\(\int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) [395]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F(-1)]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 33, antiderivative size = 374 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (2 a^2 A+A b^2+3 a b B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (3 a A+b B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{d \sqrt {\tan (c+d x)}} \]

[Out]

-1/2*(a^3*(A-B)-3*a*b^2*(A-B)-3*a^2*b*(A+B)+b^3*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/2*(a^3*
(A-B)-3*a*b^2*(A-B)-3*a^2*b*(A+B)+b^3*(A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/4*(3*a^2*b*(A-B)-b
^3*(A-B)+a^3*(A+B)-3*a*b^2*(A+B))*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)+1/4*(3*a^2*b*(A-B)-b^3*(
A-B)+a^3*(A+B)-3*a*b^2*(A+B))*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)+2*b*(2*A*a^2+A*b^2+3*B*a*b)*
tan(d*x+c)^(1/2)/d+2/3*b^2*(3*A*a+B*b)*tan(d*x+c)^(3/2)/d-2*a*A*(a+b*tan(d*x+c))^2/d/tan(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3686, 3718, 3711, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 b \left (2 a^2 A+3 a b B+A b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}-\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 b^2 (3 a A+b B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{d \sqrt {\tan (c+d x)}} \]

[In]

Int[((a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(3/2),x]

[Out]

((a^3*(A - B) - 3*a*b^2*(A - B) - 3*a^2*b*(A + B) + b^3*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt
[2]*d) - ((a^3*(A - B) - 3*a*b^2*(A - B) - 3*a^2*b*(A + B) + b^3*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]
]])/(Sqrt[2]*d) - ((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c
+ d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + ((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*Log[
1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (2*b*(2*a^2*A + A*b^2 + 3*a*b*B)*Sqrt[Tan[c +
d*x]])/d + (2*b^2*(3*a*A + b*B)*Tan[c + d*x]^(3/2))/(3*d) - (2*a*A*(a + b*Tan[c + d*x])^2)/(d*Sqrt[Tan[c + d*x
]])

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
 x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
 Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3686

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e
+ f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
 + a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

Rule 3711

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3718

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])
^(n + 1)/(d*f*(n + 2))), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
 + a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] &&  !LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a A (a+b \tan (c+d x))^2}{d \sqrt {\tan (c+d x)}}+2 \int \frac {(a+b \tan (c+d x)) \left (\frac {1}{2} a (5 A b+a B)-\frac {1}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac {1}{2} b (3 a A+b B) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 b^2 (3 a A+b B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{d \sqrt {\tan (c+d x)}}-\frac {4}{3} \int \frac {-\frac {3}{4} a^2 (5 A b+a B)+\frac {3}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)-\frac {3}{4} b \left (2 a^2 A+A b^2+3 a b B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 b \left (2 a^2 A+A b^2+3 a b B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (3 a A+b B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{d \sqrt {\tan (c+d x)}}-\frac {4}{3} \int \frac {-\frac {3}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+\frac {3}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 b \left (2 a^2 A+A b^2+3 a b B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (3 a A+b B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{d \sqrt {\tan (c+d x)}}-\frac {8 \text {Subst}\left (\int \frac {-\frac {3}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+\frac {3}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{3 d} \\ & = \frac {2 b \left (2 a^2 A+A b^2+3 a b B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (3 a A+b B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{d \sqrt {\tan (c+d x)}}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {2 b \left (2 a^2 A+A b^2+3 a b B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (3 a A+b B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{d \sqrt {\tan (c+d x)}}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d} \\ & = -\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (2 a^2 A+A b^2+3 a b B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (3 a A+b B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{d \sqrt {\tan (c+d x)}}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d} \\ & = \frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (2 a^2 A+A b^2+3 a b B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (3 a A+b B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{d \sqrt {\tan (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.86 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.71 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {8 b \left (-12 a A b-17 a^2 B+3 b^2 B\right )-3 \left (8 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},-\tan ^2(c+d x)\right )+\sqrt {2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \left (2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right ) \sqrt {\tan (c+d x)}\right )+8 b (3 A b+7 a B) (a+b \tan (c+d x))+8 b B (a+b \tan (c+d x))^2}{12 d \sqrt {\tan (c+d x)}} \]

[In]

Integrate[((a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(3/2),x]

[Out]

(8*b*(-12*a*A*b - 17*a^2*B + 3*b^2*B) - 3*(8*(a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*Hypergeometric2F1[-1/4, 1
, 3/4, -Tan[c + d*x]^2] + Sqrt[2]*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*(2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d
*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1
+ Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])*Sqrt[Tan[c + d*x]]) + 8*b*(3*A*b + 7*a*B)*(a + b*Tan[c + d*x]) +
 8*b*B*(a + b*Tan[c + d*x])^2)/(12*d*Sqrt[Tan[c + d*x]])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.76

method result size
derivativedivides \(\frac {\frac {2 B \,b^{3} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 \left (\sqrt {\tan }\left (d x +c \right )\right ) A \,b^{3}+6 \left (\sqrt {\tan }\left (d x +c \right )\right ) B a \,b^{2}-\frac {2 A \,a^{3}}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-A \,a^{3}+3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) \(286\)
default \(\frac {\frac {2 B \,b^{3} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 \left (\sqrt {\tan }\left (d x +c \right )\right ) A \,b^{3}+6 \left (\sqrt {\tan }\left (d x +c \right )\right ) B a \,b^{2}-\frac {2 A \,a^{3}}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-A \,a^{3}+3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) \(286\)
parts \(\frac {\left (A \,b^{3}+3 B a \,b^{2}\right ) \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {\left (3 A a \,b^{2}+3 B \,a^{2} b \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {A \,a^{3} \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {B \,b^{3} \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\) \(527\)

[In]

int((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(2/3*B*b^3*tan(d*x+c)^(3/2)+2*tan(d*x+c)^(1/2)*A*b^3+6*tan(d*x+c)^(1/2)*B*a*b^2-2*A*a^3/tan(d*x+c)^(1/2)+1
/4*(3*A*a^2*b-A*b^3+B*a^3-3*B*a*b^2)*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)
^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(-A*a^3+3*
A*a*b^2+3*B*a^2*b-B*b^3)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d
*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6137 vs. \(2 (338) = 676\).

Time = 1.47 (sec) , antiderivative size = 6137, normalized size of antiderivative = 16.41 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{3}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a+b*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(3/2),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**3/tan(c + d*x)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 310, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {8 \, B b^{3} \tan \left (d x + c\right )^{\frac {3}{2}} - \frac {24 \, A a^{3}}{\sqrt {\tan \left (d x + c\right )}} - 6 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 6 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 24 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{12 \, d} \]

[In]

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

1/12*(8*B*b^3*tan(d*x + c)^(3/2) - 24*A*a^3/sqrt(tan(d*x + c)) - 6*sqrt(2)*((A - B)*a^3 - 3*(A + B)*a^2*b - 3*
(A - B)*a*b^2 + (A + B)*b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) - 6*sqrt(2)*((A - B)*a^3 - 3
*(A + B)*a^2*b - 3*(A - B)*a*b^2 + (A + B)*b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + 3*sqrt
(2)*((A + B)*a^3 + 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 - (A - B)*b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x +
 c) + 1) - 3*sqrt(2)*((A + B)*a^3 + 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 - (A - B)*b^3)*log(-sqrt(2)*sqrt(tan(d*x
 + c)) + tan(d*x + c) + 1) + 24*(3*B*a*b^2 + A*b^3)*sqrt(tan(d*x + c)))/d

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 12.19 (sec) , antiderivative size = 7108, normalized size of antiderivative = 19.01 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3)/tan(c + d*x)^(3/2),x)

[Out]

atan((((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12
*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) -
(3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2
*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 2
55*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a
*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*1i - ((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((5*B^2*a^3*b^3)/d^2 - (3
0*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4
*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 + (16*ta
n(c + d*x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^
2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 3
0*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*1i)/(((8*(4*B*a^3*d^2
 - 12*B*a*b^2*d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8
*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2)
- (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*a^4*
b^2))/d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 4
52*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2
*a^5*b)/(2*d^2))^(1/2) + ((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*
b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^
4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(B^2*a^6
 - B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4
 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)
/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2) + (16*(3*B^3*a^8*b - B^3*b^9 + 6*B^3*a^4*b^5 +
 8*B^3*a^6*b^3))/d^3))*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4
*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d
^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*2i - atan((A^2*a^6*d^3*tan(c + d*x)^(1/2)*((30*A^4*a^2*b^10*d^4 - A^4*b^12*
d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1
/2)/(4*d^4) - (5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*a^5*b)/(2*d^2))^(1/2)*32i)/(16*A*b^3*(30*A^
4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4
 + 30*A^4*a^10*b^2*d^4)^(1/2) + 16*A^3*a^9*d^2 - 736*A^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4*d^2 - 288*A^3*a^7*b^2*d
^2 - 48*A*a^2*b*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4
 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 48*A^3*a*b^8*d^2) - (A^2*b^6*d^3*tan(c + d*x)^(1/2)*((30
*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*
d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*a^5*b)/(2*d^2)
)^(1/2)*32i)/(16*A*b^3*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*
b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 16*A^3*a^9*d^2 - 736*A^3*a^3*b^6*d^2 + 960*A^3*a^
5*b^4*d^2 - 288*A^3*a^7*b^2*d^2 - 48*A*a^2*b*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*
b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 48*A^3*a*b^8*d^2) + (A^2*a^
2*b^4*d^3*tan(c + d*x)^(1/2)*((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A
^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5
)/(2*d^2) + (3*A^2*a^5*b)/(2*d^2))^(1/2)*480i)/(16*A*b^3*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 -
255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 16*A^3*a^9*d^2
- 736*A^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4*d^2 - 288*A^3*a^7*b^2*d^2 - 48*A*a^2*b*(30*A^4*a^2*b^10*d^4 - A^4*b^12
*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(
1/2) + 48*A^3*a*b^8*d^2) - (A^2*a^4*b^2*d^3*tan(c + d*x)^(1/2)*((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12
*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) -
(5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*a^5*b)/(2*d^2))^(1/2)*480i)/(16*A*b^3*(30*A^4*a^2*b^10*d^
4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^1
0*b^2*d^4)^(1/2) + 16*A^3*a^9*d^2 - 736*A^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4*d^2 - 288*A^3*a^7*b^2*d^2 - 48*A*a^2
*b*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^
8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 48*A^3*a*b^8*d^2))*((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^
4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*
A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*a^5*b)/(2*d^2))^(1/2)*2i - atan((A^2*a^6*d^3*tan(c + d*x)^(1
/2)*((3*A^2*a*b^5)/(2*d^2) - (5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^
4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*A^2*a^5*b)
/(2*d^2))^(1/2)*32i)/(16*A^3*a^9*d^2 - 16*A*b^3*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a
^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) - 736*A^3*a^3*b^6*d^2 + 96
0*A^3*a^5*b^4*d^2 - 288*A^3*a^7*b^2*d^2 + 48*A*a^2*b*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*
A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 48*A^3*a*b^8*d^2) -
 (A^2*b^6*d^3*tan(c + d*x)^(1/2)*((3*A^2*a*b^5)/(2*d^2) - (5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^1
2*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^
(1/2)/(4*d^4) + (3*A^2*a^5*b)/(2*d^2))^(1/2)*32i)/(16*A^3*a^9*d^2 - 16*A*b^3*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d
^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/
2) - 736*A^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4*d^2 - 288*A^3*a^7*b^2*d^2 + 48*A*a^2*b*(30*A^4*a^2*b^10*d^4 - A^4*b
^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4
)^(1/2) + 48*A^3*a*b^8*d^2) + (A^2*a^2*b^4*d^3*tan(c + d*x)^(1/2)*((3*A^2*a*b^5)/(2*d^2) - (5*A^2*a^3*b^3)/d^2
 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^
8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*A^2*a^5*b)/(2*d^2))^(1/2)*480i)/(16*A^3*a^9*d^2 - 16*A*b^3
*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*
b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) - 736*A^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4*d^2 - 288*A^3*a^7*b^2*d^2 + 48*A*
a^2*b*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4
*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 48*A^3*a*b^8*d^2) - (A^2*a^4*b^2*d^3*tan(c + d*x)^(1/2)*((3*A^2*a*
b^5)/(2*d^2) - (5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4
+ 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*A^2*a^5*b)/(2*d^2))^(1/2
)*480i)/(16*A^3*a^9*d^2 - 16*A*b^3*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 +
452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) - 736*A^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4
*d^2 - 288*A^3*a^7*b^2*d^2 + 48*A*a^2*b*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d
^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 48*A^3*a*b^8*d^2))*((3*A^2*a*b^5
)/(2*d^2) - (5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 4
52*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*A^2*a^5*b)/(2*d^2))^(1/2)*2
i + atan((((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4
*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^
2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15
*B^2*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4
 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B
^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*1i - ((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((30*B^4*a^2*b^10*d^4
 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10
*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 + (1
6*tan(c + d*x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b
^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4
)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*1i)/(((8*(4*B*a^3
*d^2 - 12*B*a*b^2*d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6
*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d
^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*
a^4*b^2))/d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4
 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3
*B^2*a^5*b)/(2*d^2))^(1/2) + ((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^1
2*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) +
 (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(B^2
*a^6 - B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 -
255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*
a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2) + (16*(3*B^3*a^8*b - B^3*b^9 + 6*B^3*a^4*b
^5 + 8*B^3*a^6*b^3))/d^3))*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4
*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/
(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*2i - (2*A*a^3)/(d*tan(c + d*x)^(1/2)) + (2*A*b^3*tan(c + d*x)^(1/2))/d
+ (2*B*b^3*tan(c + d*x)^(3/2))/(3*d) + (6*B*a*b^2*tan(c + d*x)^(1/2))/d