3.6.92 \(\int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [592]

Optimal. Leaf size=358 \[ -\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {a^{5/2} \left (3 a^2+7 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (3 a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \]

[Out]

-a^(5/2)*(3*a^2+7*b^2)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/b^(5/2)/(a^2+b^2)^2/d+1/2*(a^2-2*a*b-b^2)*arct
an(-1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)^2/d*2^(1/2)+1/2*(a^2-2*a*b-b^2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/(
a^2+b^2)^2/d*2^(1/2)-1/4*(a^2+2*a*b-b^2)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)^2/d*2^(1/2)+1/4*(
a^2+2*a*b-b^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)^2/d*2^(1/2)+(3*a^2+2*b^2)*tan(d*x+c)^(1/2)/
b^2/(a^2+b^2)/d-a^2*tan(d*x+c)^(3/2)/b/(a^2+b^2)/d/(a+b*tan(d*x+c))

________________________________________________________________________________________

Rubi [A]
time = 0.50, antiderivative size = 358, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3646, 3728, 3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} -\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}+\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {\left (3 a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 d \left (a^2+b^2\right )}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}-\frac {a^{5/2} \left (3 a^2+7 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} d \left (a^2+b^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x])^2,x]

[Out]

-(((a^2 - 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^2*d)) + ((a^2 - 2*a*b - b^
2)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^2*d) - (a^(5/2)*(3*a^2 + 7*b^2)*ArcTan[(Sqrt[b
]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(b^(5/2)*(a^2 + b^2)^2*d) - ((a^2 + 2*a*b - b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c +
d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^2*d) + ((a^2 + 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] +
Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^2*d) + ((3*a^2 + 2*b^2)*Sqrt[Tan[c + d*x]])/(b^2*(a^2 + b^2)*d) - (a^2*
Tan[c + d*x]^(3/2))/(b*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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 211

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

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 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3728

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

Rule 3734

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

Rubi steps

\begin {align*} \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\int \frac {\sqrt {\tan (c+d x)} \left (\frac {3 a^2}{2}-a b \tan (c+d x)+\frac {1}{2} \left (3 a^2+2 b^2\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac {\left (3 a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {2 \int \frac {-\frac {1}{4} a \left (3 a^2+2 b^2\right )-\frac {1}{2} b^3 \tan (c+d x)-\frac {1}{4} a \left (3 a^2+4 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{b^2 \left (a^2+b^2\right )}\\ &=\frac {\left (3 a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {2 \int \frac {\frac {1}{2} b^2 \left (a^2-b^2\right )-a b^3 \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{b^2 \left (a^2+b^2\right )^2}-\frac {\left (a^3 \left (3 a^2+7 b^2\right )\right ) \int \frac {1+\tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=\frac {\left (3 a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {4 \text {Subst}\left (\int \frac {\frac {1}{2} b^2 \left (a^2-b^2\right )-a b^3 x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{b^2 \left (a^2+b^2\right )^2 d}-\frac {\left (a^3 \left (3 a^2+7 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{2 b^2 \left (a^2+b^2\right )^2 d}\\ &=\frac {\left (3 a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}-\frac {\left (a^3 \left (3 a^2+7 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{b^2 \left (a^2+b^2\right )^2 d}\\ &=-\frac {a^{5/2} \left (3 a^2+7 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right )^2 d}+\frac {\left (3 a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\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} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\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} \left (a^2+b^2\right )^2 d}\\ &=-\frac {a^{5/2} \left (3 a^2+7 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (3 a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}\\ &=-\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {a^{5/2} \left (3 a^2+7 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (3 a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.22, size = 375, normalized size = 1.05 \begin {gather*} \frac {-3 a^{11/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )-7 a^{7/2} b^2 \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+3 a^5 \sqrt {b} \sqrt {\tan (c+d x)}+5 a^3 b^{5/2} \sqrt {\tan (c+d x)}+2 a b^{9/2} \sqrt {\tan (c+d x)}-3 a^{9/2} b \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \tan (c+d x)-7 a^{5/2} b^3 \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \tan (c+d x)+2 a^4 b^{3/2} \tan ^{\frac {3}{2}}(c+d x)+4 a^2 b^{7/2} \tan ^{\frac {3}{2}}(c+d x)+2 b^{11/2} \tan ^{\frac {3}{2}}(c+d x)+\sqrt [4]{-1} b^{5/2} (-i a+b)^2 \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (a+b \tan (c+d x))+\sqrt [4]{-1} b^{5/2} (i a+b)^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (a+b \tan (c+d x))}{b^{5/2} \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x])^2,x]

[Out]

(-3*a^(11/2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]] - 7*a^(7/2)*b^2*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/
Sqrt[a]] + 3*a^5*Sqrt[b]*Sqrt[Tan[c + d*x]] + 5*a^3*b^(5/2)*Sqrt[Tan[c + d*x]] + 2*a*b^(9/2)*Sqrt[Tan[c + d*x]
] - 3*a^(9/2)*b*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Tan[c + d*x] - 7*a^(5/2)*b^3*ArcTan[(Sqrt[b]*Sqrt
[Tan[c + d*x]])/Sqrt[a]]*Tan[c + d*x] + 2*a^4*b^(3/2)*Tan[c + d*x]^(3/2) + 4*a^2*b^(7/2)*Tan[c + d*x]^(3/2) +
2*b^(11/2)*Tan[c + d*x]^(3/2) + (-1)^(1/4)*b^(5/2)*((-I)*a + b)^2*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*(a + b
*Tan[c + d*x]) + (-1)^(1/4)*b^(5/2)*(I*a + b)^2*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*(a + b*Tan[c + d*x]))/(
b^(5/2)*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

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Maple [A]
time = 0.14, size = 296, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{b^{2}}-\frac {2 a^{3} \left (\frac {\left (-\frac {a^{2}}{2}-\frac {b^{2}}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{a +b \tan \left (d x +c \right )}+\frac {\left (3 a^{2}+7 b^{2}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{2} b^{2}}+\frac {\frac {\left (a^{2}-b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(296\)
default	\(\frac {\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{b^{2}}-\frac {2 a^{3} \left (\frac {\left (-\frac {a^{2}}{2}-\frac {b^{2}}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{a +b \tan \left (d x +c \right )}+\frac {\left (3 a^{2}+7 b^{2}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{2} b^{2}}+\frac {\frac {\left (a^{2}-b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(296\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(2/b^2*tan(d*x+c)^(1/2)-2*a^3/(a^2+b^2)^2/b^2*((-1/2*a^2-1/2*b^2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))+1/2*(3
*a^2+7*b^2)/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2)))+2/(a^2+b^2)^2*(1/8*(a^2-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*b*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)))))

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Maxima [A]
time = 0.51, size = 296, normalized size = 0.83 \begin {gather*} \frac {\frac {4 \, a^{3} \sqrt {\tan \left (d x + c\right )}}{a^{3} b^{2} + a b^{4} + {\left (a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )} - \frac {4 \, {\left (3 \, a^{5} + 7 \, a^{3} b^{2}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a b}} + \frac {2 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {8 \, \sqrt {\tan \left (d x + c\right )}}{b^{2}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/4*(4*a^3*sqrt(tan(d*x + c))/(a^3*b^2 + a*b^4 + (a^2*b^3 + b^5)*tan(d*x + c)) - 4*(3*a^5 + 7*a^3*b^2)*arctan(
b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^4*b^2 + 2*a^2*b^4 + b^6)*sqrt(a*b)) + (2*sqrt(2)*(a^2 - 2*a*b - b^2)*arcta
n(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) -
2*sqrt(tan(d*x + c)))) + sqrt(2)*(a^2 + 2*a*b - b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqr
t(2)*(a^2 + 2*a*b - b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a^4 + 2*a^2*b^2 + b^4) + 8*sqrt
(tan(d*x + c))/b^2)/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 7100 vs. \(2 (316) = 632\).
time = 13.87, size = 14311, normalized size = 39.97 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/4*(4*sqrt(2)*((a^14*b^2 + 5*a^12*b^4 + 9*a^10*b^6 + 5*a^8*b^8 - 5*a^6*b^10 - 9*a^4*b^12 - 5*a^2*b^14 - b^16
)*d^5*cos(d*x + c)^2 + 2*(a^13*b^3 + 6*a^11*b^5 + 15*a^9*b^7 + 20*a^7*b^9 + 15*a^5*b^11 + 6*a^3*b^13 + a*b^15)
*d^5*cos(d*x + c)*sin(d*x + c) + (a^12*b^4 + 6*a^10*b^6 + 15*a^8*b^8 + 20*a^6*b^10 + 15*a^4*b^12 + 6*a^2*b^14
+ b^16)*d^5)*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^11*b + 3*a^9*b^3 + 2*a^7*b^5 - 2*a^5*b
^7 - 3*a^3*b^9 - a*b^11)*d^2*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)))/(a^8 - 12*a^6*b^2
+ 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/((a^16 + 8*a^14*b^2
+ 28*a^12*b^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^12 + 8*a^2*b^14 + b^16)*d^4))*(1/((a^8 + 4*a
^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))^(3/4)*arctan(((a^16 - 20*a^12*b^4 - 64*a^10*b^6 - 90*a^8*b^8 - 64*
a^6*b^10 - 20*a^4*b^12 + b^16)*d^4*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/((a^16 + 8*a^14*b^2
+ 28*a^12*b^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^12 + 8*a^2*b^14 + b^16)*d^4))*sqrt(1/((a^8
+ 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - sqrt(2)*((a^18 + 7*a^16*b^2 + 20*a^14*b^4 + 28*a^12*b^6 + 1
4*a^10*b^8 - 14*a^8*b^10 - 28*a^6*b^12 - 20*a^4*b^14 - 7*a^2*b^16 - b^18)*d^7*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*
b^4 - 12*a^2*b^6 + b^8)/((a^16 + 8*a^14*b^2 + 28*a^12*b^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^
12 + 8*a^2*b^14 + b^16)*d^4))*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*(a^13*b + 6*a^
11*b^3 + 15*a^9*b^5 + 20*a^7*b^7 + 15*a^5*b^9 + 6*a^3*b^11 + a*b^13)*d^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 -
12*a^2*b^6 + b^8)/((a^16 + 8*a^14*b^2 + 28*a^12*b^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^12 +
8*a^2*b^14 + b^16)*d^4)))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^11*b + 3*a^9*b^3 + 2*a^7*
b^5 - 2*a^5*b^7 - 3*a^3*b^9 - a*b^11)*d^2*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)))/(a^8
- 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(((a^12 - 10*a^10*b^2 + 15*a^8*b^4 + 52*a^6*b^6 + 15*a^4*b^
8 - 10*a^2*b^10 + b^12)*d^2*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))*cos(d*x + c) + sqrt(
2)*(2*(a^13*b - 10*a^11*b^3 + 15*a^9*b^5 + 52*a^7*b^7 + 15*a^5*b^9 - 10*a^3*b^11 + a*b^13)*d^3*sqrt(1/((a^8 +
4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))*cos(d*x + c) + (a^10 - 13*a^8*b^2 + 50*a^6*b^4 - 50*a^4*b^6 + 1
3*a^2*b^8 - b^10)*d*cos(d*x + c))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^11*b + 3*a^9*b^3
+ 2*a^7*b^5 - 2*a^5*b^7 - 3*a^3*b^9 - a*b^11)*d^2*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)
))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c))*(1/((a^8 + 4*a^6*b^2 +
6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))^(1/4) + (a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)*sin(d*x + c))/co
s(d*x + c))*(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))^(3/4) + sqrt(2)*((a^22 + a^20*b^2 - 21*a
^18*b^4 - 85*a^16*b^6 - 134*a^14*b^8 - 70*a^12*b^10 + 70*a^10*b^12 + 134*a^8*b^14 + 85*a^6*b^16 + 21*a^4*b^18
- a^2*b^20 - b^22)*d^7*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/((a^16 + 8*a^14*b^2 + 28*a^12*b
^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^12 + 8*a^2*b^14 + b^16)*d^4))*sqrt(1/((a^8 + 4*a^6*b^2
+ 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*(a^17*b - 20*a^13*b^5 - 64*a^11*b^7 - 90*a^9*b^9 - 64*a^7*b^11 - 20*a
^5*b^13 + a*b^17)*d^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/((a^16 + 8*a^14*b^2 + 28*a^12*b^
4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^12 + 8*a^2*b^14 + b^16)*d^4)))*sqrt((a^8 + 4*a^6*b^2 + 6
*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^11*b + 3*a^9*b^3 + 2*a^7*b^5 - 2*a^5*b^7 - 3*a^3*b^9 - a*b^11)*d^2*sqrt(1/((
a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt
(sin(d*x + c)/cos(d*x + c))*(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))^(3/4))/(a^8 - 12*a^6*b^2
+ 38*a^4*b^4 - 12*a^2*b^6 + b^8)) + 4*sqrt(2)*((a^14*b^2 + 5*a^12*b^4 + 9*a^10*b^6 + 5*a^8*b^8 - 5*a^6*b^10 -
9*a^4*b^12 - 5*a^2*b^14 - b^16)*d^5*cos(d*x + c)^2 + 2*(a^13*b^3 + 6*a^11*b^5 + 15*a^9*b^7 + 20*a^7*b^9 + 15*
a^5*b^11 + 6*a^3*b^13 + a*b^15)*d^5*cos(d*x + c)*sin(d*x + c) + (a^12*b^4 + 6*a^10*b^6 + 15*a^8*b^8 + 20*a^6*b
^10 + 15*a^4*b^12 + 6*a^2*b^14 + b^16)*d^5)*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^11*b +
3*a^9*b^3 + 2*a^7*b^5 - 2*a^5*b^7 - 3*a^3*b^9 - a*b^11)*d^2*sqrt(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 +
b^8)*d^4)))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*
b^6 + b^8)/((a^16 + 8*a^14*b^2 + 28*a^12*b^4 + 56*a^10*b^6 + 70*a^8*b^8 + 56*a^6*b^10 + 28*a^4*b^12 + 8*a^2*b^
14 + b^16)*d^4))*(1/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))^(3/4)*arctan(-((a^16 - 20*a^12*b^4
- 64*a^10*b^6 - 90*a^8*b^8 - 64*a^6*b^10 - 20*a^4*b^12 + b^16)*d^4*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^
2*b^6 + b^8)/((a^16 + 8*a^14*b^2 + 28*a^12*b^4 ...

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(7/2)/(a+b*tan(d*x+c))**2,x)

[Out]

Timed out

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 12.62, size = 2500, normalized size = 6.98 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(7/2)/(a + b*tan(c + d*x))^2,x)

[Out]

atan(((-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b^
4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((16*(30*a^6*b^8*d^2 - 224*a^4*b^10*d^2 - 18*a^14
*d^2 + 600*a^8*b^6*d^2 + 388*a^10*b^4*d^2 + 24*a^12*b^2*d^2))/(b^11*d^5 + 4*a^2*b^9*d^5 + 6*a^4*b^7*d^5 + 4*a^
6*b^5*d^5 + a^8*b^3*d^5) - (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(
(-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((16*(8*a*b^17*d^4 + 96*a^3*
b^15*d^4 + 360*a^5*b^13*d^4 + 640*a^7*b^11*d^4 + 600*a^9*b^9*d^4 + 288*a^11*b^7*d^4 + 56*a^13*b^5*d^4))/(b^11*
d^5 + 4*a^2*b^9*d^5 + 6*a^4*b^7*d^5 + 4*a^6*b^5*d^5 + a^8*b^3*d^5) - (16*tan(c + d*x)^(1/2)*(-1i/(4*(a^4*d^2 +
b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(32*b^20*d^4 + 160*a^2*b^18*d^4 + 288*a^4*b^16
*d^4 + 160*a^6*b^14*d^4 - 160*a^8*b^12*d^4 - 288*a^10*b^10*d^4 - 160*a^12*b^8*d^4 - 32*a^14*b^6*d^4))/(b^11*d^
4 + 4*a^2*b^9*d^4 + 6*a^4*b^7*d^4 + 4*a^6*b^5*d^4 + a^8*b^3*d^4)) + (16*tan(c + d*x)^(1/2)*(72*a^15*b*d^2 - 60
*a*b^15*d^2 - 52*a^3*b^13*d^2 + 72*a^5*b^11*d^2 + 448*a^7*b^9*d^2 + 1108*a^9*b^7*d^2 + 1132*a^11*b^5*d^2 + 480
*a^13*b^3*d^2))/(b^11*d^4 + 4*a^2*b^9*d^4 + 6*a^4*b^7*d^4 + 4*a^6*b^5*d^4 + a^8*b^3*d^4))) + (16*tan(c + d*x)^
(1/2)*(9*a^12 + 2*b^12 + 4*a^2*b^10 + 2*a^4*b^8 - 49*a^6*b^6 + 7*a^8*b^4 + 33*a^10*b^2))/(b^11*d^4 + 4*a^2*b^9
*d^4 + 6*a^4*b^7*d^4 + 4*a^6*b^5*d^4 + a^8*b^3*d^4))*1i - (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^
2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/
2)*((16*(30*a^6*b^8*d^2 - 224*a^4*b^10*d^2 - 18*a^14*d^2 + 600*a^8*b^6*d^2 + 388*a^10*b^4*d^2 + 24*a^12*b^2*d^
2))/(b^11*d^5 + 4*a^2*b^9*d^5 + 6*a^4*b^7*d^5 + 4*a^6*b^5*d^5 + a^8*b^3*d^5) - (-1i/(4*(a^4*d^2 + b^4*d^2 + a*
b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i -
6*a^2*b^2*d^2)))^(1/2)*((16*(8*a*b^17*d^4 + 96*a^3*b^15*d^4 + 360*a^5*b^13*d^4 + 640*a^7*b^11*d^4 + 600*a^9*b
^9*d^4 + 288*a^11*b^7*d^4 + 56*a^13*b^5*d^4))/(b^11*d^5 + 4*a^2*b^9*d^5 + 6*a^4*b^7*d^5 + 4*a^6*b^5*d^5 + a^8*
b^3*d^5) + (16*tan(c + d*x)^(1/2)*(-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^
(1/2)*(32*b^20*d^4 + 160*a^2*b^18*d^4 + 288*a^4*b^16*d^4 + 160*a^6*b^14*d^4 - 160*a^8*b^12*d^4 - 288*a^10*b^10
*d^4 - 160*a^12*b^8*d^4 - 32*a^14*b^6*d^4))/(b^11*d^4 + 4*a^2*b^9*d^4 + 6*a^4*b^7*d^4 + 4*a^6*b^5*d^4 + a^8*b^
3*d^4)) - (16*tan(c + d*x)^(1/2)*(72*a^15*b*d^2 - 60*a*b^15*d^2 - 52*a^3*b^13*d^2 + 72*a^5*b^11*d^2 + 448*a^7*
b^9*d^2 + 1108*a^9*b^7*d^2 + 1132*a^11*b^5*d^2 + 480*a^13*b^3*d^2))/(b^11*d^4 + 4*a^2*b^9*d^4 + 6*a^4*b^7*d^4
+ 4*a^6*b^5*d^4 + a^8*b^3*d^4))) - (16*tan(c + d*x)^(1/2)*(9*a^12 + 2*b^12 + 4*a^2*b^10 + 2*a^4*b^8 - 49*a^6*b
^6 + 7*a^8*b^4 + 33*a^10*b^2))/(b^11*d^4 + 4*a^2*b^9*d^4 + 6*a^4*b^7*d^4 + 4*a^6*b^5*d^4 + a^8*b^3*d^4))*1i)/(
(-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b^4*d^2
+ a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((16*(30*a^6*b^8*d^2 - 224*a^4*b^10*d^2 - 18*a^14*d^2 +
600*a^8*b^6*d^2 + 388*a^10*b^4*d^2 + 24*a^12*b^2*d^2))/(b^11*d^5 + 4*a^2*b^9*d^5 + 6*a^4*b^7*d^5 + 4*a^6*b^5*
d^5 + a^8*b^3*d^5) - (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(
4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((16*(8*a*b^17*d^4 + 96*a^3*b^15*d
^4 + 360*a^5*b^13*d^4 + 640*a^7*b^11*d^4 + 600*a^9*b^9*d^4 + 288*a^11*b^7*d^4 + 56*a^13*b^5*d^4))/(b^11*d^5 +
4*a^2*b^9*d^5 + 6*a^4*b^7*d^5 + 4*a^6*b^5*d^5 + a^8*b^3*d^5) - (16*tan(c + d*x)^(1/2)*(-1i/(4*(a^4*d^2 + b^4*d
^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(32*b^20*d^4 + 160*a^2*b^18*d^4 + 288*a^4*b^16*d^4 +
160*a^6*b^14*d^4 - 160*a^8*b^12*d^4 - 288*a^10*b^10*d^4 - 160*a^12*b^8*d^4 - 32*a^14*b^6*d^4))/(b^11*d^4 + 4*
a^2*b^9*d^4 + 6*a^4*b^7*d^4 + 4*a^6*b^5*d^4 + a^8*b^3*d^4)) + (16*tan(c + d*x)^(1/2)*(72*a^15*b*d^2 - 60*a*b^1
5*d^2 - 52*a^3*b^13*d^2 + 72*a^5*b^11*d^2 + 448*a^7*b^9*d^2 + 1108*a^9*b^7*d^2 + 1132*a^11*b^5*d^2 + 480*a^13*
b^3*d^2))/(b^11*d^4 + 4*a^2*b^9*d^4 + 6*a^4*b^7*d^4 + 4*a^6*b^5*d^4 + a^8*b^3*d^4))) + (16*tan(c + d*x)^(1/2)*
(9*a^12 + 2*b^12 + 4*a^2*b^10 + 2*a^4*b^8 - 49*a^6*b^6 + 7*a^8*b^4 + 33*a^10*b^2))/(b^11*d^4 + 4*a^2*b^9*d^4 +
6*a^4*b^7*d^4 + 4*a^6*b^5*d^4 + a^8*b^3*d^4)) + (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*
a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((16*(
30*a^6*b^8*d^2 - 224*a^4*b^10*d^2 - 18*a^14*d^2 + 600*a^8*b^6*d^2 + 388*a^10*b^4*d^2 + 24*a^12*b^2*d^2))/(b^11
*d^5 + 4*a^2*b^9*d^5 + 6*a^4*b^7*d^5 + 4*a^6*b^5*d^5 + a^8*b^3*d^5) - (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4
i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^
2*d^2)))^(1/2)*((16*(8*a*b^17*d^4 + 96*a^3*b^15...

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