3.5.13 \(\int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx\) [413]

Optimal. Leaf size=531 \[ -\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \left (a^2+b^2\right )^3 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]

[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))/(a^2+b^2)^3/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))/(a^2+b^2)^3/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))/(a^2+b^2)^3
/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))/(a^
2+b^2)^3/d*2^(1/2)-1/4*(15*A*a^4*b-18*A*a^2*b^3-A*b^5-3*B*a^5+26*B*a^3*b^2-3*B*a*b^4)*arctan(b^(1/2)*tan(d*x+c
)^(1/2)/a^(1/2))/a^(3/2)/(a^2+b^2)^3/d/b^(1/2)-1/2*(A*b-B*a)*tan(d*x+c)^(1/2)/(a^2+b^2)/d/(a+b*tan(d*x+c))^2-1
/4*(7*A*a^2*b-A*b^3-3*B*a^3+5*B*a*b^2)*tan(d*x+c)^(1/2)/a/(a^2+b^2)^2/d/(a+b*tan(d*x+c))

________________________________________________________________________________________

Rubi [A]
time = 0.84, antiderivative size = 531, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3689, 3730, 3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} -\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {\left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {\left (-3 a^3 B+7 a^2 A b+5 a b^2 B-A b^3\right ) \sqrt {\tan (c+d x)}}{4 a d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {\left (-\left (a^3 (A+B)\right )+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 \left (a^2+b^2\right )^3}+\frac {\left (-\left (a^3 (A+B)\right )+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 \left (a^2+b^2\right )^3}-\frac {\left (-3 a^5 B+15 a^4 A b+26 a^3 b^2 B-18 a^2 A b^3-3 a b^4 B-A b^5\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} d \left (a^2+b^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[Tan[c + d*x]]*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^3,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]]])/(Sq
rt[2]*(a^2 + b^2)^3*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]*(a^2 + b^2)^3*d) - ((15*a^4*A*b - 18*a^2*A*b^3 - A*b^5 - 3*a^5*B + 26*a^3*b^2*B
- 3*a*b^4*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*a^(3/2)*Sqrt[b]*(a^2 + b^2)^3*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*Sqr
t[2]*(a^2 + b^2)^3*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]*(a^2 + b^2)^3*d) - ((A*b - a*B)*Sqrt[Tan[c + d*x]])/(2*(a^2 + b^2)*d
*(a + b*Tan[c + d*x])^2) - ((7*a^2*A*b - A*b^3 - 3*a^3*B + 5*a*b^2*B)*Sqrt[Tan[c + d*x]])/(4*a*(a^2 + b^2)^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 3689

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

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 3730

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[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 {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\int \frac {-\frac {1}{2} b (A b-a B)-2 b (a A+b B) \tan (c+d x)+\frac {3}{2} b (A b-a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\int \frac {-\frac {1}{4} b \left (9 a^2 A b+A b^3-5 a^3 B+3 a b^2 B\right )-2 a b \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)+\frac {1}{4} b \left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 a b \left (a^2+b^2\right )^2}\\ &=-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\int \frac {-2 a b \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right )-2 a b \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}{2 a b \left (a^2+b^2\right )^3}-\frac {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \int \frac {1+\tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 a \left (a^2+b^2\right )^3}\\ &=-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\text {Subst}\left (\int \frac {-2 a b \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right )-2 a b \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 )}{a b \left (a^2+b^2\right )^3 d}-\frac {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 a \left (a^2+b^2\right )^3 d}\\ &=-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a \left (a^2+b^2\right )^3 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 {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 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 )}{\left (a^2+b^2\right )^3 d}\\ &=-\frac {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \left (a^2+b^2\right )^3 d}-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \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} \left (a^2+b^2\right )^3 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} \left (a^2+b^2\right )^3 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 \left (a^2+b^2\right )^3 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 \left (a^2+b^2\right )^3 d}\\ &=-\frac {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \left (a^2+b^2\right )^3 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} \left (a^2+b^2\right )^3 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} \left (a^2+b^2\right )^3 d}-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \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} \left (a^2+b^2\right )^3 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} \left (a^2+b^2\right )^3 d}\\ &=-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \left (a^2+b^2\right )^3 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} \left (a^2+b^2\right )^3 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} \left (a^2+b^2\right )^3 d}-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 4.01, size = 344, normalized size = 0.65 \begin {gather*} \frac {b (A b-a B) \tan ^{\frac {3}{2}}(c+d x)-(A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))+\frac {2 (a+b \tan (c+d x)) \left (\frac {1}{4} a^{3/2} b^{3/2} \left (a^2+b^2\right ) \left (-7 a^2 A b+A b^3+3 a^3 B-5 a b^2 B\right ) \sqrt {\tan (c+d x)}-\left (-\frac {1}{4} a b \left (-15 a^4 A b+18 a^2 A b^3+A b^5+3 a^5 B-26 a^3 b^2 B+3 a b^4 B\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+\sqrt [4]{-1} a^{5/2} b^{3/2} \left ((i a-b)^3 (A-i B) \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-(i a+b)^3 (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right )\right ) (a+b \tan (c+d x))\right )}{a^{3/2} b^{3/2} \left (a^2+b^2\right )^2}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(A*b - a*B)*Tan[c + d*x]^(3/2) - (A*b - a*B)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x]) + (2*(a + b*Tan[c + d*
x])*((a^(3/2)*b^(3/2)*(a^2 + b^2)*(-7*a^2*A*b + A*b^3 + 3*a^3*B - 5*a*b^2*B)*Sqrt[Tan[c + d*x]])/4 - (-1/4*(a*
b*(-15*a^4*A*b + 18*a^2*A*b^3 + A*b^5 + 3*a^5*B - 26*a^3*b^2*B + 3*a*b^4*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]]
)/Sqrt[a]]) + (-1)^(1/4)*a^(5/2)*b^(3/2)*((I*a - b)^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - (I*a +
b)^3*(A + I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]))*(a + b*Tan[c + d*x])))/(a^(3/2)*b^(3/2)*(a^2 + b^2)^2
))/(2*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2)

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Maple [A]
time = 0.12, size = 451, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (3 A \,a^{2} b -A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}-\frac {2 \left (\frac {\frac {b \left (7 A \,a^{4} b +6 A \,a^{2} b^{3}-A \,b^{5}-3 B \,a^{5}+2 B \,a^{3} b^{2}+5 B a \,b^{4}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{8 a}+\left (\frac {9}{8} A \,a^{4} b +\frac {5}{4} A \,a^{2} b^{3}-\frac {5}{8} B \,a^{5}+\frac {3}{8} B a \,b^{4}+\frac {1}{8} A \,b^{5}-\frac {1}{4} B \,a^{3} b^{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (15 A \,a^{4} b -18 A \,a^{2} b^{3}-A \,b^{5}-3 B \,a^{5}+26 B \,a^{3} b^{2}-3 B a \,b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{8 a \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(451\)
default	\(\frac {\frac {\frac {\left (3 A \,a^{2} b -A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}-\frac {2 \left (\frac {\frac {b \left (7 A \,a^{4} b +6 A \,a^{2} b^{3}-A \,b^{5}-3 B \,a^{5}+2 B \,a^{3} b^{2}+5 B a \,b^{4}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{8 a}+\left (\frac {9}{8} A \,a^{4} b +\frac {5}{4} A \,a^{2} b^{3}-\frac {5}{8} B \,a^{5}+\frac {3}{8} B a \,b^{4}+\frac {1}{8} A \,b^{5}-\frac {1}{4} B \,a^{3} b^{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (15 A \,a^{4} b -18 A \,a^{2} b^{3}-A \,b^{5}-3 B \,a^{5}+26 B \,a^{3} b^{2}-3 B a \,b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{8 a \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(451\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(2/(a^2+b^2)^3*(1/8*(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)))+1/8*(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)*ta
n(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))))-2/(a^
2+b^2)^3*((1/8*b*(7*A*a^4*b+6*A*a^2*b^3-A*b^5-3*B*a^5+2*B*a^3*b^2+5*B*a*b^4)/a*tan(d*x+c)^(3/2)+(9/8*A*a^4*b+5
/4*A*a^2*b^3-5/8*B*a^5+3/8*B*a*b^4+1/8*A*b^5-1/4*B*a^3*b^2)*tan(d*x+c)^(1/2))/(a+b*tan(d*x+c))^2+1/8*(15*A*a^4
*b-18*A*a^2*b^3-A*b^5-3*B*a^5+26*B*a^3*b^2-3*B*a*b^4)/a/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))))

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Maxima [A]
time = 0.53, size = 537, normalized size = 1.01 \begin {gather*} \frac {\frac {{\left (3 \, B a^{5} - 15 \, A a^{4} b - 26 \, B a^{3} b^{2} + 18 \, A a^{2} b^{3} + 3 \, B a b^{4} + A b^{5}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \sqrt {a b}} + \frac {2 \, \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 ) + 2 \, \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 ) - \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 ) + \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 )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (3 \, B a^{3} b - 7 \, A a^{2} b^{2} - 5 \, B a b^{3} + A b^{4}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + {\left (5 \, B a^{4} - 9 \, A a^{3} b - 3 \, B a^{2} b^{2} - A a b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4} + {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \tan \left (d x + c\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*((3*B*a^5 - 15*A*a^4*b - 26*B*a^3*b^2 + 18*A*a^2*b^3 + 3*B*a*b^4 + A*b^5)*arctan(b*sqrt(tan(d*x + c))/sqrt
(a*b))/((a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*sqrt(a*b)) + (2*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)))) + 2*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)))) - 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
) + 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))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + ((3*B*a^3*b - 7*A*a^2*b^2 - 5*B*a*b^3 + A*b^4)*tan(d*
x + c)^(3/2) + (5*B*a^4 - 9*A*a^3*b - 3*B*a^2*b^2 - A*a*b^3)*sqrt(tan(d*x + c)))/(a^7 + 2*a^5*b^2 + a^3*b^4 +
(a^5*b^2 + 2*a^3*b^4 + a*b^6)*tan(d*x + c)^2 + 2*(a^6*b + 2*a^4*b^3 + a^2*b^5)*tan(d*x + c)))/d

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Fricas [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)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

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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)**(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**3,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)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 52.40, size = 2500, normalized size = 4.71 \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)^(1/2)*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^3,x)

[Out]

(log((((((((((64*A*b^3*(b^4 - 10*a^4 + 15*a^2*b^2))/(a*d) + 128*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)
^2*((4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) + 80*A^2*a^3*b^3*d^2 - 24*A^2*a*b^5*d^2 - 24*A
^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))*((4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) + 80*A^
2*a^3*b^3*d^2 - 24*A^2*a*b^5*d^2 - 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))/4 + (8*A^2*b^2*tan(c + d*x)^(
1/2)*(8*a^10 + b^10 - 148*a^2*b^8 + 902*a^4*b^6 - 812*a^6*b^4 + 193*a^8*b^2))/(a*d^2*(a^2 + b^2)^4))*((4*(-A^4
*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) + 80*A^2*a^3*b^3*d^2 - 24*A^2*a*b^5*d^2 - 24*A^2*a^5*b*d^2
)/(d^4*(a^2 + b^2)^6))^(1/2))/4 - (2*A^3*b^2*(16*a^12 + b^12 - 71*a^2*b^10 - 1382*a^4*b^8 + 5266*a^6*b^6 - 453
9*a^8*b^4 + 1189*a^10*b^2))/(a^2*d^3*(a^2 + b^2)^6))*((4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1
/2) + 80*A^2*a^3*b^3*d^2 - 24*A^2*a*b^5*d^2 - 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))/4 - (A^4*b^3*tan(c
+ d*x)^(1/2)*(2*a^2*b^10 - b^12 - 225*a^12 + 49*a^4*b^8 + 2460*a^6*b^6 - 3631*a^8*b^4 + 1922*a^10*b^2))/(a^2*
d^4*(a^2 + b^2)^8))*((4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) + 80*A^2*a^3*b^3*d^2 - 24*A^2
*a*b^5*d^2 - 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))/4 + (A^5*b^3*(7*b^8 - 225*a^8 + 116*a^2*b^6 - 270*a
^4*b^4 + 420*a^6*b^2))/(2*a*d^5*(a^2 + b^2)^8))*(((480*A^4*a^2*b^10*d^4 - 16*A^4*b^12*d^4 - 16*A^4*a^12*d^4 -
4080*A^4*a^4*b^8*d^4 + 7232*A^4*a^6*b^6*d^4 - 4080*A^4*a^8*b^4*d^4 + 480*A^4*a^10*b^2*d^4)^(1/2) + 80*A^2*a^3*
b^3*d^2 - 24*A^2*a*b^5*d^2 - 24*A^2*a^5*b*d^2)/(a^12*d^4 + b^12*d^4 + 6*a^2*b^10*d^4 + 15*a^4*b^8*d^4 + 20*a^6
*b^6*d^4 + 15*a^8*b^4*d^4 + 6*a^10*b^2*d^4))^(1/2))/4 - ((tan(c + d*x)^(1/2)*(A*b^3 + 9*A*a^2*b))/(4*(a^4 + b^
4 + 2*a^2*b^2)) - (tan(c + d*x)^(3/2)*(A*b^4 - 7*A*a^2*b^2))/(4*a*(a^4 + b^4 + 2*a^2*b^2)))/(a^2*d + b^2*d*tan
(c + d*x)^2 + 2*a*b*d*tan(c + d*x)) - ((tan(c + d*x)^(3/2)*(5*B*b^3 - 3*B*a^2*b))/(4*(a^4 + b^4 + 2*a^2*b^2))
- (a*tan(c + d*x)^(1/2)*(5*B*a^2 - 3*B*b^2))/(4*(a^4 + b^4 + 2*a^2*b^2)))/(a^2*d + b^2*d*tan(c + d*x)^2 + 2*a*
b*d*tan(c + d*x)) + (log((((((((((64*A*b^3*(b^4 - 10*a^4 + 15*a^2*b^2))/(a*d) + 128*b^3*tan(c + d*x)^(1/2)*(a^
2 - b^2)*(a^2 + b^2)^2*(-(4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) - 80*A^2*a^3*b^3*d^2 + 24
*A^2*a*b^5*d^2 + 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))*(-(4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4
*b^2)^2)^(1/2) - 80*A^2*a^3*b^3*d^2 + 24*A^2*a*b^5*d^2 + 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))/4 + (8*
A^2*b^2*tan(c + d*x)^(1/2)*(8*a^10 + b^10 - 148*a^2*b^8 + 902*a^4*b^6 - 812*a^6*b^4 + 193*a^8*b^2))/(a*d^2*(a^
2 + b^2)^4))*(-(4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) - 80*A^2*a^3*b^3*d^2 + 24*A^2*a*b^5
*d^2 + 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))/4 - (2*A^3*b^2*(16*a^12 + b^12 - 71*a^2*b^10 - 1382*a^4*b
^8 + 5266*a^6*b^6 - 4539*a^8*b^4 + 1189*a^10*b^2))/(a^2*d^3*(a^2 + b^2)^6))*(-(4*(-A^4*d^4*(a^6 - b^6 + 15*a^2
*b^4 - 15*a^4*b^2)^2)^(1/2) - 80*A^2*a^3*b^3*d^2 + 24*A^2*a*b^5*d^2 + 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(
1/2))/4 - (A^4*b^3*tan(c + d*x)^(1/2)*(2*a^2*b^10 - b^12 - 225*a^12 + 49*a^4*b^8 + 2460*a^6*b^6 - 3631*a^8*b^4
+ 1922*a^10*b^2))/(a^2*d^4*(a^2 + b^2)^8))*(-(4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) - 80
*A^2*a^3*b^3*d^2 + 24*A^2*a*b^5*d^2 + 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))/4 + (A^5*b^3*(7*b^8 - 225*
a^8 + 116*a^2*b^6 - 270*a^4*b^4 + 420*a^6*b^2))/(2*a*d^5*(a^2 + b^2)^8))*(-((480*A^4*a^2*b^10*d^4 - 16*A^4*b^1
2*d^4 - 16*A^4*a^12*d^4 - 4080*A^4*a^4*b^8*d^4 + 7232*A^4*a^6*b^6*d^4 - 4080*A^4*a^8*b^4*d^4 + 480*A^4*a^10*b^
2*d^4)^(1/2) - 80*A^2*a^3*b^3*d^2 + 24*A^2*a*b^5*d^2 + 24*A^2*a^5*b*d^2)/(a^12*d^4 + b^12*d^4 + 6*a^2*b^10*d^4
+ 15*a^4*b^8*d^4 + 20*a^6*b^6*d^4 + 15*a^8*b^4*d^4 + 6*a^10*b^2*d^4))^(1/2))/4 - log((((((((((64*A*b^3*(b^4 -
10*a^4 + 15*a^2*b^2))/(a*d) - 128*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*((4*(-A^4*d^4*(a^6 - b^6 +
15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) + 80*A^2*a^3*b^3*d^2 - 24*A^2*a*b^5*d^2 - 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2
)^6))^(1/2))*((4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) + 80*A^2*a^3*b^3*d^2 - 24*A^2*a*b^5*
d^2 - 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))/4 - (8*A^2*b^2*tan(c + d*x)^(1/2)*(8*a^10 + b^10 - 148*a^2
*b^8 + 902*a^4*b^6 - 812*a^6*b^4 + 193*a^8*b^2))/(a*d^2*(a^2 + b^2)^4))*((4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4
- 15*a^4*b^2)^2)^(1/2) + 80*A^2*a^3*b^3*d^2 - 24*A^2*a*b^5*d^2 - 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))
/4 - (2*A^3*b^2*(16*a^12 + b^12 - 71*a^2*b^10 - 1382*a^4*b^8 + 5266*a^6*b^6 - 4539*a^8*b^4 + 1189*a^10*b^2))/(
a^2*d^3*(a^2 + b^2)^6))*((4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) + 80*A^2*a^3*b^3*d^2 - 24
*A^2*a*b^5*d^2 - 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))/4 + (A^4*b^3*tan(c + d*x)^(1/2)*(2*a^2*b^10 - b
^12 - 225*a^12 + 49*a^4*b^8 + 2460*a^6*b^6 - 3631*a^8*b^4 + 1922*a^10*b^2))/(a^2*d^4*(a^2 + b^2)^8))*((4*(-A^4
*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(...

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