3.132 \(\int \frac{(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=144 \[ \frac{8 \sqrt [4]{-1} a^3 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{16 a^3 (6 A-5 i B)}{15 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]

[Out]

(8*(-1)^(1/4)*a^3*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (16*a^3*(6*A - (5*I)*B))/(15*d*Sqrt[Tan
[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^2)/(5*d*Tan[c + d*x]^(5/2)) - (2*((9*I)*A + 5*B)*(a^3 + I*a^3*Tan[
c + d*x]))/(15*d*Tan[c + d*x]^(3/2))

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Rubi [A]  time = 0.376675, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3593, 3591, 3533, 205} \[ \frac{8 \sqrt [4]{-1} a^3 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{16 a^3 (6 A-5 i B)}{15 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

(8*(-1)^(1/4)*a^3*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (16*a^3*(6*A - (5*I)*B))/(15*d*Sqrt[Tan
[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^2)/(5*d*Tan[c + d*x]^(5/2)) - (2*((9*I)*A + 5*B)*(a^3 + I*a^3*Tan[
c + d*x]))/(15*d*Tan[c + d*x]^(3/2))

Rule 3593

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^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^
(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -
1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3591

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

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+i a \tan (c+d x))^2 \left (\frac{1}{2} a (9 i A+5 B)-\frac{1}{2} a (A-5 i B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (9 i A+5 B) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4}{15} \int \frac{(a+i a \tan (c+d x)) \left (-2 a^2 (6 A-5 i B)-a^2 (3 i A+5 B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{16 a^3 (6 A-5 i B)}{15 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (9 i A+5 B) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4}{15} \int \frac{-15 a^3 (i A+B)+15 a^3 (A-i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{16 a^3 (6 A-5 i B)}{15 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (9 i A+5 B) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{\left (120 a^6 (i A+B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-15 a^3 (i A+B)-15 a^3 (A-i B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{8 \sqrt [4]{-1} a^3 (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{16 a^3 (6 A-5 i B)}{15 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (9 i A+5 B) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [B]  time = 10.0278, size = 449, normalized size = 3.12 \[ \frac{\cos ^4(c+d x) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \left (\csc (c) \left (-\frac{2}{15} \cos (3 c)+\frac{2}{15} i \sin (3 c)\right ) \csc ^2(c+d x) (15 i A \sin (c)+3 A \cos (c)+5 B \sin (c))+\csc (c) \left (-\frac{6}{5} \cos (3 c)+\frac{6}{5} i \sin (3 c)\right ) \csc (c+d x) (7 A \sin (d x)-5 i B \sin (d x))+\csc (c) \left (\frac{2}{15} \cos (3 c)-\frac{2}{15} i \sin (3 c)\right ) (15 i A \sin (c)+63 A \cos (c)+5 B \sin (c)-45 i B \cos (c))+A \csc (c) \left (\frac{2}{5} \cos (3 c)-\frac{2}{5} i \sin (3 c)\right ) \sin (d x) \csc ^3(c+d x)\right )}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}-\frac{8 i e^{-3 i c} (A-i B) \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{d \sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

((-8*I)*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[Sqrt[(-1 + E^((2*I
)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*E^(
(3*I)*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x]
+ B*Sin[c + d*x])) + (Cos[c + d*x]^4*(Csc[c]*(63*A*Cos[c] - (45*I)*B*Cos[c] + (15*I)*A*Sin[c] + 5*B*Sin[c])*((
2*Cos[3*c])/15 - ((2*I)/15)*Sin[3*c]) + Csc[c]*Csc[c + d*x]^2*(3*A*Cos[c] + (15*I)*A*Sin[c] + 5*B*Sin[c])*((-2
*Cos[3*c])/15 + ((2*I)/15)*Sin[3*c]) + A*Csc[c]*Csc[c + d*x]^3*((2*Cos[3*c])/5 - ((2*I)/5)*Sin[3*c])*Sin[d*x]
+ Csc[c]*Csc[c + d*x]*((-6*Cos[3*c])/5 + ((6*I)/5)*Sin[3*c])*(7*A*Sin[d*x] - (5*I)*B*Sin[d*x]))*Sqrt[Tan[c + d
*x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d
*x]))

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Maple [B]  time = 0.017, size = 538, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x)

[Out]

-2/5/d*a^3*A/tan(d*x+c)^(5/2)-I/d*a^3*A*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)))-2/3/d*a^3/tan(d*x+c)^(3/2)*B-6*I/d*a^3/tan(d*x+c)^(1/2)*B+8/d*a^3*A/tan(d*x+c)^(1/2)-I/d
*a^3*B*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^(1/2)-2*I/d*a^3*A
*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-2*I/d*a^3*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-2/d*a^3*B*a
rctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-2/d*a^3*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/d*a^3*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*I/d*a^3*B*arctan(-1+
2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-2*I/d*a^3/tan(d*x+c)^(3/2)*A-2*I/d*a^3*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))
*2^(1/2)+1/d*a^3*A*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^(1/2)
+2/d*a^3*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+2/d*a^3*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)

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Maxima [A]  time = 2.28586, size = 262, normalized size = 1.82 \begin{align*} \frac{15 \,{\left (\sqrt{2}{\left (-\left (2 i - 2\right ) \, A - \left (2 i + 2\right ) \, B\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 (2 i - 2\right ) \, A - \left (2 i + 2\right ) \, B\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 (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt{2}{\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{3} + \frac{2 \,{\left ({\left (60 \, A - 45 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} + 5 \,{\left (-3 i \, A - B\right )} a^{3} \tan \left (d x + c\right ) - 3 \, A a^{3}\right )}}{\tan \left (d x + c\right )^{\frac{5}{2}}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/15*(15*(sqrt(2)*(-(2*I - 2)*A - (2*I + 2)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + sqrt(2)*
(-(2*I - 2)*A - (2*I + 2)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*(-(I + 1)*A + (I
- 1)*B)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(-sqrt(2)*sqr
t(tan(d*x + c)) + tan(d*x + c) + 1))*a^3 + 2*((60*A - 45*I*B)*a^3*tan(d*x + c)^2 + 5*(-3*I*A - B)*a^3*tan(d*x
+ c) - 3*A*a^3)/tan(d*x + c)^(5/2))/d

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Fricas [B]  time = 1.87915, size = 1366, normalized size = 9.49 \begin{align*} -\frac{15 \, \sqrt{\frac{{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac{{\left (8 \,{\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 15 \, \sqrt{\frac{{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac{{\left (8 \,{\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) -{\left ({\left (624 i \, A + 400 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (-288 i \, A - 320 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-528 i \, A - 400 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (384 i \, A + 320 \, B\right )} a^{3}\right )} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{60 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/60*(15*sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d
*e^(2*I*d*x + 2*I*c) - d)*log((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/
d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2
*I*c)/((4*I*A + 4*B)*a^3)) - 15*sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(
4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*log((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqrt((64*I*A^2 + 12
8*A*B - 64*I*B^2)*a^6/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c)
+ 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^3)) - ((624*I*A + 400*B)*a^3*e^(6*I*d*x + 6*I*c) + (-288*I*A - 32
0*B)*a^3*e^(4*I*d*x + 4*I*c) + (-528*I*A - 400*B)*a^3*e^(2*I*d*x + 2*I*c) + (384*I*A + 320*B)*a^3)*sqrt((-I*e^
(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2
*I*d*x + 2*I*c) - d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.36458, size = 146, normalized size = 1.01 \begin{align*} \frac{\left (i + 1\right ) \, \sqrt{2}{\left (-16 i \, A a^{3} - 16 \, B a^{3}\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{4 \, d} + \frac{120 \, A a^{3} \tan \left (d x + c\right )^{2} - 90 i \, B a^{3} \tan \left (d x + c\right )^{2} - 30 i \, A a^{3} \tan \left (d x + c\right ) - 10 \, B a^{3} \tan \left (d x + c\right ) - 6 \, A a^{3}}{15 \, d \tan \left (d x + c\right )^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

(1/4*I + 1/4)*sqrt(2)*(-16*I*A*a^3 - 16*B*a^3)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d + 1/15*(120
*A*a^3*tan(d*x + c)^2 - 90*I*B*a^3*tan(d*x + c)^2 - 30*I*A*a^3*tan(d*x + c) - 10*B*a^3*tan(d*x + c) - 6*A*a^3)
/(d*tan(d*x + c)^(5/2))