3.119 \(\int \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=183 \[ -\frac{2 a^2 (9 A-11 i B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{4 a^2 (B+i A) \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{4 a^2 (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{4 a^2 (B+i A) \sqrt{\tan (c+d x)}}{d}+\frac{2 i B \tan ^{\frac{7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d} \]

[Out]

(-4*(-1)^(1/4)*a^2*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (4*a^2*(I*A + B)*Sqrt[Tan[c + d*x]])/d
+ (4*a^2*(A - I*B)*Tan[c + d*x]^(3/2))/(3*d) + (4*a^2*(I*A + B)*Tan[c + d*x]^(5/2))/(5*d) - (2*a^2*(9*A - (11
*I)*B)*Tan[c + d*x]^(7/2))/(63*d) + (((2*I)/9)*B*Tan[c + d*x]^(7/2)*(a^2 + I*a^2*Tan[c + d*x]))/d

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Rubi [A]  time = 0.3568, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {3594, 3592, 3528, 3533, 205} \[ -\frac{2 a^2 (9 A-11 i B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{4 a^2 (B+i A) \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{4 a^2 (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{4 a^2 (B+i A) \sqrt{\tan (c+d x)}}{d}+\frac{2 i B \tan ^{\frac{7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

(-4*(-1)^(1/4)*a^2*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (4*a^2*(I*A + B)*Sqrt[Tan[c + d*x]])/d
+ (4*a^2*(A - I*B)*Tan[c + d*x]^(3/2))/(3*d) + (4*a^2*(I*A + B)*Tan[c + d*x]^(5/2))/(5*d) - (2*a^2*(9*A - (11
*I)*B)*Tan[c + d*x]^(7/2))/(63*d) + (((2*I)/9)*B*Tan[c + d*x]^(7/2)*(a^2 + I*a^2*Tan[c + d*x]))/d

Rule 3594

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

Rule 3592

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 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 \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\frac{2 i B \tan ^{\frac{7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac{2}{9} \int \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x)) \left (\frac{1}{2} a (9 A-7 i B)+\frac{1}{2} a (9 i A+11 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{2 a^2 (9 A-11 i B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 i B \tan ^{\frac{7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac{2}{9} \int \tan ^{\frac{5}{2}}(c+d x) \left (9 a^2 (A-i B)+9 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac{4 a^2 (i A+B) \tan ^{\frac{5}{2}}(c+d x)}{5 d}-\frac{2 a^2 (9 A-11 i B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 i B \tan ^{\frac{7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac{2}{9} \int \tan ^{\frac{3}{2}}(c+d x) \left (-9 a^2 (i A+B)+9 a^2 (A-i B) \tan (c+d x)\right ) \, dx\\ &=\frac{4 a^2 (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a^2 (i A+B) \tan ^{\frac{5}{2}}(c+d x)}{5 d}-\frac{2 a^2 (9 A-11 i B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 i B \tan ^{\frac{7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac{2}{9} \int \sqrt{\tan (c+d x)} \left (-9 a^2 (A-i B)-9 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=-\frac{4 a^2 (i A+B) \sqrt{\tan (c+d x)}}{d}+\frac{4 a^2 (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a^2 (i A+B) \tan ^{\frac{5}{2}}(c+d x)}{5 d}-\frac{2 a^2 (9 A-11 i B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 i B \tan ^{\frac{7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac{2}{9} \int \frac{9 a^2 (i A+B)-9 a^2 (A-i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{4 a^2 (i A+B) \sqrt{\tan (c+d x)}}{d}+\frac{4 a^2 (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a^2 (i A+B) \tan ^{\frac{5}{2}}(c+d x)}{5 d}-\frac{2 a^2 (9 A-11 i B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 i B \tan ^{\frac{7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac{\left (36 a^4 (i A+B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{9 a^2 (i A+B)+9 a^2 (A-i B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{4 \sqrt [4]{-1} a^2 (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{4 a^2 (i A+B) \sqrt{\tan (c+d x)}}{d}+\frac{4 a^2 (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a^2 (i A+B) \tan ^{\frac{5}{2}}(c+d x)}{5 d}-\frac{2 a^2 (9 A-11 i B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 i B \tan ^{\frac{7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}\\ \end{align*}

Mathematica [A]  time = 6.12168, size = 315, normalized size = 1.72 \[ \frac{\cos ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left (\frac{4 e^{-2 i c} (B+i A) \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac{i (\cos (2 c)-i \sin (2 c)) \sqrt{\tan (c+d x)} \sec ^4(c+d x) (30 (8 B+11 i A) \sin (2 (c+d x))+15 (20 B+17 i A) \sin (4 (c+d x))+140 (18 A-17 i B) \cos (2 (c+d x))+(756 A-791 i B) \cos (4 (c+d x))+21 (84 A-89 i B))}{1260}\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

(Cos[c + d*x]^3*((4*(I*A + 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)))]])/(E^((2*I)*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2
*I)*(c + d*x)))]) - (I/1260)*Sec[c + d*x]^4*(Cos[2*c] - I*Sin[2*c])*(21*(84*A - (89*I)*B) + 140*(18*A - (17*I)
*B)*Cos[2*(c + d*x)] + (756*A - (791*I)*B)*Cos[4*(c + d*x)] + 30*((11*I)*A + 8*B)*Sin[2*(c + d*x)] + 15*((17*I
)*A + 20*B)*Sin[4*(c + d*x)])*Sqrt[Tan[c + d*x]])*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/(d*(Cos[d*x]
+ I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B]  time = 0.015, size = 607, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

-2/9/d*a^2*B*tan(d*x+c)^(9/2)+1/2*I/d*a^2*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)-2/7/d*a^2*A*tan(d*x+c)^(7/2)+I/d*a^2*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+4/
5/d*a^2*B*tan(d*x+c)^(5/2)+I/d*a^2*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+4/3/d*a^2*A*tan(d*x+c)^(3/2)+4
/7*I/d*a^2*B*tan(d*x+c)^(7/2)-4/d*a^2*B*tan(d*x+c)^(1/2)+4/5*I/d*a^2*A*tan(d*x+c)^(5/2)-4/3*I/d*a^2*B*tan(d*x+
c)^(3/2)+1/2*I/d*a^2*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)))+1/d*a^2*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+1/2/d*a^2*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)))+1/d*a^2*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+I/d*
a^2*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+I/d*a^2*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-4*I/d*
a^2*A*tan(d*x+c)^(1/2)-1/2/d*a^2*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)-1/d*a^2*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/d*a^2*A*2^(1/2)*arctan(1+2^(1/2)*tan(
d*x+c)^(1/2))

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Maxima [A]  time = 1.84791, size = 316, normalized size = 1.73 \begin{align*} -\frac{140 \, B a^{2} \tan \left (d x + c\right )^{\frac{9}{2}} + 4 \,{\left (45 \, A - 90 i \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac{7}{2}} + 504 \,{\left (-i \, A - B\right )} a^{2} \tan \left (d x + c\right )^{\frac{5}{2}} - 4 \,{\left (210 \, A - 210 i \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac{3}{2}} + 2520 \,{\left (i \, A + B\right )} a^{2} \sqrt{\tan \left (d x + c\right )} - 315 \,{\left (2 \, \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\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 (i - 1\right ) \, A + \left (i + 1\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^{2}}{630 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/630*(140*B*a^2*tan(d*x + c)^(9/2) + 4*(45*A - 90*I*B)*a^2*tan(d*x + c)^(7/2) + 504*(-I*A - B)*a^2*tan(d*x +
c)^(5/2) - 4*(210*A - 210*I*B)*a^2*tan(d*x + c)^(3/2) + 2520*(I*A + B)*a^2*sqrt(tan(d*x + c)) - 315*(2*sqrt(2
)*((I - 1)*A + (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*((I - 1)*A + (I + 1
)*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)*sqrt(tan(d*x + c)) + tan(d*
x + c) + 1))*a^2)/d

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Fricas [B]  time = 2.65759, size = 1555, normalized size = 8.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/1260*(315*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*
d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log((4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + sqrt((16*I*A^2
+ 32*A*B - 16*I*B^2)*a^4/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)/((2*I*A + 2*B)*a^2)) - 315*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(
8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log((4*(A
- I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/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)/((2*I*A + 2*B)*a^2)) + ((-8088*
I*A - 8728*B)*a^2*e^(8*I*d*x + 8*I*c) + (-22800*I*A - 20960*B)*a^2*e^(6*I*d*x + 6*I*c) + (-28224*I*A - 29904*B
)*a^2*e^(4*I*d*x + 4*I*c) + (-17520*I*A - 17120*B)*a^2*e^(2*I*d*x + 2*I*c) + (-4008*I*A - 3928*B)*a^2)*sqrt((-
I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*
e^(4*I*d*x + 4*I*c) + 4*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(tan(d*x+c)**(5/2)*(a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.28495, size = 263, normalized size = 1.44 \begin{align*} \frac{\left (i - 1\right ) \, \sqrt{2}{\left (8 \, A a^{2} - 8 i \, B a^{2}\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{70 \, B a^{2} d^{8} \tan \left (d x + c\right )^{\frac{9}{2}} + 90 \, A a^{2} d^{8} \tan \left (d x + c\right )^{\frac{7}{2}} - 180 i \, B a^{2} d^{8} \tan \left (d x + c\right )^{\frac{7}{2}} - 252 i \, A a^{2} d^{8} \tan \left (d x + c\right )^{\frac{5}{2}} - 252 \, B a^{2} d^{8} \tan \left (d x + c\right )^{\frac{5}{2}} - 420 \, A a^{2} d^{8} \tan \left (d x + c\right )^{\frac{3}{2}} + 420 i \, B a^{2} d^{8} \tan \left (d x + c\right )^{\frac{3}{2}} + 1260 i \, A a^{2} d^{8} \sqrt{\tan \left (d x + c\right )} + 1260 \, B a^{2} d^{8} \sqrt{\tan \left (d x + c\right )}}{315 \, d^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

(1/4*I - 1/4)*sqrt(2)*(8*A*a^2 - 8*I*B*a^2)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 1/315*(70*B*
a^2*d^8*tan(d*x + c)^(9/2) + 90*A*a^2*d^8*tan(d*x + c)^(7/2) - 180*I*B*a^2*d^8*tan(d*x + c)^(7/2) - 252*I*A*a^
2*d^8*tan(d*x + c)^(5/2) - 252*B*a^2*d^8*tan(d*x + c)^(5/2) - 420*A*a^2*d^8*tan(d*x + c)^(3/2) + 420*I*B*a^2*d
^8*tan(d*x + c)^(3/2) + 1260*I*A*a^2*d^8*sqrt(tan(d*x + c)) + 1260*B*a^2*d^8*sqrt(tan(d*x + c)))/d^9