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

Optimal. Leaf size=198 \[ -\frac{16 a^3 (18 A-19 i B) \tan ^{\frac{5}{2}}(c+d x)}{315 d}+\frac{8 a^3 (B+i A) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 (9 A-13 i B) \tan ^{\frac{5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac{8 a^3 (A-i B) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d} \]

[Out]

(8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (8*a^3*(A - I*B)*Sqrt[Tan[c + d*x]])/d
+ (8*a^3*(I*A + B)*Tan[c + d*x]^(3/2))/(3*d) - (16*a^3*(18*A - (19*I)*B)*Tan[c + d*x]^(5/2))/(315*d) + (((2*I)
/9)*a*B*Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^2)/d - (2*(9*A - (13*I)*B)*Tan[c + d*x]^(5/2)*(a^3 + I*a^3*T
an[c + d*x]))/(63*d)

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Rubi [A]  time = 0.467445, antiderivative size = 198, 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} \[ -\frac{16 a^3 (18 A-19 i B) \tan ^{\frac{5}{2}}(c+d x)}{315 d}+\frac{8 a^3 (B+i A) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 (9 A-13 i B) \tan ^{\frac{5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac{8 a^3 (A-i B) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (8*a^3*(A - I*B)*Sqrt[Tan[c + d*x]])/d
+ (8*a^3*(I*A + B)*Tan[c + d*x]^(3/2))/(3*d) - (16*a^3*(18*A - (19*I)*B)*Tan[c + d*x]^(5/2))/(315*d) + (((2*I)
/9)*a*B*Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^2)/d - (2*(9*A - (13*I)*B)*Tan[c + d*x]^(5/2)*(a^3 + I*a^3*T
an[c + d*x]))/(63*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{3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\frac{2 i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}+\frac{2}{9} \int \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^2 \left (\frac{1}{2} a (9 A-5 i B)+\frac{1}{2} a (9 i A+13 B) \tan (c+d x)\right ) \, dx\\ &=\frac{2 i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac{2 (9 A-13 i B) \tan ^{\frac{5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac{4}{63} \int \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x)) \left (a^2 (27 A-25 i B)+2 a^2 (18 i A+19 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{16 a^3 (18 A-19 i B) \tan ^{\frac{5}{2}}(c+d x)}{315 d}+\frac{2 i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac{2 (9 A-13 i B) \tan ^{\frac{5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac{4}{63} \int \tan ^{\frac{3}{2}}(c+d x) \left (63 a^3 (A-i B)+63 a^3 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac{8 a^3 (i A+B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{16 a^3 (18 A-19 i B) \tan ^{\frac{5}{2}}(c+d x)}{315 d}+\frac{2 i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac{2 (9 A-13 i B) \tan ^{\frac{5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac{4}{63} \int \sqrt{\tan (c+d x)} \left (-63 a^3 (i A+B)+63 a^3 (A-i B) \tan (c+d x)\right ) \, dx\\ &=\frac{8 a^3 (A-i B) \sqrt{\tan (c+d x)}}{d}+\frac{8 a^3 (i A+B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{16 a^3 (18 A-19 i B) \tan ^{\frac{5}{2}}(c+d x)}{315 d}+\frac{2 i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac{2 (9 A-13 i B) \tan ^{\frac{5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac{4}{63} \int \frac{-63 a^3 (A-i B)-63 a^3 (i A+B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{8 a^3 (A-i B) \sqrt{\tan (c+d x)}}{d}+\frac{8 a^3 (i A+B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{16 a^3 (18 A-19 i B) \tan ^{\frac{5}{2}}(c+d x)}{315 d}+\frac{2 i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac{2 (9 A-13 i B) \tan ^{\frac{5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac{\left (504 a^6 (A-i B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-63 a^3 (A-i B)+63 a^3 (i A+B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{8 a^3 (A-i B) \sqrt{\tan (c+d x)}}{d}+\frac{8 a^3 (i A+B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{16 a^3 (18 A-19 i B) \tan ^{\frac{5}{2}}(c+d x)}{315 d}+\frac{2 i a B \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac{2 (9 A-13 i B) \tan ^{\frac{5}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}\\ \end{align*}

Mathematica [B]  time = 10.0018, size = 496, normalized size = 2.51 \[ \frac{\cos ^4(c+d x) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \left (\sec (c) \left (\frac{2}{7} \cos (3 c)-\frac{2}{7} i \sin (3 c)\right ) \sec ^3(c+d x) (-3 B \sin (d x)-i A \sin (d x))+\sec (c) \left (-\frac{2}{315} \cos (3 c)+\frac{2}{315} i \sin (3 c)\right ) \sec ^2(c+d x) (45 i A \sin (c)+189 A \cos (c)+135 B \sin (c)-322 i B \cos (c))+\sec (c) \left (\frac{2}{21} \cos (3 c)-\frac{2}{21} i \sin (3 c)\right ) \sec (c+d x) (37 B \sin (d x)+31 i A \sin (d x))+\sec (c) \left (\frac{2}{315} \cos (3 c)-\frac{2}{315} i \sin (3 c)\right ) (465 i A \sin (c)+1449 A \cos (c)+555 B \sin (c)-1547 i B \cos (c))+\left (-\frac{2}{9} B \sin (3 c)-\frac{2}{9} i B \cos (3 c)\right ) \sec ^4(c+d x)\right )}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}-\frac{8 e^{-3 i c} (A-i B) \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{d \sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-8*(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*(Sec[c]*(1449*A*Cos[c] - (1547*I)*B*Cos[c] + (465*I)*A*Sin[c] + 555*B*Sin[c])
*((2*Cos[3*c])/315 - ((2*I)/315)*Sin[3*c]) + Sec[c]*Sec[c + d*x]^2*(189*A*Cos[c] - (322*I)*B*Cos[c] + (45*I)*A
*Sin[c] + 135*B*Sin[c])*((-2*Cos[3*c])/315 + ((2*I)/315)*Sin[3*c]) + Sec[c + d*x]^4*(((-2*I)/9)*B*Cos[3*c] - (
2*B*Sin[3*c])/9) + Sec[c]*Sec[c + d*x]^3*((2*Cos[3*c])/7 - ((2*I)/7)*Sin[3*c])*((-I)*A*Sin[d*x] - 3*B*Sin[d*x]
) + Sec[c]*Sec[c + d*x]*((2*Cos[3*c])/21 - ((2*I)/21)*Sin[3*c])*((31*I)*A*Sin[d*x] + 37*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 = 0.014, size = 610, normalized size = 3.1 \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)^(3/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

-2*I/d*a^3*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+8/3*I/d*a^3*A*tan(d*x+c)^(3/2)-6/7/d*a^3*B*tan(d*x+c)^
(7/2)+2*I/d*a^3*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-6/5/d*a^3*A*tan(d*x+c)^(5/2)+2*I/d*a^3*B*arctan(-
1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+8/3/d*a^3*B*tan(d*x+c)^(3/2)-2/9*I/d*a^3*B*tan(d*x+c)^(9/2)+8*a^3*A*tan(d*
x+c)^(1/2)/d+I/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)))+8/5*I/d*a^3*B*tan(d*x+c)^(5/2)-I/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)*t
an(d*x+c)^(1/2))*2^(1/2)-1/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)))-2/7*I/d*a^3*A*tan(d*x+c)^(7/2)-8*I/d*a^3*B*tan(d*x+c)^(1/2)-2*I/d*a^3*A*arctan(-1+2^(1/2)*tan
(d*x+c)^(1/2))*2^(1/2)-1/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/d*a^3*B*arctan(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)

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Maxima [A]  time = 1.89566, size = 316, normalized size = 1.6 \begin{align*} -\frac{70 i \, B a^{3} \tan \left (d x + c\right )^{\frac{9}{2}} + 90 \,{\left (i \, A + 3 \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac{7}{2}} + 2 \,{\left (189 \, A - 252 i \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac{5}{2}} + 840 \,{\left (-i \, A - B\right )} a^{3} \tan \left (d x + c\right )^{\frac{3}{2}} - 2 \,{\left (1260 \, A - 1260 i \, B\right )} a^{3} \sqrt{\tan \left (d x + c\right )} - 315 \,{\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}}{315 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/315*(70*I*B*a^3*tan(d*x + c)^(9/2) + 90*(I*A + 3*B)*a^3*tan(d*x + c)^(7/2) + 2*(189*A - 252*I*B)*a^3*tan(d*
x + c)^(5/2) + 840*(-I*A - B)*a^3*tan(d*x + c)^(3/2) - 2*(1260*A - 1260*I*B)*a^3*sqrt(tan(d*x + c)) - 315*(sqr
t(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(sq
rt(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^3)/d

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Fricas [B]  time = 2.41392, size = 1569, normalized size = 7.92 \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)^(3/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/1260*(315*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/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((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)*(I*d*e^(2*I*d*x + 2*I*c) + I*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)) - 315*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/
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((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)*(-I*d*e^(2*I*d*x +
2*I*c) - I*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)) - 16*((957*A - 1051*I*B)*a^3*e^(8*I*d*x + 8*I*c) + 5*(579*A - 547*I*B)*a^3*e^(6*I*d*x + 6*I*c) + 21*(1
71*A - 173*I*B)*a^3*e^(4*I*d*x + 4*I*c) + 5*(429*A - 433*I*B)*a^3*e^(2*I*d*x + 2*I*c) + 4*(123*A - 124*I*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^(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)**(3/2)*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.34052, size = 262, normalized size = 1.32 \begin{align*} \frac{\left (4 i - 4\right ) \, \sqrt{2}{\left (i \, A a^{3} + 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 )}{d} - \frac{70 i \, B a^{3} d^{8} \tan \left (d x + c\right )^{\frac{9}{2}} + 90 i \, A a^{3} d^{8} \tan \left (d x + c\right )^{\frac{7}{2}} + 270 \, B a^{3} d^{8} \tan \left (d x + c\right )^{\frac{7}{2}} + 378 \, A a^{3} d^{8} \tan \left (d x + c\right )^{\frac{5}{2}} - 504 i \, B a^{3} d^{8} \tan \left (d x + c\right )^{\frac{5}{2}} - 840 i \, A a^{3} d^{8} \tan \left (d x + c\right )^{\frac{3}{2}} - 840 \, B a^{3} d^{8} \tan \left (d x + c\right )^{\frac{3}{2}} - 2520 \, A a^{3} d^{8} \sqrt{\tan \left (d x + c\right )} + 2520 i \, B a^{3} 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)^(3/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

(4*I - 4)*sqrt(2)*(I*A*a^3 + B*a^3)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 1/315*(70*I*B*a^3*d^
8*tan(d*x + c)^(9/2) + 90*I*A*a^3*d^8*tan(d*x + c)^(7/2) + 270*B*a^3*d^8*tan(d*x + c)^(7/2) + 378*A*a^3*d^8*ta
n(d*x + c)^(5/2) - 504*I*B*a^3*d^8*tan(d*x + c)^(5/2) - 840*I*A*a^3*d^8*tan(d*x + c)^(3/2) - 840*B*a^3*d^8*tan
(d*x + c)^(3/2) - 2520*A*a^3*d^8*sqrt(tan(d*x + c)) + 2520*I*B*a^3*d^8*sqrt(tan(d*x + c)))/d^9