3.128 \(\int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=171 \[ -\frac{8 a^3 (21 A-23 i B) \tan ^{\frac{3}{2}}(c+d x)}{105 d}+\frac{8 \sqrt [4]{-1} a^3 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 (7 A-11 i B) \tan ^{\frac{3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac{8 a^3 (B+i A) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d} \]

[Out]

(8*(-1)^(1/4)*a^3*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (8*a^3*(I*A + B)*Sqrt[Tan[c + d*x]])/d
- (8*a^3*(21*A - (23*I)*B)*Tan[c + d*x]^(3/2))/(105*d) + (((2*I)/7)*a*B*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*
x])^2)/d - (2*(7*A - (11*I)*B)*Tan[c + d*x]^(3/2)*(a^3 + I*a^3*Tan[c + d*x]))/(35*d)

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Rubi [A]  time = 0.421445, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{8 a^3 (21 A-23 i B) \tan ^{\frac{3}{2}}(c+d x)}{105 d}+\frac{8 \sqrt [4]{-1} a^3 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 (7 A-11 i B) \tan ^{\frac{3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac{8 a^3 (B+i A) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(-1)^(1/4)*a^3*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (8*a^3*(I*A + B)*Sqrt[Tan[c + d*x]])/d
- (8*a^3*(21*A - (23*I)*B)*Tan[c + d*x]^(3/2))/(105*d) + (((2*I)/7)*a*B*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*
x])^2)/d - (2*(7*A - (11*I)*B)*Tan[c + d*x]^(3/2)*(a^3 + I*a^3*Tan[c + d*x]))/(35*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 \sqrt{\tan (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{3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}+\frac{2}{7} \int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2 \left (\frac{1}{2} a (7 A-3 i B)+\frac{1}{2} a (7 i A+11 B) \tan (c+d x)\right ) \, dx\\ &=\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac{2 (7 A-11 i B) \tan ^{\frac{3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac{4}{35} \int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x)) \left (2 a^2 (7 A-6 i B)+a^2 (21 i A+23 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{8 a^3 (21 A-23 i B) \tan ^{\frac{3}{2}}(c+d x)}{105 d}+\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac{2 (7 A-11 i B) \tan ^{\frac{3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac{4}{35} \int \sqrt{\tan (c+d x)} \left (35 a^3 (A-i B)+35 a^3 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac{8 a^3 (i A+B) \sqrt{\tan (c+d x)}}{d}-\frac{8 a^3 (21 A-23 i B) \tan ^{\frac{3}{2}}(c+d x)}{105 d}+\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac{2 (7 A-11 i B) \tan ^{\frac{3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac{4}{35} \int \frac{-35 a^3 (i A+B)+35 a^3 (A-i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{8 a^3 (i A+B) \sqrt{\tan (c+d x)}}{d}-\frac{8 a^3 (21 A-23 i B) \tan ^{\frac{3}{2}}(c+d x)}{105 d}+\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac{2 (7 A-11 i B) \tan ^{\frac{3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac{\left (280 a^6 (i A+B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-35 a^3 (i A+B)-35 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{8 a^3 (i A+B) \sqrt{\tan (c+d x)}}{d}-\frac{8 a^3 (21 A-23 i B) \tan ^{\frac{3}{2}}(c+d x)}{105 d}+\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac{2 (7 A-11 i B) \tan ^{\frac{3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}\\ \end{align*}

Mathematica [B]  time = 9.84344, size = 452, normalized size = 2.64 \[ \frac{\cos ^4(c+d x) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \left (\sec (c) \left (-\frac{2}{35} \sin (3 c)-\frac{2}{35} i \cos (3 c)\right ) \sec ^2(c+d x) (7 A \cos (c)+5 B \sin (c)-21 i B \cos (c))+\sec (c) \left (-\frac{2}{21} \cos (3 c)+\frac{2}{21} i \sin (3 c)\right ) \sec (c+d x) (21 A \sin (d x)-31 i B \sin (d x))+\sec (c) \left (\frac{2}{105} \cos (3 c)-\frac{2}{105} i \sin (3 c)\right ) (-105 A \sin (c)+441 i A \cos (c)+155 i B \sin (c)+483 B \cos (c))-i B \sec (c) \left (\frac{2}{7} \cos (3 c)-\frac{2}{7} i \sin (3 c)\right ) \sin (d x) \sec ^3(c+d x)\right )}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}-\frac{8 i e^{-3 i c} (A-i B) \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{d \sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),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*(Sec[c]*Sec[c + d*x]^2*(7*A*Cos[c] - (21*I)*B*Cos[c] + 5*B*Sin[c])*(((-2*
I)/35)*Cos[3*c] - (2*Sin[3*c])/35) + Sec[c]*((441*I)*A*Cos[c] + 483*B*Cos[c] - 105*A*Sin[c] + (155*I)*B*Sin[c]
)*((2*Cos[3*c])/105 - ((2*I)/105)*Sin[3*c]) - I*B*Sec[c]*Sec[c + d*x]^3*((2*Cos[3*c])/7 - ((2*I)/7)*Sin[3*c])*
Sin[d*x] + Sec[c]*Sec[c + d*x]*((-2*Cos[3*c])/21 + ((2*I)/21)*Sin[3*c])*(21*A*Sin[d*x] - (31*I)*B*Sin[d*x]))*S
qrt[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 = 574, normalized size = 3.4 \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)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

-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)))-2/7*I/d
*a^3*B*tan(d*x+c)^(7/2)-6/5/d*a^3*B*tan(d*x+c)^(5/2)-2/5*I/d*a^3*A*tan(d*x+c)^(5/2)-2/d*a^3*A*tan(d*x+c)^(3/2)
-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)+8*a^3*B
*tan(d*x+c)^(1/2)/d-2*I/d*a^3*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+8*I/d*a^3*A*tan(d*x+c)^(1/2)+8/3*I/
d*a^3*B*tan(d*x+c)^(3/2)-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)-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*B*arctan(-1+2^(1/2)*tan(d*x+c)^
(1/2))*2^(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*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 = 2.10206, size = 289, normalized size = 1.69 \begin{align*} -\frac{30 i \, B a^{3} \tan \left (d x + c\right )^{\frac{7}{2}} + 42 \,{\left (i \, A + 3 \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac{5}{2}} + 2 \,{\left (105 \, A - 140 i \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac{3}{2}} + 840 \,{\left (-i \, A - B\right )} a^{3} \sqrt{\tan \left (d x + c\right )} - 105 \,{\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}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(30*I*B*a^3*tan(d*x + c)^(7/2) + 42*(I*A + 3*B)*a^3*tan(d*x + c)^(5/2) + 2*(105*A - 140*I*B)*a^3*tan(d*
x + c)^(3/2) + 840*(-I*A - B)*a^3*sqrt(tan(d*x + c)) - 105*(sqrt(2)*(-(2*I - 2)*A - (2*I + 2)*B)*arctan(1/2*sq
rt(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)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))*a^3)/d

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Fricas [B]  time = 2.08939, size = 1381, normalized size = 8.08 \begin{align*} -\frac{105 \, \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 ) - 105 \, \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 (4368 i \, A + 5104 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (10752 i \, A + 10336 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (9072 i \, A + 8816 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (2688 i \, A + 2624 \, 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}}}{420 \,{\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(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/420*(105*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)) - 105*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)) - ((4368*I*A + 5104*B)*a^3*e^(6*I*d*x + 6*I*c) + (10752*I*
A + 10336*B)*a^3*e^(4*I*d*x + 4*I*c) + (9072*I*A + 8816*B)*a^3*e^(2*I*d*x + 2*I*c) + (2688*I*A + 2624*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(tan(d*x+c)**(1/2)*(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 = 1.27522, size = 217, normalized size = 1.27 \begin{align*} -\frac{\left (i - 1\right ) \, \sqrt{2}{\left (16 \, A a^{3} - 16 i \, 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{30 i \, B a^{3} d^{6} \tan \left (d x + c\right )^{\frac{7}{2}} + 42 i \, A a^{3} d^{6} \tan \left (d x + c\right )^{\frac{5}{2}} + 126 \, B a^{3} d^{6} \tan \left (d x + c\right )^{\frac{5}{2}} + 210 \, A a^{3} d^{6} \tan \left (d x + c\right )^{\frac{3}{2}} - 280 i \, B a^{3} d^{6} \tan \left (d x + c\right )^{\frac{3}{2}} - 840 i \, A a^{3} d^{6} \sqrt{\tan \left (d x + c\right )} - 840 \, B a^{3} d^{6} \sqrt{\tan \left (d x + c\right )}}{105 \, d^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(1/4*I - 1/4)*sqrt(2)*(16*A*a^3 - 16*I*B*a^3)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 1/105*(30
*I*B*a^3*d^6*tan(d*x + c)^(7/2) + 42*I*A*a^3*d^6*tan(d*x + c)^(5/2) + 126*B*a^3*d^6*tan(d*x + c)^(5/2) + 210*A
*a^3*d^6*tan(d*x + c)^(3/2) - 280*I*B*a^3*d^6*tan(d*x + c)^(3/2) - 840*I*A*a^3*d^6*sqrt(tan(d*x + c)) - 840*B*
a^3*d^6*sqrt(tan(d*x + c)))/d^7