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

Optimal. Leaf size=105 \[ \frac {2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 a (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3592, 3528, 3533, 205} \[ \frac {2 a (B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [B]  time = 3.10, size = 266, normalized size = 2.53 \[ \frac {\cos ^2(c+d x) (\cos (d x)-i \sin (d x)) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \left (\frac {2 e^{-i c} (B+i A) \sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+\frac {1}{15} (\cos (c)-i \sin (c)) \sqrt {\tan (c+d x)} \sec ^2(c+d x) (5 (B+i A) \sin (2 (c+d x))+3 (5 A-6 i B) \cos (2 (c+d x))+3 (5 A-4 i B))\right )}{d (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^2*(Cos[d*x] - I*Sin[d*x])*((2*(I*A + B)*Sqrt[(-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^(I*c)*Sqrt[((-I)*(-1 + E^((2*I)*(c
+ d*x))))/(1 + E^((2*I)*(c + d*x)))]) + (Sec[c + d*x]^2*(Cos[c] - I*Sin[c])*(3*(5*A - (4*I)*B) + 3*(5*A - (6*I
)*B)*Cos[2*(c + d*x)] + 5*(I*A + B)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]])/15)*(a + I*a*Tan[c + d*x])*(A + B*Ta
n[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [B]  time = 0.63, size = 428, normalized size = 4.08 \[ -\frac {15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 8 \, {\left ({\left (20 \, A - 23 i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, {\left (5 \, A - 4 i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (10 \, A - 13 i \, B\right )} a\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{60 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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))*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(15*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-4*I*A^2 - 8*A*B + 4*I*B^2)*a^2/d^2)*log
((2*(A - I*B)*a*e^(2*I*d*x + 2*I*c) + (I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((-4*I*A^2 - 8*A*B + 4*I*B^2)*a^2/d^
2)*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)/((I*A + B)*a)) - 15*(d*e
^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-4*I*A^2 - 8*A*B + 4*I*B^2)*a^2/d^2)*log((2*(A - I*B)*
a*e^(2*I*d*x + 2*I*c) + (-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt((-4*I*A^2 - 8*A*B + 4*I*B^2)*a^2/d^2)*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)/((I*A + B)*a)) - 8*((20*A - 23*I*B)*a
*e^(4*I*d*x + 4*I*c) + 6*(5*A - 4*I*B)*a*e^(2*I*d*x + 2*I*c) + (10*A - 13*I*B)*a)*sqrt((-I*e^(2*I*d*x + 2*I*c)
+ I)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )} \tan \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

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))*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.10, size = 506, normalized size = 4.82 \[ -\frac {i a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {i a A \sqrt {2}\, \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 )}{4 d}+\frac {2 a B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {2 i a A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {2 a A \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}+\frac {i a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {i a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {i a B \sqrt {2}\, \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 )}{4 d}-\frac {a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a A \sqrt {2}\, \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 )}{4 d}-\frac {2 i a B \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}+\frac {2 i a B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5 d}+\frac {i a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a B \sqrt {2}\, \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 )}{4 d} \]

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))*(A+B*tan(d*x+c)),x)

[Out]

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

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maxima [B]  time = 0.81, size = 186, normalized size = 1.77 \[ -\frac {-24 i \, B a \tan \left (d x + c\right )^{\frac {5}{2}} - 8 \, {\left (5 i \, A + 5 \, B\right )} a \tan \left (d x + c\right )^{\frac {3}{2}} - 120 \, {\left (A - i \, B\right )} a \sqrt {\tan \left (d x + c\right )} - 15 \, {\left (2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a}{60 \, d} \]

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))*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(-24*I*B*a*tan(d*x + c)^(5/2) - 8*(5*I*A + 5*B)*a*tan(d*x + c)^(3/2) - 120*(A - I*B)*a*sqrt(tan(d*x + c)
) - 15*(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)/d

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mupad [B]  time = 8.40, size = 130, normalized size = 1.24 \[ \frac {2\,A\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {A\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}-\frac {B\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{d}+\frac {2\,B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}+\frac {B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,2{}\mathrm {i}}{5\,d}+\frac {\sqrt {2}\,A\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-1-\mathrm {i}\right )}{d}-\frac {{\left (-1\right )}^{1/4}\,B\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(3/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i),x)

[Out]

(2*A*a*tan(c + d*x)^(1/2))/d + (A*a*tan(c + d*x)^(3/2)*2i)/(3*d) - (B*a*tan(c + d*x)^(1/2)*2i)/d + (2*B*a*tan(
c + d*x)^(3/2))/(3*d) + (B*a*tan(c + d*x)^(5/2)*2i)/(5*d) - (2^(1/2)*A*a*atan(2^(1/2)*tan(c + d*x)^(1/2)*(1/2
- 1i/2))*(1 + 1i))/d - ((-1)^(1/4)*B*a*atan((-1)^(1/4)*tan(c + d*x)^(1/2)*1i)*2i)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int A \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx + \int B \tan ^{\frac {7}{2}}{\left (c + d x \right )}\, dx + \int \left (- i A \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int \left (- i B \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx\right ) \]

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))*(A+B*tan(d*x+c)),x)

[Out]

I*a*(Integral(A*tan(c + d*x)**(5/2), x) + Integral(B*tan(c + d*x)**(7/2), x) + Integral(-I*A*tan(c + d*x)**(3/
2), x) + Integral(-I*B*tan(c + d*x)**(5/2), x))

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