\(\int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx\) [113]

Optimal result

Integrand size = 34, antiderivative size = 105 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 \sqrt [4]{-1} a (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 a (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x)}{5 d} \]

[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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3673, 3609, 3614, 211} \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 \sqrt [4]{-1} a (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 a (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x)}{5 d} \]

[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 211

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

Rule 3609

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 3614

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 3673

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*} \text {integral}& = \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 ) \text {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) \arctan \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*}

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.85 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 a \left (15 \sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\sqrt {\tan (c+d x)} \left (15 (A-i B)+5 (i A+B) \tan (c+d x)+3 i B \tan ^2(c+d x)\right )\right )}{15 d} \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (85 ) = 170\).

Time = 0.04 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.39

method	result	size
derivativedivides	\(\frac {a \left (\frac {2 i B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 i A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {2 B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-2 i B \left (\sqrt {\tan }\left (d x +c \right )\right )+2 A \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (i B -A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-i A -B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(251\)
default	\(\frac {a \left (\frac {2 i B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 i A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {2 B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-2 i B \left (\sqrt {\tan }\left (d x +c \right )\right )+2 A \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (i B -A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-i A -B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(251\)
parts	\(\frac {\left (i a A +B a \right ) \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {a A \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {i a B \left (\frac {2 \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-2 \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(327\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 432 vs. \(2 (81) = 162\).

Time = 0.26 (sec) , antiderivative size = 432, normalized size of antiderivative = 4.11 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 4 \, {\left ({\left (20 \, A - 23 i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, {\left (5 \, A - 4 i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (10 \, A - 13 i \, B\right )} a\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{30 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[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]

-1/30*(15*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(I*A^2 + 2*A*B - 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(-(I*A^2 + 2*A*B - 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(-(I*A^2 + 2*A*B - 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(-(I*A^2 + 2*A*B - 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)) - 4*((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)

Sympy [F]

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

[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))

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (81) = 162\).

Time = 0.31 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.77 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {-24 i \, B a \tan \left (d x + c\right )^{\frac {5}{2}} + 40 \, {\left (-i \, A - B\right )} a \tan \left (d x + c\right )^{\frac {3}{2}} - 120 \, {\left (A - i \, B\right )} a \sqrt {\tan \left (d x + c\right )} - 15 \, {\left (2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a}{60 \, d} \]

[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) + 40*(-I*A - 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)*l
og(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

Giac [A] (verification not implemented)

none

Time = 0.56 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {\left (i - 1\right ) \, \sqrt {2} {\left (i \, A a + B a\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} - \frac {2 \, {\left (-3 i \, B a d^{4} \tan \left (d x + c\right )^{\frac {5}{2}} - 5 i \, A a d^{4} \tan \left (d x + c\right )^{\frac {3}{2}} - 5 \, B a d^{4} \tan \left (d x + c\right )^{\frac {3}{2}} - 15 \, A a d^{4} \sqrt {\tan \left (d x + c\right )} + 15 i \, B a d^{4} \sqrt {\tan \left (d x + c\right )}\right )}}{15 \, d^{5}} \]

[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]

(I - 1)*sqrt(2)*(I*A*a + B*a)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 2/15*(-3*I*B*a*d^4*tan(d*x
+ c)^(5/2) - 5*I*A*a*d^4*tan(d*x + c)^(3/2) - 5*B*a*d^4*tan(d*x + c)^(3/2) - 15*A*a*d^4*sqrt(tan(d*x + c)) +
15*I*B*a*d^4*sqrt(tan(d*x + c)))/d^5

Mupad [B] (verification not implemented)

Time = 10.24 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.24 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2\,A\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {A\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}-\frac {B\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{d}+\frac {2\,B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}+\frac {B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,2{}\mathrm {i}}{5\,d}+\frac {\sqrt {2}\,A\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-1-\mathrm {i}\right )}{d}-\frac {{\left (-1\right )}^{1/4}\,B\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d} \]

[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