3.761 \(\int \frac{(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=140 \[ \frac{2 a^3 (5 B+i A) \sqrt{c-i c \tan (e+f x)}}{c^2 f}+\frac{8 a^3 (2 B+i A)}{c f \sqrt{c-i c \tan (e+f x)}}-\frac{8 a^3 (B+i A)}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac{2 a^3 B (c-i c \tan (e+f x))^{3/2}}{3 c^3 f} \]

[Out]

(-8*a^3*(I*A + B))/(3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (8*a^3*(I*A + 2*B))/(c*f*Sqrt[c - I*c*Tan[e + f*x]]) +
(2*a^3*(I*A + 5*B)*Sqrt[c - I*c*Tan[e + f*x]])/(c^2*f) - (2*a^3*B*(c - I*c*Tan[e + f*x])^(3/2))/(3*c^3*f)

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Rubi [A]  time = 0.198587, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.047, Rules used = {3588, 77} \[ \frac{2 a^3 (5 B+i A) \sqrt{c-i c \tan (e+f x)}}{c^2 f}+\frac{8 a^3 (2 B+i A)}{c f \sqrt{c-i c \tan (e+f x)}}-\frac{8 a^3 (B+i A)}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac{2 a^3 B (c-i c \tan (e+f x))^{3/2}}{3 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^3*(I*A + B))/(3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (8*a^3*(I*A + 2*B))/(c*f*Sqrt[c - I*c*Tan[e + f*x]]) +
(2*a^3*(I*A + 5*B)*Sqrt[c - I*c*Tan[e + f*x]])/(c^2*f) - (2*a^3*B*(c - I*c*Tan[e + f*x])^(3/2))/(3*c^3*f)

Rule 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(a*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^2 (A+B x)}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{4 a^2 (A-i B)}{(c-i c x)^{5/2}}-\frac{4 a^2 (A-2 i B)}{c (c-i c x)^{3/2}}+\frac{a^2 (A-5 i B)}{c^2 \sqrt{c-i c x}}+\frac{i a^2 B \sqrt{c-i c x}}{c^3}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{8 a^3 (i A+B)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{8 a^3 (i A+2 B)}{c f \sqrt{c-i c \tan (e+f x)}}+\frac{2 a^3 (i A+5 B) \sqrt{c-i c \tan (e+f x)}}{c^2 f}-\frac{2 a^3 B (c-i c \tan (e+f x))^{3/2}}{3 c^3 f}\\ \end{align*}

Mathematica [A]  time = 12.1703, size = 168, normalized size = 1.2 \[ \frac{a^3 \sqrt{c-i c \tan (e+f x)} (\cos (2 e+5 f x)+i \sin (2 e+5 f x)) (A+B \tan (e+f x)) (15 (3 B+i A) \cos (e+f x)+(23 B+7 i A) \cos (3 (e+f x))+2 \sin (e+f x) ((9 A-25 i B) \cos (2 (e+f x))+9 A-26 i B))}{3 c^2 f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(15*(I*A + 3*B)*Cos[e + f*x] + ((7*I)*A + 23*B)*Cos[3*(e + f*x)] + 2*(9*A - (26*I)*B + (9*A - (25*I)*B)*C
os[2*(e + f*x)])*Sin[e + f*x])*(Cos[2*e + 5*f*x] + I*Sin[2*e + 5*f*x])*(A + B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e
+ f*x]])/(3*c^2*f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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Maple [A]  time = 0.076, size = 118, normalized size = 0.8 \begin{align*}{\frac{2\,i{a}^{3}}{f{c}^{3}} \left ({\frac{i}{3}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}-5\,iBc\sqrt{c-ic\tan \left ( fx+e \right ) }+Ac\sqrt{c-ic\tan \left ( fx+e \right ) }+4\,{\frac{{c}^{2} \left ( A-2\,iB \right ) }{\sqrt{c-ic\tan \left ( fx+e \right ) }}}-{\frac{4\,{c}^{3} \left ( A-iB \right ) }{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.11399, size = 146, normalized size = 1.04 \begin{align*} \frac{2 i \,{\left (\frac{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}{\left (12 \, A - 24 i \, B\right )} a^{3} -{\left (4 \, A - 4 i \, B\right )} a^{3} c}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}} + \frac{i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} B a^{3} + \sqrt{-i \, c \tan \left (f x + e\right ) + c}{\left (3 \, A - 15 i \, B\right )} a^{3} c}{c^{2}}\right )}}{3 \, c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*I*(((-I*c*tan(f*x + e) + c)*(12*A - 24*I*B)*a^3 - (4*A - 4*I*B)*a^3*c)/(-I*c*tan(f*x + e) + c)^(3/2) + (I*
(-I*c*tan(f*x + e) + c)^(3/2)*B*a^3 + sqrt(-I*c*tan(f*x + e) + c)*(3*A - 15*I*B)*a^3*c)/c^2)/(c*f)

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Fricas [A]  time = 1.18663, size = 309, normalized size = 2.21 \begin{align*} \frac{\sqrt{2}{\left ({\left (-2 i \, A - 2 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (6 i \, A + 18 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (24 i \, A + 72 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (16 i \, A + 48 \, B\right )} a^{3}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \,{\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/3*sqrt(2)*((-2*I*A - 2*B)*a^3*e^(6*I*f*x + 6*I*e) + (6*I*A + 18*B)*a^3*e^(4*I*f*x + 4*I*e) + (24*I*A + 72*B)
*a^3*e^(2*I*f*x + 2*I*e) + (16*I*A + 48*B)*a^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^2*f*e^(2*I*f*x + 2*I*e) +
c^2*f)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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