3.749 \(\int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=105 \[ -\frac{2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{7/2}}{7 c f}+\frac{4 a^2 (B+i A) (c-i c \tan (e+f x))^{5/2}}{5 f}+\frac{2 a^2 B (c-i c \tan (e+f x))^{9/2}}{9 c^2 f} \]

[Out]

(4*a^2*(I*A + B)*(c - I*c*Tan[e + f*x])^(5/2))/(5*f) - (2*a^2*(I*A + 3*B)*(c - I*c*Tan[e + f*x])^(7/2))/(7*c*f
) + (2*a^2*B*(c - I*c*Tan[e + f*x])^(9/2))/(9*c^2*f)

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Rubi [A]  time = 0.180887, antiderivative size = 105, 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} \[ -\frac{2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{7/2}}{7 c f}+\frac{4 a^2 (B+i A) (c-i c \tan (e+f x))^{5/2}}{5 f}+\frac{2 a^2 B (c-i c \tan (e+f x))^{9/2}}{9 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(I*A + B)*(c - I*c*Tan[e + f*x])^(5/2))/(5*f) - (2*a^2*(I*A + 3*B)*(c - I*c*Tan[e + f*x])^(7/2))/(7*c*f
) + (2*a^2*B*(c - I*c*Tan[e + f*x])^(9/2))/(9*c^2*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 (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int (a+i a x) (A+B x) (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (2 a (A-i B) (c-i c x)^{3/2}-\frac{a (A-3 i B) (c-i c x)^{5/2}}{c}-\frac{i a B (c-i c x)^{7/2}}{c^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{4 a^2 (i A+B) (c-i c \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (i A+3 B) (c-i c \tan (e+f x))^{7/2}}{7 c f}+\frac{2 a^2 B (c-i c \tan (e+f x))^{9/2}}{9 c^2 f}\\ \end{align*}

Mathematica [A]  time = 7.13579, size = 112, normalized size = 1.07 \[ \frac{a^2 c^2 (\sin (2 e)+i \cos (2 e)) \sec ^4(e+f x) \sqrt{c-i c \tan (e+f x)} (5 (13 B+9 i A) \sin (2 (e+f x))+(81 A-61 i B) \cos (2 (e+f x))+81 A+9 i B)}{315 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c^2*Sec[e + f*x]^4*(I*Cos[2*e] + Sin[2*e])*(81*A + (9*I)*B + (81*A - (61*I)*B)*Cos[2*(e + f*x)] + 5*((9*I
)*A + 13*B)*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(315*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [A]  time = 0.07, size = 83, normalized size = 0.8 \begin{align*}{\frac{-2\,i{a}^{2}}{f{c}^{2}} \left ({\frac{i}{9}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{9}{2}}}+{\frac{-3\,iBc+Ac}{7} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}}-{\frac{ \left ( -2\,iBc+2\,Ac \right ) c}{5} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

-2*I/f*a^2/c^2*(1/9*I*B*(c-I*c*tan(f*x+e))^(9/2)+1/7*(-3*I*B*c+A*c)*(c-I*c*tan(f*x+e))^(7/2)-2/5*(-I*B*c+A*c)*
c*(c-I*c*tan(f*x+e))^(5/2))

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Maxima [A]  time = 1.19194, size = 109, normalized size = 1.04 \begin{align*} -\frac{2 i \,{\left (35 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{9}{2}} B a^{2} +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}}{\left (45 \, A - 135 i \, B\right )} a^{2} c -{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}{\left (126 \, A - 126 i \, B\right )} a^{2} c^{2}\right )}}{315 \, c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-2/315*I*(35*I*(-I*c*tan(f*x + e) + c)^(9/2)*B*a^2 + (-I*c*tan(f*x + e) + c)^(7/2)*(45*A - 135*I*B)*a^2*c - (-
I*c*tan(f*x + e) + c)^(5/2)*(126*A - 126*I*B)*a^2*c^2)/(c^2*f)

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Fricas [A]  time = 1.61934, size = 379, normalized size = 3.61 \begin{align*} \frac{\sqrt{2}{\left ({\left (1008 i \, A + 1008 \, B\right )} a^{2} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (1296 i \, A - 144 \, B\right )} a^{2} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (288 i \, A - 32 \, B\right )} a^{2} c^{2}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{315 \,{\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/315*sqrt(2)*((1008*I*A + 1008*B)*a^2*c^2*e^(4*I*f*x + 4*I*e) + (1296*I*A - 144*B)*a^2*c^2*e^(2*I*f*x + 2*I*e
) + (288*I*A - 32*B)*a^2*c^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*
e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)

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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((a+I*a*tan(f*x+e))**2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [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((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out