3.748 \(\int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=105 \[ -\frac{2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{9/2}}{9 c f}+\frac{4 a^2 (B+i A) (c-i c \tan (e+f x))^{7/2}}{7 f}+\frac{2 a^2 B (c-i c \tan (e+f x))^{11/2}}{11 c^2 f} \]

[Out]

(4*a^2*(I*A + B)*(c - I*c*Tan[e + f*x])^(7/2))/(7*f) - (2*a^2*(I*A + 3*B)*(c - I*c*Tan[e + f*x])^(9/2))/(9*c*f
) + (2*a^2*B*(c - I*c*Tan[e + f*x])^(11/2))/(11*c^2*f)

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Rubi [A]  time = 0.183202, 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} \[ -\frac{2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{9/2}}{9 c f}+\frac{4 a^2 (B+i A) (c-i c \tan (e+f x))^{7/2}}{7 f}+\frac{2 a^2 B (c-i c \tan (e+f x))^{11/2}}{11 c^2 f} \]

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])^(7/2),x]

[Out]

(4*a^2*(I*A + B)*(c - I*c*Tan[e + f*x])^(7/2))/(7*f) - (2*a^2*(I*A + 3*B)*(c - I*c*Tan[e + f*x])^(9/2))/(9*c*f
) + (2*a^2*B*(c - I*c*Tan[e + f*x])^(11/2))/(11*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))^{7/2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int (a+i a x) (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 (2 a (A-i B) (c-i c x)^{5/2}-\frac{a (A-3 i B) (c-i c x)^{7/2}}{c}-\frac{i a B (c-i c x)^{9/2}}{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))^{7/2}}{7 f}-\frac{2 a^2 (i A+3 B) (c-i c \tan (e+f x))^{9/2}}{9 c f}+\frac{2 a^2 B (c-i c \tan (e+f x))^{11/2}}{11 c^2 f}\\ \end{align*}

Mathematica [A]  time = 11.1353, size = 119, normalized size = 1.13 \[ \frac{a^2 c^3 \sec ^5(e+f x) \sqrt{c-i c \tan (e+f x)} (\cos (3 e+f x)-i \sin (3 e+f x)) ((-77 A+105 i B) \sin (2 (e+f x))+(93 B+121 i A) \cos (2 (e+f x))+121 i A-33 B)}{693 f (\cos (f x)+i \sin (f x))^2} \]

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])^(7/2),x]

[Out]

(a^2*c^3*Sec[e + f*x]^5*((121*I)*A - 33*B + ((121*I)*A + 93*B)*Cos[2*(e + f*x)] + (-77*A + (105*I)*B)*Sin[2*(e
+ f*x)])*(Cos[3*e + f*x] - I*Sin[3*e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(693*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [A]  time = 0.067, size = 83, normalized size = 0.8 \begin{align*}{\frac{-2\,i{a}^{2}}{f{c}^{2}} \left ({\frac{i}{11}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{11}{2}}}+{\frac{-3\,iBc+Ac}{9} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{9}{2}}}}-{\frac{ \left ( -2\,iBc+2\,Ac \right ) c}{7} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{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))^(7/2),x)

[Out]

-2*I/f*a^2/c^2*(1/11*I*B*(c-I*c*tan(f*x+e))^(11/2)+1/9*(-3*I*B*c+A*c)*(c-I*c*tan(f*x+e))^(9/2)-2/7*(-I*B*c+A*c
)*c*(c-I*c*tan(f*x+e))^(7/2))

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Maxima [A]  time = 1.12907, size = 109, normalized size = 1.04 \begin{align*} -\frac{2 i \,{\left (63 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{11}{2}} B a^{2} +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{9}{2}}{\left (77 \, A - 231 i \, B\right )} a^{2} c -{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}}{\left (198 \, A - 198 i \, B\right )} a^{2} c^{2}\right )}}{693 \, 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))^(7/2),x, algorithm="maxima")

[Out]

-2/693*I*(63*I*(-I*c*tan(f*x + e) + c)^(11/2)*B*a^2 + (-I*c*tan(f*x + e) + c)^(9/2)*(77*A - 231*I*B)*a^2*c - (
-I*c*tan(f*x + e) + c)^(7/2)*(198*A - 198*I*B)*a^2*c^2)/(c^2*f)

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Fricas [A]  time = 2.336, size = 423, normalized size = 4.03 \begin{align*} \frac{\sqrt{2}{\left ({\left (3168 i \, A + 3168 \, B\right )} a^{2} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (3872 i \, A - 1056 \, B\right )} a^{2} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (704 i \, A - 192 \, B\right )} a^{2} c^{3}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{693 \,{\left (f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, 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))^(7/2),x, algorithm="fricas")

[Out]

1/693*sqrt(2)*((3168*I*A + 3168*B)*a^2*c^3*e^(4*I*f*x + 4*I*e) + (3872*I*A - 1056*B)*a^2*c^3*e^(2*I*f*x + 2*I*
e) + (704*I*A - 192*B)*a^2*c^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(10*I*f*x + 10*I*e) + 5*f*e^(8*I*f*x +
8*I*e) + 10*f*e^(6*I*f*x + 6*I*e) + 10*f*e^(4*I*f*x + 4*I*e) + 5*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))**(7/2),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))^(7/2),x, algorithm="giac")

[Out]

Timed out