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

Optimal. Leaf size=62 \[ \frac{2 a (B+i A) (c-i c \tan (e+f x))^{5/2}}{5 f}-\frac{2 a B (c-i c \tan (e+f x))^{7/2}}{7 c f} \]

[Out]

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

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Rubi [A]  time = 0.107791, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.049, Rules used = {3588, 43} \[ \frac{2 a (B+i A) (c-i c \tan (e+f x))^{5/2}}{5 f}-\frac{2 a B (c-i c \tan (e+f x))^{7/2}}{7 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(I*A + B)*(c - I*c*Tan[e + f*x])^(5/2))/(5*f) - (2*a*B*(c - I*c*Tan[e + f*x])^(7/2))/(7*c*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 4.41823, size = 88, normalized size = 1.42 \[ \frac{2 a c^2 \sec ^2(e+f x) (\cos (f x)-i \sin (f x)) \sqrt{c-i c \tan (e+f x)} (\sin (2 e+f x)+i \cos (2 e+f x)) (7 A+5 B \tan (e+f x)-2 i B)}{35 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*c^2*Sec[e + f*x]^2*(Cos[f*x] - I*Sin[f*x])*(I*Cos[2*e + f*x] + Sin[2*e + f*x])*(7*A - (2*I)*B + 5*B*Tan[e
+ f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(35*f)

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Maple [A]  time = 0.059, size = 55, normalized size = 0.9 \begin{align*}{\frac{2\,ia}{cf} \left ({\frac{i}{7}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}+{\frac{-iBc+Ac}{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))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.50115, size = 66, normalized size = 1.06 \begin{align*} \frac{2 i \,{\left (5 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}} B a +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}{\left (7 \, A - 7 i \, B\right )} a c\right )}}{35 \, c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.40985, size = 265, normalized size = 4.27 \begin{align*} \frac{\sqrt{2}{\left ({\left (56 i \, A + 56 \, B\right )} a c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (56 i \, A - 24 \, B\right )} a c^{2}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{35 \,{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, 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))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/35*sqrt(2)*((56*I*A + 56*B)*a*c^2*e^(2*I*f*x + 2*I*e) + (56*I*A - 24*B)*a*c^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) +
1))/(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*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))*(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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