3.968 \(\int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=94 \[ \frac{2 i a^3 (c-i c \tan (e+f x))^{9/2}}{9 c^2 f}-\frac{8 i a^3 (c-i c \tan (e+f x))^{7/2}}{7 c f}+\frac{8 i a^3 (c-i c \tan (e+f x))^{5/2}}{5 f} \]

[Out]

(((8*I)/5)*a^3*(c - I*c*Tan[e + f*x])^(5/2))/f - (((8*I)/7)*a^3*(c - I*c*Tan[e + f*x])^(7/2))/(c*f) + (((2*I)/
9)*a^3*(c - I*c*Tan[e + f*x])^(9/2))/(c^2*f)

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Rubi [A]  time = 0.157665, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3522, 3487, 43} \[ \frac{2 i a^3 (c-i c \tan (e+f x))^{9/2}}{9 c^2 f}-\frac{8 i a^3 (c-i c \tan (e+f x))^{7/2}}{7 c f}+\frac{8 i a^3 (c-i c \tan (e+f x))^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(((8*I)/5)*a^3*(c - I*c*Tan[e + f*x])^(5/2))/f - (((8*I)/7)*a^3*(c - I*c*Tan[e + f*x])^(7/2))/(c*f) + (((2*I)/
9)*a^3*(c - I*c*Tan[e + f*x])^(9/2))/(c^2*f)

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3487

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

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))^3 (c-i c \tan (e+f x))^{5/2} \, dx &=\left (a^3 c^3\right ) \int \frac{\sec ^6(e+f x)}{\sqrt{c-i c \tan (e+f x)}} \, dx\\ &=\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int (c-x)^2 (c+x)^{3/2} \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int \left (4 c^2 (c+x)^{3/2}-4 c (c+x)^{5/2}+(c+x)^{7/2}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=\frac{8 i a^3 (c-i c \tan (e+f x))^{5/2}}{5 f}-\frac{8 i a^3 (c-i c \tan (e+f x))^{7/2}}{7 c f}+\frac{2 i a^3 (c-i c \tan (e+f x))^{9/2}}{9 c^2 f}\\ \end{align*}

Mathematica [A]  time = 4.47043, size = 100, normalized size = 1.06 \[ \frac{2 a^3 c^2 \sec ^4(e+f x) \sqrt{c-i c \tan (e+f x)} (\sin (2 e-f x)+i \cos (2 e-f x)) (55 i \sin (2 (e+f x))+71 \cos (2 (e+f x))+36)}{315 f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(2*a^3*c^2*Sec[e + f*x]^4*(I*Cos[2*e - f*x] + Sin[2*e - f*x])*(36 + 71*Cos[2*(e + f*x)] + (55*I)*Sin[2*(e + f*
x)])*Sqrt[c - I*c*Tan[e + f*x]])/(315*f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]  time = 0.019, size = 66, normalized size = 0.7 \begin{align*}{\frac{2\,i{a}^{3}}{f{c}^{2}} \left ({\frac{1}{9} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{9}{2}}}}-{\frac{4\,c}{7} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}}+{\frac{4\,{c}^{2}}{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))^3*(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

2*I/f*a^3/c^2*(1/9*(c-I*c*tan(f*x+e))^(9/2)-4/7*c*(c-I*c*tan(f*x+e))^(7/2)+4/5*c^2*(c-I*c*tan(f*x+e))^(5/2))

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Maxima [A]  time = 1.34322, size = 90, normalized size = 0.96 \begin{align*} \frac{2 i \,{\left (35 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{9}{2}} a^{3} - 180 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}} a^{3} c + 252 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} a^{3} 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))^3*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

2/315*I*(35*(-I*c*tan(f*x + e) + c)^(9/2)*a^3 - 180*(-I*c*tan(f*x + e) + c)^(7/2)*a^3*c + 252*(-I*c*tan(f*x +
e) + c)^(5/2)*a^3*c^2)/(c^2*f)

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Fricas [A]  time = 1.5681, size = 331, normalized size = 3.52 \begin{align*} \frac{\sqrt{2}{\left (2016 i \, a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 1152 i \, a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 256 i \, a^{3} 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))^3*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/315*sqrt(2)*(2016*I*a^3*c^2*e^(4*I*f*x + 4*I*e) + 1152*I*a^3*c^2*e^(2*I*f*x + 2*I*e) + 256*I*a^3*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))**3*(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 (i \, a \tan \left (f x + e\right ) + a\right )}^{3}{\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))^3*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^3*(-I*c*tan(f*x + e) + c)^(5/2), x)