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

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

[Out]

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

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Rubi [A]  time = 0.169365, 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} \[ \frac{2 i a^3 (c-i c \tan (e+f x))^{7/2}}{7 c^2 f}-\frac{8 i a^3 (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac{8 i a^3 (c-i c \tan (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 3.0481, size = 94, normalized size = 1. \[ \frac{2 a^3 c \sec ^3(e+f x) \sqrt{c-i c \tan (e+f x)} (\sin (e-2 f x)+i \cos (e-2 f x)) (27 i \sin (2 (e+f x))+43 \cos (2 (e+f x))+28)}{105 f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*c*Sec[e + f*x]^3*(I*Cos[e - 2*f*x] + Sin[e - 2*f*x])*(28 + 43*Cos[2*(e + f*x)] + (27*I)*Sin[2*(e + f*x)
])*Sqrt[c - I*c*Tan[e + f*x]])/(105*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}{7} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}}-{\frac{4\,c}{5} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{4\,{c}^{2}}{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*(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.47507, size = 90, normalized size = 0.96 \begin{align*} \frac{2 i \,{\left (15 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}} a^{3} - 84 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} a^{3} c + 140 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{3} c^{2}\right )}}{105 \, 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))^(3/2),x, algorithm="maxima")

[Out]

2/105*I*(15*(-I*c*tan(f*x + e) + c)^(7/2)*a^3 - 84*(-I*c*tan(f*x + e) + c)^(5/2)*a^3*c + 140*(-I*c*tan(f*x + e
) + c)^(3/2)*a^3*c^2)/(c^2*f)

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Fricas [A]  time = 1.4764, size = 285, normalized size = 3.03 \begin{align*} \frac{\sqrt{2}{\left (560 i \, a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} + 448 i \, a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} + 128 i \, a^{3} c\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{105 \,{\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))^3*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/105*sqrt(2)*(560*I*a^3*c*e^(4*I*f*x + 4*I*e) + 448*I*a^3*c*e^(2*I*f*x + 2*I*e) + 128*I*a^3*c)*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]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int c \sqrt{- i c \tan{\left (e + f x \right )} + c}\, dx + \int - c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\, dx + \int 2 i c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan{\left (e + f x \right )}\, dx + \int 2 i c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\, dx\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))**(3/2),x)

[Out]

a**3*(Integral(c*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4, x)
+ Integral(2*I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(2*I*c*sqrt(-I*c*tan(e + f*x) + c)*ta
n(e + f*x)**3, x))

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

[Out]

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