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

Optimal. Leaf size=92 \[ -\frac{2 i a^3}{c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{8 i a^3}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{8 i a^3}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.163778, antiderivative size = 92, 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} \[ -\frac{2 i a^3}{c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{8 i a^3}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{8 i a^3}{5 f (c-i c \tan (e+f x))^{5/2}} \]

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

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 \frac{(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)}{(c-i c \tan (e+f x))^{11/2}} \, dx\\ &=\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int \frac{(c-x)^2}{(c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int \left (\frac{4 c^2}{(c+x)^{7/2}}-\frac{4 c}{(c+x)^{5/2}}+\frac{1}{(c+x)^{3/2}}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=-\frac{8 i a^3}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{8 i a^3}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i a^3}{c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 8.25086, size = 98, normalized size = 1.07 \[ \frac{2 a^3 \cos (e+f x) \sqrt{c-i c \tan (e+f x)} (-5 i \sin (2 (e+f x))+11 \cos (2 (e+f x))-4) (\sin (3 (e+2 f x))-i \cos (3 (e+2 f x)))}{15 c^3 f (\cos (f x)+i \sin (f x))^3} \]

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*Cos[e + f*x]*(-4 + 11*Cos[2*(e + f*x)] - (5*I)*Sin[2*(e + f*x)])*((-I)*Cos[3*(e + 2*f*x)] + Sin[3*(e +
2*f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(15*c^3*f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]  time = 0.024, size = 66, normalized size = 0.7 \begin{align*}{\frac{2\,i{a}^{3}}{f{c}^{2}} \left ( -{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}+{\frac{4\,c}{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{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/(c-I*c*tan(f*x+e))^(1/2)+4/3*c/(c-I*c*tan(f*x+e))^(3/2)-4/5*c^2/(c-I*c*tan(f*x+e))^(5/2))

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Maxima [A]  time = 1.07673, size = 88, normalized size = 0.96 \begin{align*} -\frac{2 i \,{\left (15 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} - 20 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c + 12 \, a^{3} c^{2}\right )}}{15 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} 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]

-2/15*I*(15*(-I*c*tan(f*x + e) + c)^2*a^3 - 20*(-I*c*tan(f*x + e) + c)*a^3*c + 12*a^3*c^2)/((-I*c*tan(f*x + e)
+ c)^(5/2)*c^2*f)

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Fricas [A]  time = 1.35031, size = 209, normalized size = 2.27 \begin{align*} \frac{\sqrt{2}{\left (-3 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 4 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{3}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, c^{3} 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="fricas")

[Out]

1/15*sqrt(2)*(-3*I*a^3*e^(6*I*f*x + 6*I*e) + I*a^3*e^(4*I*f*x + 4*I*e) - 4*I*a^3*e^(2*I*f*x + 2*I*e) - 8*I*a^3
)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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)