3.974 \(\int \frac{(a+i a \tan (e+f x))^3}{\sqrt{c-i c \tan (e+f x)}} \, dx\)

Optimal. Leaf size=90 \[ \frac{2 i a^3 (c-i c \tan (e+f x))^{3/2}}{3 c^2 f}-\frac{8 i a^3 \sqrt{c-i c \tan (e+f x)}}{c f}-\frac{8 i a^3}{f \sqrt{c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.160996, antiderivative size = 90, 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} \[ \frac{2 i a^3 (c-i c \tan (e+f x))^{3/2}}{3 c^2 f}-\frac{8 i a^3 \sqrt{c-i c \tan (e+f x)}}{c f}-\frac{8 i a^3}{f \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 2.74785, size = 94, normalized size = 1.04 \[ \frac{2 a^3 \sec (e+f x) \sqrt{c-i c \tan (e+f x)} (-5 i \sin (2 (e+f x))+11 \cos (2 (e+f x))+12) (\sin (e+4 f x)-i \cos (e+4 f x))}{3 c f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*Sec[e + f*x]*(12 + 11*Cos[2*(e + f*x)] - (5*I)*Sin[2*(e + f*x)])*((-I)*Cos[e + 4*f*x] + Sin[e + 4*f*x])
*Sqrt[c - I*c*Tan[e + f*x]])/(3*c*f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]  time = 0.05, size = 66, normalized size = 0.7 \begin{align*}{\frac{2\,i{a}^{3}}{f{c}^{2}} \left ({\frac{1}{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-4\,c\sqrt{c-ic\tan \left ( fx+e \right ) }-4\,{\frac{{c}^{2}}{\sqrt{c-ic\tan \left ( fx+e \right ) }}} \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))^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.22833, size = 95, normalized size = 1.06 \begin{align*} -\frac{2 i \,{\left (\frac{12 \, a^{3} c}{\sqrt{-i \, c \tan \left (f x + e\right ) + c}} - \frac{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{3} - 12 \, \sqrt{-i \, c \tan \left (f x + e\right ) + c} a^{3} c}{c}\right )}}{3 \, c 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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.46089, size = 207, normalized size = 2.3 \begin{align*} \frac{\sqrt{2}{\left (-12 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 48 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 32 i \, a^{3}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \,{\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c 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))^(1/2),x, algorithm="fricas")

[Out]

1/3*sqrt(2)*(-12*I*a^3*e^(4*I*f*x + 4*I*e) - 48*I*a^3*e^(2*I*f*x + 2*I*e) - 32*I*a^3)*sqrt(c/(e^(2*I*f*x + 2*I
*e) + 1))/(c*f*e^(2*I*f*x + 2*I*e) + c*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int - \frac{3 \tan ^{2}{\left (e + f x \right )}}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, dx + \int \frac{3 i \tan{\left (e + f x \right )}}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, dx + \int - \frac{i \tan ^{3}{\left (e + f x \right )}}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, dx + \int \frac{1}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, 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))**(1/2),x)

[Out]

a**3*(Integral(-3*tan(e + f*x)**2/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(3*I*tan(e + f*x)/sqrt(-I*c*tan(e
+ f*x) + c), x) + Integral(-I*tan(e + f*x)**3/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(1/sqrt(-I*c*tan(e + f
*x) + c), x))

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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}}{\sqrt{-i \, c \tan \left (f x + e\right ) + c}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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