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

Optimal. Leaf size=62 \[ \frac{4 i a^2 (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac{2 i a^2 (c-i c \tan (e+f x))^{5/2}}{5 c f} \]

[Out]

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

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Rubi [A]  time = 0.150155, antiderivative size = 62, 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} \[ \frac{4 i a^2 (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac{2 i a^2 (c-i c \tan (e+f x))^{5/2}}{5 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 2.1536, size = 80, normalized size = 1.29 \[ -\frac{2 a^2 c (3 \tan (e+f x)-7 i) \sec (e+f x) \sqrt{c-i c \tan (e+f x)} (\cos (e-f x)-i \sin (e-f x))}{15 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*c*Sec[e + f*x]*(Cos[e - f*x] - I*Sin[e - f*x])*(-7*I + 3*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(15
*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [A]  time = 0.018, size = 47, normalized size = 0.8 \begin{align*}{\frac{-2\,i{a}^{2}}{cf} \left ({\frac{1}{5} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{2\,c}{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))^2*(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.3811, size = 62, normalized size = 1. \begin{align*} -\frac{2 i \,{\left (3 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} a^{2} - 10 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{2} c\right )}}{15 \, c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*I*(3*(-I*c*tan(f*x + e) + c)^(5/2)*a^2 - 10*(-I*c*tan(f*x + e) + c)^(3/2)*a^2*c)/(c*f)

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Fricas [A]  time = 1.4106, size = 200, normalized size = 3.23 \begin{align*} \frac{\sqrt{2}{\left (40 i \, a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} + 16 i \, a^{2} c\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \,{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, 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))^2*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/15*sqrt(2)*(40*I*a^2*c*e^(2*I*f*x + 2*I*e) + 16*I*a^2*c)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(4*I*f*x + 4
*I*e) + 2*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^{2} \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 ^{2}{\left (e + f x \right )}\, dx + \int i c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan{\left (e + f x \right )}\, dx + \int 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))**2*(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

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

[Out]

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