3.160 \(\int \frac{(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=80 \[ -\frac{8 (-1)^{3/4} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{3/2} f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt{d \tan (e+f x)}} \]

[Out]

(-8*(-1)^(3/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(3/2)*f) - (2*(a^3 + I*a^3*Tan[e + f*
x]))/(d*f*Sqrt[d*Tan[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.111515, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3553, 12, 3533, 205} \[ -\frac{8 (-1)^{3/4} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{3/2} f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(-1)^(3/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(3/2)*f) - (2*(a^3 + I*a^3*Tan[e + f*
x]))/(d*f*Sqrt[d*Tan[e + f*x]])

Rule 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(a^2*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] +
Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(m
- 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; Fr
eeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 2.29109, size = 156, normalized size = 1.95 \[ \frac{2 a^3 e^{-3 i (e+f x)} (\sin (3 (e+f x))-i \cos (3 (e+f x))) \left (\sqrt{i \tan (e+f x)} (\tan (e+f x)-i)-4 \tan (e+f x) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{d f \sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*((-I)*Cos[3*(e + f*x)] + Sin[3*(e + f*x)])*(-4*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e
+ f*x)))]]*Tan[e + f*x] + Sqrt[I*Tan[e + f*x]]*(-I + Tan[e + f*x])))/(d*E^((3*I)*(e + f*x))*Sqrt[(-1 + E^((2*
I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]*f*Sqrt[d*Tan[e + f*x]])

________________________________________________________________________________________

Maple [B]  time = 0.023, size = 394, normalized size = 4.9 \begin{align*}{\frac{-2\,i{a}^{3}}{f{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }}-2\,{\frac{{a}^{3}}{fd\sqrt{d\tan \left ( fx+e \right ) }}}+{\frac{i{a}^{3}\sqrt{2}}{f{d}^{2}}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }+{\frac{2\,i{a}^{3}\sqrt{2}}{f{d}^{2}}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{2\,i{a}^{3}\sqrt{2}}{f{d}^{2}}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{{a}^{3}\sqrt{2}}{fd}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-2\,{\frac{{a}^{3}\sqrt{2}}{fd\sqrt [4]{{d}^{2}}}\arctan \left ({\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }+2\,{\frac{{a}^{3}\sqrt{2}}{fd\sqrt [4]{{d}^{2}}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(3/2),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.39304, size = 895, normalized size = 11.19 \begin{align*} \frac{-16 i \, a^{3} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (2 i \, f x + 2 i \, e\right )} -{\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt{\frac{64 i \, a^{6}}{d^{3} f^{2}}} \log \left (\frac{{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d^{2} f\right )} \sqrt{\frac{64 i \, a^{6}}{d^{3} f^{2}}} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) +{\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt{\frac{64 i \, a^{6}}{d^{3} f^{2}}} \log \left (\frac{{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-i \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, d^{2} f\right )} \sqrt{\frac{64 i \, a^{6}}{d^{3} f^{2}}} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right )}{4 \,{\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/4*(-16*I*a^3*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*e^(2*I*f*x + 2*I*e) - (d^2*f*e
^(2*I*f*x + 2*I*e) - d^2*f)*sqrt(64*I*a^6/(d^3*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (I*d^2*f*e^(2*I
*f*x + 2*I*e) + I*d^2*f)*sqrt(64*I*a^6/(d^3*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) +
1)))*e^(-2*I*f*x - 2*I*e)/a^3) + (d^2*f*e^(2*I*f*x + 2*I*e) - d^2*f)*sqrt(64*I*a^6/(d^3*f^2))*log(1/4*(-8*I*a
^3*d*e^(2*I*f*x + 2*I*e) + (-I*d^2*f*e^(2*I*f*x + 2*I*e) - I*d^2*f)*sqrt(64*I*a^6/(d^3*f^2))*sqrt((-I*d*e^(2*I
*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/a^3))/(d^2*f*e^(2*I*f*x + 2*I*e) - d^2*f
)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**3/(d*tan(f*x+e))**(3/2),x)

[Out]

a**3*(Integral((d*tan(e + f*x))**(-3/2), x) + Integral(-3*tan(e + f*x)**2/(d*tan(e + f*x))**(3/2), x) + Integr
al(3*I*tan(e + f*x)/(d*tan(e + f*x))**(3/2), x) + Integral(-I*tan(e + f*x)**3/(d*tan(e + f*x))**(3/2), x))

________________________________________________________________________________________

Giac [A]  time = 1.27003, size = 155, normalized size = 1.94 \begin{align*} \frac{\frac{8 i \, \sqrt{2} a^{3} \arctan \left (-\frac{16 i \, \sqrt{d^{2}} \sqrt{d \tan \left (f x + e\right )}}{-8 i \, \sqrt{2} d^{\frac{3}{2}} + 8 \, \sqrt{2} \sqrt{d^{2}} \sqrt{d}}\right )}{\sqrt{d} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{2 \, a^{3}}{\sqrt{d \tan \left (f x + e\right )} f} - \frac{2 i \, \sqrt{d \tan \left (f x + e\right )} a^{3}}{d f}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

(8*I*sqrt(2)*a^3*arctan(-16*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(
d)))/(sqrt(d)*f*(-I*d/sqrt(d^2) + 1)) - 2*a^3/(sqrt(d*tan(f*x + e))*f) - 2*I*sqrt(d*tan(f*x + e))*a^3/(d*f))/d