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

Optimal. Leaf size=109 \[ \frac{8 \sqrt [4]{-1} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{5/2} f}-\frac{16 i a^3}{3 d^2 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}} \]

[Out]

(8*(-1)^(1/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(5/2)*f) - (((16*I)/3)*a^3)/(d^2*f*Sqr
t[d*Tan[e + f*x]]) - (2*(a^3 + I*a^3*Tan[e + f*x]))/(3*d*f*(d*Tan[e + f*x])^(3/2))

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Rubi [A]  time = 0.20379, antiderivative size = 109, 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, 3591, 3533, 205} \[ \frac{8 \sqrt [4]{-1} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{5/2} f}-\frac{16 i a^3}{3 d^2 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(5/2),x]

[Out]

(8*(-1)^(1/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(5/2)*f) - (((16*I)/3)*a^3)/(d^2*f*Sqr
t[d*Tan[e + f*x]]) - (2*(a^3 + I*a^3*Tan[e + f*x]))/(3*d*f*(d*Tan[e + f*x])^(3/2))

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 3591

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2
+ b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& LtQ[m, -1] && NeQ[a^2 + b^2, 0]

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))^{5/2}} \, dx &=-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}-\frac{2 \int \frac{(a+i a \tan (e+f x)) \left (-4 i a^2 d+2 a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{3/2}} \, dx}{3 d^2}\\ &=-\frac{16 i a^3}{3 d^2 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}-\frac{2 \int \frac{6 a^3 d^2+6 i a^3 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{3 d^4}\\ &=-\frac{16 i a^3}{3 d^2 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}-\frac{\left (48 a^6\right ) \operatorname{Subst}\left (\int \frac{1}{6 a^3 d^3-6 i a^3 d^2 x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=\frac{8 \sqrt [4]{-1} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{5/2} f}-\frac{16 i a^3}{3 d^2 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 3.30157, size = 87, normalized size = 0.8 \[ -\frac{2 a^3 \left (\cot (e+f x)-12 i \sqrt{i \tan (e+f x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+9 i\right )}{3 d^2 f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(5/2),x]

[Out]

(-2*a^3*(9*I + Cot[e + f*x] - (12*I)*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]]*Sqrt[
I*Tan[e + f*x]]))/(3*d^2*f*Sqrt[d*Tan[e + f*x]])

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Maple [B]  time = 0.026, size = 394, normalized size = 3.6 \begin{align*}{\frac{-6\,i{a}^{3}}{f{d}^{2}}{\frac{1}{\sqrt{d\tan \left ( fx+e \right ) }}}}-{\frac{2\,{a}^{3}}{3\,fd} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{{a}^{3}\sqrt{2}}{f{d}^{3}}\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 ) }-2\,{\frac{{a}^{3}\sqrt [4]{{d}^{2}}\sqrt{2}}{f{d}^{3}}\arctan \left ({\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }+2\,{\frac{{a}^{3}\sqrt [4]{{d}^{2}}\sqrt{2}}{f{d}^{3}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }-{\frac{i{a}^{3}\sqrt{2}}{f{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{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{2\,i{a}^{3}\sqrt{2}}{f{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{2\,i{a}^{3}\sqrt{2}}{f{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}} \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))^(5/2),x)

[Out]

-6*I/f*a^3/d^2/(d*tan(f*x+e))^(1/2)-2/3/f*a^3/d/(d*tan(f*x+e))^(3/2)-1/f*a^3/d^3*(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^3*(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^3*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-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)

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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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.35729, size = 1065, normalized size = 9.77 \begin{align*} -\frac{3 \,{\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt{-\frac{64 i \, a^{6}}{d^{5} f^{2}}} \log \left (\frac{{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \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}} \sqrt{-\frac{64 i \, a^{6}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 3 \,{\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt{-\frac{64 i \, a^{6}}{d^{5} f^{2}}} \log \left (\frac{{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} -{\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \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}} \sqrt{-\frac{64 i \, a^{6}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 16 \,{\left (5 \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, a^{3}\right )} \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}}}{12 \,{\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

-1/12*(3*(d^3*f*e^(4*I*f*x + 4*I*e) - 2*d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f)*sqrt(-64*I*a^6/(d^5*f^2))*log(1/4*(
-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^
(2*I*f*x + 2*I*e) + 1))*sqrt(-64*I*a^6/(d^5*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) - 3*(d^3*f*e^(4*I*f*x + 4*I*e) -
2*d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f)*sqrt(-64*I*a^6/(d^5*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) - (d^3*
f*e^(2*I*f*x + 2*I*e) + d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-64*I*a^6
/(d^5*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) - 16*(5*a^3*e^(4*I*f*x + 4*I*e) + a^3*e^(2*I*f*x + 2*I*e) - 4*a^3)*sqrt
((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d^3*f*e^(4*I*f*x + 4*I*e) - 2*d^3*f*e^(2*I*f*x
+ 2*I*e) + d^3*f)

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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{5}{2}}}\, dx + \int - \frac{3 \tan ^{2}{\left (e + f x \right )}}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx + \int \frac{3 i \tan{\left (e + f x \right )}}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx + \int - \frac{i \tan ^{3}{\left (e + f x \right )}}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{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))**(5/2),x)

[Out]

a**3*(Integral((d*tan(e + f*x))**(-5/2), x) + Integral(-3*tan(e + f*x)**2/(d*tan(e + f*x))**(5/2), x) + Integr
al(3*I*tan(e + f*x)/(d*tan(e + f*x))**(5/2), x) + Integral(-I*tan(e + f*x)**3/(d*tan(e + f*x))**(5/2), x))

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Giac [A]  time = 1.2821, size = 158, normalized size = 1.45 \begin{align*} -\frac{8 i \, \sqrt{2} a^{3} \arctan \left (\frac{16 \, \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 )}{d^{\frac{5}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{2 \,{\left (9 i \, a^{3} d \tan \left (f x + e\right ) + a^{3} d\right )}}{3 \, \sqrt{d \tan \left (f x + e\right )} d^{3} f \tan \left (f x + e\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))^(5/2),x, algorithm="giac")

[Out]

-8*I*sqrt(2)*a^3*arctan(16*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(d)))
/(d^(5/2)*f*(I*d/sqrt(d^2) + 1)) - 2/3*(9*I*a^3*d*tan(f*x + e) + a^3*d)/(sqrt(d*tan(f*x + e))*d^3*f*tan(f*x +
e))