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

Optimal. Leaf size=93 \[ \frac{4 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{5/2} f}-\frac{4 i a^2}{d^2 f \sqrt{d \tan (e+f x)}}-\frac{2 a^2}{3 d f (d \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.15475, antiderivative size = 93, 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 = {3542, 3529, 3533, 205} \[ \frac{4 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{5/2} f}-\frac{4 i a^2}{d^2 f \sqrt{d \tan (e+f x)}}-\frac{2 a^2}{3 d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3542

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

Rule 3529

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

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

Mathematica [A]  time = 1.96359, size = 87, normalized size = 0.94 \[ -\frac{2 a^2 \left (\cot (e+f x)-6 i \sqrt{i \tan (e+f x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+6 i\right )}{3 d^2 f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(6*I + Cot[e + f*x] - (6*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 = 0.022, size = 393, normalized size = 4.2 \begin{align*} -{\frac{2\,{a}^{2}}{3\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{4\,i{a}^{2}}{{d}^{2}f}{\frac{1}{\sqrt{d\tan \left ( fx+e \right ) }}}}-{\frac{{a}^{2}\sqrt{2}}{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 ) }-{\frac{{a}^{2}\sqrt{2}}{f{d}^{3}}\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}^{2}\sqrt{2}}{f{d}^{3}}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{{\frac{i}{2}}{a}^{2}\sqrt{2}}{{d}^{2}f}\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{i{a}^{2}\sqrt{2}}{{d}^{2}f}\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{i{a}^{2}\sqrt{2}}{{d}^{2}f}\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))^2/(d*tan(f*x+e))^(5/2),x)

[Out]

-2/3*a^2/d/f/(d*tan(f*x+e))^(3/2)-4*I*a^2/d^2/f/(d*tan(f*x+e))^(1/2)-1/2/f*a^2/d^3*(d^2)^(1/4)*2^(1/2)*ln((d*t
an(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)))-1/f*a^2/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)+1/f*
a^2/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)-1/2*I/f*a^2/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)))-I/f*a^2/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)+I/f*a^2/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))^2/(d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.13909, size = 1067, normalized size = 11.47 \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{16 i \, a^{4}}{d^{5} f^{2}}} \log \left (\frac{{\left (-4 i \, a^{2} 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{16 i \, a^{4}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\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{16 i \, a^{4}}{d^{5} f^{2}}} \log \left (\frac{{\left (-4 i \, a^{2} 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{16 i \, a^{4}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 8 \,{\left (7 \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 5 \, a^{2}\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))^2/(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(-16*I*a^4/(d^5*f^2))*log(1/2*(
-4*I*a^2*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(-16*I*a^4/(d^5*f^2)))*e^(-2*I*f*x - 2*I*e)/a^2) - 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(-16*I*a^4/(d^5*f^2))*log(1/2*(-4*I*a^2*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(-16*I*a^4
/(d^5*f^2)))*e^(-2*I*f*x - 2*I*e)/a^2) - 8*(7*a^2*e^(4*I*f*x + 4*I*e) + 2*a^2*e^(2*I*f*x + 2*I*e) - 5*a^2)*sqr
t((-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^{2} \left (\int \frac{1}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx + \int - \frac{\tan ^{2}{\left (e + f x \right )}}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx + \int \frac{2 i \tan{\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))**2/(d*tan(f*x+e))**(5/2),x)

[Out]

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

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Giac [A]  time = 1.22296, size = 158, normalized size = 1.7 \begin{align*} -\frac{4 i \, \sqrt{2} a^{2} \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 (6 i \, a^{2} d \tan \left (f x + e\right ) + a^{2} 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))^2/(d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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