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

Optimal. Leaf size=66 \[ -\frac{4 (-1)^{3/4} a^2 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{3/2} f}-\frac{2 a^2}{d f \sqrt{d \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.110197, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3542, 3533, 205} \[ -\frac{4 (-1)^{3/4} a^2 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{3/2} f}-\frac{2 a^2}{d f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 1.45099, size = 147, normalized size = 2.23 \[ -\frac{2 a^2 e^{-2 i (e+f x)} (\cos (2 (e+f x))+i \sin (2 (e+f x))) \left (\sqrt{i \tan (e+f x)}-2 i \tan (e+f x) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{d f \sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.02, size = 371, normalized size = 5.6 \begin{align*} -2\,{\frac{{a}^{2}}{df\sqrt{d\tan \left ( fx+e \right ) }}}+{\frac{{\frac{i}{2}}{a}^{2}\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{i{a}^{2}\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{i{a}^{2}\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}^{2}\sqrt{2}}{2\,df}\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{{a}^{2}\sqrt{2}}{df}\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{{a}^{2}\sqrt{2}}{df}\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))^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.01644, size = 911, normalized size = 13.8 \begin{align*} -\frac{{\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt{\frac{16 i \, a^{4}}{d^{3} f^{2}}} \log \left (\frac{{\left (-4 i \, a^{2} 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{-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^{3} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) -{\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt{\frac{16 i \, a^{4}}{d^{3} f^{2}}} \log \left (\frac{{\left (-4 i \, a^{2} 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{-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^{3} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) -{\left (-8 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, 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}}}{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))^2/(d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/4*((d^2*f*e^(2*I*f*x + 2*I*e) - d^2*f)*sqrt(16*I*a^4/(d^3*f^2))*log(1/2*(-4*I*a^2*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((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1
6*I*a^4/(d^3*f^2)))*e^(-2*I*f*x - 2*I*e)/a^2) - (d^2*f*e^(2*I*f*x + 2*I*e) - d^2*f)*sqrt(16*I*a^4/(d^3*f^2))*l
og(1/2*(-4*I*a^2*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((-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^3*f^2)))*e^(-2*I*f*x - 2*I*e)/a^2) - (-8*I*a^2*e^(2*I*f*
x + 2*I*e) - 8*I*a^2)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d^2*f*e^(2*I*f*x + 2*
I*e) - d^2*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{3}{2}}}\, dx + \int - \frac{\tan ^{2}{\left (e + f x \right )}}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx + \int \frac{2 i \tan{\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))**2/(d*tan(f*x+e))**(3/2),x)

[Out]

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

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Giac [C]  time = 1.20306, size = 126, normalized size = 1.91 \begin{align*} \frac{\frac{4 i \, \sqrt{2} a^{2} \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^{2}}{\sqrt{d \tan \left (f x + e\right )} f}}{d} \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))^(3/2),x, algorithm="giac")

[Out]

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