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

Optimal. Leaf size=62 \[ \frac{2 (-1)^{3/4} a \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{3/2} f}-\frac{2 a}{d f \sqrt{d \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0777917, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3529, 3533, 208} \[ \frac{2 (-1)^{3/4} a \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{3/2} f}-\frac{2 a}{d f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 208

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

Rubi steps

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

Mathematica [C]  time = 0.109179, size = 39, normalized size = 0.63 \[ -\frac{2 a \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-i \tan (e+f x)\right )}{d f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*Hypergeometric2F1[-1/2, 1, 1/2, (-I)*Tan[e + f*x]])/(d*f*Sqrt[d*Tan[e + f*x]])

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Maple [C]  time = 0.019, size = 358, normalized size = 5.8 \begin{align*} -2\,{\frac{a}{df\sqrt{d\tan \left ( fx+e \right ) }}}-{\frac{{\frac{i}{4}}a\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{{\frac{i}{2}}a\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{{\frac{i}{2}}a\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\sqrt{2}}{4\,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\sqrt{2}}{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\sqrt{2}}{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))/(d*tan(f*x+e))^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.37847, size = 805, normalized size = 12.98 \begin{align*} \frac{{\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt{-\frac{4 i \, a^{2}}{d^{3} f^{2}}} \log \left (\frac{{\left ({\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d 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{4 i \, a^{2}}{d^{3} f^{2}}} + 2 \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d f}\right ) -{\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt{-\frac{4 i \, a^{2}}{d^{3} f^{2}}} \log \left (-\frac{{\left ({\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d 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{4 i \, a^{2}}{d^{3} f^{2}}} - 2 \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d f}\right ) +{\left (-8 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a\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))/(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(-4*I*a^2/(d^3*f^2))*log(((d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt((-
I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-4*I*a^2/(d^3*f^2)) + 2*a)*e^(-2*I*f*x - 2*I*e)
/(d*f)) - (d^2*f*e^(2*I*f*x + 2*I*e) - d^2*f)*sqrt(-4*I*a^2/(d^3*f^2))*log(-((d*f*e^(2*I*f*x + 2*I*e) + d*f)*s
qrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-4*I*a^2/(d^3*f^2)) - 2*a)*e^(-2*I*f*x -
2*I*e)/(d*f)) + (-8*I*a*e^(2*I*f*x + 2*I*e) - 8*I*a)*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 \left (\int - \frac{1}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx + \int \frac{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))/(d*tan(f*x+e))**(3/2),x)

[Out]

-a*(Integral(-1/(d*tan(e + f*x))**(3/2), x) + Integral(I*tan(e + f*x)/(d*tan(e + f*x))**(3/2), x))

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Giac [C]  time = 1.18, size = 120, normalized size = 1.94 \begin{align*} 2 \, a{\left (-\frac{i \, \sqrt{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 )}{d^{\frac{3}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{1}{\sqrt{d \tan \left (f x + e\right )} d f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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