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

Optimal. Leaf size=87 \[ \frac{2 i a}{d^2 f \sqrt{d \tan (e+f x)}}+\frac{2 \sqrt [4]{-1} a \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{5/2} f}-\frac{2 a}{3 d f (d \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.125183, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3529, 3533, 208} \[ \frac{2 i a}{d^2 f \sqrt{d \tan (e+f x)}}+\frac{2 \sqrt [4]{-1} a \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{5/2} f}-\frac{2 a}{3 d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.114292, size = 41, normalized size = 0.47 \[ -\frac{2 a \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-i \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*Hypergeometric2F1[-3/2, 1, -1/2, (-I)*Tan[e + f*x]])/(3*d*f*(d*Tan[e + f*x])^(3/2))

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Maple [B]  time = 0.022, size = 378, normalized size = 4.3 \begin{align*}{\frac{2\,ia}{{d}^{2}f}{\frac{1}{\sqrt{d\tan \left ( fx+e \right ) }}}}-{\frac{2\,a}{3\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{a\sqrt{2}}{4\,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\sqrt{2}}{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\sqrt{2}}{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}{4}}a\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{{\frac{i}{2}}a\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{{\frac{i}{2}}a\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))/(d*tan(f*x+e))^(5/2),x)

[Out]

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

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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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.30662, size = 977, normalized size = 11.23 \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{4 i \, a^{2}}{d^{5} f^{2}}} \log \left (-\frac{{\left ({\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 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{4 i \, a^{2}}{d^{5} f^{2}}} + 2 i \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2} f}\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{4 i \, a^{2}}{d^{5} f^{2}}} \log \left (\frac{{\left ({\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 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{4 i \, a^{2}}{d^{5} f^{2}}} - 2 i \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2} f}\right ) + 16 \,{\left (a e^{\left (4 i \, f x + 4 i \, e\right )} - a e^{\left (2 i \, f x + 2 i \, e\right )} - 2 \, 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}}}{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))/(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(4*I*a^2/(d^5*f^2))*log(-((d^2*
f*e^(2*I*f*x + 2*I*e) + 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(4*I*a^2/(
d^5*f^2)) + 2*I*a)*e^(-2*I*f*x - 2*I*e)/(d^2*f)) - 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(4*I*a^2/(d^5*f^2))*log(((d^2*f*e^(2*I*f*x + 2*I*e) + 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(4*I*a^2/(d^5*f^2)) - 2*I*a)*e^(-2*I*f*x - 2*I*e)/(d^2*f)) + 16*(a*e^(4*I*f*x
+ 4*I*e) - a*e^(2*I*f*x + 2*I*e) - 2*a)*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 \left (\int - \frac{1}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx + \int \frac{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))/(d*tan(f*x+e))**(5/2),x)

[Out]

-a*(Integral(-1/(d*tan(e + f*x))**(5/2), x) + Integral(I*tan(e + f*x)/(d*tan(e + f*x))**(5/2), x))

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Giac [A]  time = 1.19358, size = 147, normalized size = 1.69 \begin{align*} -\frac{2}{3} \, a{\left (-\frac{3 i \, \sqrt{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{-3 i \, d \tan \left (f x + e\right ) + d}{\sqrt{d \tan \left (f x + e\right )} d^{3} f \tan \left (f x + e\right )}\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))^(5/2),x, algorithm="giac")

[Out]

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