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

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Rubi [A]  time = 0.08, antiderivative size = 62, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.11, 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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fricas [C]  time = 0.50, size = 313, normalized size = 5.05 \[ \frac {{\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{3} f^{2}}} \log \left (\frac {{\left ({\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i \, a^{2}}{d^{3} f^{2}}} + 2 \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d f}\right ) - {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{3} f^{2}}} \log \left (-\frac {{\left ({\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i \, a^{2}}{d^{3} f^{2}}} - 2 \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d f}\right ) + {\left (-8 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )}} \]

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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giac [C]  time = 0.78, size = 89, normalized size = 1.44 \[ \frac {2 \, a {\left (-\frac {i \, \sqrt {2} \arctan \left (\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{\sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {1}{\sqrt {d \tan \left (f x + e\right )} f}\right )}}{d} \]

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

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maple [C]  time = 0.17, size = 358, normalized size = 5.77 \[ -\frac {i a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f \,d^{2}}-\frac {i a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \,d^{2}}+\frac {i a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \,d^{2}}-\frac {a \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f d \left (d^{2}\right )^{\frac {1}{4}}}-\frac {a \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f d \left (d^{2}\right )^{\frac {1}{4}}}+\frac {a \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f d \left (d^{2}\right )^{\frac {1}{4}}}-\frac {2 a}{d f \sqrt {d \tan \left (f x +e \right )}} \]

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]

-1/4*I*a/f/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*t
an(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))-1/2*I*a/f/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*a/f/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*a/f/d*2^(1/2)/(d^2)^(1/4)*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*a/f/d*2^(1/2)/(d^2)^(1/4)*arct
an(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/2*a/f/d*2^(1/2)/(d^2)^(1/4)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*ta
n(f*x+e))^(1/2)+1)-2*a/d/f/(d*tan(f*x+e))^(1/2)

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maxima [C]  time = 0.72, size = 172, normalized size = 2.77 \[ \frac {a {\left (-\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - \frac {8 \, a}{\sqrt {d \tan \left (f x + e\right )}}}{4 \, d f} \]

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]

1/4*(a*(-(2*I + 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - (2
*I + 2)*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - (I - 1)*sqrt
(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + (I - 1)*sqrt(2)*log(d*tan(f*x + e
) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d)) - 8*a/sqrt(d*tan(f*x + e)))/(d*f)

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mupad [B]  time = 4.30, size = 50, normalized size = 0.81 \[ -\frac {2\,a}{d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {2\,{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{d^{3/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - a*tan(e + f*x)*1i)/(d*tan(e + f*x))^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i a \left (\int \frac {i}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \]

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]

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

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