3.152 \(\int (e x)^m \tan ^3(a+i \log (x)) \, dx\)

Optimal. Leaf size=184 \[ -\frac {i \left (m^2+2 m+3\right ) x (e x)^m \, _2F_1\left (1,\frac {1}{2} (-m-1);\frac {1-m}{2};-\frac {e^{2 i a}}{x^2}\right )}{m+1}+\frac {i e^{-2 i a} x \left (\frac {e^{4 i a} (1-m)}{x^2}+e^{2 i a} (m+3)\right ) (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )}+\frac {i x \left (1-\frac {e^{2 i a}}{x^2}\right )^2 (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}-\frac {i (1-m) m x (e x)^m}{2 (m+1)} \]

[Out]

-1/2*I*(1-m)*m*x*(e*x)^m/(1+m)+1/2*I*(1-exp(2*I*a)/x^2)^2*x*(e*x)^m/(1+exp(2*I*a)/x^2)^2+1/2*I*(exp(2*I*a)*(3+
m)+exp(4*I*a)*(1-m)/x^2)*x*(e*x)^m/exp(2*I*a)/(1+exp(2*I*a)/x^2)-I*(m^2+2*m+3)*x*(e*x)^m*hypergeom([1, -1/2-1/
2*m],[1/2-1/2*m],-exp(2*I*a)/x^2)/(1+m)

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Rubi [F]  time = 0.09, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[(e*x)^m*Tan[a + I*Log[x]]^3,x]

[Out]

Defer[Int][(e*x)^m*Tan[a + I*Log[x]]^3, x]

Rubi steps

\begin {align*} \int (e x)^m \tan ^3(a+i \log (x)) \, dx &=\int (e x)^m \tan ^3(a+i \log (x)) \, dx\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 125, normalized size = 0.68 \[ \frac {i x (e x)^m \left (6 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-x^2 (\cos (2 a)-i \sin (2 a))\right )-12 \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-x^2 (\cos (2 a)-i \sin (2 a))\right )+8 \, _2F_1\left (3,\frac {m+1}{2};\frac {m+3}{2};-x^2 (\cos (2 a)-i \sin (2 a))\right )-1\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Tan[a + I*Log[x]]^3,x]

[Out]

(I*x*(e*x)^m*(-1 + 6*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -(x^2*(Cos[2*a] - I*Sin[2*a]))] - 12*Hypergeom
etric2F1[2, (1 + m)/2, (3 + m)/2, -(x^2*(Cos[2*a] - I*Sin[2*a]))] + 8*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/
2, -(x^2*(Cos[2*a] - I*Sin[2*a]))]))/(1 + m)

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-i \, x^{6} + 3 i \, x^{4} e^{\left (2 i \, a\right )} - 3 i \, x^{2} e^{\left (4 i \, a\right )} + i \, e^{\left (6 i \, a\right )}\right )} e^{\left (m \log \relax (e) + m \log \relax (x)\right )}}{x^{6} + 3 \, x^{4} e^{\left (2 i \, a\right )} + 3 \, x^{2} e^{\left (4 i \, a\right )} + e^{\left (6 i \, a\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tan(a+I*log(x))^3,x, algorithm="fricas")

[Out]

integral((-I*x^6 + 3*I*x^4*e^(2*I*a) - 3*I*x^2*e^(4*I*a) + I*e^(6*I*a))*e^(m*log(e) + m*log(x))/(x^6 + 3*x^4*e
^(2*I*a) + 3*x^2*e^(4*I*a) + e^(6*I*a)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{m} \tan \left (a + i \, \log \relax (x)\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tan(a+I*log(x))^3,x, algorithm="giac")

[Out]

integrate((e*x)^m*tan(a + I*log(x))^3, x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (\tan ^{3}\left (a +i \ln \relax (x )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*tan(a+I*ln(x))^3,x)

[Out]

int((e*x)^m*tan(a+I*ln(x))^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{m} \tan \left (a + i \, \log \relax (x)\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tan(a+I*log(x))^3,x, algorithm="maxima")

[Out]

integrate((e*x)^m*tan(a + I*log(x))^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (a+\ln \relax (x)\,1{}\mathrm {i}\right )}^3\,{\left (e\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a + log(x)*1i)^3*(e*x)^m,x)

[Out]

int(tan(a + log(x)*1i)^3*(e*x)^m, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{m} \tan ^{3}{\left (a + i \log {\relax (x )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*tan(a+I*ln(x))**3,x)

[Out]

Integral((e*x)**m*tan(a + I*log(x))**3, x)

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