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

Optimal. Leaf size=184 \[ -\frac{i \left (m^2+2 m+3\right ) x (e x)^m \text{Hypergeometric2F1}\left (1,\frac{1}{2} (-m-1),\frac{1-m}{2},-\frac{e^{2 i a}}{x^2}\right )}{m+1}+\frac{i e^{-2 i a} x \left (\frac{e^{4 i a} (1-m)}{x^2}+e^{2 i a} (m+3)\right ) (e x)^m}{2 \left (1+\frac{e^{2 i a}}{x^2}\right )}+\frac{i x \left (1-\frac{e^{2 i a}}{x^2}\right )^2 (e x)^m}{2 \left (1+\frac{e^{2 i a}}{x^2}\right )^2}-\frac{i (1-m) m x (e x)^m}{2 (m+1)} \]

[Out]

((-I/2)*(1 - m)*m*x*(e*x)^m)/(1 + m) + ((I/2)*(1 - E^((2*I)*a)/x^2)^2*x*(e*x)^m)/(1 + E^((2*I)*a)/x^2)^2 + ((I
/2)*(E^((2*I)*a)*(3 + m) + (E^((4*I)*a)*(1 - m))/x^2)*x*(e*x)^m)/(E^((2*I)*a)*(1 + E^((2*I)*a)/x^2)) - (I*(3 +
2*m + m^2)*x*(e*x)^m*Hypergeometric2F1[1, (-1 - m)/2, (1 - m)/2, -(E^((2*I)*a)/x^2)])/(1 + m)

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Rubi [F]  time = 0.0906369, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., 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*}

Mathematica [A]  time = 0.929593, size = 255, normalized size = 1.39 \[ \frac{x (e x)^m \left (-\frac{i x^4 (\cos (2 a)-i \sin (2 a)) \left ((m+5) x^2 (\cos (a)-i \sin (a)) \text{Hypergeometric2F1}\left (3,\frac{m+7}{2},\frac{m+9}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right )-3 (m+7) (\cos (a)+i \sin (a)) \text{Hypergeometric2F1}\left (3,\frac{m+5}{2},\frac{m+7}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right )\right )}{(m+5) (m+7)}+\frac{3 x^2 (\sin (a)-i \cos (a)) \text{Hypergeometric2F1}\left (3,\frac{m+3}{2},\frac{m+5}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right )}{m+3}+\frac{i (\cos (a)+i \sin (a))^3 \text{Hypergeometric2F1}\left (3,\frac{m+1}{2},\frac{m+3}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right )}{m+1}\right )}{(\cos (a)+i \sin (a))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*((I*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -(x^2*(Cos[2*a] - I*Sin[2*a]))]*(Cos[a] + I*Sin[a])^
3)/(1 + m) + (3*x^2*Hypergeometric2F1[3, (3 + m)/2, (5 + m)/2, -(x^2*(Cos[2*a] - I*Sin[2*a]))]*((-I)*Cos[a] +
Sin[a]))/(3 + m) - (I*x^4*((5 + m)*x^2*Hypergeometric2F1[3, (7 + m)/2, (9 + m)/2, -(x^2*(Cos[2*a] - I*Sin[2*a]
))]*(Cos[a] - I*Sin[a]) - 3*(7 + m)*Hypergeometric2F1[3, (5 + m)/2, (7 + m)/2, -(x^2*(Cos[2*a] - I*Sin[2*a]))]
*(Cos[a] + I*Sin[a]))*(Cos[2*a] - I*Sin[2*a]))/((5 + m)*(7 + m))))/(Cos[a] + I*Sin[a])^3

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Maple [F]  time = 0.067, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( \tan \left ( a+i\ln \left ( x \right ) \right ) \right ) ^{3}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \left (e x\right )^{m} \tan \left (a + i \, \log \left (x\right )\right )^{3}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left ({\left (i \, m - i\right )} x e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} +{\left (i \, m + i\right )} x\right )} \left (e x\right )^{m} +{\left (e^{\left (4 i \, a - 4 \, \log \left (x\right )\right )} + 2 \, e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + 1\right )}{\rm integral}\left (\frac{{\left (-i \, m^{2} - 2 i \, m + i \, e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} - 2 i\right )} \left (e x\right )^{m}}{e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + 1}, x\right )}{e^{\left (4 i \, a - 4 \, \log \left (x\right )\right )} + 2 \, e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + 1} \end{align*}

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]

(((I*m - I)*x*e^(2*I*a - 2*log(x)) + (I*m + I)*x)*(e*x)^m + (e^(4*I*a - 4*log(x)) + 2*e^(2*I*a - 2*log(x)) + 1
)*integral((-I*m^2 - 2*I*m + I*e^(2*I*a - 2*log(x)) - 2*I)*(e*x)^m/(e^(2*I*a - 2*log(x)) + 1), x))/(e^(4*I*a -
4*log(x)) + 2*e^(2*I*a - 2*log(x)) + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e x\right )^{m} \tan \left (a + i \, \log \left (x\right )\right )^{3}\,{d x} \end{align*}

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)