3.162 \(\int (e x)^m \tanh ^3(a+2 \log (x)) \, dx\)

Optimal. Leaf size=176 \[ -\frac {\left (m^2+2 m+9\right ) (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{4};\frac {m+5}{4};-e^{2 a} x^4\right )}{4 e (m+1)}-\frac {e^{-2 a} \left (e^{4 a} (m+5) x^4+e^{2 a} (3-m)\right ) (e x)^{m+1}}{8 e \left (e^{2 a} x^4+1\right )}-\frac {\left (1-e^{2 a} x^4\right )^2 (e x)^{m+1}}{4 e \left (e^{2 a} x^4+1\right )^2}+\frac {(m+3) (m+5) (e x)^{m+1}}{8 e (m+1)} \]

[Out]

1/8*(3+m)*(5+m)*(e*x)^(1+m)/e/(1+m)-1/4*(e*x)^(1+m)*(1-exp(2*a)*x^4)^2/e/(1+exp(2*a)*x^4)^2-1/8*(e*x)^(1+m)*(e
xp(2*a)*(3-m)+exp(4*a)*(5+m)*x^4)/e/exp(2*a)/(1+exp(2*a)*x^4)-1/4*(m^2+2*m+9)*(e*x)^(1+m)*hypergeom([1, 1/4+1/
4*m],[5/4+1/4*m],-exp(2*a)*x^4)/e/(1+m)

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Rubi [F]  time = 0.07, 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 = {} \[ \int (e x)^m \tanh ^3(a+2 \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[(e*x)^m*Tanh[a + 2*Log[x]]^3,x]

[Out]

Defer[Int][(e*x)^m*Tanh[a + 2*Log[x]]^3, x]

Rubi steps

\begin {align*} \int (e x)^m \tanh ^3(a+2 \log (x)) \, dx &=\int (e x)^m \tanh ^3(a+2 \log (x)) \, dx\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 111, normalized size = 0.63 \[ -\frac {x (e x)^m \left (6 \, _2F_1\left (1,\frac {m+1}{4};\frac {m+5}{4};-x^4 (\cosh (2 a)+\sinh (2 a))\right )-12 \, _2F_1\left (2,\frac {m+1}{4};\frac {m+5}{4};-x^4 (\cosh (2 a)+\sinh (2 a))\right )+8 \, _2F_1\left (3,\frac {m+1}{4};\frac {m+5}{4};-x^4 (\cosh (2 a)+\sinh (2 a))\right )-1\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Tanh[a + 2*Log[x]]^3,x]

[Out]

-((x*(e*x)^m*(-1 + 6*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, -(x^4*(Cosh[2*a] + Sinh[2*a]))] - 12*Hypergeom
etric2F1[2, (1 + m)/4, (5 + m)/4, -(x^4*(Cosh[2*a] + Sinh[2*a]))] + 8*Hypergeometric2F1[3, (1 + m)/4, (5 + m)/
4, -(x^4*(Cosh[2*a] + Sinh[2*a]))]))/(1 + m))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tanh(a+2*log(x))^3,x, algorithm="fricas")

[Out]

integral((e*x)^m*tanh(a + 2*log(x))^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{m} \tanh \left (a + 2 \, \log \relax (x)\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tanh(a+2*log(x))^3,x, algorithm="giac")

[Out]

integrate((e*x)^m*tanh(a + 2*log(x))^3, x)

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (\tanh ^{3}\left (a +2 \ln \relax (x )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*tanh(a+2*ln(x))^3,x)

[Out]

int((e*x)^m*tanh(a+2*ln(x))^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{m} \tanh \left (a + 2 \, \log \relax (x)\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*tanh(a+2*log(x))^3,x, algorithm="maxima")

[Out]

integrate((e*x)^m*tanh(a + 2*log(x))^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tanh}\left (a+2\,\ln \relax (x)\right )}^3\,{\left (e\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + 2*log(x))^3*(e*x)^m,x)

[Out]

int(tanh(a + 2*log(x))^3*(e*x)^m, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{m} \tanh ^{3}{\left (a + 2 \log {\relax (x )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*tanh(a+2*ln(x))**3,x)

[Out]

Integral((e*x)**m*tanh(a + 2*log(x))**3, x)

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