3.167 \(\int \tanh ^p(a+\frac {\log (x)}{6}) \, dx\)

Optimal. Leaf size=158 \[ \frac {e^{-6 a} 2^{-p} \left (2 p^2+1\right ) \left (e^{2 a} \sqrt [3]{x}-1\right )^{p+1} \, _2F_1\left (p,p+1;p+2;\frac {1}{2} \left (1-e^{2 a} \sqrt [3]{x}\right )\right )}{p+1}-e^{-6 a} p \left (e^{2 a} \sqrt [3]{x}-1\right )^{p+1} \left (e^{2 a} \sqrt [3]{x}+1\right )^{1-p}+e^{-4 a} \sqrt [3]{x} \left (e^{2 a} \sqrt [3]{x}-1\right )^{p+1} \left (e^{2 a} \sqrt [3]{x}+1\right )^{1-p} \]

[Out]

-p*(-1+exp(2*a)*x^(1/3))^(1+p)*(1+exp(2*a)*x^(1/3))^(1-p)/exp(6*a)+(-1+exp(2*a)*x^(1/3))^(1+p)*(1+exp(2*a)*x^(
1/3))^(1-p)*x^(1/3)/exp(4*a)+(2*p^2+1)*(-1+exp(2*a)*x^(1/3))^(1+p)*hypergeom([p, 1+p],[2+p],1/2-1/2*exp(2*a)*x
^(1/3))/(2^p)/exp(6*a)/(1+p)

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Rubi [F]  time = 0.05, 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 = {} \[ \int \tanh ^p\left (a+\frac {\log (x)}{6}\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Int[Tanh[a + Log[x]/6]^p,x]

[Out]

Defer[Int][Tanh[(6*a + Log[x])/6]^p, x]

Rubi steps

\begin {align*} \int \tanh ^p\left (a+\frac {\log (x)}{6}\right ) \, dx &=\int \tanh ^p\left (\frac {1}{6} (6 a+\log (x))\right ) \, dx\\ \end {align*}

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Mathematica [C]  time = 4.28, size = 177, normalized size = 1.12 \[ \frac {4 x \left (\frac {e^{2 a} \sqrt [3]{x}-1}{e^{2 a} \sqrt [3]{x}+1}\right )^p F_1\left (3;-p,p;4;e^{2 a} \sqrt [3]{x},-e^{2 a} \sqrt [3]{x}\right )}{4 F_1\left (3;-p,p;4;e^{2 a} \sqrt [3]{x},-e^{2 a} \sqrt [3]{x}\right )-e^{2 a} p \sqrt [3]{x} \left (F_1\left (4;1-p,p;5;e^{2 a} \sqrt [3]{x},-e^{2 a} \sqrt [3]{x}\right )+F_1\left (4;-p,p+1;5;e^{2 a} \sqrt [3]{x},-e^{2 a} \sqrt [3]{x}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tanh[a + Log[x]/6]^p,x]

[Out]

(4*((-1 + E^(2*a)*x^(1/3))/(1 + E^(2*a)*x^(1/3)))^p*x*AppellF1[3, -p, p, 4, E^(2*a)*x^(1/3), -(E^(2*a)*x^(1/3)
)])/(4*AppellF1[3, -p, p, 4, E^(2*a)*x^(1/3), -(E^(2*a)*x^(1/3))] - E^(2*a)*p*x^(1/3)*(AppellF1[4, 1 - p, p, 5
, E^(2*a)*x^(1/3), -(E^(2*a)*x^(1/3))] + AppellF1[4, -p, 1 + p, 5, E^(2*a)*x^(1/3), -(E^(2*a)*x^(1/3))]))

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fricas [F]  time = 1.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\tanh \left (a + \frac {1}{6} \, \log \relax (x)\right )^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/6*log(x))^p,x, algorithm="fricas")

[Out]

integral(tanh(a + 1/6*log(x))^p, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \tanh \left (a + \frac {1}{6} \, \log \relax (x)\right )^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/6*log(x))^p,x, algorithm="giac")

[Out]

integrate(tanh(a + 1/6*log(x))^p, x)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \tanh ^{p}\left (a +\frac {\ln \relax (x )}{6}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a+1/6*ln(x))^p,x)

[Out]

int(tanh(a+1/6*ln(x))^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \tanh \left (a + \frac {1}{6} \, \log \relax (x)\right )^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/6*log(x))^p,x, algorithm="maxima")

[Out]

integrate(tanh(a + 1/6*log(x))^p, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tanh}\left (a+\frac {\ln \relax (x)}{6}\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + log(x)/6)^p,x)

[Out]

int(tanh(a + log(x)/6)^p, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \tanh ^{p}{\left (a + \frac {\log {\relax (x )}}{6} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/6*ln(x))**p,x)

[Out]

Integral(tanh(a + log(x)/6)**p, x)

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