3.168 \(\int \tanh ^p(a+\frac{\log (x)}{8}) \, dx\)

Optimal. Leaf size=190 \[ -\frac{e^{-8 a} 2^{2-p} p \left (p^2+2\right ) \left (e^{2 a} \sqrt [4]{x}-1\right )^{p+1} \, _2F_1\left (p,p+1;p+2;\frac{1}{2} \left (1-e^{2 a} \sqrt [4]{x}\right )\right )}{3 (p+1)}+\frac{1}{3} e^{-12 a} \left (e^{2 a} \sqrt [4]{x}-1\right )^{p+1} \left (e^{4 a} \left (2 p^2+3\right )-2 e^{6 a} p \sqrt [4]{x}\right ) \left (e^{2 a} \sqrt [4]{x}+1\right )^{1-p}+e^{-4 a} \sqrt{x} \left (e^{2 a} \sqrt [4]{x}-1\right )^{p+1} \left (e^{2 a} \sqrt [4]{x}+1\right )^{1-p} \]

[Out]

((-1 + E^(2*a)*x^(1/4))^(1 + p)*(1 + E^(2*a)*x^(1/4))^(1 - p)*(E^(4*a)*(3 + 2*p^2) - 2*E^(6*a)*p*x^(1/4)))/(3*
E^(12*a)) + ((-1 + E^(2*a)*x^(1/4))^(1 + p)*(1 + E^(2*a)*x^(1/4))^(1 - p)*Sqrt[x])/E^(4*a) - (2^(2 - p)*p*(2 +
p^2)*(-1 + E^(2*a)*x^(1/4))^(1 + p)*Hypergeometric2F1[p, 1 + p, 2 + p, (1 - E^(2*a)*x^(1/4))/2])/(3*E^(8*a)*(
1 + p))

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Rubi [F]  time = 0.0519402, 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 = {} \[ \int \tanh ^p\left (a+\frac{\log (x)}{8}\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Int[Tanh[a + Log[x]/8]^p,x]

[Out]

Defer[Int][Tanh[(8*a + Log[x])/8]^p, x]

Rubi steps

\begin{align*} \int \tanh ^p\left (a+\frac{\log (x)}{8}\right ) \, dx &=\int \tanh ^p\left (\frac{1}{8} (8 a+\log (x))\right ) \, dx\\ \end{align*}

Mathematica [C]  time = 3.17013, size = 177, normalized size = 0.93 \[ \frac{5 x \left (\frac{e^{2 a} \sqrt [4]{x}-1}{e^{2 a} \sqrt [4]{x}+1}\right )^p F_1\left (4;-p,p;5;e^{2 a} \sqrt [4]{x},-e^{2 a} \sqrt [4]{x}\right )}{5 F_1\left (4;-p,p;5;e^{2 a} \sqrt [4]{x},-e^{2 a} \sqrt [4]{x}\right )-e^{2 a} p \sqrt [4]{x} \left (F_1\left (5;1-p,p;6;e^{2 a} \sqrt [4]{x},-e^{2 a} \sqrt [4]{x}\right )+F_1\left (5;-p,p+1;6;e^{2 a} \sqrt [4]{x},-e^{2 a} \sqrt [4]{x}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tanh[a + Log[x]/8]^p,x]

[Out]

(5*((-1 + E^(2*a)*x^(1/4))/(1 + E^(2*a)*x^(1/4)))^p*x*AppellF1[4, -p, p, 5, E^(2*a)*x^(1/4), -(E^(2*a)*x^(1/4)
)])/(5*AppellF1[4, -p, p, 5, E^(2*a)*x^(1/4), -(E^(2*a)*x^(1/4))] - E^(2*a)*p*x^(1/4)*(AppellF1[5, 1 - p, p, 6
, E^(2*a)*x^(1/4), -(E^(2*a)*x^(1/4))] + AppellF1[5, -p, 1 + p, 6, E^(2*a)*x^(1/4), -(E^(2*a)*x^(1/4))]))

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Maple [F]  time = 0.029, size = 0, normalized size = 0. \begin{align*} \int \left ( \tanh \left ( a+{\frac{\ln \left ( x \right ) }{8}} \right ) \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a+1/8*ln(x))^p,x)

[Out]

int(tanh(a+1/8*ln(x))^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \tanh \left (a + \frac{1}{8} \, \log \left (x\right )\right )^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/8*log(x))^p,x, algorithm="maxima")

[Out]

integrate(tanh(a + 1/8*log(x))^p, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\tanh \left (a + \frac{1}{8} \, \log \left (x\right )\right )^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/8*log(x))^p,x, algorithm="fricas")

[Out]

integral(tanh(a + 1/8*log(x))^p, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \tanh ^{p}{\left (a + \frac{\log{\left (x \right )}}{8} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/8*ln(x))**p,x)

[Out]

Integral(tanh(a + log(x)/8)**p, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \tanh \left (a + \frac{1}{8} \, \log \left (x\right )\right )^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/8*log(x))^p,x, algorithm="giac")

[Out]

integrate(tanh(a + 1/8*log(x))^p, x)