3.2.66 \(\int \tanh ^p(a+\frac {\log (x)}{4}) \, dx\) [166]

Optimal. Leaf size=106 \[ e^{-4 a} \left (-1+e^{2 a} \sqrt {x}\right )^{1+p} \left (1+e^{2 a} \sqrt {x}\right )^{1-p}-\frac {2^{1-p} e^{-4 a} p \left (-1+e^{2 a} \sqrt {x}\right )^{1+p} \, _2F_1\left (p,1+p;2+p;\frac {1}{2} \left (1-e^{2 a} \sqrt {x}\right )\right )}{1+p} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5652, 383, 81, 71} \begin {gather*} e^{-4 a} \left (e^{2 a} \sqrt {x}-1\right )^{p+1} \left (e^{2 a} \sqrt {x}+1\right )^{1-p}-\frac {e^{-4 a} 2^{1-p} p \left (e^{2 a} \sqrt {x}-1\right )^{p+1} \, _2F_1\left (p,p+1;p+2;\frac {1}{2} \left (1-e^{2 a} \sqrt {x}\right )\right )}{p+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tanh[a + Log[x]/4]^p,x]

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 383

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, Dis
t[g, Subst[Int[x^(g - 1)*(a + b*x^(g*n))^p*(c + d*x^(g*n))^q, x], x, x^(1/g)], x]] /; FreeQ[{a, b, c, d, p, q}
, x] && NeQ[b*c - a*d, 0] && FractionQ[n]

Rule 5652

Int[Tanh[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[(-1 + E^(2*a*d)*x^(2*b*d))^p/(1 + E^(2*a*d)*x^
(2*b*d))^p, x] /; FreeQ[{a, b, d, p}, x]

Rubi steps

\begin {align*} \int \tanh ^p\left (a+\frac {\log (x)}{4}\right ) \, dx &=\int \tanh ^p\left (\frac {1}{4} (4 a+\log (x))\right ) \, dx\\ \end {align*}

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Mathematica [A]
time = 2.00, size = 121, normalized size = 1.14 \begin {gather*} \frac {e^{-4 a} \left (-1+e^{2 a} \sqrt {x}\right ) \left (\frac {-1+e^{2 a} \sqrt {x}}{2+2 e^{2 a} \sqrt {x}}\right )^p \left (2^p (1+p) \left (1+e^{2 a} \sqrt {x}\right )-2 p \left (1+e^{2 a} \sqrt {x}\right )^p \, _2F_1\left (p,1+p;2+p;\frac {1}{2} \left (1-e^{2 a} \sqrt {x}\right )\right )\right )}{1+p} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tanh[a + Log[x]/4]^p,x]

[Out]

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

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Maple [F]
time = 0.62, size = 0, normalized size = 0.00 \[\int \tanh ^{p}\left (a +\frac {\ln \left (x \right )}{4}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a+1/4*ln(x))^p,x)

[Out]

int(tanh(a+1/4*ln(x))^p,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/4*log(x))^p,x, algorithm="maxima")

[Out]

integrate(tanh(a + 1/4*log(x))^p, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/4*log(x))^p,x, algorithm="fricas")

[Out]

integral(tanh(a + 1/4*log(x))^p, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \tanh ^{p}{\left (a + \frac {\log {\left (x \right )}}{4} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/4*ln(x))**p,x)

[Out]

Integral(tanh(a + log(x)/4)**p, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+1/4*log(x))^p,x, algorithm="giac")

[Out]

integrate(tanh(a + 1/4*log(x))^p, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {tanh}\left (a+\frac {\ln \left (x\right )}{4}\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + log(x)/4)^p,x)

[Out]

int(tanh(a + log(x)/4)^p, x)

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