3.2.72 \(\int \coth ^p(a+\frac {\log (x)}{6}) \, dx\) [172]

Optimal. Leaf size=162 \[ e^{-6 a} p \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p}+e^{-4 a} \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p} \sqrt [3]{x}-\frac {2^{-p} e^{-6 a} \left (1+2 p^2\right ) \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \, _2F_1\left (p,1+p;2+p;\frac {1}{2} \left (1+e^{2 a} \sqrt [3]{x}\right )\right )}{1+p} \]

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {5653, 383, 92, 81, 71} \begin {gather*} -\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 (e^{2 a} \sqrt [3]{x}+1\right )\right )}{p+1}+e^{-6 a} p \left (-e^{2 a} \sqrt [3]{x}-1\right )^{p+1} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p}+e^{-4 a} \sqrt [3]{x} \left (-e^{2 a} \sqrt [3]{x}-1\right )^{p+1} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Coth[a + Log[x]/6]^p,x]

[Out]

(p*(-1 - E^(2*a)*x^(1/3))^(1 + p)*(1 - E^(2*a)*x^(1/3))^(1 - p))/E^(6*a) + ((-1 - E^(2*a)*x^(1/3))^(1 + p)*(1
- E^(2*a)*x^(1/3))^(1 - p)*x^(1/3))/E^(4*a) - ((1 + 2*p^2)*(-1 - E^(2*a)*x^(1/3))^(1 + p)*Hypergeometric2F1[p,
1 + p, 2 + p, (1 + E^(2*a)*x^(1/3))/2])/(2^p*E^(6*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 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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 5653

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

________________________________________________________________________________________

Mathematica [A]
time = 0.50, size = 142, normalized size = 0.88 \begin {gather*} \frac {e^{-6 a} \left (1+e^{2 a} \sqrt [3]{x}\right )^{1-p} \left (\frac {1+e^{2 a} \sqrt [3]{x}}{-1+e^{2 a} \sqrt [3]{x}}\right )^{-1+p} \left ((-1+p) \left (1+e^{2 a} \sqrt [3]{x}\right )^{1+p} \left (p+e^{2 a} \sqrt [3]{x}\right )-2^p \left (1+2 p^2\right ) \, _2F_1\left (1-p,-p;2-p;\frac {1}{2}-\frac {1}{2} e^{2 a} \sqrt [3]{x}\right )\right )}{-1+p} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Coth[a + Log[x]/6]^p,x]

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.85, size = 0, normalized size = 0.00 \[\int \coth ^{p}\left (a +\frac {\ln \left (x \right )}{6}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

integrate(coth(a + 1/6*log(x))^p, x)

________________________________________________________________________________________

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

[Out]

integral(coth(a + 1/6*log(x))^p, x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \coth ^{p}{\left (a + \frac {\log {\left (x \right )}}{6} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(coth(a + log(x)/6)**p, x)

________________________________________________________________________________________

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

[Out]

integrate(coth(a + 1/6*log(x))^p, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {coth}\left (a+\frac {\ln \left (x\right )}{6}\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a + log(x)/6)^p,x)

[Out]

int(coth(a + log(x)/6)^p, x)

________________________________________________________________________________________