3.2.86 \(\int x \coth ^2(d (a+b \log (c x^n))) \, dx\) [186]

Optimal. Leaf size=130 \[ \frac {1}{2} \left (1+\frac {2}{b d n}\right ) x^2+\frac {x^2 \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {2 x^2 \, _2F_1\left (1,\frac {1}{b d n};1+\frac {1}{b d n};e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n} \]

[Out]

1/2*(1+2/b/d/n)*x^2+x^2*(1+exp(2*a*d)*(c*x^n)^(2*b*d))/b/d/n/(1-exp(2*a*d)*(c*x^n)^(2*b*d))-2*x^2*hypergeom([1
, 1/b/d/n],[1+1/b/d/n],exp(2*a*d)*(c*x^n)^(2*b*d))/b/d/n

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5659, 5657, 516, 470, 371} \begin {gather*} -\frac {2 x^2 \, _2F_1\left (1,\frac {1}{b d n};1+\frac {1}{b d n};e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n}+\frac {x^2 \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}{b d n \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}+\frac {1}{2} x^2 \left (\frac {2}{b d n}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Coth[d*(a + b*Log[c*x^n])]^2,x]

[Out]

((1 + 2/(b*d*n))*x^2)/2 + (x^2*(1 + E^(2*a*d)*(c*x^n)^(2*b*d)))/(b*d*n*(1 - E^(2*a*d)*(c*x^n)^(2*b*d))) - (2*x
^2*Hypergeometric2F1[1, 1/(b*d*n), 1 + 1/(b*d*n), E^(2*a*d)*(c*x^n)^(2*b*d)])/(b*d*n)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rule 470

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

Rule 516

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

Rule 5657

Int[Coth[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*((-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, e, m, p}, x]

Rule 5659

Int[Coth[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1
)/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Coth[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[
{a, b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])

Rubi steps

\begin {align*} \int x \coth ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx &=\int x \coth ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 4.50, size = 151, normalized size = 1.16 \begin {gather*} \frac {x^2 \left (-2 e^{2 d \left (a+b \log \left (c x^n\right )\right )} \, _2F_1\left (1,1+\frac {1}{b d n};2+\frac {1}{b d n};e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )+(1+b d n) \left (b d n-2 \coth \left (d \left (a+b \log \left (c x^n\right )\right )\right )-2 \, _2F_1\left (1,\frac {1}{b d n};1+\frac {1}{b d n};e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )\right )\right )}{2 b d n (1+b d n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Coth[d*(a + b*Log[c*x^n])]^2,x]

[Out]

(x^2*(-2*E^(2*d*(a + b*Log[c*x^n]))*Hypergeometric2F1[1, 1 + 1/(b*d*n), 2 + 1/(b*d*n), E^(2*d*(a + b*Log[c*x^n
]))] + (1 + b*d*n)*(b*d*n - 2*Coth[d*(a + b*Log[c*x^n])] - 2*Hypergeometric2F1[1, 1/(b*d*n), 1 + 1/(b*d*n), E^
(2*d*(a + b*Log[c*x^n]))])))/(2*b*d*n*(1 + b*d*n))

________________________________________________________________________________________

Maple [F]
time = 0.57, size = 0, normalized size = 0.00 \[\int x \left (\coth ^{2}\left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*coth(d*(a+b*ln(c*x^n)))^2,x)

[Out]

int(x*coth(d*(a+b*ln(c*x^n)))^2,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(x*coth(d*(a+b*log(c*x^n)))^2,x, algorithm="maxima")

[Out]

1/2*(b*c^(2*b*d)*d*n*x^2*e^(2*b*d*log(x^n) + 2*a*d) - (b*d*n + 4)*x^2)/(b*c^(2*b*d)*d*n*e^(2*b*d*log(x^n) + 2*
a*d) - b*d*n) - 2*integrate(x/(b*c^(b*d)*d*n*e^(b*d*log(x^n) + a*d) + b*d*n), x) + 2*integrate(x/(b*c^(b*d)*d*
n*e^(b*d*log(x^n) + a*d) - b*d*n), 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(x*coth(d*(a+b*log(c*x^n)))^2,x, algorithm="fricas")

[Out]

integral(x*coth(b*d*log(c*x^n) + a*d)^2, x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \coth ^{2}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*coth(d*(a+b*ln(c*x**n)))**2,x)

[Out]

Integral(x*coth(a*d + b*d*log(c*x**n))**2, 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(x*coth(d*(a+b*log(c*x^n)))^2,x, algorithm="giac")

[Out]

integrate(x*coth((b*log(c*x^n) + a)*d)^2, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\mathrm {coth}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*coth(d*(a + b*log(c*x^n)))^2,x)

[Out]

int(x*coth(d*(a + b*log(c*x^n)))^2, x)

________________________________________________________________________________________