3.2.85 \(\int x^2 \coth ^2(d (a+b \log (c x^n))) \, dx\) [185]

Optimal. Leaf size=136 \[ \frac {1}{3} \left (1+\frac {3}{b d n}\right ) x^3+\frac {x^3 \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {2 x^3 \, _2F_1\left (1,\frac {3}{2 b d n};1+\frac {3}{2 b d n};e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n} \]

[Out]

1/3*(1+3/b/d/n)*x^3+x^3*(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^3*hypergeom([1
, 3/2/b/d/n],[1+3/2/b/d/n],exp(2*a*d)*(c*x^n)^(2*b*d))/b/d/n

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Rubi [A]
time = 0.12, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5659, 5657, 516, 470, 371} \begin {gather*} -\frac {2 x^3 \, _2F_1\left (1,\frac {3}{2 b d n};1+\frac {3}{2 b d n};e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n}+\frac {x^3 \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}{3} x^3 \left (\frac {3}{b d n}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 3/(b*d*n))*x^3)/3 + (x^3*(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
^3*Hypergeometric2F1[1, 3/(2*b*d*n), 1 + 3/(2*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^2 \coth ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx &=\int x^2 \coth ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\\ \end {align*}

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Mathematica [A]
time = 3.09, size = 165, normalized size = 1.21 \begin {gather*} \frac {x^3 \left (-9 e^{2 d \left (a+b \log \left (c x^n\right )\right )} \, _2F_1\left (1,1+\frac {3}{2 b d n};2+\frac {3}{2 b d n};e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )+(3+2 b d n) \left (b d n-3 \coth \left (d \left (a+b \log \left (c x^n\right )\right )\right )-3 \, _2F_1\left (1,\frac {3}{2 b d n};1+\frac {3}{2 b d n};e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )\right )\right )}{3 b d n (3+2 b d n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(-9*E^(2*d*(a + b*Log[c*x^n]))*Hypergeometric2F1[1, 1 + 3/(2*b*d*n), 2 + 3/(2*b*d*n), E^(2*d*(a + b*Log[c
*x^n]))] + (3 + 2*b*d*n)*(b*d*n - 3*Coth[d*(a + b*Log[c*x^n])] - 3*Hypergeometric2F1[1, 3/(2*b*d*n), 1 + 3/(2*
b*d*n), E^(2*d*(a + b*Log[c*x^n]))])))/(3*b*d*n*(3 + 2*b*d*n))

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Maple [F]
time = 0.55, size = 0, normalized size = 0.00 \[\int x^{2} \left (\coth ^{2}\left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

1/3*(b*c^(2*b*d)*d*n*x^3*e^(2*b*d*log(x^n) + 2*a*d) - (b*d*n + 6)*x^3)/(b*c^(2*b*d)*d*n*e^(2*b*d*log(x^n) + 2*
a*d) - b*d*n) - 3*integrate(x^2/(b*c^(b*d)*d*n*e^(b*d*log(x^n) + a*d) + b*d*n), x) + 3*integrate(x^2/(b*c^(b*d
)*d*n*e^(b*d*log(x^n) + a*d) - b*d*n), 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(x^2*coth(d*(a+b*log(c*x^n)))^2,x, algorithm="fricas")

[Out]

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

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

[Out]

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

[Out]

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

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

[Out]

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

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