\(\int x \tanh ^2(d (a+b \log (c x^n))) \, dx\) [181]

Optimal result

Integrand size = 17, antiderivative size = 131 \[ \int x \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {1}{2} \left (1+\frac {2}{b d n}\right ) x^2+\frac {x^2 \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {2 x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{b d n},1+\frac {1}{b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n} \]

[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] (verified)

Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5658, 5656, 516, 470, 371} \[ \int x \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {2 x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{b d n},1+\frac {1}{b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n}+\frac {x^2 \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}+\frac {1}{2} x^2 \left (\frac {2}{b d n}+1\right ) \]

[In]

Int[x*Tanh[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 5656

Int[((e_.)*(x_))^(m_.)*Tanh[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), 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 5658

Int[((e_.)*(x_))^(m_.)*Tanh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(e*x)^(m + 1
)/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Tanh[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*} \text {integral}& = \frac {\left (x^2 \left (c x^n\right )^{-2/n}\right ) \text {Subst}\left (\int x^{-1+\frac {2}{n}} \tanh ^2(d (a+b \log (x))) \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (x^2 \left (c x^n\right )^{-2/n}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {2}{n}} \left (-1+e^{2 a d} x^{2 b d}\right )^2}{\left (1+e^{2 a d} x^{2 b d}\right )^2} \, dx,x,c x^n\right )}{n} \\ & = \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 {\left (e^{-2 a d} x^2 \left (c x^n\right )^{-2/n}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {2}{n}} \left (\frac {2 e^{2 a d} (2-b d n)}{n}-\frac {2 e^{4 a d} (2+b d n) x^{2 b d}}{n}\right )}{1+e^{2 a d} x^{2 b d}} \, dx,x,c x^n\right )}{2 b d n} \\ & = \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 {\left (4 x^2 \left (c x^n\right )^{-2/n}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {2}{n}}}{1+e^{2 a d} x^{2 b d}} \, dx,x,c x^n\right )}{b d n^2} \\ & = \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 \operatorname {Hypergeometric2F1}\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 8.30 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.18 \[ \int x \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x^2 \left (2 e^{2 d \left (a+b \log \left (c x^n\right )\right )} \operatorname {Hypergeometric2F1}\left (1,1+\frac {1}{b d n},2+\frac {1}{b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )+(1+b d n) \left (b d n-2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{b d n},1+\frac {1}{b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )-2 \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )\right )}{2 b d n (1+b d n)} \]

[In]

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

Maple [F]

\[\int x {\tanh \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{2}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int x \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]

[In]

integrate(x*tanh(d*(a+b*log(c*x^n)))^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int x \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int x \tanh ^{2}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \]

[In]

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

[Out]

Integral(x*tanh(a*d + b*d*log(c*x**n))**2, x)

Maxima [F]

\[ \int x \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]

[In]

integrate(x*tanh(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) - 4*integrate(x/(b*c^(2*b*d)*d*n*e^(2*b*d*log(x^n) + 2*a*d) + b*d*n), x)

Giac [F]

\[ \int x \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int x\,{\mathrm {tanh}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2 \,d x \]

[In]

int(x*tanh(d*(a + b*log(c*x^n)))^2,x)

[Out]

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