Integrand size = 19, antiderivative size = 133 \[ \int x^3 \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {1}{4} \left (1+\frac {4}{b d n}\right ) x^4+\frac {x^4 \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {2 x^4 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{b d n},1+\frac {2}{b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n} \]
1/4*(1+4/b/d/n)*x^4+x^4*(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^4*hypergeom([1, 2/b/d/n],[1+2/b/d/n],-exp(2*a*d)*(c* x^n)^(2*b*d))/b/d/n
Time = 8.36 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.20 \[ \int x^3 \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x^4 \left (8 e^{2 d \left (a+b \log \left (c x^n\right )\right )} \operatorname {Hypergeometric2F1}\left (1,1+\frac {2}{b d n},2+\frac {2}{b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )+(2+b d n) \left (b d n-4 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{b d n},1+\frac {2}{b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )-4 \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )\right )}{4 b d n (2+b d n)} \]
(x^4*(8*E^(2*d*(a + b*Log[c*x^n]))*Hypergeometric2F1[1, 1 + 2/(b*d*n), 2 + 2/(b*d*n), -E^(2*d*(a + b*Log[c*x^n]))] + (2 + b*d*n)*(b*d*n - 4*Hypergeo metric2F1[1, 2/(b*d*n), 1 + 2/(b*d*n), -E^(2*d*(a + b*Log[c*x^n]))] - 4*Ta nh[d*(a + b*Log[c*x^n])])))/(4*b*d*n*(2 + b*d*n))
Time = 0.49 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.38, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {6073, 6071, 1004, 27, 959, 888}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x^3 \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\) |
| \(\Big \downarrow \) 6073 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \int \left (c x^n\right )^{\frac {4}{n}-1} \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )d\left (c x^n\right )}{n}\) |
| \(\Big \downarrow \) 6071 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \int \frac {\left (c x^n\right )^{\frac {4}{n}-1} \left (e^{2 a d} \left (c x^n\right )^{2 b d}-1\right )^2}{\left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^2}d\left (c x^n\right )}{n}\) |
| \(\Big \downarrow \) 1004 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \left (\frac {\left (c x^n\right )^{4/n} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}-\frac {e^{-2 a d} \int \frac {2 \left (c x^n\right )^{\frac {4}{n}-1} \left (\frac {e^{2 a d} (4-b d n)}{n}-\frac {e^{4 a d} (b d n+4) \left (c x^n\right )^{2 b d}}{n}\right )}{e^{2 a d} \left (c x^n\right )^{2 b d}+1}d\left (c x^n\right )}{2 b d}\right )}{n}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \left (\frac {\left (c x^n\right )^{4/n} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}-\frac {e^{-2 a d} \int \frac {\left (c x^n\right )^{\frac {4}{n}-1} \left (\frac {e^{2 a d} (4-b d n)}{n}-\frac {e^{4 a d} (b d n+4) \left (c x^n\right )^{2 b d}}{n}\right )}{e^{2 a d} \left (c x^n\right )^{2 b d}+1}d\left (c x^n\right )}{b d}\right )}{n}\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \left (\frac {\left (c x^n\right )^{4/n} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}-\frac {e^{-2 a d} \left (\frac {8 e^{2 a d} \int \frac {\left (c x^n\right )^{\frac {4}{n}-1}}{e^{2 a d} \left (c x^n\right )^{2 b d}+1}d\left (c x^n\right )}{n}-\frac {1}{4} e^{2 a d} (b d n+4) \left (c x^n\right )^{4/n}\right )}{b d}\right )}{n}\) |
| \(\Big \downarrow \) 888 |
| \(\displaystyle \frac {x^4 \left (c x^n\right )^{-4/n} \left (\frac {\left (c x^n\right )^{4/n} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}-\frac {e^{-2 a d} \left (2 e^{2 a d} \left (c x^n\right )^{4/n} \operatorname {Hypergeometric2F1}\left (1,\frac {2}{b d n},1+\frac {2}{b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )-\frac {1}{4} e^{2 a d} (b d n+4) \left (c x^n\right )^{4/n}\right )}{b d}\right )}{n}\) |
(x^4*(((c*x^n)^(4/n)*(1 - E^(2*a*d)*(c*x^n)^(2*b*d)))/(b*d*(1 + E^(2*a*d)* (c*x^n)^(2*b*d))) - (-1/4*(E^(2*a*d)*(4 + b*d*n)*(c*x^n)^(4/n)) + 2*E^(2*a *d)*(c*x^n)^(4/n)*Hypergeometric2F1[1, 2/(b*d*n), 1 + 2/(b*d*n), -(E^(2*a* d)*(c*x^n)^(2*b*d))])/(b*d*E^(2*a*d))))/(n*(c*x^n)^(4/n))
3.2.79.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
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] - Simp[(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]
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] + Simp[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] && Lt Q[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
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]
Int[((e_.)*(x_))^(m_.)*Tanh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p _.), x_Symbol] :> Simp[(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])
\[\int x^{3} {\tanh \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{2}d x\]
\[ \int x^3 \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x^{3} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]
Timed out. \[ \int x^3 \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \]
\[ \int x^3 \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x^{3} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]
1/4*(b*c^(2*b*d)*d*n*x^4*e^(2*b*d*log(x^n) + 2*a*d) + (b*d*n + 8)*x^4)/(b* c^(2*b*d)*d*n*e^(2*b*d*log(x^n) + 2*a*d) + b*d*n) - 8*integrate(x^3/(b*c^( 2*b*d)*d*n*e^(2*b*d*log(x^n) + 2*a*d) + b*d*n), x)
\[ \int x^3 \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x^{3} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \]
Timed out. \[ \int x^3 \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int x^3\,{\mathrm {tanh}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2 \,d x \]