Integrand size = 21, antiderivative size = 178 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{x} \, dx=\frac {b \left (6 c^2 d-e\right ) e \sqrt {1+\frac {1}{c^2 x^2}} x}{6 c^3}+\frac {b e^2 \sqrt {1+\frac {1}{c^2 x^2}} x^3}{12 c}+\frac {1}{2} b d^2 \text {csch}^{-1}(c x)^2+d e x^2 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \text {csch}^{-1}(c x)\right )-b d^2 \text {csch}^{-1}(c x) \log \left (1-e^{2 \text {csch}^{-1}(c x)}\right )+b d^2 \text {csch}^{-1}(c x) \log \left (\frac {1}{x}\right )-d^2 \left (a+b \text {csch}^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {1}{2} b d^2 \operatorname {PolyLog}\left (2,e^{2 \text {csch}^{-1}(c x)}\right ) \]
1/2*b*d^2*arccsch(c*x)^2+d*e*x^2*(a+b*arccsch(c*x))+1/4*e^2*x^4*(a+b*arccs ch(c*x))-b*d^2*arccsch(c*x)*ln(1-(1/c/x+(1+1/c^2/x^2)^(1/2))^2)+b*d^2*arcc sch(c*x)*ln(1/x)-d^2*(a+b*arccsch(c*x))*ln(1/x)-1/2*b*d^2*polylog(2,(1/c/x +(1+1/c^2/x^2)^(1/2))^2)+1/6*b*(6*c^2*d-e)*e*x*(1+1/c^2/x^2)^(1/2)/c^3+1/1 2*b*e^2*x^3*(1+1/c^2/x^2)^(1/2)/c
Time = 0.34 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.84 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{x} \, dx=a d e x^2+\frac {1}{4} a e^2 x^4+\frac {b e^2 \sqrt {1+\frac {1}{c^2 x^2}} x \left (-2+c^2 x^2\right )}{12 c^3}+\frac {1}{4} b e^2 x^4 \text {csch}^{-1}(c x)+\frac {b d e x \left (\sqrt {1+\frac {1}{c^2 x^2}}+c x \text {csch}^{-1}(c x)\right )}{c}+a d^2 \log (x)+\frac {1}{2} b d^2 \left (\text {csch}^{-1}(c x) \left (\text {csch}^{-1}(c x)-2 \log \left (1-e^{2 \text {csch}^{-1}(c x)}\right )\right )-\operatorname {PolyLog}\left (2,e^{2 \text {csch}^{-1}(c x)}\right )\right ) \]
a*d*e*x^2 + (a*e^2*x^4)/4 + (b*e^2*Sqrt[1 + 1/(c^2*x^2)]*x*(-2 + c^2*x^2)) /(12*c^3) + (b*e^2*x^4*ArcCsch[c*x])/4 + (b*d*e*x*(Sqrt[1 + 1/(c^2*x^2)] + c*x*ArcCsch[c*x]))/c + a*d^2*Log[x] + (b*d^2*(ArcCsch[c*x]*(ArcCsch[c*x] - 2*Log[1 - E^(2*ArcCsch[c*x])]) - PolyLog[2, E^(2*ArcCsch[c*x])]))/2
Time = 0.78 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6858, 6237, 27, 7293, 2009}
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.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{x} \, dx\) |
| \(\Big \downarrow \) 6858 |
| \(\displaystyle -\int \left (\frac {d}{x^2}+e\right )^2 x^5 \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )d\frac {1}{x}\) |
| \(\Big \downarrow \) 6237 |
| \(\displaystyle \frac {b \int -\frac {e \left (\frac {4 d}{x^2}+e\right ) x^4-4 d^2 \log \left (\frac {1}{x}\right )}{4 \sqrt {1+\frac {1}{c^2 x^2}}}d\frac {1}{x}}{c}-d^2 \log \left (\frac {1}{x}\right ) \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )+d e x^2 \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )+\frac {1}{4} e^2 x^4 \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b \int \frac {e \left (\frac {4 d}{x^2}+e\right ) x^4-4 d^2 \log \left (\frac {1}{x}\right )}{\sqrt {1+\frac {1}{c^2 x^2}}}d\frac {1}{x}}{4 c}-d^2 \log \left (\frac {1}{x}\right ) \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )+d e x^2 \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )+\frac {1}{4} e^2 x^4 \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {b \int \left (\frac {e \left (\frac {4 d}{x^2}+e\right ) x^4}{\sqrt {1+\frac {1}{c^2 x^2}}}-\frac {4 d^2 \log \left (\frac {1}{x}\right )}{\sqrt {1+\frac {1}{c^2 x^2}}}\right )d\frac {1}{x}}{4 c}-d^2 \log \left (\frac {1}{x}\right ) \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )+d e x^2 \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )+\frac {1}{4} e^2 x^4 \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -d^2 \log \left (\frac {1}{x}\right ) \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )+d e x^2 \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )+\frac {1}{4} e^2 x^4 \left (a+b \text {arcsinh}\left (\frac {1}{c x}\right )\right )-\frac {b \left (2 c d^2 \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {1}{c x}\right )}\right )-2 c d^2 \text {arcsinh}\left (\frac {1}{c x}\right )^2+4 c d^2 \text {arcsinh}\left (\frac {1}{c x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\frac {1}{c x}\right )}\right )-4 c d^2 \log \left (\frac {1}{x}\right ) \text {arcsinh}\left (\frac {1}{c x}\right )-\frac {2}{3} e x \sqrt {\frac {1}{c^2 x^2}+1} \left (6 d-\frac {e}{c^2}\right )-\frac {1}{3} e^2 x^3 \sqrt {\frac {1}{c^2 x^2}+1}\right )}{4 c}\) |
d*e*x^2*(a + b*ArcSinh[1/(c*x)]) + (e^2*x^4*(a + b*ArcSinh[1/(c*x)]))/4 - d^2*(a + b*ArcSinh[1/(c*x)])*Log[x^(-1)] - (b*((-2*e*(6*d - e/c^2)*Sqrt[1 + 1/(c^2*x^2)]*x)/3 - (e^2*Sqrt[1 + 1/(c^2*x^2)]*x^3)/3 - 2*c*d^2*ArcSinh[ 1/(c*x)]^2 + 4*c*d^2*ArcSinh[1/(c*x)]*Log[1 - E^(2*ArcSinh[1/(c*x)])] - 4* c*d^2*ArcSinh[1/(c*x)]*Log[x^(-1)] + 2*c*d^2*PolyLog[2, E^(2*ArcSinh[1/(c* x)])]))/(4*c)
3.1.96.3.1 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[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x _)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Sim p[(a + b*ArcSinh[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/Sqrt[1 + c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[e, c^2*d] && IntegerQ[p] && (GtQ[p, 0] || (IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))
Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_.) + (e_.)*(x_ )^2)^(p_.), x_Symbol] :> -Subst[Int[(e + d*x^2)^p*((a + b*ArcSinh[x/c])^n/x ^(m + 2*(p + 1))), x], x, 1/x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[n, 0 ] && IntegersQ[m, p]
\[\int \frac {\left (e \,x^{2}+d \right )^{2} \left (a +b \,\operatorname {arccsch}\left (c x \right )\right )}{x}d x\]
\[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsch}\left (c x\right ) + a\right )}}{x} \,d x } \]
integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d ^2)*arccsch(c*x))/x, x)
\[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{x} \, dx=\int \frac {\left (a + b \operatorname {acsch}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x}\, dx \]
\[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsch}\left (c x\right ) + a\right )}}{x} \,d x } \]
1/4*a*e^2*x^4 + 4*b*c^2*d^2*integrate(1/4*x*log(x)/(sqrt(c^2*x^2 + 1)*c^2* x^2 + c^2*x^2 + sqrt(c^2*x^2 + 1) + 1), x) + a*d*e*x^2 - b*d^2*log(c)*log( x) - 1/4*(2*log(c^2*x^2 + 1)*log(x) + dilog(-c^2*x^2))*b*d^2 + a*d^2*log(x ) + 1/2*b*d*e*(2*sqrt(c^2*x^2 + 1) - log(c^2*x^2 + 1))/c^2 - 1/24*(3*c^2*x ^2 - 2*(c^2*x^2 + 1)^(3/2) + 6*sqrt(c^2*x^2 + 1) - 3*log(c^2*x^2 + 1) + 3) *b*e^2/c^4 - 1/8*(2*b*c^2*e^2*x^4*log(c) + 4*b*c^2*d^2*log(x)^2 + (8*c^2*d *e*log(c) - e^2)*b*x^2 + 2*(b*c^2*e^2*x^4 + 4*b*c^2*d*e*x^2)*log(x) - 2*(b *c^2*e^2*x^4 + 4*b*c^2*d*e*x^2 + 4*b*c^2*d^2*log(x))*log(sqrt(c^2*x^2 + 1) + 1))/c^2 + 1/8*(4*c^2*d*e - e^2)*b*log(c^2*x^2 + 1)/c^4
\[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsch}\left (c x\right ) + a\right )}}{x} \,d x } \]
Timed out. \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{x} \, dx=\int \frac {{\left (e\,x^2+d\right )}^2\,\left (a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )\right )}{x} \,d x \]