Integrand size = 25, antiderivative size = 464 \[ \int x^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {11 b d^2 n x \sqrt {d+e x^2}}{192 e}-\frac {23}{288} b d n x^3 \sqrt {d+e x^2}-\frac {1}{36} b e n x^5 \sqrt {d+e x^2}-\frac {b d^{5/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{192 e^{3/2} \sqrt {1+\frac {e x^2}{d}}}-\frac {b d^{5/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{32 e^{3/2} \sqrt {1+\frac {e x^2}{d}}}+\frac {b d^{5/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{16 e^{3/2} \sqrt {1+\frac {e x^2}{d}}}+\frac {d^2 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e}+\frac {1}{8} d x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac {1}{6} x^3 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )-\frac {d^{5/2} \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{3/2} \sqrt {1+\frac {e x^2}{d}}}+\frac {b d^{5/2} n \sqrt {d+e x^2} \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{32 e^{3/2} \sqrt {1+\frac {e x^2}{d}}} \]
1/6*x^3*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n))-11/192*b*d^2*n*x*(e*x^2+d)^(1/2)/e -23/288*b*d*n*x^3*(e*x^2+d)^(1/2)-1/36*b*e*n*x^5*(e*x^2+d)^(1/2)+1/16*d^2* x*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e+1/8*d*x^3*(a+b*ln(c*x^n))*(e*x^2+d)^(1 /2)-1/192*b*d^(5/2)*n*arcsinh(x*e^(1/2)/d^(1/2))*(e*x^2+d)^(1/2)/e^(3/2)/( 1+e*x^2/d)^(1/2)-1/32*b*d^(5/2)*n*arcsinh(x*e^(1/2)/d^(1/2))^2*(e*x^2+d)^( 1/2)/e^(3/2)/(1+e*x^2/d)^(1/2)+1/16*b*d^(5/2)*n*arcsinh(x*e^(1/2)/d^(1/2)) *ln(1-(x*e^(1/2)/d^(1/2)+(1+e*x^2/d)^(1/2))^2)*(e*x^2+d)^(1/2)/e^(3/2)/(1+ e*x^2/d)^(1/2)-1/16*d^(5/2)*arcsinh(x*e^(1/2)/d^(1/2))*(a+b*ln(c*x^n))*(e* x^2+d)^(1/2)/e^(3/2)/(1+e*x^2/d)^(1/2)+1/32*b*d^(5/2)*n*polylog(2,(x*e^(1/ 2)/d^(1/2)+(1+e*x^2/d)^(1/2))^2)*(e*x^2+d)^(1/2)/e^(3/2)/(1+e*x^2/d)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.63 (sec) , antiderivative size = 331, normalized size of antiderivative = 0.71 \[ \int x^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {-400 b d e^{3/2} n x^3 \sqrt {d+e x^2} \, _3F_2\left (-\frac {1}{2},\frac {3}{2},\frac {3}{2};\frac {5}{2},\frac {5}{2};-\frac {e x^2}{d}\right )-144 b e^{5/2} n x^5 \sqrt {d+e x^2} \, _3F_2\left (-\frac {1}{2},\frac {5}{2},\frac {5}{2};\frac {7}{2},\frac {7}{2};-\frac {e x^2}{d}\right )-75 \left (3 b d^{5/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)+\sqrt {1+\frac {e x^2}{d}} \left (-a \sqrt {e} x \sqrt {d+e x^2} \left (3 d^2+14 d e x^2+8 e^2 x^4\right )+3 d^3 (a-b n \log (x)) \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )-b \log \left (c x^n\right ) \left (\sqrt {e} x \sqrt {d+e x^2} \left (3 d^2+14 d e x^2+8 e^2 x^4\right )-3 d^3 \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )\right )\right )\right )}{3600 e^{3/2} \sqrt {1+\frac {e x^2}{d}}} \]
(-400*b*d*e^(3/2)*n*x^3*Sqrt[d + e*x^2]*HypergeometricPFQ[{-1/2, 3/2, 3/2} , {5/2, 5/2}, -((e*x^2)/d)] - 144*b*e^(5/2)*n*x^5*Sqrt[d + e*x^2]*Hypergeo metricPFQ[{-1/2, 5/2, 5/2}, {7/2, 7/2}, -((e*x^2)/d)] - 75*(3*b*d^(5/2)*n* Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*Log[x] + Sqrt[1 + (e*x^2)/d]* (-(a*Sqrt[e]*x*Sqrt[d + e*x^2]*(3*d^2 + 14*d*e*x^2 + 8*e^2*x^4)) + 3*d^3*( a - b*n*Log[x])*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]] - b*Log[c*x^n]*(Sqrt[e] *x*Sqrt[d + e*x^2]*(3*d^2 + 14*d*e*x^2 + 8*e^2*x^4) - 3*d^3*Log[e*x + Sqrt [e]*Sqrt[d + e*x^2]]))))/(3600*e^(3/2)*Sqrt[1 + (e*x^2)/d])
Time = 0.88 (sec) , antiderivative size = 371, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2786, 2792, 27, 2010, 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.
| \(\displaystyle \int x^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2786 |
| \(\displaystyle \frac {d \sqrt {d+e x^2} \int x^2 \left (\frac {e x^2}{d}+1\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )dx}{\sqrt {\frac {e x^2}{d}+1}}\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle \frac {d \sqrt {d+e x^2} \left (-b n \int \frac {\sqrt {e} x \sqrt {\frac {e x^2}{d}+1} \left (8 e^2 x^4+14 d e x^2+3 d^2\right )-3 d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{48 d e^{3/2} x}dx-\frac {d^{3/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{3/2}}+\frac {d x \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )}{16 e}+\frac {1}{6} x^3 \left (\frac {e x^2}{d}+1\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )+\frac {1}{8} x^3 \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )\right )}{\sqrt {\frac {e x^2}{d}+1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d \sqrt {d+e x^2} \left (-\frac {b n \int \frac {\sqrt {e} x \sqrt {\frac {e x^2}{d}+1} \left (8 e^2 x^4+14 d e x^2+3 d^2\right )-3 d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x}dx}{48 d e^{3/2}}-\frac {d^{3/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{3/2}}+\frac {d x \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )}{16 e}+\frac {1}{6} x^3 \left (\frac {e x^2}{d}+1\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )+\frac {1}{8} x^3 \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )\right )}{\sqrt {\frac {e x^2}{d}+1}}\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle \frac {d \sqrt {d+e x^2} \left (-\frac {b n \int \left (8 e^{5/2} \sqrt {\frac {e x^2}{d}+1} x^4+14 d e^{3/2} \sqrt {\frac {e x^2}{d}+1} x^2+3 d^2 \sqrt {e} \sqrt {\frac {e x^2}{d}+1}-\frac {3 d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x}\right )dx}{48 d e^{3/2}}-\frac {d^{3/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{3/2}}+\frac {d x \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )}{16 e}+\frac {1}{6} x^3 \left (\frac {e x^2}{d}+1\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )+\frac {1}{8} x^3 \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )\right )}{\sqrt {\frac {e x^2}{d}+1}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d \sqrt {d+e x^2} \left (-\frac {d^{3/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{3/2}}+\frac {d x \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )}{16 e}+\frac {1}{6} x^3 \left (\frac {e x^2}{d}+1\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )+\frac {1}{8} x^3 \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )-\frac {b n \left (-\frac {3}{2} d^{5/2} \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )+\frac {3}{2} d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2+\frac {1}{4} d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )-3 d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )+\frac {11}{4} d^2 \sqrt {e} x \sqrt {\frac {e x^2}{d}+1}+\frac {4}{3} e^{5/2} x^5 \sqrt {\frac {e x^2}{d}+1}+\frac {23}{6} d e^{3/2} x^3 \sqrt {\frac {e x^2}{d}+1}\right )}{48 d e^{3/2}}\right )}{\sqrt {\frac {e x^2}{d}+1}}\) |
(d*Sqrt[d + e*x^2]*((d*x*Sqrt[1 + (e*x^2)/d]*(a + b*Log[c*x^n]))/(16*e) + (x^3*Sqrt[1 + (e*x^2)/d]*(a + b*Log[c*x^n]))/8 + (x^3*(1 + (e*x^2)/d)^(3/2 )*(a + b*Log[c*x^n]))/6 - (d^(3/2)*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*(a + b*Log [c*x^n]))/(16*e^(3/2)) - (b*n*((11*d^2*Sqrt[e]*x*Sqrt[1 + (e*x^2)/d])/4 + (23*d*e^(3/2)*x^3*Sqrt[1 + (e*x^2)/d])/6 + (4*e^(5/2)*x^5*Sqrt[1 + (e*x^2) /d])/3 + (d^(5/2)*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])/4 + (3*d^(5/2)*ArcSinh[(Sq rt[e]*x)/Sqrt[d]]^2)/2 - 3*d^(5/2)*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*Log[1 - E^ (2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])] - (3*d^(5/2)*PolyLog[2, E^(2*ArcSinh[(Sq rt[e]*x)/Sqrt[d]])])/2))/(48*d*e^(3/2))))/Sqrt[1 + (e*x^2)/d]
3.3.68.3.1 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[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^ (q_), x_Symbol] :> Simp[d^IntPart[q]*((d + e*x^2)^FracPart[q]/(1 + (e/d)*x^ 2)^FracPart[q]) Int[x^m*(1 + (e/d)*x^2)^q*(a + b*Log[c*x^n]), x], x] /; F reeQ[{a, b, c, d, e, n}, x] && IntegerQ[m/2] && IntegerQ[q - 1/2] && !(LtQ [m + 2*q, -2] || GtQ[d, 0])
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x] }, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[SimplifyIntegrand[u/x, x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2] ) || InverseFunctionFreeQ[u, x]] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x ] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])
\[\int x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]
\[ \int x^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \,d x } \]
integral((b*e*x^4 + b*d*x^2)*sqrt(e*x^2 + d)*log(c*x^n) + (a*e*x^4 + a*d*x ^2)*sqrt(e*x^2 + d), x)
Timed out. \[ \int x^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \]
Exception generated. \[ \int x^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int x^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \,d x } \]
Timed out. \[ \int x^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^2\,{\left (e\,x^2+d\right )}^{3/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]