Integrand size = 25, antiderivative size = 409 \[ \int x^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {3 b d n x \sqrt {d+e x^2}}{32 e}-\frac {1}{16} b n x^3 \sqrt {d+e x^2}-\frac {b d^{3/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{32 e^{3/2} \sqrt {1+\frac {e x^2}{d}}}-\frac {b d^{3/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{16 e^{3/2} \sqrt {1+\frac {e x^2}{d}}}+\frac {b d^{3/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{8 e^{3/2} \sqrt {1+\frac {e x^2}{d}}}+\frac {d x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{4} x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )-\frac {d^{3/2} \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 e^{3/2} \sqrt {1+\frac {e x^2}{d}}}+\frac {b d^{3/2} n \sqrt {d+e x^2} \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{16 e^{3/2} \sqrt {1+\frac {e x^2}{d}}} \]
-3/32*b*d*n*x*(e*x^2+d)^(1/2)/e-1/16*b*n*x^3*(e*x^2+d)^(1/2)+1/8*d*x*(a+b* ln(c*x^n))*(e*x^2+d)^(1/2)/e+1/4*x^3*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)-1/32* b*d^(3/2)*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)-1/16*b*d^(3/2)*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)+1/8*b*d^(3/2)*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)-1/8*d^(3/2)*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)+1/16*b*d^(3/2)*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 = 0.27 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.61 \[ \int x^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {-8 b e^{3/2} n x^3 \sqrt {d+e x^2} \, _3F_2\left (-\frac {1}{2},\frac {3}{2},\frac {3}{2};\frac {5}{2},\frac {5}{2};-\frac {e x^2}{d}\right )-9 b d^{3/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)+9 \sqrt {1+\frac {e x^2}{d}} \left (a \sqrt {e} x \sqrt {d+e x^2} \left (d+2 e x^2\right )+d^2 (-a+b n \log (x)) \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )+b \log \left (c x^n\right ) \left (\sqrt {e} x \sqrt {d+e x^2} \left (d+2 e x^2\right )-d^2 \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )\right )\right )}{72 e^{3/2} \sqrt {1+\frac {e x^2}{d}}} \]
(-8*b*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)] - 9*b*d^(3/2)*n*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x )/Sqrt[d]]*Log[x] + 9*Sqrt[1 + (e*x^2)/d]*(a*Sqrt[e]*x*Sqrt[d + e*x^2]*(d + 2*e*x^2) + d^2*(-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]*(d + 2*e*x^2) - d^2*Log[e*x + Sqrt[e ]*Sqrt[d + e*x^2]])))/(72*e^(3/2)*Sqrt[1 + (e*x^2)/d])
Time = 0.65 (sec) , antiderivative size = 317, normalized size of antiderivative = 0.78, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2786, 2792, 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.
\(\displaystyle \int x^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2786 |
\(\displaystyle \frac {\sqrt {d+e x^2} \int x^2 \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )dx}{\sqrt {\frac {e x^2}{d}+1}}\) |
\(\Big \downarrow \) 2792 |
\(\displaystyle \frac {\sqrt {d+e x^2} \left (-b n \int \left (\frac {\left (2 e x^2+d\right ) \sqrt {\frac {e x^2}{d}+1}}{8 e}-\frac {d^{3/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 e^{3/2} x}\right )dx-\frac {d^{3/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 e^{3/2}}+\frac {d x \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{4} x^3 \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )\right )}{\sqrt {\frac {e x^2}{d}+1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {d+e x^2} \left (-\frac {d^{3/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 e^{3/2}}+\frac {d x \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{4} x^3 \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )-b n \left (-\frac {d^{3/2} \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{16 e^{3/2}}+\frac {d^{3/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{16 e^{3/2}}+\frac {d^{3/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{32 e^{3/2}}-\frac {d^{3/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{8 e^{3/2}}+\frac {d x \left (\frac {e x^2}{d}+1\right )^{3/2}}{16 e}+\frac {d x \sqrt {\frac {e x^2}{d}+1}}{32 e}\right )\right )}{\sqrt {\frac {e x^2}{d}+1}}\) |
(Sqrt[d + e*x^2]*((d*x*Sqrt[1 + (e*x^2)/d]*(a + b*Log[c*x^n]))/(8*e) + (x^ 3*Sqrt[1 + (e*x^2)/d]*(a + b*Log[c*x^n]))/4 - (d^(3/2)*ArcSinh[(Sqrt[e]*x) /Sqrt[d]]*(a + b*Log[c*x^n]))/(8*e^(3/2)) - b*n*((d*x*Sqrt[1 + (e*x^2)/d]) /(32*e) + (d*x*(1 + (e*x^2)/d)^(3/2))/(16*e) + (d^(3/2)*ArcSinh[(Sqrt[e]*x )/Sqrt[d]])/(32*e^(3/2)) + (d^(3/2)*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]^2)/(16*e^ (3/2)) - (d^(3/2)*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*Log[1 - E^(2*ArcSinh[(Sqrt[ e]*x)/Sqrt[d]])])/(8*e^(3/2)) - (d^(3/2)*PolyLog[2, E^(2*ArcSinh[(Sqrt[e]* x)/Sqrt[d]])])/(16*e^(3/2)))))/Sqrt[1 + (e*x^2)/d]
3.3.57.3.1 Defintions of rubi rules used
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])
\[\int x^{2} \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e \,x^{2}+d}d x\]
\[ \int x^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \sqrt {e x^{2} + d} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \,d x } \]
\[ \int x^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^{2} \left (a + b \log {\left (c x^{n} \right )}\right ) \sqrt {d + e x^{2}}\, dx \]
Exception generated. \[ \int x^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]
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
\[ \int x^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \sqrt {e x^{2} + d} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \,d x } \]
Timed out. \[ \int x^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^2\,\sqrt {e\,x^2+d}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]