Integrand size = 25, antiderivative size = 383 \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=-\frac {b n x}{3 e^2 \sqrt {d+e x^2}}+\frac {4 b \sqrt {d} n \sqrt {1+\frac {e x^2}{d}} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{5/2} \sqrt {d+e x^2}}+\frac {b \sqrt {d} n \sqrt {1+\frac {e x^2}{d}} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 e^{5/2} \sqrt {d+e x^2}}-\frac {b \sqrt {d} n \sqrt {1+\frac {e x^2}{d}} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{e^{5/2} \sqrt {d+e x^2}}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^{5/2} \sqrt {d+e x^2}}-\frac {b \sqrt {d} n \sqrt {1+\frac {e x^2}{d}} \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 e^{5/2} \sqrt {d+e x^2}} \]
-1/3*x^3*(a+b*ln(c*x^n))/e/(e*x^2+d)^(3/2)-1/3*b*n*x/e^2/(e*x^2+d)^(1/2)-x *(a+b*ln(c*x^n))/e^2/(e*x^2+d)^(1/2)+4/3*b*n*arcsinh(x*e^(1/2)/d^(1/2))*d^ (1/2)*(1+e*x^2/d)^(1/2)/e^(5/2)/(e*x^2+d)^(1/2)+1/2*b*n*arcsinh(x*e^(1/2)/ d^(1/2))^2*d^(1/2)*(1+e*x^2/d)^(1/2)/e^(5/2)/(e*x^2+d)^(1/2)-b*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)*d^(1/2)*(1 +e*x^2/d)^(1/2)/e^(5/2)/(e*x^2+d)^(1/2)+arcsinh(x*e^(1/2)/d^(1/2))*(a+b*ln (c*x^n))*d^(1/2)*(1+e*x^2/d)^(1/2)/e^(5/2)/(e*x^2+d)^(1/2)-1/2*b*n*polylog (2,(x*e^(1/2)/d^(1/2)+(1+e*x^2/d)^(1/2))^2)*d^(1/2)*(1+e*x^2/d)^(1/2)/e^(5 /2)/(e*x^2+d)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.59 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.64 \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=-\frac {b n \sqrt {1+\frac {e x^2}{d}} \left (3 e^{5/2} x^5 \left (d+e x^2\right )^2 \, _3F_2\left (\frac {5}{2},\frac {5}{2},\frac {5}{2};\frac {7}{2},\frac {7}{2};-\frac {e x^2}{d}\right )+25 d^3 \sqrt {e} x \left (3 d+4 e x^2\right ) \sqrt {1+\frac {e x^2}{d}} \log (x)-75 d^{5/2} \left (d+e x^2\right )^2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)\right )}{75 d^2 e^{5/2} \left (d+e x^2\right )^{5/2}}-\frac {x \left (3 d+4 e x^2\right ) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}+\frac {\left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{e^{5/2}} \]
-1/75*(b*n*Sqrt[1 + (e*x^2)/d]*(3*e^(5/2)*x^5*(d + e*x^2)^2*Hypergeometric PFQ[{5/2, 5/2, 5/2}, {7/2, 7/2}, -((e*x^2)/d)] + 25*d^3*Sqrt[e]*x*(3*d + 4 *e*x^2)*Sqrt[1 + (e*x^2)/d]*Log[x] - 75*d^(5/2)*(d + e*x^2)^2*ArcSinh[(Sqr t[e]*x)/Sqrt[d]]*Log[x]))/(d^2*e^(5/2)*(d + e*x^2)^(5/2)) - (x*(3*d + 4*e* x^2)*(a - b*n*Log[x] + b*Log[c*x^n]))/(3*e^2*(d + e*x^2)^(3/2)) + ((a - b* n*Log[x] + b*Log[c*x^n])*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/e^(5/2)
Time = 0.75 (sec) , antiderivative size = 297, 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.
| \(\displaystyle \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2786 |
| \(\displaystyle \frac {\sqrt {\frac {e x^2}{d}+1} \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (\frac {e x^2}{d}+1\right )^{5/2}}dx}{d^2 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle \frac {\sqrt {\frac {e x^2}{d}+1} \left (-b n \int \left (\frac {d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{e^{5/2} x}-\frac {d^3 \left (4 e x^2+3 d\right ) \sqrt {\frac {e x^2}{d}+1}}{3 e^2 \left (e x^2+d\right )^2}\right )dx+\frac {d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^{5/2}}-\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {\frac {e x^2}{d}+1}}-\frac {d x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (\frac {e x^2}{d}+1\right )^{3/2}}\right )}{d^2 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {\frac {e x^2}{d}+1} \left (\frac {d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^{5/2}}-\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {\frac {e x^2}{d}+1}}-\frac {d x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (\frac {e x^2}{d}+1\right )^{3/2}}-b n \left (\frac {d^{5/2} \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 e^{5/2}}-\frac {d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 e^{5/2}}-\frac {4 d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{5/2}}+\frac {d^{5/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{e^{5/2}}+\frac {d^2 x}{3 e^2 \sqrt {\frac {e x^2}{d}+1}}\right )\right )}{d^2 \sqrt {d+e x^2}}\) |
(Sqrt[1 + (e*x^2)/d]*(-1/3*(d*x^3*(a + b*Log[c*x^n]))/(e*(1 + (e*x^2)/d)^( 3/2)) - (d^2*x*(a + b*Log[c*x^n]))/(e^2*Sqrt[1 + (e*x^2)/d]) + (d^(5/2)*Ar cSinh[(Sqrt[e]*x)/Sqrt[d]]*(a + b*Log[c*x^n]))/e^(5/2) - b*n*((d^2*x)/(3*e ^2*Sqrt[1 + (e*x^2)/d]) - (4*d^(5/2)*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])/(3*e^(5 /2)) - (d^(5/2)*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]^2)/(2*e^(5/2)) + (d^(5/2)*Arc Sinh[(Sqrt[e]*x)/Sqrt[d]]*Log[1 - E^(2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])])/e^( 5/2) + (d^(5/2)*PolyLog[2, E^(2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])])/(2*e^(5/2) ))))/(d^2*Sqrt[d + e*x^2])
3.4.4.3.1 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])
\[\int \frac {x^{4} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
integral((sqrt(e*x^2 + d)*b*x^4*log(c*x^n) + sqrt(e*x^2 + d)*a*x^4)/(e^3*x ^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {x^{4} \left (a + b \log {\left (c x^{n} \right )}\right )}{\left (d + e x^{2}\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {x^4\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]