Integrand size = 25, antiderivative size = 443 \[ \int \frac {x^6 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {b d n x}{3 e^3 \sqrt {d+e x^2}}-\frac {b n x \sqrt {d+e x^2}}{4 e^3}-\frac {31 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{12 e^{7/2} \sqrt {d+e x^2}}-\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{4 e^{7/2} \sqrt {d+e x^2}}+\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {d+e x^2}}+\frac {5 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 d^{3/2} \sqrt {1+\frac {e x^2}{d}} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}+\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{4 e^{7/2} \sqrt {d+e x^2}} \]
-1/3*x^5*(a+b*ln(c*x^n))/e/(e*x^2+d)^(3/2)+5/6*b*d*n*x/e^3/(e*x^2+d)^(1/2) +1/2*b*n*x^3/e^2/(e*x^2+d)^(1/2)-5/3*x^3*(a+b*ln(c*x^n))/e^2/(e*x^2+d)^(1/ 2)-3/4*b*n*x*(e*x^2+d)^(1/2)/e^3+5/2*x*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e^3 -31/12*b*d^(3/2)*n*arcsinh(x*e^(1/2)/d^(1/2))*(1+e*x^2/d)^(1/2)/e^(7/2)/(e *x^2+d)^(1/2)-5/4*b*d^(3/2)*n*arcsinh(x*e^(1/2)/d^(1/2))^2*(1+e*x^2/d)^(1/ 2)/e^(7/2)/(e*x^2+d)^(1/2)-5*b*d^(3/2)*n*arcsinh(x*e^(1/2)/d^(1/2))*arctan h((x*e^(1/2)/d^(1/2)+(1+e*x^2/d)^(1/2))^2)*(1+e*x^2/d)^(1/2)/e^(7/2)/(e*x^ 2+d)^(1/2)+5/2*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)*(1+e*x^2/d)^(1/2)/e^(7/2)/(e*x^2+d)^(1/2)-5/2*d^ (3/2)*arcsinh(x*e^(1/2)/d^(1/2))*(a+b*ln(c*x^n))*(1+e*x^2/d)^(1/2)/e^(7/2) /(e*x^2+d)^(1/2)+5/4*b*d^(3/2)*n*polylog(2,(x*e^(1/2)/d^(1/2)+(1+e*x^2/d)^ (1/2))^2)*(1+e*x^2/d)^(1/2)/e^(7/2)/(e*x^2+d)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.21 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.45 \[ \int \frac {x^6 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {b n x^7 \sqrt {1+\frac {e x^2}{d}} \left (5 \, _3F_2\left (\frac {7}{2},\frac {7}{2},\frac {7}{2};\frac {9}{2},\frac {9}{2};-\frac {e x^2}{d}\right )+7 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {7}{2},\frac {9}{2},-\frac {e x^2}{d}\right ) (-1+2 \log (x))\right )}{98 d^2 \sqrt {d+e x^2}}+\frac {x \left (15 d^2+20 d e x^2+3 e^2 x^4\right ) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{6 e^3 \left (d+e x^2\right )^{3/2}}-\frac {5 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{2 e^{7/2}} \]
(b*n*x^7*Sqrt[1 + (e*x^2)/d]*(5*HypergeometricPFQ[{7/2, 7/2, 7/2}, {9/2, 9 /2}, -((e*x^2)/d)] + 7*Hypergeometric2F1[5/2, 7/2, 9/2, -((e*x^2)/d)]*(-1 + 2*Log[x])))/(98*d^2*Sqrt[d + e*x^2]) + (x*(15*d^2 + 20*d*e*x^2 + 3*e^2*x ^4)*(a - b*n*Log[x] + b*Log[c*x^n]))/(6*e^3*(d + e*x^2)^(3/2)) - (5*d*(a - b*n*Log[x] + b*Log[c*x^n])*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(2*e^(7/2) )
Time = 0.77 (sec) , antiderivative size = 367, normalized size of antiderivative = 0.83, 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.
| \(\displaystyle \int \frac {x^6 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2786 |
| \(\displaystyle \frac {\sqrt {\frac {e x^2}{d}+1} \int \frac {x^6 \left (a+b \log \left (c x^n\right )\right )}{\left (\frac {e x^2}{d}+1\right )^{5/2}}dx}{d^2 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle \frac {\sqrt {\frac {e x^2}{d}+1} \left (-b n \int \left (\frac {d^3 \sqrt {\frac {e x^2}{d}+1} \left (3 e^2 x^4+20 d e x^2+15 d^2\right )}{6 e^3 \left (e x^2+d\right )^2}-\frac {5 d^{7/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 e^{7/2} x}\right )dx-\frac {5 d^{7/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2}}+\frac {5 d^3 x \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 d^2 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {\frac {e x^2}{d}+1}}-\frac {d x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (\frac {e x^2}{d}+1\right )^{3/2}}\right )}{d^2 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {\frac {e x^2}{d}+1} \left (-\frac {5 d^{7/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2}}+\frac {5 d^3 x \sqrt {\frac {e x^2}{d}+1} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 d^2 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {\frac {e x^2}{d}+1}}-\frac {d x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (\frac {e x^2}{d}+1\right )^{3/2}}-b n \left (-\frac {5 d^{7/2} \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{4 e^{7/2}}+\frac {5 d^{7/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{4 e^{7/2}}+\frac {31 d^{7/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{12 e^{7/2}}-\frac {5 d^{7/2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 e^{7/2}}+\frac {d^3 x \sqrt {\frac {e x^2}{d}+1}}{4 e^3}-\frac {d^3 x}{3 e^3 \sqrt {\frac {e x^2}{d}+1}}\right )\right )}{d^2 \sqrt {d+e x^2}}\) |
(Sqrt[1 + (e*x^2)/d]*(-1/3*(d*x^5*(a + b*Log[c*x^n]))/(e*(1 + (e*x^2)/d)^( 3/2)) - (5*d^2*x^3*(a + b*Log[c*x^n]))/(3*e^2*Sqrt[1 + (e*x^2)/d]) + (5*d^ 3*x*Sqrt[1 + (e*x^2)/d]*(a + b*Log[c*x^n]))/(2*e^3) - (5*d^(7/2)*ArcSinh[( Sqrt[e]*x)/Sqrt[d]]*(a + b*Log[c*x^n]))/(2*e^(7/2)) - b*n*(-1/3*(d^3*x)/(e ^3*Sqrt[1 + (e*x^2)/d]) + (d^3*x*Sqrt[1 + (e*x^2)/d])/(4*e^3) + (31*d^(7/2 )*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])/(12*e^(7/2)) + (5*d^(7/2)*ArcSinh[(Sqrt[e] *x)/Sqrt[d]]^2)/(4*e^(7/2)) - (5*d^(7/2)*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*Log[ 1 - E^(2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])])/(2*e^(7/2)) - (5*d^(7/2)*PolyLog[ 2, E^(2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])])/(4*e^(7/2)))))/(d^2*Sqrt[d + e*x^2 ])
3.4.3.3.1 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])
\[\int \frac {x^{6} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {x^6 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{6}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
integral((sqrt(e*x^2 + d)*b*x^6*log(c*x^n) + sqrt(e*x^2 + d)*a*x^6)/(e^3*x ^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)
Timed out. \[ \int \frac {x^6 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x^6 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^6 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{6}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^6 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {x^6\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]