Integrand size = 28, antiderivative size = 403 \[ \int x^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {7 b e^5 k n \sqrt {x}}{9 f^5}+\frac {2 b e^4 k n x}{9 f^4}-\frac {b e^3 k n x^{3/2}}{9 f^3}+\frac {5 b e^2 k n x^2}{72 f^2}-\frac {11 b e k n x^{5/2}}{225 f}+\frac {1}{27} b k n x^3+\frac {b e^6 k n \log \left (e+f \sqrt {x}\right )}{9 f^6}-\frac {1}{9} b n x^3 \log \left (d \left (e+f \sqrt {x}\right )^k\right )+\frac {2 b e^6 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{3 f^6}+\frac {e^5 k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{3 f^5}-\frac {e^4 k x \left (a+b \log \left (c x^n\right )\right )}{6 f^4}+\frac {e^3 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{9 f^3}-\frac {e^2 k x^2 \left (a+b \log \left (c x^n\right )\right )}{12 f^2}+\frac {e k x^{5/2} \left (a+b \log \left (c x^n\right )\right )}{15 f}-\frac {1}{18} k x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {e^6 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 f^6}+\frac {1}{3} x^3 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {2 b e^6 k n \operatorname {PolyLog}\left (2,1+\frac {f \sqrt {x}}{e}\right )}{3 f^6} \]
2/9*b*e^4*k*n*x/f^4-1/9*b*e^3*k*n*x^(3/2)/f^3+5/72*b*e^2*k*n*x^2/f^2-11/22 5*b*e*k*n*x^(5/2)/f+1/27*b*k*n*x^3-1/6*e^4*k*x*(a+b*ln(c*x^n))/f^4+1/9*e^3 *k*x^(3/2)*(a+b*ln(c*x^n))/f^3-1/12*e^2*k*x^2*(a+b*ln(c*x^n))/f^2+1/15*e*k *x^(5/2)*(a+b*ln(c*x^n))/f-1/18*k*x^3*(a+b*ln(c*x^n))+1/9*b*e^6*k*n*ln(e+f *x^(1/2))/f^6-1/3*e^6*k*(a+b*ln(c*x^n))*ln(e+f*x^(1/2))/f^6+2/3*b*e^6*k*n* ln(-f*x^(1/2)/e)*ln(e+f*x^(1/2))/f^6-1/9*b*n*x^3*ln(d*(e+f*x^(1/2))^k)+1/3 *x^3*(a+b*ln(c*x^n))*ln(d*(e+f*x^(1/2))^k)+2/3*b*e^6*k*n*polylog(2,1+f*x^( 1/2)/e)/f^6-7/9*b*e^5*k*n*x^(1/2)/f^5+1/3*e^5*k*(a+b*ln(c*x^n))*x^(1/2)/f^ 5
Time = 0.36 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.08 \[ \int x^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {-1800 a e^5 f k \sqrt {x}+4200 b e^5 f k n \sqrt {x}+900 a e^4 f^2 k x-1200 b e^4 f^2 k n x-600 a e^3 f^3 k x^{3/2}+600 b e^3 f^3 k n x^{3/2}+450 a e^2 f^4 k x^2-375 b e^2 f^4 k n x^2-360 a e f^5 k x^{5/2}+264 b e f^5 k n x^{5/2}+300 a f^6 k x^3-200 b f^6 k n x^3-1800 a f^6 x^3 \log \left (d \left (e+f \sqrt {x}\right )^k\right )+600 b f^6 n x^3 \log \left (d \left (e+f \sqrt {x}\right )^k\right )+1800 b e^6 k n \log \left (1+\frac {f \sqrt {x}}{e}\right ) \log (x)-1800 b e^5 f k \sqrt {x} \log \left (c x^n\right )+900 b e^4 f^2 k x \log \left (c x^n\right )-600 b e^3 f^3 k x^{3/2} \log \left (c x^n\right )+450 b e^2 f^4 k x^2 \log \left (c x^n\right )-360 b e f^5 k x^{5/2} \log \left (c x^n\right )+300 b f^6 k x^3 \log \left (c x^n\right )-1800 b f^6 x^3 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log \left (c x^n\right )+600 e^6 k \log \left (e+f \sqrt {x}\right ) \left (3 a-b n-3 b n \log (x)+3 b \log \left (c x^n\right )\right )+3600 b e^6 k n \operatorname {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right )}{5400 f^6} \]
-1/5400*(-1800*a*e^5*f*k*Sqrt[x] + 4200*b*e^5*f*k*n*Sqrt[x] + 900*a*e^4*f^ 2*k*x - 1200*b*e^4*f^2*k*n*x - 600*a*e^3*f^3*k*x^(3/2) + 600*b*e^3*f^3*k*n *x^(3/2) + 450*a*e^2*f^4*k*x^2 - 375*b*e^2*f^4*k*n*x^2 - 360*a*e*f^5*k*x^( 5/2) + 264*b*e*f^5*k*n*x^(5/2) + 300*a*f^6*k*x^3 - 200*b*f^6*k*n*x^3 - 180 0*a*f^6*x^3*Log[d*(e + f*Sqrt[x])^k] + 600*b*f^6*n*x^3*Log[d*(e + f*Sqrt[x ])^k] + 1800*b*e^6*k*n*Log[1 + (f*Sqrt[x])/e]*Log[x] - 1800*b*e^5*f*k*Sqrt [x]*Log[c*x^n] + 900*b*e^4*f^2*k*x*Log[c*x^n] - 600*b*e^3*f^3*k*x^(3/2)*Lo g[c*x^n] + 450*b*e^2*f^4*k*x^2*Log[c*x^n] - 360*b*e*f^5*k*x^(5/2)*Log[c*x^ n] + 300*b*f^6*k*x^3*Log[c*x^n] - 1800*b*f^6*x^3*Log[d*(e + f*Sqrt[x])^k]* Log[c*x^n] + 600*e^6*k*Log[e + f*Sqrt[x]]*(3*a - b*n - 3*b*n*Log[x] + 3*b* Log[c*x^n]) + 3600*b*e^6*k*n*PolyLog[2, -((f*Sqrt[x])/e)])/f^6
Time = 0.60 (sec) , antiderivative size = 388, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2823, 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.
| \(\displaystyle \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \, dx\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle -b n \int \left (-\frac {k \log \left (e+f \sqrt {x}\right ) e^6}{3 f^6 x}+\frac {k e^5}{3 f^5 \sqrt {x}}-\frac {k e^4}{6 f^4}+\frac {k \sqrt {x} e^3}{9 f^3}-\frac {k x e^2}{12 f^2}+\frac {k x^{3/2} e}{15 f}-\frac {k x^2}{18}+\frac {1}{3} x^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right )\right )dx+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {e^6 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 f^6}+\frac {e^5 k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{3 f^5}-\frac {e^4 k x \left (a+b \log \left (c x^n\right )\right )}{6 f^4}+\frac {e^3 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{9 f^3}-\frac {e^2 k x^2 \left (a+b \log \left (c x^n\right )\right )}{12 f^2}+\frac {e k x^{5/2} \left (a+b \log \left (c x^n\right )\right )}{15 f}-\frac {1}{18} k x^3 \left (a+b \log \left (c x^n\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {e^6 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 f^6}+\frac {e^5 k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{3 f^5}-\frac {e^4 k x \left (a+b \log \left (c x^n\right )\right )}{6 f^4}+\frac {e^3 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{9 f^3}-\frac {e^2 k x^2 \left (a+b \log \left (c x^n\right )\right )}{12 f^2}+\frac {e k x^{5/2} \left (a+b \log \left (c x^n\right )\right )}{15 f}-\frac {1}{18} k x^3 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {1}{9} x^3 \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {2 e^6 k \operatorname {PolyLog}\left (2,\frac {\sqrt {x} f}{e}+1\right )}{3 f^6}-\frac {e^6 k \log \left (e+f \sqrt {x}\right )}{9 f^6}-\frac {2 e^6 k \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{3 f^6}+\frac {7 e^5 k \sqrt {x}}{9 f^5}-\frac {2 e^4 k x}{9 f^4}+\frac {e^3 k x^{3/2}}{9 f^3}-\frac {5 e^2 k x^2}{72 f^2}+\frac {11 e k x^{5/2}}{225 f}-\frac {k x^3}{27}\right )\) |
(e^5*k*Sqrt[x]*(a + b*Log[c*x^n]))/(3*f^5) - (e^4*k*x*(a + b*Log[c*x^n]))/ (6*f^4) + (e^3*k*x^(3/2)*(a + b*Log[c*x^n]))/(9*f^3) - (e^2*k*x^2*(a + b*L og[c*x^n]))/(12*f^2) + (e*k*x^(5/2)*(a + b*Log[c*x^n]))/(15*f) - (k*x^3*(a + b*Log[c*x^n]))/18 - (e^6*k*Log[e + f*Sqrt[x]]*(a + b*Log[c*x^n]))/(3*f^ 6) + (x^3*Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]))/3 - b*n*((7*e^5*k*S qrt[x])/(9*f^5) - (2*e^4*k*x)/(9*f^4) + (e^3*k*x^(3/2))/(9*f^3) - (5*e^2*k *x^2)/(72*f^2) + (11*e*k*x^(5/2))/(225*f) - (k*x^3)/27 - (e^6*k*Log[e + f* Sqrt[x]])/(9*f^6) + (x^3*Log[d*(e + f*Sqrt[x])^k])/9 - (2*e^6*k*Log[e + f* Sqrt[x]]*Log[-((f*Sqrt[x])/e)])/(3*f^6) - (2*e^6*k*PolyLog[2, 1 + (f*Sqrt[ x])/e])/(3*f^6))
3.2.15.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* (e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[1/x u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] && RationalQ[q])) && NeQ[q, -1]
\[\int x^{2} \left (a +b \ln \left (c \,x^{n}\right )\right ) \ln \left (d \left (e +f \sqrt {x}\right )^{k}\right )d x\]
\[ \int x^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \]
Timed out. \[ \int x^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \]
\[ \int x^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \]
1/441*(147*b*e*x^3*log(d)*log(x^n) + 49*(3*a*e*log(d) - (e*n*log(d) - 3*e* log(c)*log(d))*b)*x^3 + 49*(3*b*e*x^3*log(x^n) - ((e*n - 3*e*log(c))*b - 3 *a*e)*x^3)*log((f*sqrt(x) + e)^k) - (21*b*f*k*x^4*log(x^n) + (21*a*f*k - ( 13*f*k*n - 21*f*k*log(c))*b)*x^4)/sqrt(x))/e + integrate(1/18*(3*b*f^2*k*x ^3*log(x^n) + (3*a*f^2*k - (f^2*k*n - 3*f^2*k*log(c))*b)*x^3)/(e*f*sqrt(x) + e^2), x)
\[ \int x^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \]
Timed out. \[ \int x^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^2\,\ln \left (d\,{\left (e+f\,\sqrt {x}\right )}^k\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]