Integrand size = 24, antiderivative size = 148 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\frac {1}{2} b m n x^2-\frac {1}{2} m x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {b n \left (e+f x^2\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 f}-\frac {b e n \log \left (-\frac {f x^2}{e}\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 f}+\frac {\left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{2 f}-\frac {b e m n \operatorname {PolyLog}\left (2,1+\frac {f x^2}{e}\right )}{4 f} \]
1/2*b*m*n*x^2-1/2*m*x^2*(a+b*ln(c*x^n))-1/4*b*n*(f*x^2+e)*ln(d*(f*x^2+e)^m )/f-1/4*b*e*n*ln(-f*x^2/e)*ln(d*(f*x^2+e)^m)/f+1/2*(f*x^2+e)*(a+b*ln(c*x^n ))*ln(d*(f*x^2+e)^m)/f-1/4*b*e*m*n*polylog(2,1+f*x^2/e)/f
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.80 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\frac {-2 a f m x^2+2 b f m n x^2-2 b f m x^2 \log \left (c x^n\right )+2 b e m n \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 b e m n \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-b e m n \log \left (e+f x^2\right )-2 b e m n \log (x) \log \left (e+f x^2\right )+2 b e m \log \left (c x^n\right ) \log \left (e+f x^2\right )+2 a e \log \left (d \left (e+f x^2\right )^m\right )+2 a f x^2 \log \left (d \left (e+f x^2\right )^m\right )-b f n x^2 \log \left (d \left (e+f x^2\right )^m\right )+2 b f x^2 \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+2 b e m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 b e m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{4 f} \]
(-2*a*f*m*x^2 + 2*b*f*m*n*x^2 - 2*b*f*m*x^2*Log[c*x^n] + 2*b*e*m*n*Log[x]* Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 2*b*e*m*n*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sq rt[e]] - b*e*m*n*Log[e + f*x^2] - 2*b*e*m*n*Log[x]*Log[e + f*x^2] + 2*b*e* m*Log[c*x^n]*Log[e + f*x^2] + 2*a*e*Log[d*(e + f*x^2)^m] + 2*a*f*x^2*Log[d *(e + f*x^2)^m] - b*f*n*x^2*Log[d*(e + f*x^2)^m] + 2*b*f*x^2*Log[c*x^n]*Lo g[d*(e + f*x^2)^m] + 2*b*e*m*n*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]] + 2*b* e*m*n*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/(4*f)
Time = 0.44 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.
\(\displaystyle \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx\) |
\(\Big \downarrow \) 2823 |
\(\displaystyle -b n \int \left (\frac {\left (f x^2+e\right ) \log \left (d \left (f x^2+e\right )^m\right )}{2 f x}-\frac {m x}{2}\right )dx+\frac {\left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{2 f}-\frac {1}{2} m x^2 \left (a+b \log \left (c x^n\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{2 f}-\frac {1}{2} m x^2 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {\left (e+f x^2\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 f}+\frac {e \log \left (-\frac {f x^2}{e}\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 f}+\frac {e m \operatorname {PolyLog}\left (2,\frac {f x^2}{e}+1\right )}{4 f}-\frac {m x^2}{2}\right )\) |
-1/2*(m*x^2*(a + b*Log[c*x^n])) + ((e + f*x^2)*(a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/(2*f) - b*n*(-1/2*(m*x^2) + ((e + f*x^2)*Log[d*(e + f*x^2)^m ])/(4*f) + (e*Log[-((f*x^2)/e)]*Log[d*(e + f*x^2)^m])/(4*f) + (e*m*PolyLog [2, 1 + (f*x^2)/e])/(4*f))
3.1.91.3.1 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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 39.64 (sec) , antiderivative size = 828, normalized size of antiderivative = 5.59
(1/2*b*x^2*ln(x^n)+1/4*x^2*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I* b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*c sgn(I*c*x^n)^3+2*b*ln(c)-b*n+2*a))*ln((f*x^2+e)^m)+(1/4*I*Pi*csgn(I*(f*x^2 +e)^m)*csgn(I*d*(f*x^2+e)^m)^2-1/4*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^ 2+e)^m)*csgn(I*d)-1/4*I*Pi*csgn(I*d*(f*x^2+e)^m)^3+1/4*I*Pi*csgn(I*d*(f*x^ 2+e)^m)^2*csgn(I*d)+1/2*ln(d))*(1/2*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I* c*x^n)+I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2 -I*b*Pi*csgn(I*c*x^n)^3+2*b*ln(c)+2*a)*x^2+b*x^2*ln(x^n)-1/2*b*n*x^2)-1/4* I*m/f*e*ln(f*x^2+e)*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/4*I*m*x^2*b *Pi*csgn(I*c*x^n)^3-1/4*I*m/f*e*ln(f*x^2+e)*b*Pi*csgn(I*c*x^n)^3-1/4*I*m*x ^2*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1/2*m*x^2*b*ln(c)+1/2*b*m*n*x^2-1/2*x^2* a*m+1/4*I*m/f*e*ln(f*x^2+e)*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1/4*I*m*x^2*b*P i*csgn(I*x^n)*csgn(I*c*x^n)^2+1/4*I*m*x^2*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn( I*c*x^n)+1/4*I*m/f*e*ln(f*x^2+e)*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/2*m/f* e*ln(f*x^2+e)*b*ln(c)-1/4*m/f*b*n*e*ln(f*x^2+e)+1/2*m/f*e*ln(f*x^2+e)*a-1/ 2*m*b*ln(x^n)*x^2+1/2*m/f*b*ln(x^n)*e*ln(f*x^2+e)-1/2*m/f*b*n*e*ln(x)*ln(f *x^2+e)+1/2*m/f*b*n*e*ln(x)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/2*m/f*b *n*e*ln(x)*ln((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/2*m/f*b*n*e*dilog((-f*x+( -e*f)^(1/2))/(-e*f)^(1/2))+1/2*m/f*b*n*e*dilog((f*x+(-e*f)^(1/2))/(-e*f)^( 1/2))
\[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\text {Timed out} \]
\[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
1/4*(2*b*x^2*log(x^n) - (b*(n - 2*log(c)) - 2*a)*x^2)*log((f*x^2 + e)^m) + integrate(-1/2*((2*(f*m - f*log(d))*a - (f*m*n - 2*(f*m - f*log(d))*log(c ))*b)*x^3 - 2*(b*e*log(c)*log(d) + a*e*log(d))*x + 2*((f*m - f*log(d))*b*x ^3 - b*e*x*log(d))*log(x^n))/(f*x^2 + e), x)
\[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int x\,\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]