Integrand size = 23, antiderivative size = 194 \[ \int \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=4 b m n x-\frac {2 b \sqrt {e} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {f}}-2 m x \left (a+b \log \left (c x^n\right )\right )+\frac {2 \sqrt {e} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {f}}-b n x \log \left (d \left (e+f x^2\right )^m\right )+x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac {i b \sqrt {e} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {f}}+\frac {i b \sqrt {e} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {f}} \]
4*b*m*n*x-2*m*x*(a+b*ln(c*x^n))-b*n*x*ln(d*(f*x^2+e)^m)+x*(a+b*ln(c*x^n))* ln(d*(f*x^2+e)^m)-2*b*m*n*arctan(x*f^(1/2)/e^(1/2))*e^(1/2)/f^(1/2)+2*m*ar ctan(x*f^(1/2)/e^(1/2))*(a+b*ln(c*x^n))*e^(1/2)/f^(1/2)-I*b*m*n*polylog(2, -I*x*f^(1/2)/e^(1/2))*e^(1/2)/f^(1/2)+I*b*m*n*polylog(2,I*x*f^(1/2)/e^(1/2 ))*e^(1/2)/f^(1/2)
Time = 0.08 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.71 \[ \int \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\frac {-2 a \sqrt {f} m x+4 b \sqrt {f} m n x+2 a \sqrt {e} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )-2 b \sqrt {e} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )-2 b \sqrt {e} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log (x)-2 b \sqrt {f} m x \log \left (c x^n\right )+2 b \sqrt {e} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log \left (c x^n\right )+i b \sqrt {e} m n \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-i b \sqrt {e} m n \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+a \sqrt {f} x \log \left (d \left (e+f x^2\right )^m\right )-b \sqrt {f} n x \log \left (d \left (e+f x^2\right )^m\right )+b \sqrt {f} x \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )-i b \sqrt {e} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+i b \sqrt {e} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {f}} \]
(-2*a*Sqrt[f]*m*x + 4*b*Sqrt[f]*m*n*x + 2*a*Sqrt[e]*m*ArcTan[(Sqrt[f]*x)/S qrt[e]] - 2*b*Sqrt[e]*m*n*ArcTan[(Sqrt[f]*x)/Sqrt[e]] - 2*b*Sqrt[e]*m*n*Ar cTan[(Sqrt[f]*x)/Sqrt[e]]*Log[x] - 2*b*Sqrt[f]*m*x*Log[c*x^n] + 2*b*Sqrt[e ]*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[c*x^n] + I*b*Sqrt[e]*m*n*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - I*b*Sqrt[e]*m*n*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sq rt[e]] + a*Sqrt[f]*x*Log[d*(e + f*x^2)^m] - b*Sqrt[f]*n*x*Log[d*(e + f*x^2 )^m] + b*Sqrt[f]*x*Log[c*x^n]*Log[d*(e + f*x^2)^m] - I*b*Sqrt[e]*m*n*PolyL og[2, ((-I)*Sqrt[f]*x)/Sqrt[e]] + I*b*Sqrt[e]*m*n*PolyLog[2, (I*Sqrt[f]*x) /Sqrt[e]])/Sqrt[f]
Time = 0.32 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2817, 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.
\(\displaystyle \int \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx\) |
\(\Big \downarrow \) 2817 |
\(\displaystyle -b n \int \left (\frac {2 \sqrt {e} \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) m}{\sqrt {f} x}-2 m+\log \left (d \left (f x^2+e\right )^m\right )\right )dx+\frac {2 \sqrt {e} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {f}}+x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-2 m x \left (a+b \log \left (c x^n\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \sqrt {e} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {f}}+x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-2 m x \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {2 \sqrt {e} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {f}}+x \log \left (d \left (e+f x^2\right )^m\right )+\frac {i \sqrt {e} m \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {f}}-\frac {i \sqrt {e} m \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {f}}-4 m x\right )\) |
-2*m*x*(a + b*Log[c*x^n]) + (2*Sqrt[e]*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*(a + b*Log[c*x^n]))/Sqrt[f] + x*(a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m] - b*n*( -4*m*x + (2*Sqrt[e]*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/Sqrt[f] + x*Log[d*(e + f*x^2)^m] + (I*Sqrt[e]*m*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]])/Sqrt[f] - ( I*Sqrt[e]*m*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/Sqrt[f])
3.1.96.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))^(p_.), x_Symbol] :> With[{u = IntHide[Log[d*(e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n])^p u, x] - Simp[b*n*p Int[(a + b*Log[c*x^n])^(p - 1)/x u, x], x]] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p, 0] && RationalQ[m] && (EqQ[p, 1] || (FractionQ[m] && IntegerQ[1/m]) || (EqQ[r, 1] && EqQ[m, 1] && EqQ[d*e, 1]))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 16.92 (sec) , antiderivative size = 841, normalized size of antiderivative = 4.34
(b*x*ln(x^n)+1/2*x*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csg n(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*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))*(I*Pi*b*x*csgn(I*c)*csgn(I*c*x^n)^2+I*Pi*b*x*csgn (I*x^n)*csgn(I*c*x^n)^2+2*a*x+2*ln(c)*b*x+2*b*x*ln(x^n)-2*b*n*x-I*Pi*b*x*c sgn(I*c*x^n)^3-I*Pi*b*x*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n))-I*m*x*b*Pi*cs gn(I*c)*csgn(I*c*x^n)^2+I*m*x*b*Pi*csgn(I*c*x^n)^3+I*m*e/(e*f)^(1/2)*arcta n(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2-I*m*e/(e*f)^(1/2)*arctan (x*f/(e*f)^(1/2))*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-2*m*x*b*ln(c)+4 *b*m*n*x-2*x*a*m-I*m*x*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*m*e/(e*f)^(1/2)* arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c*x^n)^3+I*m*x*b*Pi*csgn(I*c)*csgn(I*x ^n)*csgn(I*c*x^n)+I*m*e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*x^ n)*csgn(I*c*x^n)^2+2*m*e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*ln(c)-2*m*e /(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*n+2*a*m*e/(e*f)^(1/2)*arctan(x*f/(e *f)^(1/2))-2*m*b*ln(x^n)*x-2*m*b*e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*n*l n(x)+2*m*b*e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*ln(x^n)+m*b*n*e*ln(x)/(-e *f)^(1/2)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-m*b*n*e*ln(x)/(-e*f)^(1/2)* ln((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+m*b*n*e/(-e*f)^(1/2)*dilog((-f*x+(-...
\[ \int \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\text {Timed out} \]
Exception generated. \[ \int \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int \ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]