Integrand size = 26, antiderivative size = 227 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx=-\frac {8 b f m n}{9 e x}-\frac {2 b f^{3/2} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 e^{3/2}}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}-\frac {2 f^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}+\frac {i b f^{3/2} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}-\frac {i b f^{3/2} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}} \]
-8/9*b*f*m*n/e/x-2/9*b*f^(3/2)*m*n*arctan(x*f^(1/2)/e^(1/2))/e^(3/2)-2/3*f *m*(a+b*ln(c*x^n))/e/x-2/3*f^(3/2)*m*arctan(x*f^(1/2)/e^(1/2))*(a+b*ln(c*x ^n))/e^(3/2)-1/9*b*n*ln(d*(f*x^2+e)^m)/x^3-1/3*(a+b*ln(c*x^n))*ln(d*(f*x^2 +e)^m)/x^3+1/3*I*b*f^(3/2)*m*n*polylog(2,-I*x*f^(1/2)/e^(1/2))/e^(3/2)-1/3 *I*b*f^(3/2)*m*n*polylog(2,I*x*f^(1/2)/e^(1/2))/e^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.10 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx=\frac {-8 b \sqrt {e} f m n x^2-2 b f^{3/2} m n x^3 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )-6 a \sqrt {e} f m x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {f x^2}{e}\right )+6 b f^{3/2} m n x^3 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log (x)-6 b \sqrt {e} f m x^2 \log \left (c x^n\right )-6 b f^{3/2} m x^3 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log \left (c x^n\right )-3 i b f^{3/2} m n x^3 \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+3 i b f^{3/2} m n x^3 \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-3 a e^{3/2} \log \left (d \left (e+f x^2\right )^m\right )-b e^{3/2} n \log \left (d \left (e+f x^2\right )^m\right )-3 b e^{3/2} \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+3 i b f^{3/2} m n x^3 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-3 i b f^{3/2} m n x^3 \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{9 e^{3/2} x^3} \]
(-8*b*Sqrt[e]*f*m*n*x^2 - 2*b*f^(3/2)*m*n*x^3*ArcTan[(Sqrt[f]*x)/Sqrt[e]] - 6*a*Sqrt[e]*f*m*x^2*Hypergeometric2F1[-1/2, 1, 1/2, -((f*x^2)/e)] + 6*b* f^(3/2)*m*n*x^3*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[x] - 6*b*Sqrt[e]*f*m*x^2*L og[c*x^n] - 6*b*f^(3/2)*m*x^3*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[c*x^n] - (3* I)*b*f^(3/2)*m*n*x^3*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + (3*I)*b*f^(3/ 2)*m*n*x^3*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - 3*a*e^(3/2)*Log[d*(e + f*x^2)^m] - b*e^(3/2)*n*Log[d*(e + f*x^2)^m] - 3*b*e^(3/2)*Log[c*x^n]*Log[ d*(e + f*x^2)^m] + (3*I)*b*f^(3/2)*m*n*x^3*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqr t[e]] - (3*I)*b*f^(3/2)*m*n*x^3*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/(9*e^(3 /2)*x^3)
Time = 0.37 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, 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.
\(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 2823 |
\(\displaystyle -b n \int \left (-\frac {2 m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) f^{3/2}}{3 e^{3/2} x}-\frac {2 m f}{3 e x^2}-\frac {\log \left (d \left (f x^2+e\right )^m\right )}{3 x^4}\right )dx-\frac {2 f^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 f^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}-b n \left (\frac {2 f^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 e^{3/2}}+\frac {\log \left (d \left (e+f x^2\right )^m\right )}{9 x^3}-\frac {i f^{3/2} m \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}+\frac {i f^{3/2} m \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}+\frac {8 f m}{9 e x}\right )\) |
(-2*f*m*(a + b*Log[c*x^n]))/(3*e*x) - (2*f^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt [e]]*(a + b*Log[c*x^n]))/(3*e^(3/2)) - ((a + b*Log[c*x^n])*Log[d*(e + f*x^ 2)^m])/(3*x^3) - b*n*((8*f*m)/(9*e*x) + (2*f^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sq rt[e]])/(9*e^(3/2)) + Log[d*(e + f*x^2)^m]/(9*x^3) - ((I/3)*f^(3/2)*m*Poly Log[2, ((-I)*Sqrt[f]*x)/Sqrt[e]])/e^(3/2) + ((I/3)*f^(3/2)*m*PolyLog[2, (I *Sqrt[f]*x)/Sqrt[e]])/e^(3/2))
3.1.98.3.1 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 = 35.93 (sec) , antiderivative size = 965, normalized size of antiderivative = 4.25
(-1/3*b/x^3*ln(x^n)-1/18*(-3*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+3* I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+3*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-3*I* b*Pi*csgn(I*c*x^n)^3+6*b*ln(c)+2*b*n+6*a)/x^3)*ln((f*x^2+e)^m)+(1/4*I*Pi*c sgn(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)*cs gn(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*cs gn(I*d*(f*x^2+e)^m)^2*csgn(I*d)+1/2*ln(d))*(-1/3*(-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)*cs gn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x^n)^3+2*b*ln(c)+2*a)/x^3-2/3*b/x^3*ln(x^n)- 2/9*b/x^3*n)-1/3*I*m*f/e/x*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/3*I*m*f^2/e/ (e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/3*I*m *f^2/e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c*x^n)^3-1/3*I*m*f^ 2/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 /3*m*f^2/e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*ln(c)-2/9*m*f^2/e/(e*f)^( 1/2)*arctan(x*f/(e*f)^(1/2))*b*n-2/3*m*f^2/e/(e*f)^(1/2)*arctan(x*f/(e*f)^ (1/2))*a+1/3*I*m*f^2/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)-1/3*I*m*f/e/x*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/3 *I*m*f/e/x*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/3*I*m*f/e/x*b*Pi*csg n(I*c*x^n)^3-2/3*m*f/e/x*b*ln(c)-8/9*b*f*m*n/e/x-2/3*m*f/e/x*a+2/3*m*f^2*b /e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*n*ln(x)-2/3*m*f^2*b/e/(e*f)^(1/2)*a rctan(x*f/(e*f)^(1/2))*ln(x^n)-2/3*m*f*b*ln(x^n)/e/x-1/3*m*f^2*b*n/e*ln...
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx=\int \frac {\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^4} \,d x \]