Integrand size = 28, antiderivative size = 451 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x^3} \, dx=\frac {3 b^3 f m n^3 \log (x)}{4 e}-\frac {3 b^2 f m n^2 \log \left (1+\frac {e}{f x^2}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac {3 b f m n \log \left (1+\frac {e}{f x^2}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{4 e}-\frac {f m \log \left (1+\frac {e}{f x^2}\right ) \left (a+b \log \left (c x^n\right )\right )^3}{2 e}-\frac {3 b^3 f m n^3 \log \left (e+f x^2\right )}{8 e}-\frac {3 b^3 n^3 \log \left (d \left (e+f x^2\right )^m\right )}{8 x^2}-\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 x^2}-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{2 x^2}+\frac {3 b^3 f m n^3 \operatorname {PolyLog}\left (2,-\frac {e}{f x^2}\right )}{8 e}+\frac {3 b^2 f m n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e}{f x^2}\right )}{4 e}+\frac {3 b f m n \left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}\left (2,-\frac {e}{f x^2}\right )}{4 e}+\frac {3 b^3 f m n^3 \operatorname {PolyLog}\left (3,-\frac {e}{f x^2}\right )}{8 e}+\frac {3 b^2 f m n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (3,-\frac {e}{f x^2}\right )}{4 e}+\frac {3 b^3 f m n^3 \operatorname {PolyLog}\left (4,-\frac {e}{f x^2}\right )}{8 e} \]
3/4*b^3*f*m*n^3*ln(x)/e-3/4*b^2*f*m*n^2*ln(1+e/f/x^2)*(a+b*ln(c*x^n))/e-3/ 4*b*f*m*n*ln(1+e/f/x^2)*(a+b*ln(c*x^n))^2/e-1/2*f*m*ln(1+e/f/x^2)*(a+b*ln( c*x^n))^3/e-3/8*b^3*f*m*n^3*ln(f*x^2+e)/e-3/8*b^3*n^3*ln(d*(f*x^2+e)^m)/x^ 2-3/4*b^2*n^2*(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)/x^2-3/4*b*n*(a+b*ln(c*x^n) )^2*ln(d*(f*x^2+e)^m)/x^2-1/2*(a+b*ln(c*x^n))^3*ln(d*(f*x^2+e)^m)/x^2+3/8* b^3*f*m*n^3*polylog(2,-e/f/x^2)/e+3/4*b^2*f*m*n^2*(a+b*ln(c*x^n))*polylog( 2,-e/f/x^2)/e+3/4*b*f*m*n*(a+b*ln(c*x^n))^2*polylog(2,-e/f/x^2)/e+3/8*b^3* f*m*n^3*polylog(3,-e/f/x^2)/e+3/4*b^2*f*m*n^2*(a+b*ln(c*x^n))*polylog(3,-e /f/x^2)/e+3/8*b^3*f*m*n^3*polylog(4,-e/f/x^2)/e
Result contains complex when optimal does not.
Time = 0.58 (sec) , antiderivative size = 2248, normalized size of antiderivative = 4.98 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x^3} \, dx=\text {Result too large to show} \]
-1/8*(-8*a^3*f*m*x^2*Log[x] - 12*a^2*b*f*m*n*x^2*Log[x] - 12*a*b^2*f*m*n^2 *x^2*Log[x] - 6*b^3*f*m*n^3*x^2*Log[x] + 12*a^2*b*f*m*n*x^2*Log[x]^2 + 12* a*b^2*f*m*n^2*x^2*Log[x]^2 + 6*b^3*f*m*n^3*x^2*Log[x]^2 - 8*a*b^2*f*m*n^2* x^2*Log[x]^3 - 4*b^3*f*m*n^3*x^2*Log[x]^3 + 2*b^3*f*m*n^3*x^2*Log[x]^4 - 2 4*a^2*b*f*m*x^2*Log[x]*Log[c*x^n] - 24*a*b^2*f*m*n*x^2*Log[x]*Log[c*x^n] - 12*b^3*f*m*n^2*x^2*Log[x]*Log[c*x^n] + 24*a*b^2*f*m*n*x^2*Log[x]^2*Log[c* x^n] + 12*b^3*f*m*n^2*x^2*Log[x]^2*Log[c*x^n] - 8*b^3*f*m*n^2*x^2*Log[x]^3 *Log[c*x^n] - 24*a*b^2*f*m*x^2*Log[x]*Log[c*x^n]^2 - 12*b^3*f*m*n*x^2*Log[ x]*Log[c*x^n]^2 + 12*b^3*f*m*n*x^2*Log[x]^2*Log[c*x^n]^2 - 8*b^3*f*m*x^2*L og[x]*Log[c*x^n]^3 + 12*a^2*b*f*m*n*x^2*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[ e]] + 12*a*b^2*f*m*n^2*x^2*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 6*b^3*f *m*n^3*x^2*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - 12*a*b^2*f*m*n^2*x^2*Lo g[x]^2*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - 6*b^3*f*m*n^3*x^2*Log[x]^2*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 4*b^3*f*m*n^3*x^2*Log[x]^3*Log[1 - (I*Sqrt[f]*x) /Sqrt[e]] + 24*a*b^2*f*m*n*x^2*Log[x]*Log[c*x^n]*Log[1 - (I*Sqrt[f]*x)/Sqr t[e]] + 12*b^3*f*m*n^2*x^2*Log[x]*Log[c*x^n]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e] ] - 12*b^3*f*m*n^2*x^2*Log[x]^2*Log[c*x^n]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 12*b^3*f*m*n*x^2*Log[x]*Log[c*x^n]^2*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 12 *a^2*b*f*m*n*x^2*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + 12*a*b^2*f*m*n^2* x^2*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + 6*b^3*f*m*n^3*x^2*Log[x]*Lo...
Time = 0.80 (sec) , antiderivative size = 434, 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 = {2825, 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.
\(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2825 |
\(\displaystyle -2 f m \int \left (-\frac {3 b^3 n^3}{8 x \left (f x^2+e\right )}-\frac {3 b^2 \left (a+b \log \left (c x^n\right )\right ) n^2}{4 x \left (f x^2+e\right )}-\frac {3 b \left (a+b \log \left (c x^n\right )\right )^2 n}{4 x \left (f x^2+e\right )}-\frac {\left (a+b \log \left (c x^n\right )\right )^3}{2 x \left (f x^2+e\right )}\right )dx-\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 x^2}-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{2 x^2}-\frac {3 b^3 n^3 \log \left (d \left (e+f x^2\right )^m\right )}{8 x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 x^2}-2 f m \left (-\frac {3 b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e}{f x^2}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {3 b^2 n^2 \operatorname {PolyLog}\left (3,-\frac {e}{f x^2}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {3 b^2 n^2 \log \left (\frac {e}{f x^2}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {3 b n \operatorname {PolyLog}\left (2,-\frac {e}{f x^2}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{8 e}+\frac {3 b n \log \left (\frac {e}{f x^2}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{8 e}+\frac {\log \left (\frac {e}{f x^2}+1\right ) \left (a+b \log \left (c x^n\right )\right )^3}{4 e}-\frac {3 b^3 n^3 \operatorname {PolyLog}\left (2,-\frac {e}{f x^2}\right )}{16 e}-\frac {3 b^3 n^3 \operatorname {PolyLog}\left (3,-\frac {e}{f x^2}\right )}{16 e}-\frac {3 b^3 n^3 \operatorname {PolyLog}\left (4,-\frac {e}{f x^2}\right )}{16 e}+\frac {3 b^3 n^3 \log \left (e+f x^2\right )}{16 e}-\frac {3 b^3 n^3 \log (x)}{8 e}\right )-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{2 x^2}-\frac {3 b^3 n^3 \log \left (d \left (e+f x^2\right )^m\right )}{8 x^2}\) |
(-3*b^3*n^3*Log[d*(e + f*x^2)^m])/(8*x^2) - (3*b^2*n^2*(a + b*Log[c*x^n])* Log[d*(e + f*x^2)^m])/(4*x^2) - (3*b*n*(a + b*Log[c*x^n])^2*Log[d*(e + f*x ^2)^m])/(4*x^2) - ((a + b*Log[c*x^n])^3*Log[d*(e + f*x^2)^m])/(2*x^2) - 2* f*m*((-3*b^3*n^3*Log[x])/(8*e) + (3*b^2*n^2*Log[1 + e/(f*x^2)]*(a + b*Log[ c*x^n]))/(8*e) + (3*b*n*Log[1 + e/(f*x^2)]*(a + b*Log[c*x^n])^2)/(8*e) + ( Log[1 + e/(f*x^2)]*(a + b*Log[c*x^n])^3)/(4*e) + (3*b^3*n^3*Log[e + f*x^2] )/(16*e) - (3*b^3*n^3*PolyLog[2, -(e/(f*x^2))])/(16*e) - (3*b^2*n^2*(a + b *Log[c*x^n])*PolyLog[2, -(e/(f*x^2))])/(8*e) - (3*b*n*(a + b*Log[c*x^n])^2 *PolyLog[2, -(e/(f*x^2))])/(8*e) - (3*b^3*n^3*PolyLog[3, -(e/(f*x^2))])/(1 6*e) - (3*b^2*n^2*(a + b*Log[c*x^n])*PolyLog[3, -(e/(f*x^2))])/(8*e) - (3* b^3*n^3*PolyLog[4, -(e/(f*x^2))])/(16*e))
3.2.10.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))^(p_.)*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q* (a + b*Log[c*x^n])^p, x]}, Simp[Log[d*(e + f*x^m)^r] u, x] - Simp[f*m*r Int[x^(m - 1)/(e + f*x^m) u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m , n, q}, x] && IGtQ[p, 0] && RationalQ[m] && RationalQ[q]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 291.58 (sec) , antiderivative size = 22905, normalized size of antiderivative = 50.79
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{3}} \,d x } \]
integral((b^3*log(c*x^n)^3 + 3*a*b^2*log(c*x^n)^2 + 3*a^2*b*log(c*x^n) + a ^3)*log((f*x^2 + e)^m*d)/x^3, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x^3} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{3}} \,d x } \]
-1/8*(4*b^3*log(x^n)^3 + 6*(n^2 + 2*n*log(c) + 2*log(c)^2)*a*b^2 + (3*n^3 + 6*n^2*log(c) + 6*n*log(c)^2 + 4*log(c)^3)*b^3 + 6*a^2*b*(n + 2*log(c)) + 4*a^3 + 6*(b^3*(n + 2*log(c)) + 2*a*b^2)*log(x^n)^2 + 6*((n^2 + 2*n*log(c ) + 2*log(c)^2)*b^3 + 2*a*b^2*(n + 2*log(c)) + 2*a^2*b)*log(x^n))*log((f*x ^2 + e)^m)/x^2 + integrate(1/4*(4*b^3*e*log(c)^3*log(d) + 12*a*b^2*e*log(c )^2*log(d) + 12*a^2*b*e*log(c)*log(d) + 4*a^3*e*log(d) + 4*((f*m + f*log(d ))*b^3*x^2 + b^3*e*log(d))*log(x^n)^3 + (4*(f*m + f*log(d))*a^3 + 6*(f*m*n + 2*(f*m + f*log(d))*log(c))*a^2*b + 6*(f*m*n^2 + 2*f*m*n*log(c) + 2*(f*m + f*log(d))*log(c)^2)*a*b^2 + (3*f*m*n^3 + 6*f*m*n^2*log(c) + 6*f*m*n*log (c)^2 + 4*(f*m + f*log(d))*log(c)^3)*b^3)*x^2 + 6*(2*b^3*e*log(c)*log(d) + 2*a*b^2*e*log(d) + (2*(f*m + f*log(d))*a*b^2 + (f*m*n + 2*(f*m + f*log(d) )*log(c))*b^3)*x^2)*log(x^n)^2 + 6*(2*b^3*e*log(c)^2*log(d) + 4*a*b^2*e*lo g(c)*log(d) + 2*a^2*b*e*log(d) + (2*(f*m + f*log(d))*a^2*b + 2*(f*m*n + 2* (f*m + f*log(d))*log(c))*a*b^2 + (f*m*n^2 + 2*f*m*n*log(c) + 2*(f*m + f*lo g(d))*log(c)^2)*b^3)*x^2)*log(x^n))/(f*x^5 + e*x^3), x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x^3} \, dx=\int \frac {\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3}{x^3} \,d x \]