Integrand size = 28, antiderivative size = 181 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx=\frac {\left (a+b \log \left (c x^n\right )\right )^4 \log \left (d \left (e+f x^2\right )^m\right )}{4 b n}-\frac {m \left (a+b \log \left (c x^n\right )\right )^4 \log \left (1+\frac {f x^2}{e}\right )}{4 b n}-\frac {1}{2} m \left (a+b \log \left (c x^n\right )\right )^3 \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right )+\frac {3}{4} b m n \left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}\left (3,-\frac {f x^2}{e}\right )-\frac {3}{4} b^2 m n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (4,-\frac {f x^2}{e}\right )+\frac {3}{8} b^3 m n^3 \operatorname {PolyLog}\left (5,-\frac {f x^2}{e}\right ) \]
1/4*(a+b*ln(c*x^n))^4*ln(d*(f*x^2+e)^m)/b/n-1/4*m*(a+b*ln(c*x^n))^4*ln(1+f *x^2/e)/b/n-1/2*m*(a+b*ln(c*x^n))^3*polylog(2,-f*x^2/e)+3/4*b*m*n*(a+b*ln( c*x^n))^2*polylog(3,-f*x^2/e)-3/4*b^2*m*n^2*(a+b*ln(c*x^n))*polylog(4,-f*x ^2/e)+3/8*b^3*m*n^3*polylog(5,-f*x^2/e)
Result contains complex when optimal does not.
Time = 0.22 (sec) , antiderivative size = 1348, normalized size of antiderivative = 7.45 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx =\text {Too large to display} \]
-(a^3*m*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]]) + (3*a^2*b*m*n*Log[x]^2*Log [1 - (I*Sqrt[f]*x)/Sqrt[e]])/2 - a*b^2*m*n^2*Log[x]^3*Log[1 - (I*Sqrt[f]*x )/Sqrt[e]] + (b^3*m*n^3*Log[x]^4*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]])/4 - 3*a^2 *b*m*Log[x]*Log[c*x^n]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 3*a*b^2*m*n*Log[x] ^2*Log[c*x^n]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - b^3*m*n^2*Log[x]^3*Log[c*x^ n]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - 3*a*b^2*m*Log[x]*Log[c*x^n]^2*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + (3*b^3*m*n*Log[x]^2*Log[c*x^n]^2*Log[1 - (I*Sqrt[ f]*x)/Sqrt[e]])/2 - b^3*m*Log[x]*Log[c*x^n]^3*Log[1 - (I*Sqrt[f]*x)/Sqrt[e ]] - a^3*m*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + (3*a^2*b*m*n*Log[x]^2*L og[1 + (I*Sqrt[f]*x)/Sqrt[e]])/2 - a*b^2*m*n^2*Log[x]^3*Log[1 + (I*Sqrt[f] *x)/Sqrt[e]] + (b^3*m*n^3*Log[x]^4*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]])/4 - 3*a ^2*b*m*Log[x]*Log[c*x^n]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + 3*a*b^2*m*n*Log[ x]^2*Log[c*x^n]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - b^3*m*n^2*Log[x]^3*Log[c* x^n]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - 3*a*b^2*m*Log[x]*Log[c*x^n]^2*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + (3*b^3*m*n*Log[x]^2*Log[c*x^n]^2*Log[1 + (I*Sqr t[f]*x)/Sqrt[e]])/2 - b^3*m*Log[x]*Log[c*x^n]^3*Log[1 + (I*Sqrt[f]*x)/Sqrt [e]] + a^3*Log[x]*Log[d*(e + f*x^2)^m] - (3*a^2*b*n*Log[x]^2*Log[d*(e + f* x^2)^m])/2 + a*b^2*n^2*Log[x]^3*Log[d*(e + f*x^2)^m] - (b^3*n^3*Log[x]^4*L og[d*(e + f*x^2)^m])/4 + 3*a^2*b*Log[x]*Log[c*x^n]*Log[d*(e + f*x^2)^m] - 3*a*b^2*n*Log[x]^2*Log[c*x^n]*Log[d*(e + f*x^2)^m] + b^3*n^2*Log[x]^3*L...
Time = 0.68 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2822, 2775, 2821, 2830, 2830, 7143}
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.
\(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx\) |
\(\Big \downarrow \) 2822 |
\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^4 \log \left (d \left (e+f x^2\right )^m\right )}{4 b n}-\frac {f m \int \frac {x \left (a+b \log \left (c x^n\right )\right )^4}{f x^2+e}dx}{2 b n}\) |
\(\Big \downarrow \) 2775 |
\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^4 \log \left (d \left (e+f x^2\right )^m\right )}{4 b n}-\frac {f m \left (\frac {\log \left (\frac {f x^2}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^4}{2 f}-\frac {2 b n \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (\frac {f x^2}{e}+1\right )}{x}dx}{f}\right )}{2 b n}\) |
\(\Big \downarrow \) 2821 |
\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^4 \log \left (d \left (e+f x^2\right )^m\right )}{4 b n}-\frac {f m \left (\frac {\log \left (\frac {f x^2}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^4}{2 f}-\frac {2 b n \left (\frac {3}{2} b n \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right )}{x}dx-\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^3\right )}{f}\right )}{2 b n}\) |
\(\Big \downarrow \) 2830 |
\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^4 \log \left (d \left (e+f x^2\right )^m\right )}{4 b n}-\frac {f m \left (\frac {\log \left (\frac {f x^2}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^4}{2 f}-\frac {2 b n \left (\frac {3}{2} b n \left (\frac {1}{2} \operatorname {PolyLog}\left (3,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^2-b n \int \frac {\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (3,-\frac {f x^2}{e}\right )}{x}dx\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^3\right )}{f}\right )}{2 b n}\) |
\(\Big \downarrow \) 2830 |
\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^4 \log \left (d \left (e+f x^2\right )^m\right )}{4 b n}-\frac {f m \left (\frac {\log \left (\frac {f x^2}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^4}{2 f}-\frac {2 b n \left (\frac {3}{2} b n \left (\frac {1}{2} \operatorname {PolyLog}\left (3,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^2-b n \left (\frac {1}{2} \operatorname {PolyLog}\left (4,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} b n \int \frac {\operatorname {PolyLog}\left (4,-\frac {f x^2}{e}\right )}{x}dx\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^3\right )}{f}\right )}{2 b n}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^4 \log \left (d \left (e+f x^2\right )^m\right )}{4 b n}-\frac {f m \left (\frac {\log \left (\frac {f x^2}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^4}{2 f}-\frac {2 b n \left (\frac {3}{2} b n \left (\frac {1}{2} \operatorname {PolyLog}\left (3,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^2-b n \left (\frac {1}{2} \operatorname {PolyLog}\left (4,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b n \operatorname {PolyLog}\left (5,-\frac {f x^2}{e}\right )\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^3\right )}{f}\right )}{2 b n}\) |
((a + b*Log[c*x^n])^4*Log[d*(e + f*x^2)^m])/(4*b*n) - (f*m*(((a + b*Log[c* x^n])^4*Log[1 + (f*x^2)/e])/(2*f) - (2*b*n*(-1/2*((a + b*Log[c*x^n])^3*Pol yLog[2, -((f*x^2)/e)]) + (3*b*n*(((a + b*Log[c*x^n])^2*PolyLog[3, -((f*x^2 )/e)])/2 - b*n*(((a + b*Log[c*x^n])*PolyLog[4, -((f*x^2)/e)])/2 - (b*n*Pol yLog[5, -((f*x^2)/e)])/4)))/2))/f))/(2*b*n)
3.2.9.3.1 Defintions of rubi rules used
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Simp[f^m*Log[1 + e*(x^r/d)]*((a + b*Log[c* x^n])^p/(e*r)), x] - Simp[b*f^m*n*(p/(e*r)) Int[Log[1 + e*(x^r/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] & & EqQ[m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]
Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b _.))^(p_.))/(x_), x_Symbol] :> Simp[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c *x^n])^p/m), x] + Simp[b*n*(p/m) Int[PolyLog[2, (-d)*f*x^m]*((a + b*Log[c *x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]
Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_ .)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp[Log[d*(e + f*x^m)^r]*((a + b*Log[ c*x^n])^(p + 1)/(b*n*(p + 1))), x] - Simp[f*m*(r/(b*n*(p + 1))) Int[x^(m - 1)*((a + b*Log[c*x^n])^(p + 1)/(e + f*x^m)), x], x] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p, 0] && NeQ[d*e, 1]
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_ .)])/(x_), x_Symbol] :> Simp[PolyLog[k + 1, e*x^q]*((a + b*Log[c*x^n])^p/q) , x] - Simp[b*n*(p/q) Int[PolyLog[k + 1, e*x^q]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, e, k, n, q}, x] && GtQ[p, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 308.50 (sec) , antiderivative size = 5816, normalized size of antiderivative = 32.13
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x} \,d x } \]
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, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x} \,d x } \]
-1/4*(b^3*n^3*log(x)^4 - 4*b^3*log(x)*log(x^n)^3 - 4*(b^3*n^2*log(c) + a*b ^2*n^2)*log(x)^3 + 6*(b^3*n*log(c)^2 + 2*a*b^2*n*log(c) + a^2*b*n)*log(x)^ 2 + 6*(b^3*n*log(x)^2 - 2*(b^3*log(c) + a*b^2)*log(x))*log(x^n)^2 - 4*(b^3 *n^2*log(x)^3 - 3*(b^3*n*log(c) + a*b^2*n)*log(x)^2 + 3*(b^3*log(c)^2 + 2* a*b^2*log(c) + a^2*b)*log(x))*log(x^n) - 4*(b^3*log(c)^3 + 3*a*b^2*log(c)^ 2 + 3*a^2*b*log(c) + a^3)*log(x))*log((f*x^2 + e)^m) - integrate(-1/2*(b^3 *f*m*n^3*x^2*log(x)^4 + 2*b^3*e*log(c)^3*log(d) + 6*a*b^2*e*log(c)^2*log(d ) + 6*a^2*b*e*log(c)*log(d) - 4*(b^3*f*m*n^2*log(c) + a*b^2*f*m*n^2)*x^2*l og(x)^3 + 2*a^3*e*log(d) + 6*(b^3*f*m*n*log(c)^2 + 2*a*b^2*f*m*n*log(c) + a^2*b*f*m*n)*x^2*log(x)^2 - 4*(b^3*f*m*log(c)^3 + 3*a*b^2*f*m*log(c)^2 + 3 *a^2*b*f*m*log(c) + a^3*f*m)*x^2*log(x) - 2*(2*b^3*f*m*x^2*log(x) - b^3*f* x^2*log(d) - b^3*e*log(d))*log(x^n)^3 + 2*(b^3*f*log(c)^3*log(d) + 3*a*b^2 *f*log(c)^2*log(d) + 3*a^2*b*f*log(c)*log(d) + a^3*f*log(d))*x^2 + 6*(b^3* f*m*n*x^2*log(x)^2 + b^3*e*log(c)*log(d) + a*b^2*e*log(d) - 2*(b^3*f*m*log (c) + a*b^2*f*m)*x^2*log(x) + (b^3*f*log(c)*log(d) + a*b^2*f*log(d))*x^2)* log(x^n)^2 - 2*(2*b^3*f*m*n^2*x^2*log(x)^3 - 3*b^3*e*log(c)^2*log(d) - 6*a *b^2*e*log(c)*log(d) - 3*a^2*b*e*log(d) - 6*(b^3*f*m*n*log(c) + a*b^2*f*m* n)*x^2*log(x)^2 + 6*(b^3*f*m*log(c)^2 + 2*a*b^2*f*m*log(c) + a^2*b*f*m)*x^ 2*log(x) - 3*(b^3*f*log(c)^2*log(d) + 2*a*b^2*f*log(c)*log(d) + a^2*b*f*lo g(d))*x^2)*log(x^n))/(f*x^3 + e*x), x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx=\int \frac {\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3}{x} \,d x \]