Integrand size = 26, antiderivative size = 514 \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\frac {3}{2} b^3 m n^3 x^2-\frac {9}{4} b^2 m n^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {3}{2} b m n x^2 \left (a+b \log \left (c x^n\right )\right )^2-\frac {1}{2} m x^2 \left (a+b \log \left (c x^n\right )\right )^3-\frac {3 b^3 e m n^3 \log \left (e+f x^2\right )}{8 f}-\frac {3}{8} b^3 n^3 x^2 \log \left (d \left (e+f x^2\right )^m\right )+\frac {3}{4} b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac {3}{4} b n x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )+\frac {3 b^2 e m n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {f x^2}{e}\right )}{4 f}-\frac {3 b e m n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {f x^2}{e}\right )}{4 f}+\frac {e m \left (a+b \log \left (c x^n\right )\right )^3 \log \left (1+\frac {f x^2}{e}\right )}{2 f}+\frac {3 b^3 e m n^3 \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right )}{8 f}-\frac {3 b^2 e m n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right )}{4 f}+\frac {3 b e m n \left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right )}{4 f}+\frac {3 b^3 e m n^3 \operatorname {PolyLog}\left (3,-\frac {f x^2}{e}\right )}{8 f}-\frac {3 b^2 e m n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (3,-\frac {f x^2}{e}\right )}{4 f}+\frac {3 b^3 e m n^3 \operatorname {PolyLog}\left (4,-\frac {f x^2}{e}\right )}{8 f} \]
3/2*b^3*m*n^3*x^2-9/4*b^2*m*n^2*x^2*(a+b*ln(c*x^n))+3/2*b*m*n*x^2*(a+b*ln( c*x^n))^2-1/2*m*x^2*(a+b*ln(c*x^n))^3-3/8*b^3*e*m*n^3*ln(f*x^2+e)/f-3/8*b^ 3*n^3*x^2*ln(d*(f*x^2+e)^m)+3/4*b^2*n^2*x^2*(a+b*ln(c*x^n))*ln(d*(f*x^2+e) ^m)-3/4*b*n*x^2*(a+b*ln(c*x^n))^2*ln(d*(f*x^2+e)^m)+1/2*x^2*(a+b*ln(c*x^n) )^3*ln(d*(f*x^2+e)^m)+3/4*b^2*e*m*n^2*(a+b*ln(c*x^n))*ln(1+f*x^2/e)/f-3/4* b*e*m*n*(a+b*ln(c*x^n))^2*ln(1+f*x^2/e)/f+1/2*e*m*(a+b*ln(c*x^n))^3*ln(1+f *x^2/e)/f+3/8*b^3*e*m*n^3*polylog(2,-f*x^2/e)/f-3/4*b^2*e*m*n^2*(a+b*ln(c* x^n))*polylog(2,-f*x^2/e)/f+3/4*b*e*m*n*(a+b*ln(c*x^n))^2*polylog(2,-f*x^2 /e)/f+3/8*b^3*e*m*n^3*polylog(3,-f*x^2/e)/f-3/4*b^2*e*m*n^2*(a+b*ln(c*x^n) )*polylog(3,-f*x^2/e)/f+3/8*b^3*e*m*n^3*polylog(4,-f*x^2/e)/f
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 1911, normalized size of antiderivative = 3.72 \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right ) \, dx =\text {Too large to display} \]
(-4*a^3*f*m*x^2 + 12*a^2*b*f*m*n*x^2 - 18*a*b^2*f*m*n^2*x^2 + 12*b^3*f*m*n ^3*x^2 - 12*a^2*b*f*m*x^2*Log[c*x^n] + 24*a*b^2*f*m*n*x^2*Log[c*x^n] - 18* b^3*f*m*n^2*x^2*Log[c*x^n] - 12*a*b^2*f*m*x^2*Log[c*x^n]^2 + 12*b^3*f*m*n* x^2*Log[c*x^n]^2 - 4*b^3*f*m*x^2*Log[c*x^n]^3 + 12*a^2*b*e*m*n*Log[x]*Log[ 1 - (I*Sqrt[f]*x)/Sqrt[e]] - 12*a*b^2*e*m*n^2*Log[x]*Log[1 - (I*Sqrt[f]*x) /Sqrt[e]] + 6*b^3*e*m*n^3*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - 12*a*b^2 *e*m*n^2*Log[x]^2*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 6*b^3*e*m*n^3*Log[x]^2* Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 4*b^3*e*m*n^3*Log[x]^3*Log[1 - (I*Sqrt[f] *x)/Sqrt[e]] + 24*a*b^2*e*m*n*Log[x]*Log[c*x^n]*Log[1 - (I*Sqrt[f]*x)/Sqrt [e]] - 12*b^3*e*m*n^2*Log[x]*Log[c*x^n]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - 1 2*b^3*e*m*n^2*Log[x]^2*Log[c*x^n]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 12*b^3* e*m*n*Log[x]*Log[c*x^n]^2*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 12*a^2*b*e*m*n* Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - 12*a*b^2*e*m*n^2*Log[x]*Log[1 + (I *Sqrt[f]*x)/Sqrt[e]] + 6*b^3*e*m*n^3*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - 12*a*b^2*e*m*n^2*Log[x]^2*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + 6*b^3*e*m*n^ 3*Log[x]^2*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + 4*b^3*e*m*n^3*Log[x]^3*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + 24*a*b^2*e*m*n*Log[x]*Log[c*x^n]*Log[1 + (I*Sqrt [f]*x)/Sqrt[e]] - 12*b^3*e*m*n^2*Log[x]*Log[c*x^n]*Log[1 + (I*Sqrt[f]*x)/S qrt[e]] - 12*b^3*e*m*n^2*Log[x]^2*Log[c*x^n]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e] ] + 12*b^3*e*m*n*Log[x]*Log[c*x^n]^2*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + 4...
Time = 1.06 (sec) , antiderivative size = 517, normalized size of antiderivative = 1.01, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, 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.
\(\displaystyle \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right ) \, dx\) |
\(\Big \downarrow \) 2825 |
\(\displaystyle -2 f m \int \left (\frac {\left (a+b \log \left (c x^n\right )\right )^3 x^3}{2 \left (f x^2+e\right )}-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 x^3}{4 \left (f x^2+e\right )}+\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) x^3}{4 \left (f x^2+e\right )}-\frac {3 b^3 n^3 x^3}{8 \left (f x^2+e\right )}\right )dx+\frac {3}{4} b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac {3}{4} b n x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )-\frac {3}{8} b^3 n^3 x^2 \log \left (d \left (e+f x^2\right )^m\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{4} b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-2 f m \left (\frac {3 b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 f^2}+\frac {3 b^2 e n^2 \operatorname {PolyLog}\left (3,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 f^2}-\frac {3 b^2 e n^2 \log \left (\frac {f x^2}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{8 f^2}+\frac {9 b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right )}{8 f}-\frac {3 b e n \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{8 f^2}+\frac {3 b e n \log \left (\frac {f x^2}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{8 f^2}-\frac {e \log \left (\frac {f x^2}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^3}{4 f^2}-\frac {3 b n x^2 \left (a+b \log \left (c x^n\right )\right )^2}{4 f}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^3}{4 f}-\frac {3 b^3 e n^3 \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right )}{16 f^2}-\frac {3 b^3 e n^3 \operatorname {PolyLog}\left (3,-\frac {f x^2}{e}\right )}{16 f^2}-\frac {3 b^3 e n^3 \operatorname {PolyLog}\left (4,-\frac {f x^2}{e}\right )}{16 f^2}+\frac {3 b^3 e n^3 \log \left (e+f x^2\right )}{16 f^2}-\frac {3 b^3 n^3 x^2}{4 f}\right )-\frac {3}{4} b n x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )-\frac {3}{8} b^3 n^3 x^2 \log \left (d \left (e+f x^2\right )^m\right )\) |
(-3*b^3*n^3*x^2*Log[d*(e + f*x^2)^m])/8 + (3*b^2*n^2*x^2*(a + b*Log[c*x^n] )*Log[d*(e + f*x^2)^m])/4 - (3*b*n*x^2*(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 - 2*f*m*((-3 *b^3*n^3*x^2)/(4*f) + (9*b^2*n^2*x^2*(a + b*Log[c*x^n]))/(8*f) - (3*b*n*x^ 2*(a + b*Log[c*x^n])^2)/(4*f) + (x^2*(a + b*Log[c*x^n])^3)/(4*f) + (3*b^3* e*n^3*Log[e + f*x^2])/(16*f^2) - (3*b^2*e*n^2*(a + b*Log[c*x^n])*Log[1 + ( f*x^2)/e])/(8*f^2) + (3*b*e*n*(a + b*Log[c*x^n])^2*Log[1 + (f*x^2)/e])/(8* f^2) - (e*(a + b*Log[c*x^n])^3*Log[1 + (f*x^2)/e])/(4*f^2) - (3*b^3*e*n^3* PolyLog[2, -((f*x^2)/e)])/(16*f^2) + (3*b^2*e*n^2*(a + b*Log[c*x^n])*PolyL og[2, -((f*x^2)/e)])/(8*f^2) - (3*b*e*n*(a + b*Log[c*x^n])^2*PolyLog[2, -( (f*x^2)/e)])/(8*f^2) - (3*b^3*e*n^3*PolyLog[3, -((f*x^2)/e)])/(16*f^2) + ( 3*b^2*e*n^2*(a + b*Log[c*x^n])*PolyLog[3, -((f*x^2)/e)])/(8*f^2) - (3*b^3* e*n^3*PolyLog[4, -((f*x^2)/e)])/(16*f^2))
3.2.8.3.1 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 = 0.09 (sec) , antiderivative size = 21242, normalized size of antiderivative = 41.33
\[\text {output too large to display}\]
\[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{3} x \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
integral((b^3*x*log(c*x^n)^3 + 3*a*b^2*x*log(c*x^n)^2 + 3*a^2*b*x*log(c*x^ n) + a^3*x)*log((f*x^2 + e)^m*d), x)
Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\text {Timed out} \]
\[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{3} x \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
1/8*(4*b^3*x^2*log(x^n)^3 - 6*(b^3*(n - 2*log(c)) - 2*a*b^2)*x^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)*x^2*log(x^n) + (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)*x^2)*log((f*x^2 + e)^m) + integrate(-1/4*((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^3 + 4*((f*m - f*log(d))* b^3*x^3 - b^3*e*x*log(d))*log(x^n)^3 + 6*((2*(f*m - f*log(d))*a*b^2 - (f*m *n - 2*(f*m - f*log(d))*log(c))*b^3)*x^3 - 2*(b^3*e*log(c)*log(d) + a*b^2* e*log(d))*x)*log(x^n)^2 - 4*(b^3*e*log(c)^3*log(d) + 3*a*b^2*e*log(c)^2*lo g(d) + 3*a^2*b*e*log(c)*log(d) + a^3*e*log(d))*x + 6*((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*l og(c) + 2*(f*m - f*log(d))*log(c)^2)*b^3)*x^3 - 2*(b^3*e*log(c)^2*log(d) + 2*a*b^2*e*log(c)*log(d) + a^2*b*e*log(d))*x)*log(x^n))/(f*x^2 + e), x)
\[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{3} x \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \]
Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int x\,\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3 \,d x \]