Integrand size = 29, antiderivative size = 232 \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=-a g m x-b f n x+2 b g m n x-\frac {b g m (d+e x) \log \left (c (d+e x)^n\right )}{e}+\frac {g i m \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (i+j x)}{e i-d j}\right )}{j}-\frac {b g n (i+j x) \log \left (h (i+j x)^m\right )}{j}+\frac {b d n \log \left (-\frac {j (d+e x)}{e i-d j}\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )}{e}+x \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )+\frac {b g i m n \operatorname {PolyLog}\left (2,-\frac {j (d+e x)}{e i-d j}\right )}{j}+\frac {b d g m n \operatorname {PolyLog}\left (2,\frac {e (i+j x)}{e i-d j}\right )}{e} \]
-a*g*m*x-b*f*n*x+2*b*g*m*n*x-b*g*m*(e*x+d)*ln(c*(e*x+d)^n)/e+g*i*m*(a+b*ln (c*(e*x+d)^n))*ln(e*(j*x+i)/(-d*j+e*i))/j-b*g*n*(j*x+i)*ln(h*(j*x+i)^m)/j+ b*d*n*ln(-j*(e*x+d)/(-d*j+e*i))*(f+g*ln(h*(j*x+i)^m))/e+x*(a+b*ln(c*(e*x+d )^n))*(f+g*ln(h*(j*x+i)^m))+b*g*i*m*n*polylog(2,-j*(e*x+d)/(-d*j+e*i))/j+b *d*g*m*n*polylog(2,e*(j*x+i)/(-d*j+e*i))/e
Time = 0.14 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.42 \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\frac {-b d f j n+b d g j m n+a e f j x-a e g j m x-b e f j n x+2 b e g j m n x+b e f j x \log \left (c (d+e x)^n\right )-b e g j m x \log \left (c (d+e x)^n\right )+a e g i m \log (i+j x)-b e g i m n \log (i+j x)+b d g j m n \log (i+j x)+b e g i m \log \left (c (d+e x)^n\right ) \log (i+j x)-b d g j n \log \left (h (i+j x)^m\right )+a e g j x \log \left (h (i+j x)^m\right )-b e g j n x \log \left (h (i+j x)^m\right )+b e g j x \log \left (c (d+e x)^n\right ) \log \left (h (i+j x)^m\right )+b n \log (d+e x) \left (-e g i m \log (i+j x)+g (e i-d j) m \log \left (\frac {e (i+j x)}{e i-d j}\right )+d j \left (f-g m+g \log \left (h (i+j x)^m\right )\right )\right )+b g (e i-d j) m n \operatorname {PolyLog}\left (2,\frac {j (d+e x)}{-e i+d j}\right )}{e j} \]
(-(b*d*f*j*n) + b*d*g*j*m*n + a*e*f*j*x - a*e*g*j*m*x - b*e*f*j*n*x + 2*b* e*g*j*m*n*x + b*e*f*j*x*Log[c*(d + e*x)^n] - b*e*g*j*m*x*Log[c*(d + e*x)^n ] + a*e*g*i*m*Log[i + j*x] - b*e*g*i*m*n*Log[i + j*x] + b*d*g*j*m*n*Log[i + j*x] + b*e*g*i*m*Log[c*(d + e*x)^n]*Log[i + j*x] - b*d*g*j*n*Log[h*(i + j*x)^m] + a*e*g*j*x*Log[h*(i + j*x)^m] - b*e*g*j*n*x*Log[h*(i + j*x)^m] + b*e*g*j*x*Log[c*(d + e*x)^n]*Log[h*(i + j*x)^m] + b*n*Log[d + e*x]*(-(e*g* i*m*Log[i + j*x]) + g*(e*i - d*j)*m*Log[(e*(i + j*x))/(e*i - d*j)] + d*j*( f - g*m + g*Log[h*(i + j*x)^m])) + b*g*(e*i - d*j)*m*n*PolyLog[2, (j*(d + e*x))/(-(e*i) + d*j)])/(e*j)
Time = 0.56 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2879, 2863, 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.
\(\displaystyle \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx\) |
\(\Big \downarrow \) 2879 |
\(\displaystyle -g j m \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{i+j x}dx-b e n \int \frac {x \left (f+g \log \left (h (i+j x)^m\right )\right )}{d+e x}dx+x \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle -g j m \int \left (\frac {a+b \log \left (c (d+e x)^n\right )}{j}-\frac {i \left (a+b \log \left (c (d+e x)^n\right )\right )}{j (i+j x)}\right )dx-b e n \int \left (\frac {f+g \log \left (h (i+j x)^m\right )}{e}-\frac {d \left (f+g \log \left (h (i+j x)^m\right )\right )}{e (d+e x)}\right )dx+x \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )-g j m \left (-\frac {i \log \left (\frac {e (i+j x)}{e i-d j}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{j^2}+\frac {a x}{j}+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e j}-\frac {b i n \operatorname {PolyLog}\left (2,-\frac {j (d+e x)}{e i-d j}\right )}{j^2}-\frac {b n x}{j}\right )-b e n \left (-\frac {d \log \left (-\frac {j (d+e x)}{e i-d j}\right ) \left (f+g \log \left (h (i+j x)^m\right )\right )}{e^2}-\frac {d g m \operatorname {PolyLog}\left (2,\frac {e (i+j x)}{e i-d j}\right )}{e^2}+\frac {f x}{e}+\frac {g (i+j x) \log \left (h (i+j x)^m\right )}{e j}-\frac {g m x}{e}\right )\) |
x*(a + b*Log[c*(d + e*x)^n])*(f + g*Log[h*(i + j*x)^m]) - g*j*m*((a*x)/j - (b*n*x)/j + (b*(d + e*x)*Log[c*(d + e*x)^n])/(e*j) - (i*(a + b*Log[c*(d + e*x)^n])*Log[(e*(i + j*x))/(e*i - d*j)])/j^2 - (b*i*n*PolyLog[2, -((j*(d + e*x))/(e*i - d*j))])/j^2) - b*e*n*((f*x)/e - (g*m*x)/e + (g*(i + j*x)*Lo g[h*(i + j*x)^m])/(e*j) - (d*Log[-((j*(d + e*x))/(e*i - d*j))]*(f + g*Log[ h*(i + j*x)^m]))/e^2 - (d*g*m*PolyLog[2, (e*(i + j*x))/(e*i - d*j)])/e^2)
3.4.89.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) ^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c , d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log [(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*(g_.)), x_Symbol] :> Simp[x*(a + b*Log[c *(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]), x] + (-Simp[g*j*m Int[x*((a + b*Log[c*(d + e*x)^n])^p/(i + j*x)), x], x] - Simp[b*e*n*p Int[x*(a + b* Log[c*(d + e*x)^n])^(p - 1)*((f + g*Log[h*(i + j*x)^m])/(d + e*x)), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 49.44 (sec) , antiderivative size = 1012, normalized size of antiderivative = 4.36
(x*b*g*ln((j*x+i)^m)+1/2*b*(I*Pi*g*j*x*csgn(I*(j*x+i)^m)*csgn(I*h*(j*x+i)^ m)^2-I*Pi*g*j*x*csgn(I*(j*x+i)^m)*csgn(I*h*(j*x+i)^m)*csgn(I*h)-I*Pi*g*j*x *csgn(I*h*(j*x+i)^m)^3+I*Pi*g*j*x*csgn(I*h*(j*x+i)^m)^2*csgn(I*h)+2*j*x*ln (h)*g+2*g*i*m*ln(j*x+i)-2*x*g*m*j+2*f*j*x)/j)*ln((e*x+d)^n)+(-1/4*I*b*Pi*c sgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/4*I*b*Pi*csgn(I*c)*csgn(I *c*(e*x+d)^n)^2+1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/4*I*b *Pi*csgn(I*c*(e*x+d)^n)^3+1/2*b*ln(c)+1/2*a)*(I*Pi*g*x*csgn(I*(j*x+i)^m)*c sgn(I*h*(j*x+i)^m)^2+I*Pi*g*x*csgn(I*h*(j*x+i)^m)^2*csgn(I*h)+2*f*x+2*ln(h )*g*x+2*g*ln((j*x+i)^m)*x-2*g*m*x+2*g*m/j*i*ln(j*x+i)-I*Pi*g*x*csgn(I*h*(j *x+i)^m)^3-I*Pi*g*x*csgn(I*(j*x+i)^m)*csgn(I*h*(j*x+i)^m)*csgn(I*h))+1/2*I *n*b*x*g*Pi*csgn(I*(j*x+i)^m)*csgn(I*h*(j*x+i)^m)*csgn(I*h)+1/2*I*n*b*x*g* Pi*csgn(I*h*(j*x+i)^m)^3-1/2*I*n*b*x*g*Pi*csgn(I*h*(j*x+i)^m)^2*csgn(I*h)+ 1/2*I/e*n*b*d*ln(e*x+d)*g*Pi*csgn(I*h*(j*x+i)^m)^2*csgn(I*h)-ln(h)*x*b*g*n +2*b*g*m*n*x-b*f*n*x+1/2*I/e*n*b*d*ln(e*x+d)*g*Pi*csgn(I*(j*x+i)^m)*csgn(I *h*(j*x+i)^m)^2-1/2*I*n*b*x*g*Pi*csgn(I*(j*x+i)^m)*csgn(I*h*(j*x+i)^m)^2-1 /2*I/e*n*b*d*ln(e*x+d)*g*Pi*csgn(I*h*(j*x+i)^m)^3-1/2*I/e*n*b*d*ln(e*x+d)* g*Pi*csgn(I*(j*x+i)^m)*csgn(I*h*(j*x+i)^m)*csgn(I*h)+1/e*n*b*d*ln(e*x+d)*g *ln(h)-1/e*n*b*d*ln(e*x+d)*g*m+b*f/e*n*d*ln(e*x+d)-n*b*g*ln((j*x+i)^m)*x+1 /e*n*b*g*ln((j*x+i)^m)*d*ln(e*x+d)+b*d*g*m*n/e-n*b*g*m/j*i*ln((e*x+d)*j-d* j+e*i)-1/e*n*b*g*m*d*dilog(((e*x+d)*j-d*j+e*i)/(-d*j+e*i))-1/e*n*b*g*m*...
\[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} {\left (g \log \left ({\left (j x + i\right )}^{m} h\right ) + f\right )} \,d x } \]
integral(b*f*log((e*x + d)^n*c) + a*f + (b*g*log((e*x + d)^n*c) + a*g)*log ((j*x + i)^m*h), x)
Timed out. \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\text {Timed out} \]
\[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} {\left (g \log \left ({\left (j x + i\right )}^{m} h\right ) + f\right )} \,d x } \]
-b*e*f*n*(x/e - d*log(e*x + d)/e^2) - a*g*j*m*(x/j - i*log(j*x + i)/j^2) + b*f*x*log((e*x + d)^n*c) + a*g*x*log((j*x + i)^m*h) + a*f*x - b*g*((e*i*m *n*log(e*x + d)*log(j*x + i) - (e*i*m*log(j*x + i) - (j*m - j*log(h))*e*x) *log((e*x + d)^n) - (d*j*n*log(e*x + d) + e*j*x*log((e*x + d)^n) - (e*j*n - e*j*log(c))*x)*log((j*x + i)^m))/(e*j) + integrate(-(d*e*i*log(c)*log(h) - ((j*m - j*log(h))*e^2*log(c) - (2*j*m*n - j*n*log(h))*e^2)*x^2 + (d*e*j *m*n + (i*m*n - i*n*log(h))*e^2 + (e^2*i*log(h) - (j*m - j*log(h))*d*e)*lo g(c))*x + (d*e*i*m*n - d^2*j*m*n + (e^2*i*m*n - d*e*j*m*n)*x)*log(e*x + d) )/(e^2*j*x^2 + d*e*i + (e^2*i + d*e*j)*x), x))
\[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\int { {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} {\left (g \log \left ({\left (j x + i\right )}^{m} h\right ) + f\right )} \,d x } \]
Timed out. \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (h (i+j x)^m\right )\right ) \, dx=\int \left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )\,\left (f+g\,\ln \left (h\,{\left (i+j\,x\right )}^m\right )\right ) \,d x \]