Integrand size = 20, antiderivative size = 210 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=-\frac {5 b n x}{16 e^3}+\frac {3 b n x^2}{32 e^2}-\frac {7 b n x^3}{144 e}+\frac {1}{32} b n x^4+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{8 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1+e x)}{16 e^4}-\frac {1}{16} b n x^4 \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{4 e^4}+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {b n \operatorname {PolyLog}(2,-e x)}{4 e^4} \] Output:
-5/16*b*n*x/e^3+3/32*b*n*x^2/e^2-7/144*b*n*x^3/e+1/32*b*n*x^4+1/4*x*(a+b*l n(c*x^n))/e^3-1/8*x^2*(a+b*ln(c*x^n))/e^2+1/12*x^3*(a+b*ln(c*x^n))/e-1/16* x^4*(a+b*ln(c*x^n))+1/16*b*n*ln(e*x+1)/e^4-1/16*b*n*x^4*ln(e*x+1)-1/4*(a+b *ln(c*x^n))*ln(e*x+1)/e^4+1/4*x^4*(a+b*ln(c*x^n))*ln(e*x+1)-1/4*b*n*polylo g(2,-e*x)/e^4
Time = 0.15 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.90 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\frac {72 a e x-90 b e n x-36 a e^2 x^2+27 b e^2 n x^2+24 a e^3 x^3-14 b e^3 n x^3-18 a e^4 x^4+9 b e^4 n x^4-72 a \log (1+e x)+18 b n \log (1+e x)+72 a e^4 x^4 \log (1+e x)-18 b e^4 n x^4 \log (1+e x)+6 b \log \left (c x^n\right ) \left (e x \left (12-6 e x+4 e^2 x^2-3 e^3 x^3\right )+12 \left (-1+e^4 x^4\right ) \log (1+e x)\right )-72 b n \operatorname {PolyLog}(2,-e x)}{288 e^4} \] Input:
Integrate[x^3*(a + b*Log[c*x^n])*Log[1 + e*x],x]
Output:
(72*a*e*x - 90*b*e*n*x - 36*a*e^2*x^2 + 27*b*e^2*n*x^2 + 24*a*e^3*x^3 - 14 *b*e^3*n*x^3 - 18*a*e^4*x^4 + 9*b*e^4*n*x^4 - 72*a*Log[1 + e*x] + 18*b*n*L og[1 + e*x] + 72*a*e^4*x^4*Log[1 + e*x] - 18*b*e^4*n*x^4*Log[1 + e*x] + 6* b*Log[c*x^n]*(e*x*(12 - 6*e*x + 4*e^2*x^2 - 3*e^3*x^3) + 12*(-1 + e^4*x^4) *Log[1 + e*x]) - 72*b*n*PolyLog[2, -(e*x)])/(288*e^4)
Time = 0.40 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
| \(\displaystyle \int x^3 \log (e x+1) \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle -b n \int \left (\frac {1}{4} \log (e x+1) x^3-\frac {x^3}{16}+\frac {x^2}{12 e}-\frac {x}{8 e^2}+\frac {1}{4 e^3}-\frac {\log (e x+1)}{4 e^4 x}\right )dx-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{4 e^4}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{8 e^2}+\frac {1}{4} x^4 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{4 e^4}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{8 e^2}+\frac {1}{4} x^4 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {\operatorname {PolyLog}(2,-e x)}{4 e^4}-\frac {\log (e x+1)}{16 e^4}+\frac {5 x}{16 e^3}-\frac {3 x^2}{32 e^2}+\frac {1}{16} x^4 \log (e x+1)+\frac {7 x^3}{144 e}-\frac {x^4}{32}\right )\) |
Input:
Int[x^3*(a + b*Log[c*x^n])*Log[1 + e*x],x]
Output:
(x*(a + b*Log[c*x^n]))/(4*e^3) - (x^2*(a + b*Log[c*x^n]))/(8*e^2) + (x^3*( a + b*Log[c*x^n]))/(12*e) - (x^4*(a + b*Log[c*x^n]))/16 - ((a + b*Log[c*x^ n])*Log[1 + e*x])/(4*e^4) + (x^4*(a + b*Log[c*x^n])*Log[1 + e*x])/4 - b*n* ((5*x)/(16*e^3) - (3*x^2)/(32*e^2) + (7*x^3)/(144*e) - x^4/32 - Log[1 + e* x]/(16*e^4) + (x^4*Log[1 + e*x])/16 + PolyLog[2, -(e*x)]/(4*e^4))
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 = 12.16 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.55
| method | result | size |
| risch | \(\left (\frac {b \,x^{4} \ln \left (e x +1\right )}{4}-\frac {b \left (3 e^{4} x^{4}-4 e^{3} x^{3}+6 e^{2} x^{2}-12 e x +12 \ln \left (e x +1\right )\right )}{48 e^{4}}\right ) \ln \left (x^{n}\right )+\frac {\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\left (e x +1\right )^{4} \ln \left (e x +1\right )}{4}-\frac {\left (e x +1\right )^{4}}{16}-\ln \left (e x +1\right ) \left (e x +1\right )^{3}+\frac {\left (e x +1\right )^{3}}{3}+\frac {3 \ln \left (e x +1\right ) \left (e x +1\right )^{2}}{2}-\frac {3 \left (e x +1\right )^{2}}{4}-\ln \left (e x +1\right ) \left (e x +1\right )+e x +1\right )}{e^{4}}+\frac {b n \,x^{4}}{32}-\frac {7 b n \,x^{3}}{144 e}+\frac {3 b n \,x^{2}}{32 e^{2}}-\frac {5 b n x}{16 e^{3}}-\frac {35 n b}{72 e^{4}}-\frac {b n \,x^{4} \ln \left (e x +1\right )}{16}+\frac {b n \ln \left (e x +1\right )}{16 e^{4}}-\frac {n b \operatorname {dilog}\left (e x +1\right )}{4 e^{4}}\) | \(326\) |
Input:
int(x^3*(a+b*ln(c*x^n))*ln(e*x+1),x,method=_RETURNVERBOSE)
Output:
(1/4*b*x^4*ln(e*x+1)-1/48*b*(3*e^4*x^4-4*e^3*x^3+6*e^2*x^2-12*e*x+12*ln(e* x+1))/e^4)*ln(x^n)+(1/2*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*Pi*b*csgn (I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/2*I*Pi*b*csgn(I*c*x^n)^3+1/2*I*Pi*b*csgn (I*c*x^n)^2*csgn(I*c)+b*ln(c)+a)/e^4*(1/4*(e*x+1)^4*ln(e*x+1)-1/16*(e*x+1) ^4-ln(e*x+1)*(e*x+1)^3+1/3*(e*x+1)^3+3/2*ln(e*x+1)*(e*x+1)^2-3/4*(e*x+1)^2 -ln(e*x+1)*(e*x+1)+e*x+1)+1/32*b*n*x^4-7/144*b*n*x^3/e+3/32*b*n*x^2/e^2-5/ 16*b*n*x/e^3-35/72*n*b/e^4-1/16*b*n*x^4*ln(e*x+1)+1/16*b*n*ln(e*x+1)/e^4-1 /4*n*b/e^4*dilog(e*x+1)
\[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \log \left (e x + 1\right ) \,d x } \] Input:
integrate(x^3*(a+b*log(c*x^n))*log(e*x+1),x, algorithm="fricas")
Output:
integral(b*x^3*log(c*x^n)*log(e*x + 1) + a*x^3*log(e*x + 1), x)
Timed out. \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\text {Timed out} \] Input:
integrate(x**3*(a+b*ln(c*x**n))*ln(e*x+1),x)
Output:
Timed out
Time = 0.08 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.23 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=-\frac {{\left (\log \left (e x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-e x\right )\right )} b n}{4 \, e^{4}} + \frac {{\left (b {\left (n - 4 \, \log \left (c\right )\right )} - 4 \, a\right )} \log \left (e x + 1\right )}{16 \, e^{4}} - \frac {9 \, {\left (2 \, a e^{4} - {\left (e^{4} n - 2 \, e^{4} \log \left (c\right )\right )} b\right )} x^{4} - 2 \, {\left (12 \, a e^{3} - {\left (7 \, e^{3} n - 12 \, e^{3} \log \left (c\right )\right )} b\right )} x^{3} + 9 \, {\left (4 \, a e^{2} - {\left (3 \, e^{2} n - 4 \, e^{2} \log \left (c\right )\right )} b\right )} x^{2} + 18 \, {\left ({\left (5 \, e n - 4 \, e \log \left (c\right )\right )} b - 4 \, a e\right )} x - 18 \, {\left ({\left (4 \, a e^{4} - {\left (e^{4} n - 4 \, e^{4} \log \left (c\right )\right )} b\right )} x^{4} + 4 \, b n \log \left (x\right )\right )} \log \left (e x + 1\right ) + 6 \, {\left (3 \, b e^{4} x^{4} - 4 \, b e^{3} x^{3} + 6 \, b e^{2} x^{2} - 12 \, b e x - 12 \, {\left (b e^{4} x^{4} - b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{288 \, e^{4}} \] Input:
integrate(x^3*(a+b*log(c*x^n))*log(e*x+1),x, algorithm="maxima")
Output:
-1/4*(log(e*x + 1)*log(x) + dilog(-e*x))*b*n/e^4 + 1/16*(b*(n - 4*log(c)) - 4*a)*log(e*x + 1)/e^4 - 1/288*(9*(2*a*e^4 - (e^4*n - 2*e^4*log(c))*b)*x^ 4 - 2*(12*a*e^3 - (7*e^3*n - 12*e^3*log(c))*b)*x^3 + 9*(4*a*e^2 - (3*e^2*n - 4*e^2*log(c))*b)*x^2 + 18*((5*e*n - 4*e*log(c))*b - 4*a*e)*x - 18*((4*a *e^4 - (e^4*n - 4*e^4*log(c))*b)*x^4 + 4*b*n*log(x))*log(e*x + 1) + 6*(3*b *e^4*x^4 - 4*b*e^3*x^3 + 6*b*e^2*x^2 - 12*b*e*x - 12*(b*e^4*x^4 - b)*log(e *x + 1))*log(x^n))/e^4
\[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \log \left (e x + 1\right ) \,d x } \] Input:
integrate(x^3*(a+b*log(c*x^n))*log(e*x+1),x, algorithm="giac")
Output:
integrate((b*log(c*x^n) + a)*x^3*log(e*x + 1), x)
Timed out. \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\int x^3\,\ln \left (e\,x+1\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \] Input:
int(x^3*log(e*x + 1)*(a + b*log(c*x^n)),x)
Output:
int(x^3*log(e*x + 1)*(a + b*log(c*x^n)), x)
\[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\frac {72 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e \,x^{2}+x}d x \right ) b n +72 \,\mathrm {log}\left (e x +1\right ) \mathrm {log}\left (x^{n} c \right ) b \,e^{4} n \,x^{4}+72 \,\mathrm {log}\left (e x +1\right ) a \,e^{4} n \,x^{4}-72 \,\mathrm {log}\left (e x +1\right ) a n -18 \,\mathrm {log}\left (e x +1\right ) b \,e^{4} n^{2} x^{4}+18 \,\mathrm {log}\left (e x +1\right ) b \,n^{2}-36 \mathrm {log}\left (x^{n} c \right )^{2} b -18 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{4} n \,x^{4}+24 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{3} n \,x^{3}-36 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{2} n \,x^{2}+72 \,\mathrm {log}\left (x^{n} c \right ) b e n x -18 a \,e^{4} n \,x^{4}+24 a \,e^{3} n \,x^{3}-36 a \,e^{2} n \,x^{2}+72 a e n x +9 b \,e^{4} n^{2} x^{4}-14 b \,e^{3} n^{2} x^{3}+27 b \,e^{2} n^{2} x^{2}-90 b e \,n^{2} x}{288 e^{4} n} \] Input:
int(x^3*(a+b*log(c*x^n))*log(e*x+1),x)
Output:
(72*int(log(x**n*c)/(e*x**2 + x),x)*b*n + 72*log(e*x + 1)*log(x**n*c)*b*e* *4*n*x**4 + 72*log(e*x + 1)*a*e**4*n*x**4 - 72*log(e*x + 1)*a*n - 18*log(e *x + 1)*b*e**4*n**2*x**4 + 18*log(e*x + 1)*b*n**2 - 36*log(x**n*c)**2*b - 18*log(x**n*c)*b*e**4*n*x**4 + 24*log(x**n*c)*b*e**3*n*x**3 - 36*log(x**n* c)*b*e**2*n*x**2 + 72*log(x**n*c)*b*e*n*x - 18*a*e**4*n*x**4 + 24*a*e**3*n *x**3 - 36*a*e**2*n*x**2 + 72*a*e*n*x + 9*b*e**4*n**2*x**4 - 14*b*e**3*n** 2*x**3 + 27*b*e**2*n**2*x**2 - 90*b*e*n**2*x)/(288*e**4*n)