Integrand size = 21, antiderivative size = 141 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}-\frac {x^2 \left (3 a+b n+3 b \log \left (c x^n\right )\right )}{6 e^2 (d+e x)^2}-\frac {x \left (6 a+5 b n+6 b \log \left (c x^n\right )\right )}{6 e^3 (d+e x)}+\frac {\left (6 a+11 b n+6 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{6 e^4}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4} \]
-1/3*x^3*(a+b*ln(c*x^n))/e/(e*x+d)^3-1/6*x^2*(3*a+b*n+3*b*ln(c*x^n))/e^2/( e*x+d)^2-1/6*x*(6*a+5*b*n+6*b*ln(c*x^n))/e^3/(e*x+d)+1/6*(6*a+11*b*n+6*b*l n(c*x^n))*ln(1+e*x/d)/e^4+b*n*polylog(2,-e*x/d)/e^4
Time = 0.15 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.27 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\frac {\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3}-\frac {9 d^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac {18 d \left (a+b \log \left (c x^n\right )\right )}{d+e x}-b n \left (\frac {d (3 d+2 e x)}{(d+e x)^2}+2 \log (x)-2 \log (d+e x)\right )-18 b n (\log (x)-\log (d+e x))+9 b n \left (\frac {d}{d+e x}+\log (x)-\log (d+e x)\right )+6 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+6 b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{6 e^4} \]
((2*d^3*(a + b*Log[c*x^n]))/(d + e*x)^3 - (9*d^2*(a + b*Log[c*x^n]))/(d + e*x)^2 + (18*d*(a + b*Log[c*x^n]))/(d + e*x) - b*n*((d*(3*d + 2*e*x))/(d + e*x)^2 + 2*Log[x] - 2*Log[d + e*x]) - 18*b*n*(Log[x] - Log[d + e*x]) + 9* b*n*(d/(d + e*x) + Log[x] - Log[d + e*x]) + 6*(a + b*Log[c*x^n])*Log[1 + ( e*x)/d] + 6*b*n*PolyLog[2, -((e*x)/d)])/(6*e^4)
Time = 0.54 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2784, 2784, 2784, 2754, 2838}
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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {\int \frac {x^2 \left (3 a+b n+3 b \log \left (c x^n\right )\right )}{(d+e x)^3}dx}{3 e}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {\frac {\int \frac {x \left (6 a+5 b n+6 b \log \left (c x^n\right )\right )}{(d+e x)^2}dx}{2 e}-\frac {x^2 \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {6 a+11 b n+6 b \log \left (c x^n\right )}{d+e x}dx}{e}-\frac {x \left (6 a+6 b \log \left (c x^n\right )+5 b n\right )}{e (d+e x)}}{2 e}-\frac {x^2 \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2754 |
| \(\displaystyle \frac {\frac {\frac {\frac {\log \left (\frac {e x}{d}+1\right ) \left (6 a+6 b \log \left (c x^n\right )+11 b n\right )}{e}-\frac {6 b n \int \frac {\log \left (\frac {e x}{d}+1\right )}{x}dx}{e}}{e}-\frac {x \left (6 a+6 b \log \left (c x^n\right )+5 b n\right )}{e (d+e x)}}{2 e}-\frac {x^2 \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {\frac {\frac {\frac {\log \left (\frac {e x}{d}+1\right ) \left (6 a+6 b \log \left (c x^n\right )+11 b n\right )}{e}+\frac {6 b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e}}{e}-\frac {x \left (6 a+6 b \log \left (c x^n\right )+5 b n\right )}{e (d+e x)}}{2 e}-\frac {x^2 \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
-1/3*(x^3*(a + b*Log[c*x^n]))/(e*(d + e*x)^3) + (-1/2*(x^2*(3*a + b*n + 3* b*Log[c*x^n]))/(e*(d + e*x)^2) + (-((x*(6*a + 5*b*n + 6*b*Log[c*x^n]))/(e* (d + e*x))) + (((6*a + 11*b*n + 6*b*Log[c*x^n])*Log[1 + (e*x)/d])/e + (6*b *n*PolyLog[2, -((e*x)/d)])/e)/e)/(2*e))/(3*e)
3.1.55.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symb ol] :> Simp[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^p/e), x] - Simp[b*n*(p/e) Int[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_))^(q_.), x_Symbol] :> Simp[(f*x)^m*(d + e*x)^(q + 1)*((a + b*Log[c*x^n] )/(e*(q + 1))), x] - Simp[f/(e*(q + 1)) Int[(f*x)^(m - 1)*(d + e*x)^(q + 1)*(a*m + b*n + b*m*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && ILtQ[q, -1] && GtQ[m, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.54 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.20
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) d^{3}}{3 e^{4} \left (e x +d \right )^{3}}+\frac {b \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{e^{4}}+\frac {3 b \ln \left (x^{n}\right ) d}{e^{4} \left (e x +d \right )}-\frac {3 b \ln \left (x^{n}\right ) d^{2}}{2 e^{4} \left (e x +d \right )^{2}}+\frac {7 b n d}{6 e^{4} \left (e x +d \right )}+\frac {11 b n \ln \left (e x +d \right )}{6 e^{4}}-\frac {b n \,d^{2}}{6 e^{4} \left (e x +d \right )^{2}}-\frac {11 b n \ln \left (e x \right )}{6 e^{4}}-\frac {b n \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{4}}-\frac {b n \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{4}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {d^{3}}{3 e^{4} \left (e x +d \right )^{3}}+\frac {\ln \left (e x +d \right )}{e^{4}}+\frac {3 d}{e^{4} \left (e x +d \right )}-\frac {3 d^{2}}{2 e^{4} \left (e x +d \right )^{2}}\right )\) | \(310\) |
1/3*b*ln(x^n)/e^4*d^3/(e*x+d)^3+b*ln(x^n)/e^4*ln(e*x+d)+3*b*ln(x^n)/e^4*d/ (e*x+d)-3/2*b*ln(x^n)/e^4*d^2/(e*x+d)^2+7/6*b*n/e^4*d/(e*x+d)+11/6*b*n/e^4 *ln(e*x+d)-1/6*b*n/e^4*d^2/(e*x+d)^2-11/6*b*n/e^4*ln(e*x)-b*n/e^4*ln(e*x+d )*ln(-e*x/d)-b*n/e^4*dilog(-e*x/d)+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn (I*c*x^n)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/2*I*b*Pi*csgn(I*x^n)*csgn (I*c*x^n)^2-1/2*I*b*Pi*csgn(I*c*x^n)^3+b*ln(c)+a)*(1/3/e^4*d^3/(e*x+d)^3+1 /e^4*ln(e*x+d)+3/e^4*d/(e*x+d)-3/2/e^4*d^2/(e*x+d)^2)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{4}} \,d x } \]
integral((b*x^3*log(c*x^n) + a*x^3)/(e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4), x)
Time = 30.94 (sec) , antiderivative size = 518, normalized size of antiderivative = 3.67 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\text {Too large to display} \]
-a*d**3*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))/e**3 + 3*a*d**2*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))/e* *3 - 3*a*d*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/e**3 + a*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**3 + b*d**3*n*Piec ewise((x/d**4, Eq(e, 0)), (-3*d/(6*d**4*e + 12*d**3*e**2*x + 6*d**2*e**3*x **2) - 2*e*x/(6*d**4*e + 12*d**3*e**2*x + 6*d**2*e**3*x**2) - log(x)/(3*d* *3*e) + log(d/e + x)/(3*d**3*e), True))/e**3 - b*d**3*Piecewise((x/d**4, E q(e, 0)), (-1/(3*e*(d + e*x)**3), True))*log(c*x**n)/e**3 - 3*b*d**2*n*Pie cewise((x/d**3, Eq(e, 0)), (-1/(2*d**2*e + 2*d*e**2*x) - log(x)/(2*d**2*e) + log(d/e + x)/(2*d**2*e), True))/e**3 + 3*b*d**2*Piecewise((x/d**3, Eq(e , 0)), (-1/(2*e*(d + e*x)**2), True))*log(c*x**n)/e**3 + 3*b*d*n*Piecewise ((x/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x)/(d*e), True))/e**3 - 3* b*d*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))*log(c*x**n)/e **3 - b*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((-polylog(2, e*x*exp_polar (I*pi)/d), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d)*log(x) - polylog(2, e*x *exp_polar(I*pi)/d), Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x*exp_p olar(I*pi)/d), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log (d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x*exp_p olar(I*pi)/d), True))/e, True))/e**3 + b*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e**3
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{4}} \,d x } \]
1/6*a*((18*d*e^2*x^2 + 27*d^2*e*x + 11*d^3)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2 *e^5*x + d^3*e^4) + 6*log(e*x + d)/e^4) + b*integrate((x^3*log(c) + x^3*lo g(x^n))/(e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4), x)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^4} \,d x \]