Integrand size = 21, antiderivative size = 183 \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=-\frac {4 b n x}{e^4}+\frac {(12 a+13 b n) x}{3 e^4}+\frac {4 b x \log \left (c x^n\right )}{e^4}-\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}-\frac {x^3 \left (4 a+b n+4 b \log \left (c x^n\right )\right )}{6 e^2 (d+e x)^2}-\frac {x^2 \left (12 a+7 b n+12 b \log \left (c x^n\right )\right )}{6 e^3 (d+e x)}-\frac {d \left (12 a+13 b n+12 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{3 e^5}-\frac {4 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^5} \]
-4*b*n*x/e^4+1/3*(13*b*n+12*a)*x/e^4+4*b*x*ln(c*x^n)/e^4-1/3*x^4*(a+b*ln(c *x^n))/e/(e*x+d)^3-1/6*x^3*(4*a+b*n+4*b*ln(c*x^n))/e^2/(e*x+d)^2-1/6*x^2*( 12*a+7*b*n+12*b*ln(c*x^n))/e^3/(e*x+d)-1/3*d*(12*a+13*b*n+12*b*ln(c*x^n))* ln(1+e*x/d)/e^5-4*b*d*n*polylog(2,-e*x/d)/e^5
Time = 0.14 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.13 \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\frac {6 a e x-6 b e n x+6 b e x \log \left (c x^n\right )-\frac {2 d^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3}+\frac {12 d^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}-\frac {36 d^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}+b d n \left (\frac {d (3 d+2 e x)}{(d+e x)^2}+2 \log (x)-2 \log (d+e x)\right )+36 b d n (\log (x)-\log (d+e x))-12 b d n \left (\frac {d}{d+e x}+\log (x)-\log (d+e x)\right )-24 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-24 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{6 e^5} \]
(6*a*e*x - 6*b*e*n*x + 6*b*e*x*Log[c*x^n] - (2*d^4*(a + b*Log[c*x^n]))/(d + e*x)^3 + (12*d^3*(a + b*Log[c*x^n]))/(d + e*x)^2 - (36*d^2*(a + b*Log[c* x^n]))/(d + e*x) + b*d*n*((d*(3*d + 2*e*x))/(d + e*x)^2 + 2*Log[x] - 2*Log [d + e*x]) + 36*b*d*n*(Log[x] - Log[d + e*x]) - 12*b*d*n*(d/(d + e*x) + Lo g[x] - Log[d + e*x]) - 24*d*(a + b*Log[c*x^n])*Log[1 + (e*x)/d] - 24*b*d*n *PolyLog[2, -((e*x)/d)])/(6*e^5)
Time = 0.64 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2784, 2784, 2784, 27, 2793, 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.
| \(\displaystyle \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {\int \frac {x^3 \left (4 a+b n+4 b \log \left (c x^n\right )\right )}{(d+e x)^3}dx}{3 e}-\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {\frac {\int \frac {x^2 \left (12 a+7 b n+12 b \log \left (c x^n\right )\right )}{(d+e x)^2}dx}{2 e}-\frac {x^3 \left (4 a+4 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 x \left (12 a+13 b n+12 b \log \left (c x^n\right )\right )}{d+e x}dx}{e}-\frac {x^2 \left (12 a+12 b \log \left (c x^n\right )+7 b n\right )}{e (d+e x)}}{2 e}-\frac {x^3 \left (4 a+4 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {2 \int \frac {x \left (12 a+13 b n+12 b \log \left (c x^n\right )\right )}{d+e x}dx}{e}-\frac {x^2 \left (12 a+12 b \log \left (c x^n\right )+7 b n\right )}{e (d+e x)}}{2 e}-\frac {x^3 \left (4 a+4 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \frac {\frac {\frac {2 \int \left (\frac {12 a+13 b n+12 b \log \left (c x^n\right )}{e}-\frac {d \left (12 a+13 b n+12 b \log \left (c x^n\right )\right )}{e (d+e x)}\right )dx}{e}-\frac {x^2 \left (12 a+12 b \log \left (c x^n\right )+7 b n\right )}{e (d+e x)}}{2 e}-\frac {x^3 \left (4 a+4 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {2 \left (-\frac {d \log \left (\frac {e x}{d}+1\right ) \left (12 a+12 b \log \left (c x^n\right )+13 b n\right )}{e^2}+\frac {x (12 a+13 b n)}{e}+\frac {12 b x \log \left (c x^n\right )}{e}-\frac {12 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^2}-\frac {12 b n x}{e}\right )}{e}-\frac {x^2 \left (12 a+12 b \log \left (c x^n\right )+7 b n\right )}{e (d+e x)}}{2 e}-\frac {x^3 \left (4 a+4 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
-1/3*(x^4*(a + b*Log[c*x^n]))/(e*(d + e*x)^3) + (-1/2*(x^3*(4*a + b*n + 4* b*Log[c*x^n]))/(e*(d + e*x)^2) + (-((x^2*(12*a + 7*b*n + 12*b*Log[c*x^n])) /(e*(d + e*x))) + (2*((-12*b*n*x)/e + ((12*a + 13*b*n)*x)/e + (12*b*x*Log[ c*x^n])/e - (d*(12*a + 13*b*n + 12*b*Log[c*x^n])*Log[1 + (e*x)/d])/e^2 - ( 12*b*d*n*PolyLog[2, -((e*x)/d)])/e^2))/e)/(2*e))/(3*e)
3.1.54.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && Integer Q[r]))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.53 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.94
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) x}{e^{4}}-\frac {b \ln \left (x^{n}\right ) d^{4}}{3 e^{5} \left (e x +d \right )^{3}}-\frac {4 b \ln \left (x^{n}\right ) d \ln \left (e x +d \right )}{e^{5}}-\frac {6 b \ln \left (x^{n}\right ) d^{2}}{e^{5} \left (e x +d \right )}+\frac {2 b \ln \left (x^{n}\right ) d^{3}}{e^{5} \left (e x +d \right )^{2}}-\frac {b n x}{e^{4}}-\frac {b n d}{e^{5}}+\frac {b n \,d^{3}}{6 e^{5} \left (e x +d \right )^{2}}-\frac {13 b n d \ln \left (e x +d \right )}{3 e^{5}}-\frac {5 b n \,d^{2}}{3 e^{5} \left (e x +d \right )}+\frac {13 b n d \ln \left (e x \right )}{3 e^{5}}+\frac {4 b n d \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{5}}+\frac {4 b n d \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{5}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {x}{e^{4}}-\frac {d^{4}}{3 e^{5} \left (e x +d \right )^{3}}-\frac {4 d \ln \left (e x +d \right )}{e^{5}}-\frac {6 d^{2}}{e^{5} \left (e x +d \right )}+\frac {2 d^{3}}{e^{5} \left (e x +d \right )^{2}}\right )\) | \(355\) |
b*ln(x^n)/e^4*x-1/3*b*ln(x^n)/e^5*d^4/(e*x+d)^3-4*b*ln(x^n)/e^5*d*ln(e*x+d )-6*b*ln(x^n)/e^5*d^2/(e*x+d)+2*b*ln(x^n)/e^5*d^3/(e*x+d)^2-b*n*x/e^4-b*n/ e^5*d+1/6*b*n/e^5*d^3/(e*x+d)^2-13/3*b*n/e^5*d*ln(e*x+d)-5/3*b*n/e^5*d^2/( e*x+d)+13/3*b*n/e^5*d*ln(e*x)+4*b*n/e^5*d*ln(e*x+d)*ln(-e*x/d)+4*b*n/e^5*d *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)*(x/e^4-1/3/e^5*d^4/(e*x+d)^3-4/e^5*d*ln(e*x+ d)-6/e^5*d^2/(e*x+d)+2/e^5*d^3/(e*x+d)^2)
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{4}}{{\left (e x + d\right )}^{4}} \,d x } \]
integral((b*x^4*log(c*x^n) + a*x^4)/(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 = 31.90 (sec) , antiderivative size = 563, normalized size of antiderivative = 3.08 \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\text {Too large to display} \]
a*d**4*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))/e**4 - 4*a*d**3*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))/e** 4 + 6*a*d**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/e**4 - 4*a*d*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**4 + a*x/e** 4 - b*d**4*n*Piecewise((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**4 + b*d**4*Pie cewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))*log(c*x**n)/e**4 + 4*b*d**3*n*Piecewise((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**4 - 4*b*d**3*Piecew ise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))*log(c*x**n)/e**4 - 6*b*d**2*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x)/(d* e), True))/e**4 + 6*b*d**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x ), True))*log(c*x**n)/e**4 + 4*b*d*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_polar(I*pi)/d), 1/Abs(x) < 1), (-meijerg(((), (1 , 1)), ((0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*lo g(d) - polylog(2, e*x*exp_polar(I*pi)/d), True))/e, True))/e**4 - 4*b*d*Pi ecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e**4 - b*n...
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{4}}{{\left (e x + d\right )}^{4}} \,d x } \]
-1/3*a*((18*d^2*e^2*x^2 + 30*d^3*e*x + 13*d^4)/(e^8*x^3 + 3*d*e^7*x^2 + 3* d^2*e^6*x + d^3*e^5) - 3*x/e^4 + 12*d*log(e*x + d)/e^5) + b*integrate((x^4 *log(c) + x^4*log(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)
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{4}}{{\left (e x + d\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int \frac {x^4\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^4} \,d x \]