Integrand size = 21, antiderivative size = 229 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\frac {10 b d n x}{e^5}-\frac {d (60 a+47 b n) x}{6 e^5}-\frac {5 b n x^2}{2 e^4}-\frac {10 b d x \log \left (c x^n\right )}{e^5}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}-\frac {x^4 \left (5 a+b n+5 b \log \left (c x^n\right )\right )}{6 e^2 (d+e x)^2}-\frac {x^3 \left (20 a+9 b n+20 b \log \left (c x^n\right )\right )}{6 e^3 (d+e x)}+\frac {x^2 \left (60 a+47 b n+60 b \log \left (c x^n\right )\right )}{12 e^4}+\frac {d^2 \left (60 a+47 b n+60 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{6 e^6}+\frac {10 b d^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^6} \]
10*b*d*n*x/e^5-1/6*d*(47*b*n+60*a)*x/e^5-5/2*b*n*x^2/e^4-10*b*d*x*ln(c*x^n )/e^5-1/3*x^5*(a+b*ln(c*x^n))/e/(e*x+d)^3-1/6*x^4*(5*a+b*n+5*b*ln(c*x^n))/ e^2/(e*x+d)^2-1/6*x^3*(20*a+9*b*n+20*b*ln(c*x^n))/e^3/(e*x+d)+1/12*x^2*(60 *a+47*b*n+60*b*ln(c*x^n))/e^4+1/6*d^2*(60*a+47*b*n+60*b*ln(c*x^n))*ln(1+e* x/d)/e^6+10*b*d^2*n*polylog(2,-e*x/d)/e^6
Time = 0.18 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.09 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\frac {-48 a d e x+48 b d e n x-3 b e^2 n x^2-48 b d e x \log \left (c x^n\right )+6 e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {4 d^5 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3}-\frac {30 d^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac {120 d^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x}-2 b d^2 n \left (\frac {d (3 d+2 e x)}{(d+e x)^2}+2 \log (x)-2 \log (d+e x)\right )-120 b d^2 n (\log (x)-\log (d+e x))+30 b d^2 n \left (\frac {d}{d+e x}+\log (x)-\log (d+e x)\right )+120 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+120 b d^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{12 e^6} \]
(-48*a*d*e*x + 48*b*d*e*n*x - 3*b*e^2*n*x^2 - 48*b*d*e*x*Log[c*x^n] + 6*e^ 2*x^2*(a + b*Log[c*x^n]) + (4*d^5*(a + b*Log[c*x^n]))/(d + e*x)^3 - (30*d^ 4*(a + b*Log[c*x^n]))/(d + e*x)^2 + (120*d^3*(a + b*Log[c*x^n]))/(d + e*x) - 2*b*d^2*n*((d*(3*d + 2*e*x))/(d + e*x)^2 + 2*Log[x] - 2*Log[d + e*x]) - 120*b*d^2*n*(Log[x] - Log[d + e*x]) + 30*b*d^2*n*(d/(d + e*x) + Log[x] - Log[d + e*x]) + 120*d^2*(a + b*Log[c*x^n])*Log[1 + (e*x)/d] + 120*b*d^2*n* PolyLog[2, -((e*x)/d)])/(12*e^6)
Time = 0.70 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.05, 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, 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.
| \(\displaystyle \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {\int \frac {x^4 \left (5 a+b n+5 b \log \left (c x^n\right )\right )}{(d+e x)^3}dx}{3 e}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {\frac {\int \frac {x^3 \left (20 a+9 b n+20 b \log \left (c x^n\right )\right )}{(d+e x)^2}dx}{2 e}-\frac {x^4 \left (5 a+5 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {x^2 \left (60 a+47 b n+60 b \log \left (c x^n\right )\right )}{d+e x}dx}{e}-\frac {x^3 \left (20 a+20 b \log \left (c x^n\right )+9 b n\right )}{e (d+e x)}}{2 e}-\frac {x^4 \left (5 a+5 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \frac {\frac {\frac {\int \left (\frac {\left (60 a+47 b n+60 b \log \left (c x^n\right )\right ) d^2}{e^2 (d+e x)}-\frac {\left (60 a+47 b n+60 b \log \left (c x^n\right )\right ) d}{e^2}+\frac {x \left (60 a+47 b n+60 b \log \left (c x^n\right )\right )}{e}\right )dx}{e}-\frac {x^3 \left (20 a+20 b \log \left (c x^n\right )+9 b n\right )}{e (d+e x)}}{2 e}-\frac {x^4 \left (5 a+5 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {\frac {d^2 \log \left (\frac {e x}{d}+1\right ) \left (60 a+60 b \log \left (c x^n\right )+47 b n\right )}{e^3}+\frac {x^2 \left (60 a+60 b \log \left (c x^n\right )+47 b n\right )}{2 e}-\frac {d x (60 a+47 b n)}{e^2}-\frac {60 b d x \log \left (c x^n\right )}{e^2}+\frac {60 b d^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3}+\frac {60 b d n x}{e^2}-\frac {15 b n x^2}{e}}{e}-\frac {x^3 \left (20 a+20 b \log \left (c x^n\right )+9 b n\right )}{e (d+e x)}}{2 e}-\frac {x^4 \left (5 a+5 b \log \left (c x^n\right )+b n\right )}{2 e (d+e x)^2}}{3 e}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}\) |
-1/3*(x^5*(a + b*Log[c*x^n]))/(e*(d + e*x)^3) + (-1/2*(x^4*(5*a + b*n + 5* b*Log[c*x^n]))/(e*(d + e*x)^2) + (-((x^3*(20*a + 9*b*n + 20*b*Log[c*x^n])) /(e*(d + e*x))) + ((60*b*d*n*x)/e^2 - (d*(60*a + 47*b*n)*x)/e^2 - (15*b*n* x^2)/e - (60*b*d*x*Log[c*x^n])/e^2 + (x^2*(60*a + 47*b*n + 60*b*Log[c*x^n] ))/(2*e) + (d^2*(60*a + 47*b*n + 60*b*Log[c*x^n])*Log[1 + (e*x)/d])/e^3 + (60*b*d^2*n*PolyLog[2, -((e*x)/d)])/e^3)/e)/(2*e))/(3*e)
3.1.53.3.1 Defintions of rubi rules used
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 = 0.56 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.77
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) x^{2}}{2 e^{4}}-\frac {4 b \ln \left (x^{n}\right ) d x}{e^{5}}+\frac {b \ln \left (x^{n}\right ) d^{5}}{3 e^{6} \left (e x +d \right )^{3}}+\frac {10 b \ln \left (x^{n}\right ) d^{2} \ln \left (e x +d \right )}{e^{6}}+\frac {10 b \ln \left (x^{n}\right ) d^{3}}{e^{6} \left (e x +d \right )}-\frac {5 b \ln \left (x^{n}\right ) d^{4}}{2 e^{6} \left (e x +d \right )^{2}}-\frac {b n \,x^{2}}{4 e^{4}}+\frac {4 b d n x}{e^{5}}+\frac {17 b n \,d^{2}}{4 e^{6}}+\frac {47 b n \,d^{2} \ln \left (e x +d \right )}{6 e^{6}}+\frac {13 b n \,d^{3}}{6 e^{6} \left (e x +d \right )}-\frac {b n \,d^{4}}{6 e^{6} \left (e x +d \right )^{2}}-\frac {47 b n \,d^{2} \ln \left (e x \right )}{6 e^{6}}-\frac {10 b n \,d^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{6}}-\frac {10 b n \,d^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{6}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{2} e \,x^{2}-4 d x}{e^{5}}+\frac {d^{5}}{3 e^{6} \left (e x +d \right )^{3}}+\frac {10 d^{2} \ln \left (e x +d \right )}{e^{6}}+\frac {10 d^{3}}{e^{6} \left (e x +d \right )}-\frac {5 d^{4}}{2 e^{6} \left (e x +d \right )^{2}}\right )\) | \(405\) |
1/2*b*ln(x^n)/e^4*x^2-4*b*ln(x^n)/e^5*d*x+1/3*b*ln(x^n)*d^5/e^6/(e*x+d)^3+ 10*b*ln(x^n)/e^6*d^2*ln(e*x+d)+10*b*ln(x^n)/e^6*d^3/(e*x+d)-5/2*b*ln(x^n)/ e^6*d^4/(e*x+d)^2-1/4*b*n*x^2/e^4+4*b*d*n*x/e^5+17/4*b*n/e^6*d^2+47/6*b*n/ e^6*d^2*ln(e*x+d)+13/6*b*n/e^6*d^3/(e*x+d)-1/6*b*n/e^6*d^4/(e*x+d)^2-47/6* b*n/e^6*d^2*ln(e*x)-10*b*n/e^6*d^2*ln(e*x+d)*ln(-e*x/d)-10*b*n/e^6*d^2*dil og(-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*csg n(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*c sgn(I*c*x^n)^3+b*ln(c)+a)*(1/e^5*(1/2*e*x^2-4*d*x)+1/3*d^5/e^6/(e*x+d)^3+1 0/e^6*d^2*ln(e*x+d)+10/e^6*d^3/(e*x+d)-5/2/e^6*d^4/(e*x+d)^2)
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x + d\right )}^{4}} \,d x } \]
integral((b*x^5*log(c*x^n) + a*x^5)/(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 = 62.66 (sec) , antiderivative size = 617, normalized size of antiderivative = 2.69 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\text {Too large to display} \]
-a*d**5*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))/e**5 + 5*a*d**4*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))/e* *5 - 10*a*d**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/e* *5 + 10*a*d**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**5 - 4 *a*d*x/e**5 + a*x**2/(2*e**4) + b*d**5*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**5 - b*d**5*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)* *3), True))*log(c*x**n)/e**5 - 5*b*d**4*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), T rue))/e**5 + 5*b*d**4*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2) , True))*log(c*x**n)/e**5 + 10*b*d**3*n*Piecewise((x/d**2, Eq(e, 0)), (-lo g(x)/(d*e) + log(d/e + x)/(d*e), True))/e**5 - 10*b*d**3*Piecewise((x/d**2 , Eq(e, 0)), (-1/(d*e + e**2*x), True))*log(c*x**n)/e**5 - 10*b*d**2*n*Pie cewise((x/d, Eq(e, 0)), (Piecewise((-polylog(2, e*x*exp_polar(I*pi)/d), (A bs(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)*log(d) - polylog(2, e*x*exp_polar(I*pi)/d) , True))/e, True))/e**5 + 10*b*d**2*Piecewise((x/d, Eq(e, 0)), (log(d +...
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x + d\right )}^{4}} \,d x } \]
1/6*a*((60*d^3*e^2*x^2 + 105*d^4*e*x + 47*d^5)/(e^9*x^3 + 3*d*e^8*x^2 + 3* d^2*e^7*x + d^3*e^6) + 60*d^2*log(e*x + d)/e^6 + 3*(e*x^2 - 8*d*x)/e^5) + b*integrate((x^5*log(c) + x^5*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)
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x + d\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^4} \,d x \]