Integrand size = 21, antiderivative size = 263 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^4} \, dx=-\frac {b n}{4 d^4 x^2}+\frac {4 b e n}{d^5 x}-\frac {b e^2 n}{6 d^4 (d+e x)^2}-\frac {11 b e^2 n}{6 d^5 (d+e x)}-\frac {11 b e^2 n \log (x)}{6 d^6}-\frac {a+b \log \left (c x^n\right )}{2 d^4 x^2}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{d^5 x}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)^3}+\frac {3 e^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^4 (d+e x)^2}-\frac {6 e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^6 (d+e x)}-\frac {10 e^2 \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^6}+\frac {47 b e^2 n \log (d+e x)}{6 d^6}+\frac {10 b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^6} \]
-1/4*b*n/d^4/x^2+4*b*e*n/d^5/x-1/6*b*e^2*n/d^4/(e*x+d)^2-11/6*b*e^2*n/d^5/ (e*x+d)-11/6*b*e^2*n*ln(x)/d^6+1/2*(-a-b*ln(c*x^n))/d^4/x^2+4*e*(a+b*ln(c* x^n))/d^5/x+1/3*e^2*(a+b*ln(c*x^n))/d^3/(e*x+d)^3+3/2*e^2*(a+b*ln(c*x^n))/ d^4/(e*x+d)^2-6*e^3*x*(a+b*ln(c*x^n))/d^6/(e*x+d)-10*e^2*ln(1+d/e/x)*(a+b* ln(c*x^n))/d^6+47/6*b*e^2*n*ln(e*x+d)/d^6+10*b*e^2*n*polylog(2,-d/e/x)/d^6
Time = 0.20 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^4} \, dx=\frac {-\frac {3 b d^2 n}{x^2}+\frac {48 b d e n}{x}-\frac {18 b d e^2 n}{d+e x}-\frac {2 b d e^2 n (3 d+2 e x)}{(d+e x)^2}-22 b e^2 n \log (x)-\frac {6 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^2}+\frac {48 d e \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {4 d^3 e^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3}+\frac {18 d^2 e^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac {72 d e^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}+\frac {60 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}-72 b e^2 n (\log (x)-\log (d+e x))+22 b e^2 n \log (d+e x)-120 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-120 b e^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{12 d^6} \]
((-3*b*d^2*n)/x^2 + (48*b*d*e*n)/x - (18*b*d*e^2*n)/(d + e*x) - (2*b*d*e^2 *n*(3*d + 2*e*x))/(d + e*x)^2 - 22*b*e^2*n*Log[x] - (6*d^2*(a + b*Log[c*x^ n]))/x^2 + (48*d*e*(a + b*Log[c*x^n]))/x + (4*d^3*e^2*(a + b*Log[c*x^n]))/ (d + e*x)^3 + (18*d^2*e^2*(a + b*Log[c*x^n]))/(d + e*x)^2 + (72*d*e^2*(a + b*Log[c*x^n]))/(d + e*x) + (60*e^2*(a + b*Log[c*x^n])^2)/(b*n) - 72*b*e^2 *n*(Log[x] - Log[d + e*x]) + 22*b*e^2*n*Log[d + e*x] - 120*e^2*(a + b*Log[ c*x^n])*Log[1 + (e*x)/d] - 120*b*e^2*n*PolyLog[2, -((e*x)/d)])/(12*d^6)
Time = 0.57 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {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.
| \(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (-\frac {6 e^3 \left (a+b \log \left (c x^n\right )\right )}{d^5 (d+e x)^2}+\frac {10 e^2 \left (a+b \log \left (c x^n\right )\right )}{d^5 x (d+e x)}-\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{d^5 x^2}-\frac {3 e^3 \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)^3}+\frac {a+b \log \left (c x^n\right )}{d^4 x^3}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {6 e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^6 (d+e x)}-\frac {10 e^2 \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^6}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{d^5 x}+\frac {3 e^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^4 (d+e x)^2}-\frac {a+b \log \left (c x^n\right )}{2 d^4 x^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)^3}+\frac {10 b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^6}-\frac {11 b e^2 n \log (x)}{6 d^6}+\frac {47 b e^2 n \log (d+e x)}{6 d^6}-\frac {11 b e^2 n}{6 d^5 (d+e x)}+\frac {4 b e n}{d^5 x}-\frac {b e^2 n}{6 d^4 (d+e x)^2}-\frac {b n}{4 d^4 x^2}\) |
-1/4*(b*n)/(d^4*x^2) + (4*b*e*n)/(d^5*x) - (b*e^2*n)/(6*d^4*(d + e*x)^2) - (11*b*e^2*n)/(6*d^5*(d + e*x)) - (11*b*e^2*n*Log[x])/(6*d^6) - (a + b*Log [c*x^n])/(2*d^4*x^2) + (4*e*(a + b*Log[c*x^n]))/(d^5*x) + (e^2*(a + b*Log[ c*x^n]))/(3*d^3*(d + e*x)^3) + (3*e^2*(a + b*Log[c*x^n]))/(2*d^4*(d + e*x) ^2) - (6*e^3*x*(a + b*Log[c*x^n]))/(d^6*(d + e*x)) - (10*e^2*Log[1 + d/(e* x)]*(a + b*Log[c*x^n]))/d^6 + (47*b*e^2*n*Log[d + e*x])/(6*d^6) + (10*b*e^ 2*n*PolyLog[2, -(d/(e*x))])/d^6
3.1.61.3.1 Defintions of rubi rules used
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 = 0.92 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.67
| method | result | size |
| risch | \(-\frac {10 b \ln \left (x^{n}\right ) e^{2} \ln \left (e x +d \right )}{d^{6}}+\frac {6 b \ln \left (x^{n}\right ) e^{2}}{d^{5} \left (e x +d \right )}+\frac {3 b \ln \left (x^{n}\right ) e^{2}}{2 d^{4} \left (e x +d \right )^{2}}+\frac {b \ln \left (x^{n}\right ) e^{2}}{3 d^{3} \left (e x +d \right )^{3}}-\frac {b \ln \left (x^{n}\right )}{2 d^{4} x^{2}}+\frac {10 b \ln \left (x^{n}\right ) e^{2} \ln \left (x \right )}{d^{6}}+\frac {4 b \ln \left (x^{n}\right ) e}{d^{5} x}-\frac {11 b \,e^{2} n}{6 d^{5} \left (e x +d \right )}+\frac {47 b \,e^{2} n \ln \left (e x +d \right )}{6 d^{6}}-\frac {b \,e^{2} n}{6 d^{4} \left (e x +d \right )^{2}}-\frac {b n}{4 d^{4} x^{2}}+\frac {4 b e n}{d^{5} x}-\frac {47 b \,e^{2} n \ln \left (x \right )}{6 d^{6}}-\frac {5 b n \,e^{2} \ln \left (x \right )^{2}}{d^{6}}+\frac {10 b n \,e^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{6}}+\frac {10 b n \,e^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{6}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {10 e^{2} \ln \left (e x +d \right )}{d^{6}}+\frac {6 e^{2}}{d^{5} \left (e x +d \right )}+\frac {3 e^{2}}{2 d^{4} \left (e x +d \right )^{2}}+\frac {e^{2}}{3 d^{3} \left (e x +d \right )^{3}}-\frac {1}{2 d^{4} x^{2}}+\frac {10 e^{2} \ln \left (x \right )}{d^{6}}+\frac {4 e}{d^{5} x}\right )\) | \(438\) |
-10*b*ln(x^n)/d^6*e^2*ln(e*x+d)+6*b*ln(x^n)/d^5*e^2/(e*x+d)+3/2*b*ln(x^n)/ d^4*e^2/(e*x+d)^2+1/3*b*ln(x^n)/d^3/(e*x+d)^3*e^2-1/2*b*ln(x^n)/d^4/x^2+10 *b*ln(x^n)/d^6*e^2*ln(x)+4*b*ln(x^n)/d^5*e/x-11/6*b*e^2*n/d^5/(e*x+d)+47/6 *b*e^2*n*ln(e*x+d)/d^6-1/6*b*e^2*n/d^4/(e*x+d)^2-1/4*b*n/d^4/x^2+4*b*e*n/d ^5/x-47/6*b*e^2*n*ln(x)/d^6-5*b*n/d^6*e^2*ln(x)^2+10*b*n/d^6*e^2*ln(e*x+d) *ln(-e*x/d)+10*b*n/d^6*e^2*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)*(-10/d^6*e^2*ln(e* x+d)+6/d^5*e^2/(e*x+d)+3/2/d^4*e^2/(e*x+d)^2+1/3/d^3/(e*x+d)^3*e^2-1/2/d^4 /x^2+10/d^6*e^2*ln(x)+4/d^5*e/x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^4} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{4} x^{3}} \,d x } \]
integral((b*log(c*x^n) + a)/(e^4*x^7 + 4*d*e^3*x^6 + 6*d^2*e^2*x^5 + 4*d^3 *e*x^4 + d^4*x^3), x)
Time = 61.94 (sec) , antiderivative size = 668, normalized size of antiderivative = 2.54 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^4} \, dx=\text {Too large to display} \]
-a*e**3*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))/d**3 - 3*a*e**3*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))/d* *4 - a/(2*d**4*x**2) - 6*a*e**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e **2*x), True))/d**5 + 4*a*e/(d**5*x) - 10*a*e**3*Piecewise((x/d, Eq(e, 0)) , (log(d + e*x)/e, True))/d**6 + 10*a*e**2*log(x)/d**6 + b*e**3*n*Piecewis e((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))/d**3 - b*e**3*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))*log(c*x**n)/d**3 + 3*b*e**3*n*Piecewi se((x/d**3, Eq(e, 0)), (-1/(2*d**2*e + 2*d*e**2*x) - log(x)/(2*d**2*e) + l og(d/e + x)/(2*d**2*e), True))/d**4 - 3*b*e**3*Piecewise((x/d**3, Eq(e, 0) ), (-1/(2*e*(d + e*x)**2), True))*log(c*x**n)/d**4 - b*n/(4*d**4*x**2) - b *log(c*x**n)/(2*d**4*x**2) + 6*b*e**3*n*Piecewise((x/d**2, Eq(e, 0)), (-lo g(x)/(d*e) + log(d/e + x)/(d*e), True))/d**5 - 6*b*e**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))*log(c*x**n)/d**5 + 4*b*e*n/(d**5*x) + 4*b*e*log(c*x**n)/(d**5*x) + 10*b*e**3*n*Piecewise((x/d, Eq(e, 0)), (Pi ecewise((-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, ...
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^4} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{4} x^{3}} \,d x } \]
1/6*a*((60*e^4*x^4 + 150*d*e^3*x^3 + 110*d^2*e^2*x^2 + 15*d^3*e*x - 3*d^4) /(d^5*e^3*x^5 + 3*d^6*e^2*x^4 + 3*d^7*e*x^3 + d^8*x^2) - 60*e^2*log(e*x + d)/d^6 + 60*e^2*log(x)/d^6) + b*integrate((log(c) + log(x^n))/(e^4*x^7 + 4 *d*e^3*x^6 + 6*d^2*e^2*x^5 + 4*d^3*e*x^4 + d^4*x^3), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^4} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{4} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^4} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^3\,{\left (d+e\,x\right )}^4} \,d x \]