Integrand size = 21, antiderivative size = 171 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)^3} \, dx=-\frac {b n}{d^3 x}+\frac {b e n}{2 d^3 (d+e x)}+\frac {b e n \log (x)}{2 d^4}-\frac {a+b \log \left (c x^n\right )}{d^3 x}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 (d+e x)^2}+\frac {2 e^2 x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac {3 e \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac {5 b e n \log (d+e x)}{2 d^4}-\frac {3 b e n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^4} \] Output:
-b*n/d^3/x+1/2*b*e*n/d^3/(e*x+d)+1/2*b*e*n*ln(x)/d^4-(a+b*ln(c*x^n))/d^3/x -1/2*e*(a+b*ln(c*x^n))/d^2/(e*x+d)^2+2*e^2*x*(a+b*ln(c*x^n))/d^4/(e*x+d)+3 *e*ln(1+d/e/x)*(a+b*ln(c*x^n))/d^4-5/2*b*e*n*ln(e*x+d)/d^4-3*b*e*n*polylog (2,-d/e/x)/d^4
Time = 0.21 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.01 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)^3} \, dx=\frac {-\frac {2 b d n}{x}-\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {d^2 e \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}-\frac {4 d e \left (a+b \log \left (c x^n\right )\right )}{d+e x}-\frac {3 e \left (a+b \log \left (c x^n\right )\right )^2}{b n}+4 b e n (\log (x)-\log (d+e x))+b e n \left (\frac {d}{d+e x}+\log (x)-\log (d+e x)\right )+6 e \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+6 b e n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{2 d^4} \] Input:
Integrate[(a + b*Log[c*x^n])/(x^2*(d + e*x)^3),x]
Output:
((-2*b*d*n)/x - (2*d*(a + b*Log[c*x^n]))/x - (d^2*e*(a + b*Log[c*x^n]))/(d + e*x)^2 - (4*d*e*(a + b*Log[c*x^n]))/(d + e*x) - (3*e*(a + b*Log[c*x^n]) ^2)/(b*n) + 4*b*e*n*(Log[x] - Log[d + e*x]) + b*e*n*(d/(d + e*x) + Log[x] - Log[d + e*x]) + 6*e*(a + b*Log[c*x^n])*Log[1 + (e*x)/d] + 6*b*e*n*PolyLo g[2, -((e*x)/d)])/(2*d^4)
Time = 0.47 (sec) , antiderivative size = 171, 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.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)^3} \, dx\) |
\(\Big \downarrow \) 2793 |
\(\displaystyle \int \left (\frac {2 e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)^2}-\frac {3 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x (d+e x)}+\frac {a+b \log \left (c x^n\right )}{d^3 x^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{d^2 (d+e x)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 e^2 x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac {3 e \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac {a+b \log \left (c x^n\right )}{d^3 x}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 (d+e x)^2}-\frac {3 b e n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^4}+\frac {b e n \log (x)}{2 d^4}-\frac {5 b e n \log (d+e x)}{2 d^4}+\frac {b e n}{2 d^3 (d+e x)}-\frac {b n}{d^3 x}\) |
Input:
Int[(a + b*Log[c*x^n])/(x^2*(d + e*x)^3),x]
Output:
-((b*n)/(d^3*x)) + (b*e*n)/(2*d^3*(d + e*x)) + (b*e*n*Log[x])/(2*d^4) - (a + b*Log[c*x^n])/(d^3*x) - (e*(a + b*Log[c*x^n]))/(2*d^2*(d + e*x)^2) + (2 *e^2*x*(a + b*Log[c*x^n]))/(d^4*(d + e*x)) + (3*e*Log[1 + d/(e*x)]*(a + b* Log[c*x^n]))/d^4 - (5*b*e*n*Log[d + e*x])/(2*d^4) - (3*b*e*n*PolyLog[2, -( d/(e*x))])/d^4
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 = 324, normalized size of antiderivative = 1.89
method | result | size |
risch | \(-\frac {b \ln \left (x^{n}\right ) e}{2 d^{2} \left (e x +d \right )^{2}}+\frac {3 b \ln \left (x^{n}\right ) e \ln \left (e x +d \right )}{d^{4}}-\frac {2 b \ln \left (x^{n}\right ) e}{d^{3} \left (e x +d \right )}-\frac {b \ln \left (x^{n}\right )}{d^{3} x}-\frac {3 b \ln \left (x^{n}\right ) e \ln \left (x \right )}{d^{4}}+\frac {3 b n e \ln \left (x \right )^{2}}{2 d^{4}}-\frac {3 b n e \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{4}}-\frac {3 b n e \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{4}}+\frac {b e n}{2 d^{3} \left (e x +d \right )}-\frac {5 b e n \ln \left (e x +d \right )}{2 d^{4}}-\frac {b n}{d^{3} x}+\frac {5 b e n \ln \left (x \right )}{2 d^{4}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {e}{2 d^{2} \left (e x +d \right )^{2}}+\frac {3 e \ln \left (e x +d \right )}{d^{4}}-\frac {2 e}{d^{3} \left (e x +d \right )}-\frac {1}{d^{3} x}-\frac {3 e \ln \left (x \right )}{d^{4}}\right )\) | \(324\) |
Input:
int((a+b*ln(c*x^n))/x^2/(e*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-1/2*b*ln(x^n)/d^2/(e*x+d)^2*e+3*b*ln(x^n)/d^4*e*ln(e*x+d)-2*b*ln(x^n)/d^3 *e/(e*x+d)-b*ln(x^n)/d^3/x-3*b*ln(x^n)/d^4*e*ln(x)+3/2*b*n/d^4*e*ln(x)^2-3 *b*n/d^4*e*ln(e*x+d)*ln(-e*x/d)-3*b*n/d^4*e*dilog(-e*x/d)+1/2*b*e*n/d^3/(e *x+d)-5/2*b*e*n*ln(e*x+d)/d^4-b*n/d^3/x+5/2*b*e*n*ln(x)/d^4+(1/2*I*Pi*b*cs gn(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)* (-1/2/d^2/(e*x+d)^2*e+3/d^4*e*ln(e*x+d)-2/d^3*e/(e*x+d)-1/d^3/x-3/d^4*e*ln (x))
\[ \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{3} x^{2}} \,d x } \] Input:
integrate((a+b*log(c*x^n))/x^2/(e*x+d)^3,x, algorithm="fricas")
Output:
integral((b*log(c*x^n) + a)/(e^3*x^5 + 3*d*e^2*x^4 + 3*d^2*e*x^3 + d^3*x^2 ), x)
Time = 45.74 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.60 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)^3} \, dx =\text {Too large to display} \] Input:
integrate((a+b*ln(c*x**n))/x**2/(e*x+d)**3,x)
Output:
a*e**2*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))/d**2 + 2*a*e**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/d**3 - a/(d**3*x) + 3*a*e**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/d **4 - 3*a*e*log(x)/d**4 - b*e**2*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))/d **2 + b*e**2*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))* log(c*x**n)/d**2 - 2*b*e**2*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x)/(d*e), True))/d**3 + 2*b*e**2*Piecewise((x/d**2, Eq(e, 0)) , (-1/(d*e + e**2*x), True))*log(c*x**n)/d**3 - b*n/(d**3*x) - b*log(c*x** n)/(d**3*x) - 3*b*e**2*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) - polyl og(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) - polyl og(2, e*x*exp_polar(I*pi)/d), True))/e, True))/d**4 + 3*b*e**2*Piecewise(( x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/d**4 + 3*b*e*n*log(x)* *2/(2*d**4) - 3*b*e*log(x)*log(c*x**n)/d**4
\[ \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{3} x^{2}} \,d x } \] Input:
integrate((a+b*log(c*x^n))/x^2/(e*x+d)^3,x, algorithm="maxima")
Output:
-1/2*a*((6*e^2*x^2 + 9*d*e*x + 2*d^2)/(d^3*e^2*x^3 + 2*d^4*e*x^2 + d^5*x) - 6*e*log(e*x + d)/d^4 + 6*e*log(x)/d^4) + b*integrate((log(c) + log(x^n)) /(e^3*x^5 + 3*d*e^2*x^4 + 3*d^2*e*x^3 + d^3*x^2), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{3} x^{2}} \,d x } \] Input:
integrate((a+b*log(c*x^n))/x^2/(e*x+d)^3,x, algorithm="giac")
Output:
integrate((b*log(c*x^n) + a)/((e*x + d)^3*x^2), x)
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)^3} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^2\,{\left (d+e\,x\right )}^3} \,d x \] Input:
int((a + b*log(c*x^n))/(x^2*(d + e*x)^3),x)
Output:
int((a + b*log(c*x^n))/(x^2*(d + e*x)^3), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)^3} \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{3} x^{5}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{3}+d^{3} x^{2}}d x \right ) b \,d^{6} x +4 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{3} x^{5}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{3}+d^{3} x^{2}}d x \right ) b \,d^{5} e \,x^{2}+2 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{3} x^{5}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{3}+d^{3} x^{2}}d x \right ) b \,d^{4} e^{2} x^{3}+6 \,\mathrm {log}\left (e x +d \right ) a \,d^{2} e x +12 \,\mathrm {log}\left (e x +d \right ) a d \,e^{2} x^{2}+6 \,\mathrm {log}\left (e x +d \right ) a \,e^{3} x^{3}-6 \,\mathrm {log}\left (x \right ) a \,d^{2} e x -12 \,\mathrm {log}\left (x \right ) a d \,e^{2} x^{2}-6 \,\mathrm {log}\left (x \right ) a \,e^{3} x^{3}-2 a \,d^{3}-6 a \,d^{2} e x +3 a \,e^{3} x^{3}}{2 d^{4} x \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:
int((a+b*log(c*x^n))/x^2/(e*x+d)^3,x)
Output:
(2*int(log(x**n*c)/(d**3*x**2 + 3*d**2*e*x**3 + 3*d*e**2*x**4 + e**3*x**5) ,x)*b*d**6*x + 4*int(log(x**n*c)/(d**3*x**2 + 3*d**2*e*x**3 + 3*d*e**2*x** 4 + e**3*x**5),x)*b*d**5*e*x**2 + 2*int(log(x**n*c)/(d**3*x**2 + 3*d**2*e* x**3 + 3*d*e**2*x**4 + e**3*x**5),x)*b*d**4*e**2*x**3 + 6*log(d + e*x)*a*d **2*e*x + 12*log(d + e*x)*a*d*e**2*x**2 + 6*log(d + e*x)*a*e**3*x**3 - 6*l og(x)*a*d**2*e*x - 12*log(x)*a*d*e**2*x**2 - 6*log(x)*a*e**3*x**3 - 2*a*d* *3 - 6*a*d**2*e*x + 3*a*e**3*x**3)/(2*d**4*x*(d**2 + 2*d*e*x + e**2*x**2))