\(\int \frac {a+b \log (c x^n)}{x^3 (d+e x)^2} \, dx\) [45]

Optimal result

Integrand size = 21, antiderivative size = 154 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=-\frac {b n}{4 d^2 x^2}+\frac {2 b e n}{d^3 x}-\frac {a+b \log \left (c x^n\right )}{2 d^2 x^2}+\frac {2 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}-\frac {3 e^2 \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}+\frac {b e^2 n \log (d+e x)}{d^4}+\frac {3 b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^4} \] Output:

-1/4*b*n/d^2/x^2+2*b*e*n/d^3/x-1/2*(a+b*ln(c*x^n))/d^2/x^2+2*e*(a+b*ln(c*x 
^n))/d^3/x-e^3*x*(a+b*ln(c*x^n))/d^4/(e*x+d)-3*e^2*ln(1+d/e/x)*(a+b*ln(c*x 
^n))/d^4+b*e^2*n*ln(e*x+d)/d^4+3*b*e^2*n*polylog(2,-d/e/x)/d^4
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.07 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=-\frac {\frac {b d^2 n}{x^2}-\frac {8 b d e n}{x}+\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac {8 d e \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {4 d e^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}-\frac {6 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+4 b e^2 n (\log (x)-\log (d+e x))+12 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+12 b e^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 d^4} \] Input:

Integrate[(a + b*Log[c*x^n])/(x^3*(d + e*x)^2),x]
 

Output:

-1/4*((b*d^2*n)/x^2 - (8*b*d*e*n)/x + (2*d^2*(a + b*Log[c*x^n]))/x^2 - (8* 
d*e*(a + b*Log[c*x^n]))/x - (4*d*e^2*(a + b*Log[c*x^n]))/(d + e*x) - (6*e^ 
2*(a + b*Log[c*x^n])^2)/(b*n) + 4*b*e^2*n*(Log[x] - Log[d + e*x]) + 12*e^2 
*(a + b*Log[c*x^n])*Log[1 + (e*x)/d] + 12*b*e^2*n*PolyLog[2, -((e*x)/d)])/ 
d^4
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 154, 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.

\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx\)

\(\Big \downarrow \) 2793

\(\displaystyle \int \left (-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)^2}+\frac {3 e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3 x (d+e x)}-\frac {2 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x^2}+\frac {a+b \log \left (c x^n\right )}{d^2 x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}-\frac {3 e^2 \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}+\frac {2 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {a+b \log \left (c x^n\right )}{2 d^2 x^2}+\frac {3 b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^4}+\frac {b e^2 n \log (d+e x)}{d^4}+\frac {2 b e n}{d^3 x}-\frac {b n}{4 d^2 x^2}\)

Input:

Int[(a + b*Log[c*x^n])/(x^3*(d + e*x)^2),x]
 

Output:

-1/4*(b*n)/(d^2*x^2) + (2*b*e*n)/(d^3*x) - (a + b*Log[c*x^n])/(2*d^2*x^2) 
+ (2*e*(a + b*Log[c*x^n]))/(d^3*x) - (e^3*x*(a + b*Log[c*x^n]))/(d^4*(d + 
e*x)) - (3*e^2*Log[1 + d/(e*x)]*(a + b*Log[c*x^n]))/d^4 + (b*e^2*n*Log[d + 
 e*x])/d^4 + (3*b*e^2*n*PolyLog[2, -(d/(e*x))])/d^4
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2793

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]))
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.77 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.15

method	result	size
risch	\(-\frac {3 b \ln \left (x^{n}\right ) e^{2} \ln \left (e x +d \right )}{d^{4}}+\frac {b \ln \left (x^{n}\right ) e^{2}}{d^{3} \left (e x +d \right )}-\frac {b \ln \left (x^{n}\right )}{2 d^{2} x^{2}}+\frac {3 b \ln \left (x^{n}\right ) e^{2} \ln \left (x \right )}{d^{4}}+\frac {2 b \ln \left (x^{n}\right ) e}{d^{3} x}+\frac {b \,e^{2} n \ln \left (e x +d \right )}{d^{4}}-\frac {b n}{4 d^{2} x^{2}}+\frac {2 b e n}{d^{3} x}-\frac {b n \,e^{2} \ln \left (x \right )}{d^{4}}-\frac {3 b n \,e^{2} \ln \left (x \right )^{2}}{2 d^{4}}+\frac {3 b n \,e^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{4}}+\frac {3 b n \,e^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{4}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {3 e^{2} \ln \left (e x +d \right )}{d^{4}}+\frac {e^{2}}{d^{3} \left (e x +d \right )}-\frac {1}{2 d^{2} x^{2}}+\frac {3 e^{2} \ln \left (x \right )}{d^{4}}+\frac {2 e}{d^{3} x}\right )\)	\(331\)

Input:

int((a+b*ln(c*x^n))/x^3/(e*x+d)^2,x,method=_RETURNVERBOSE)
 

Output:

-3*b*ln(x^n)/d^4*e^2*ln(e*x+d)+b*ln(x^n)/d^3*e^2/(e*x+d)-1/2*b*ln(x^n)/d^2 
/x^2+3*b*ln(x^n)/d^4*e^2*ln(x)+2*b*ln(x^n)/d^3*e/x+b*e^2*n*ln(e*x+d)/d^4-1 
/4*b*n/d^2/x^2+2*b*e*n/d^3/x-b*n/d^4*e^2*ln(x)-3/2*b*n/d^4*e^2*ln(x)^2+3*b 
*n/d^4*e^2*ln(e*x+d)*ln(-e*x/d)+3*b*n/d^4*e^2*dilog(-e*x/d)+(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)* 
(-3/d^4*e^2*ln(e*x+d)+1/d^3*e^2/(e*x+d)-1/2/d^2/x^2+3/d^4*e^2*ln(x)+2/d^3* 
e/x)
 

Fricas [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{2} x^{3}} \,d x } \] Input:

integrate((a+b*log(c*x^n))/x^3/(e*x+d)^2,x, algorithm="fricas")
 

Output:

integral((b*log(c*x^n) + a)/(e^2*x^5 + 2*d*e*x^4 + d^2*x^3), x)
 

Sympy [A] (verification not implemented)

Time = 51.74 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.44 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx =\text {Too large to display} \] Input:

integrate((a+b*ln(c*x**n))/x**3/(e*x+d)**2,x)
 

Output:

-a/(2*d**2*x**2) - a*e**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x) 
, True))/d**3 + 2*a*e/(d**3*x) - 3*a*e**3*Piecewise((x/d, Eq(e, 0)), (log( 
d + e*x)/e, True))/d**4 + 3*a*e**2*log(x)/d**4 - b*n/(4*d**2*x**2) - b*log 
(c*x**n)/(2*d**2*x**2) + b*e**3*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/( 
d*e) + log(d/e + x)/(d*e), True))/d**3 - b*e**3*Piecewise((x/d**2, Eq(e, 0 
)), (-1/(d*e + e**2*x), True))*log(c*x**n)/d**3 + 2*b*e*n/(d**3*x) + 2*b*e 
*log(c*x**n)/(d**3*x) + 3*b*e**3*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)*log( 
d) - polylog(2, e*x*exp_polar(I*pi)/d), True))/e, True))/d**4 - 3*b*e**3*P 
iecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/d**4 - 3*b*e 
**2*n*log(x)**2/(2*d**4) + 3*b*e**2*log(x)*log(c*x**n)/d**4
 

Maxima [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{2} x^{3}} \,d x } \] Input:

integrate((a+b*log(c*x^n))/x^3/(e*x+d)^2,x, algorithm="maxima")
 

Output:

1/2*a*((6*e^2*x^2 + 3*d*e*x - d^2)/(d^3*e*x^3 + d^4*x^2) - 6*e^2*log(e*x + 
 d)/d^4 + 6*e^2*log(x)/d^4) + b*integrate((log(c) + log(x^n))/(e^2*x^5 + 2 
*d*e*x^4 + d^2*x^3), x)
 

Giac [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{2} x^{3}} \,d x } \] Input:

integrate((a+b*log(c*x^n))/x^3/(e*x+d)^2,x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)/((e*x + d)^2*x^3), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^3\,{\left (d+e\,x\right )}^2} \,d x \] Input:

int((a + b*log(c*x^n))/(x^3*(d + e*x)^2),x)
 

Output:

int((a + b*log(c*x^n))/(x^3*(d + e*x)^2), x)
 

Reduce [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{2} x^{5}+2 d e \,x^{4}+d^{2} x^{3}}d x \right ) b \,d^{5} x^{2}+2 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{2} x^{5}+2 d e \,x^{4}+d^{2} x^{3}}d x \right ) b \,d^{4} e \,x^{3}-6 \,\mathrm {log}\left (e x +d \right ) a d \,e^{2} x^{2}-6 \,\mathrm {log}\left (e x +d \right ) a \,e^{3} x^{3}+6 \,\mathrm {log}\left (x \right ) a d \,e^{2} x^{2}+6 \,\mathrm {log}\left (x \right ) a \,e^{3} x^{3}-a \,d^{3}+3 a \,d^{2} e x -6 a \,e^{3} x^{3}}{2 d^{4} x^{2} \left (e x +d \right )} \] Input:

int((a+b*log(c*x^n))/x^3/(e*x+d)^2,x)
 

Output:

(2*int(log(x**n*c)/(d**2*x**3 + 2*d*e*x**4 + e**2*x**5),x)*b*d**5*x**2 + 2 
*int(log(x**n*c)/(d**2*x**3 + 2*d*e*x**4 + e**2*x**5),x)*b*d**4*e*x**3 - 6 
*log(d + e*x)*a*d*e**2*x**2 - 6*log(d + e*x)*a*e**3*x**3 + 6*log(x)*a*d*e* 
*2*x**2 + 6*log(x)*a*e**3*x**3 - a*d**3 + 3*a*d**2*e*x - 6*a*e**3*x**3)/(2 
*d**4*x**2*(d + e*x))