\(\int \frac {a+b \log (c (d+e x)^n)}{x^3 (f+g x^2)} \, dx\) [260]

Optimal result

Integrand size = 27, antiderivative size = 331 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )} \, dx=-\frac {b e n}{2 d f x}-\frac {b e^2 n \log (x)}{2 d^2 f}+\frac {b e^2 n \log (d+e x)}{2 d^2 f}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f x^2}-\frac {g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f^2}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}+\frac {b g n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}+\frac {b g n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f^2}-\frac {b g n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.84 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )} \, dx=\frac {-\frac {b e f n (d+e x \log (x)-e x \log (d+e x))}{d^2 x}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^2}-2 g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )+g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )+b g n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+b g n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )-2 b g n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{2 f^2} \] Input:

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

Output:

(-((b*e*f*n*(d + e*x*Log[x] - e*x*Log[d + e*x]))/(d^2*x)) - (f*(a + b*Log[ 
c*(d + e*x)^n]))/x^2 - 2*g*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]) + g* 
(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d* 
Sqrt[g])] + g*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e 
*Sqrt[-f] - d*Sqrt[g])] + b*g*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[- 
f] - d*Sqrt[g]))] + b*g*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*S 
qrt[g])] - 2*b*g*n*PolyLog[2, 1 + (e*x)/d])/(2*f^2)
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2863, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2863

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) 
^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a 
 + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
 

Maple [C] (warning: unable to verify)

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

Time = 2.50 (sec) , antiderivative size = 499, normalized size of antiderivative = 1.51

method	result	size
risch	\(-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{2 f \,x^{2}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g \ln \left (x \right )}{f^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g \ln \left (g \,x^{2}+f \right )}{2 f^{2}}-\frac {b n g \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{2 f^{2}}+\frac {b n g \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-g f}-g \left (e x +d \right )+d g}{e \sqrt {-g f}+d g}\right )}{2 f^{2}}+\frac {b n g \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-g f}+g \left (e x +d \right )-d g}{e \sqrt {-g f}-d g}\right )}{2 f^{2}}+\frac {b n g \operatorname {dilog}\left (\frac {e \sqrt {-g f}-g \left (e x +d \right )+d g}{e \sqrt {-g f}+d g}\right )}{2 f^{2}}+\frac {b n g \operatorname {dilog}\left (\frac {e \sqrt {-g f}+g \left (e x +d \right )-d g}{e \sqrt {-g f}-d g}\right )}{2 f^{2}}+\frac {b \,e^{2} n \ln \left (e x +d \right )}{2 d^{2} f}-\frac {b e n}{2 d f x}-\frac {b \,e^{2} n \ln \left (x \right )}{2 d^{2} f}+\frac {b n g \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f^{2}}+\frac {b n g \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{2}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {1}{2 f \,x^{2}}-\frac {g \ln \left (x \right )}{f^{2}}+\frac {g \ln \left (g \,x^{2}+f \right )}{2 f^{2}}\right )\)	\(499\)

Input:

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

Output:

-1/2*b*ln((e*x+d)^n)/f/x^2-b*ln((e*x+d)^n)*g/f^2*ln(x)+1/2*b*ln((e*x+d)^n) 
*g/f^2*ln(g*x^2+f)-1/2*b*n*g/f^2*ln(e*x+d)*ln(g*x^2+f)+1/2*b*n*g/f^2*ln(e* 
x+d)*ln((e*(-g*f)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1/2)+d*g))+1/2*b*n*g/f^2 
*ln(e*x+d)*ln((e*(-g*f)^(1/2)+g*(e*x+d)-d*g)/(e*(-g*f)^(1/2)-d*g))+1/2*b*n 
*g/f^2*dilog((e*(-g*f)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1/2)+d*g))+1/2*b*n* 
g/f^2*dilog((e*(-g*f)^(1/2)+g*(e*x+d)-d*g)/(e*(-g*f)^(1/2)-d*g))+1/2*b*e^2 
*n*ln(e*x+d)/d^2/f-1/2*b*e*n/d/f/x-1/2*b*e^2*n*ln(x)/d^2/f+b*n*g/f^2*dilog 
((e*x+d)/d)+b*n*g/f^2*ln(x)*ln((e*x+d)/d)+(1/2*I*b*Pi*csgn(I*(e*x+d)^n)*cs 
gn(I*c*(e*x+d)^n)^2-1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*csgn( 
I*c)-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3+1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^2*csg 
n(I*c)+b*ln(c)+a)*(-1/2/f/x^2-g/f^2*ln(x)+1/2*g/f^2*ln(g*x^2+f))
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

1/2*a*(g*log(g*x^2 + f)/f^2 - 2*g*log(x)/f^2 - 1/(f*x^2)) + b*integrate((l 
og((e*x + d)^n) + log(c))/(g*x^5 + f*x^3), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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