\(\int \frac {x (a+b \log (c (d+e x)^n))}{f+g x^2} \, dx\) [258]

Optimal result

Integrand size = 25, antiderivative size = 203 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g}+\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g}+\frac {b n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.85 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \left (\log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )+\log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )\right )+b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+b n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 = 1.34 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.75

method	result	size
risch	\(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g \,x^{2}+f \right )}{2 g}-\frac {b n \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{2 g}+\frac {b n \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-g f}-g \left (e x +d \right )+d g}{e \sqrt {-g f}+d g}\right )}{2 g}+\frac {b n \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-g f}+g \left (e x +d \right )-d g}{e \sqrt {-g f}-d g}\right )}{2 g}+\frac {b n \operatorname {dilog}\left (\frac {e \sqrt {-g f}-g \left (e x +d \right )+d g}{e \sqrt {-g f}+d g}\right )}{2 g}+\frac {b n \operatorname {dilog}\left (\frac {e \sqrt {-g f}+g \left (e x +d \right )-d g}{e \sqrt {-g f}-d g}\right )}{2 g}+\frac {\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \ln \left (g \,x^{2}+f \right )}{2 g}\)	\(356\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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