\(\int \frac {x^3 (a+b \log (c (d+e x)^n))}{(f+g x^2)^2} \, dx\) [267]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (warning: unable to verify)
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 27, antiderivative size = 344 \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=-\frac {b d e \sqrt {f} n \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 g^{3/2} \left (e^2 f+d^2 g\right )}-\frac {b e^2 f n \log (d+e x)}{2 g^2 \left (e^2 f+d^2 g\right )}+\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2 \left (f+g x^2\right )}+\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g^2}+\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^2}+\frac {b e^2 f n \log \left (f+g x^2\right )}{4 g^2 \left (e^2 f+d^2 g\right )}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^2}+\frac {b n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g^2} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.19 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.32 \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\frac {\frac {2 f \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2}+2 \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right ) \log \left (f+g x^2\right )+b n \left (\frac {\sqrt {f} \left (-i \sqrt {g} (d+e x) \log (d+e x)+e \left (\sqrt {f}+i \sqrt {g} x\right ) \log \left (i \sqrt {f}-\sqrt {g} x\right )\right )}{\left (e \sqrt {f}-i d \sqrt {g}\right ) \left (\sqrt {f}+i \sqrt {g} x\right )}+\frac {\sqrt {f} \left (i \sqrt {g} (d+e x) \log (d+e x)+e \left (\sqrt {f}-i \sqrt {g} x\right ) \log \left (i \sqrt {f}+\sqrt {g} x\right )\right )}{\left (e \sqrt {f}+i d \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}+2 \left (\log (d+e x) \log \left (\frac {e \left (\sqrt {f}+i \sqrt {g} x\right )}{e \sqrt {f}-i d \sqrt {g}}\right )+\operatorname {PolyLog}\left (2,-\frac {i \sqrt {g} (d+e x)}{e \sqrt {f}-i d \sqrt {g}}\right )\right )+2 \left (\log (d+e x) \log \left (\frac {e \left (\sqrt {f}-i \sqrt {g} x\right )}{e \sqrt {f}+i d \sqrt {g}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {g} (d+e x)}{e \sqrt {f}+i d \sqrt {g}}\right )\right )\right )}{4 g^2} \] Input:

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

Output:

((2*f*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n]))/(f + g*x^2) + 2*(a - 
b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])*Log[f + g*x^2] + b*n*((Sqrt[f]*(( 
-I)*Sqrt[g]*(d + e*x)*Log[d + e*x] + e*(Sqrt[f] + I*Sqrt[g]*x)*Log[I*Sqrt[ 
f] - Sqrt[g]*x]))/((e*Sqrt[f] - I*d*Sqrt[g])*(Sqrt[f] + I*Sqrt[g]*x)) + (S 
qrt[f]*(I*Sqrt[g]*(d + e*x)*Log[d + e*x] + e*(Sqrt[f] - I*Sqrt[g]*x)*Log[I 
*Sqrt[f] + Sqrt[g]*x]))/((e*Sqrt[f] + I*d*Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x) 
) + 2*(Log[d + e*x]*Log[(e*(Sqrt[f] + I*Sqrt[g]*x))/(e*Sqrt[f] - I*d*Sqrt[ 
g])] + PolyLog[2, ((-I)*Sqrt[g]*(d + e*x))/(e*Sqrt[f] - I*d*Sqrt[g])]) + 2 
*(Log[d + e*x]*Log[(e*(Sqrt[f] - I*Sqrt[g]*x))/(e*Sqrt[f] + I*d*Sqrt[g])] 
+ PolyLog[2, (I*Sqrt[g]*(d + e*x))/(e*Sqrt[f] + I*d*Sqrt[g])])))/(4*g^2)
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 344, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/2*(b*d*e*Sqrt[f]*n*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/(g^(3/2)*(e^2*f + d^2*g 
)) - (b*e^2*f*n*Log[d + e*x])/(2*g^2*(e^2*f + d^2*g)) + (f*(a + b*Log[c*(d 
 + e*x)^n]))/(2*g^2*(f + g*x^2)) + ((a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqr 
t[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*g^2) + ((a + b*Log[c*(d 
+ e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/(2*g^ 
2) + (b*e^2*f*n*Log[f + g*x^2])/(4*g^2*(e^2*f + d^2*g)) + (b*n*PolyLog[2, 
-((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(2*g^2) + (b*n*PolyLog[2 
, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*g^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 = 2.36 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.44

method result size
risch \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g \,x^{2}+f \right )}{2 g^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f}{2 g^{2} \left (g \,x^{2}+f \right )}-\frac {b n \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{2 g^{2}}+\frac {b n \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-g f}-g \left (e x +d \right )+d g}{e \sqrt {-g f}+d g}\right )}{2 g^{2}}+\frac {b n \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-g f}+g \left (e x +d \right )-d g}{e \sqrt {-g f}-d g}\right )}{2 g^{2}}+\frac {b n \operatorname {dilog}\left (\frac {e \sqrt {-g f}-g \left (e x +d \right )+d g}{e \sqrt {-g f}+d g}\right )}{2 g^{2}}+\frac {b n \operatorname {dilog}\left (\frac {e \sqrt {-g f}+g \left (e x +d \right )-d g}{e \sqrt {-g f}-d g}\right )}{2 g^{2}}-\frac {b \,e^{2} f n \ln \left (e x +d \right )}{2 g^{2} \left (d^{2} g +f \,e^{2}\right )}+\frac {b \,e^{2} f n \ln \left (g \,x^{2}+f \right )}{4 g^{2} \left (d^{2} g +f \,e^{2}\right )}-\frac {b e n f d \arctan \left (\frac {g x}{\sqrt {g f}}\right )}{2 g \left (d^{2} g +f \,e^{2}\right ) \sqrt {g f}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\ln \left (g \,x^{2}+f \right )}{2 g^{2}}+\frac {f}{2 g^{2} \left (g \,x^{2}+f \right )}\right )\) \(496\)

Input:

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

Output:

1/2*b*ln((e*x+d)^n)/g^2*ln(g*x^2+f)+1/2*b*ln((e*x+d)^n)*f/g^2/(g*x^2+f)-1/ 
2*b*n/g^2*ln(e*x+d)*ln(g*x^2+f)+1/2*b*n/g^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^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^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^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*f*n*ln(e*x+d)/g^2/(d^2*g+e^2*f 
)+1/4*b*e^2*f*n*ln(g*x^2+f)/g^2/(d^2*g+e^2*f)-1/2*b*e*n*f/g/(d^2*g+e^2*f)* 
d/(g*f)^(1/2)*arctan(g*x/(g*f)^(1/2))+(1/2*I*b*Pi*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)*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*csgn(I* 
c)+b*ln(c)+a)*(1/2/g^2*ln(g*x^2+f)+1/2*f/g^2/(g*x^2+f))
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(6*sqrt(g)*sqrt(f)*atan((g*x)/(sqrt(g)*sqrt(f)))*b*d*e**3*f**2*n**2 + 6*sq 
rt(g)*sqrt(f)*atan((g*x)/(sqrt(g)*sqrt(f)))*b*d*e**3*f*g*n**2*x**2 + 6*int 
(log((d + e*x)**n*c)/(d*f**2 + 2*d*f*g*x**2 + d*g**2*x**4 + e*f**2*x + 2*e 
*f*g*x**3 + e*g**2*x**5),x)*b*d**4*e*f**3*g**2*n + 6*int(log((d + e*x)**n* 
c)/(d*f**2 + 2*d*f*g*x**2 + d*g**2*x**4 + e*f**2*x + 2*e*f*g*x**3 + e*g**2 
*x**5),x)*b*d**4*e*f**2*g**3*n*x**2 + 8*int(log((d + e*x)**n*c)/(d*f**2 + 
2*d*f*g*x**2 + d*g**2*x**4 + e*f**2*x + 2*e*f*g*x**3 + e*g**2*x**5),x)*b*d 
**2*e**3*f**4*g*n + 8*int(log((d + e*x)**n*c)/(d*f**2 + 2*d*f*g*x**2 + d*g 
**2*x**4 + e*f**2*x + 2*e*f*g*x**3 + e*g**2*x**5),x)*b*d**2*e**3*f**3*g**2 
*n*x**2 + 2*int(log((d + e*x)**n*c)/(d*f**2 + 2*d*f*g*x**2 + d*g**2*x**4 + 
 e*f**2*x + 2*e*f*g*x**3 + e*g**2*x**5),x)*b*e**5*f**5*n + 2*int(log((d + 
e*x)**n*c)/(d*f**2 + 2*d*f*g*x**2 + d*g**2*x**4 + e*f**2*x + 2*e*f*g*x**3 
+ e*g**2*x**5),x)*b*e**5*f**4*g*n*x**2 + 2*int((log((d + e*x)**n*c)*x**4)/ 
(d*f**2 + 2*d*f*g*x**2 + d*g**2*x**4 + e*f**2*x + 2*e*f*g*x**3 + e*g**2*x* 
*5),x)*b*d**4*e*f*g**4*n + 2*int((log((d + e*x)**n*c)*x**4)/(d*f**2 + 2*d* 
f*g*x**2 + d*g**2*x**4 + e*f**2*x + 2*e*f*g*x**3 + e*g**2*x**5),x)*b*d**4* 
e*g**5*n*x**2 + 8*int((log((d + e*x)**n*c)*x**4)/(d*f**2 + 2*d*f*g*x**2 + 
d*g**2*x**4 + e*f**2*x + 2*e*f*g*x**3 + e*g**2*x**5),x)*b*d**2*e**3*f**2*g 
**3*n + 8*int((log((d + e*x)**n*c)*x**4)/(d*f**2 + 2*d*f*g*x**2 + d*g**2*x 
**4 + e*f**2*x + 2*e*f*g*x**3 + e*g**2*x**5),x)*b*d**2*e**3*f*g**4*n*x*...