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

3.3.67.1 Optimal result
3.3.67.2 Mathematica [C] (verified)
3.3.67.3 Rubi [A] (verified)
3.3.67.4 Maple [C] (warning: unable to verify)
3.3.67.5 Fricas [F]
3.3.67.6 Sympy [F(-1)]
3.3.67.7 Maxima [F]
3.3.67.8 Giac [F]
3.3.67.9 Mupad [F(-1)]

3.3.67.1 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*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)+1/4*b*e^2*f*n*ln(g*x^2+f)/g^2/(d^2*g+e^2*f)+1/2*(a+b*ln(c*(e*x+ 
d)^n))*ln(e*((-f)^(1/2)-x*g^(1/2))/(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)+x*g^(1/2))/(e*(-f)^(1/2)-d*g^(1/2)))/g^2 
+1/2*b*n*polylog(2,-(e*x+d)*g^(1/2)/(e*(-f)^(1/2)-d*g^(1/2)))/g^2+1/2*b*n* 
polylog(2,(e*x+d)*g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2)))/g^2-1/2*b*d*e*n*arctan 
(x*g^(1/2)/f^(1/2))*f^(1/2)/g^(3/2)/(d^2*g+e^2*f)
 
3.3.67.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.73 (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)
 
3.3.67.3 Rubi [A] (verified)

Time = 0.68 (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)
 

3.3.67.3.1 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]
 
3.3.67.4 Maple [C] (warning: unable to verify)

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

Time = 1.18 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.44

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

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)*f/g^2/(g*x^2+f)+1/2*b*ln((e*x+d)^n)/g^2*ln(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*(-f*g)^(1/2)-g 
*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+d*g))+1/2*b*n/g^2*ln(e*x+d)*ln((e*(-f*g)^(1/ 
2)+g*(e*x+d)-d*g)/(e*(-f*g)^(1/2)-d*g))+1/2*b*n/g^2*dilog((e*(-f*g)^(1/2)- 
g*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+d*g))+1/2*b*n/g^2*dilog((e*(-f*g)^(1/2)+g*( 
e*x+d)-d*g)/(e*(-f*g)^(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/(f*g)^(1/2)*arctan(g*x/(f*g)^(1/2))+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d 
)^n)*csgn(I*c*(e*x+d)^n)+1/2*I*b*Pi*csgn(I*c)*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)^2-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n 
)^3+b*ln(c)+a)*(1/2*f/g^2/(g*x^2+f)+1/2/g^2*ln(g*x^2+f))
 
3.3.67.5 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 
)
 
3.3.67.6 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
 
3.3.67.7 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)
 
3.3.67.8 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)
 
3.3.67.9 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)