3.3.65 \(\int \frac {a+b \log (c (d+e x)^n)}{x^4 (f+g x^2)} \, dx\) [265]

3.3.65.1 Optimal result
3.3.65.2 Mathematica [A] (verified)
3.3.65.3 Rubi [A] (verified)
3.3.65.4 Maple [C] (warning: unable to verify)
3.3.65.5 Fricas [F]
3.3.65.6 Sympy [F(-1)]
3.3.65.7 Maxima [F]
3.3.65.8 Giac [F]
3.3.65.9 Mupad [F(-1)]

3.3.65.1 Optimal result

Integrand size = 27, antiderivative size = 388 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=-\frac {b e n}{6 d f x^2}+\frac {b e^2 n}{3 d^2 f x}+\frac {b e^3 n \log (x)}{3 d^3 f}-\frac {b e g n \log (x)}{d f^2}-\frac {b e^3 n \log (d+e x)}{3 d^3 f}+\frac {b e g n \log (d+e x)}{d f^2}-\frac {a+b \log \left (c (d+e x)^n\right )}{3 f x^3}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {g^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 (-f)^{5/2}}-\frac {g^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{5/2}}-\frac {b g^{3/2} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{5/2}}+\frac {b g^{3/2} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 (-f)^{5/2}} \]

output
-1/6*b*e*n/d/f/x^2+1/3*b*e^2*n/d^2/f/x+1/3*b*e^3*n*ln(x)/d^3/f-b*e*g*n*ln( 
x)/d/f^2-1/3*b*e^3*n*ln(e*x+d)/d^3/f+b*e*g*n*ln(e*x+d)/d/f^2+1/3*(-a-b*ln( 
c*(e*x+d)^n))/f/x^3+g*(a+b*ln(c*(e*x+d)^n))/f^2/x+1/2*g^(3/2)*(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)))/(-f)^(5/2) 
-1/2*g^(3/2)*(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)))/(-f)^(5/2)-1/2*b*g^(3/2)*n*polylog(2,-(e*x+d)*g^(1/2)/(e*(- 
f)^(1/2)-d*g^(1/2)))/(-f)^(5/2)+1/2*b*g^(3/2)*n*polylog(2,(e*x+d)*g^(1/2)/ 
(e*(-f)^(1/2)+d*g^(1/2)))/(-f)^(5/2)
 
3.3.65.2 Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 350, normalized size of antiderivative = 0.90 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\frac {1}{6} \left (-\frac {6 b e g n (\log (x)-\log (d+e x))}{d f^2}-\frac {b e n \left (d (d-2 e x)-2 e^2 x^2 \log (x)+2 e^2 x^2 \log (d+e x)\right )}{d^3 f x^2}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f x^3}+\frac {6 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {3 g^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2}}-\frac {3 g^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{5/2}}-\frac {3 b g^{3/2} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{5/2}}+\frac {3 b g^{3/2} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2}}\right ) \]

input
Integrate[(a + b*Log[c*(d + e*x)^n])/(x^4*(f + g*x^2)),x]
 
output
((-6*b*e*g*n*(Log[x] - Log[d + e*x]))/(d*f^2) - (b*e*n*(d*(d - 2*e*x) - 2* 
e^2*x^2*Log[x] + 2*e^2*x^2*Log[d + e*x]))/(d^3*f*x^2) - (2*(a + b*Log[c*(d 
 + e*x)^n]))/(f*x^3) + (6*g*(a + b*Log[c*(d + e*x)^n]))/(f^2*x) + (3*g^(3/ 
2)*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + 
 d*Sqrt[g])])/(-f)^(5/2) - (3*g^(3/2)*(a + b*Log[c*(d + e*x)^n])*Log[(e*(S 
qrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/(-f)^(5/2) - (3*b*g^(3/2) 
*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(-f)^(5/2) 
 + (3*b*g^(3/2)*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])] 
)/(-f)^(5/2))/6
 
3.3.65.3 Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 388, 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.

\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx\)

\(\Big \downarrow \) 2863

\(\displaystyle \int \left (\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 \left (f+g x^2\right )}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x^2}+\frac {a+b \log \left (c (d+e x)^n\right )}{f x^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {g^{3/2} \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{5/2}}-\frac {g^{3/2} \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{5/2}}-\frac {a+b \log \left (c (d+e x)^n\right )}{3 f x^3}+\frac {b e^3 n \log (x)}{3 d^3 f}-\frac {b e^3 n \log (d+e x)}{3 d^3 f}+\frac {b e^2 n}{3 d^2 f x}-\frac {b e g n \log (x)}{d f^2}+\frac {b e g n \log (d+e x)}{d f^2}-\frac {b g^{3/2} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{5/2}}+\frac {b g^{3/2} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 (-f)^{5/2}}-\frac {b e n}{6 d f x^2}\)

input
Int[(a + b*Log[c*(d + e*x)^n])/(x^4*(f + g*x^2)),x]
 
output
-1/6*(b*e*n)/(d*f*x^2) + (b*e^2*n)/(3*d^2*f*x) + (b*e^3*n*Log[x])/(3*d^3*f 
) - (b*e*g*n*Log[x])/(d*f^2) - (b*e^3*n*Log[d + e*x])/(3*d^3*f) + (b*e*g*n 
*Log[d + e*x])/(d*f^2) - (a + b*Log[c*(d + e*x)^n])/(3*f*x^3) + (g*(a + b* 
Log[c*(d + e*x)^n]))/(f^2*x) + (g^(3/2)*(a + b*Log[c*(d + e*x)^n])*Log[(e* 
(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*(-f)^(5/2)) - (g^(3/ 
2)*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - 
 d*Sqrt[g])])/(2*(-f)^(5/2)) - (b*g^(3/2)*n*PolyLog[2, -((Sqrt[g]*(d + e*x 
))/(e*Sqrt[-f] - d*Sqrt[g]))])/(2*(-f)^(5/2)) + (b*g^(3/2)*n*PolyLog[2, (S 
qrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*(-f)^(5/2))
 

3.3.65.3.1 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]
 
3.3.65.4 Maple [C] (warning: unable to verify)

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

Time = 1.75 (sec) , antiderivative size = 598, normalized size of antiderivative = 1.54

method result size
risch \(-\frac {b \,g^{2} \arctan \left (\frac {2 g \left (e x +d \right )-2 d g}{2 e \sqrt {f g}}\right ) n \ln \left (e x +d \right )}{f^{2} \sqrt {f g}}+\frac {b \,g^{2} \arctan \left (\frac {2 g \left (e x +d \right )-2 d g}{2 e \sqrt {f g}}\right ) \ln \left (\left (e x +d \right )^{n}\right )}{f^{2} \sqrt {f g}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{3 f \,x^{3}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g}{f^{2} x}-\frac {e b n g \ln \left (e x \right )}{f^{2} d}+\frac {b e g n \ln \left (e x +d \right )}{d \,f^{2}}-\frac {b e n}{6 d f \,x^{2}}+\frac {b \,e^{2} n}{3 d^{2} f x}+\frac {e^{3} b n \ln \left (e x \right )}{3 f \,d^{3}}-\frac {b \,e^{3} n \ln \left (e x +d \right )}{3 d^{3} f}+\frac {b n \,g^{2} \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 f^{2} \sqrt {-f g}}-\frac {b n \,g^{2} \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 f^{2} \sqrt {-f g}}+\frac {b n \,g^{2} \operatorname {dilog}\left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 f^{2} \sqrt {-f g}}-\frac {b n \,g^{2} \operatorname {dilog}\left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 f^{2} \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 {1}{3 f \,x^{3}}+\frac {g}{f^{2} x}+\frac {g^{2} \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{f^{2} \sqrt {f g}}\right )\) \(598\)

input
int((a+b*ln(c*(e*x+d)^n))/x^4/(g*x^2+f),x,method=_RETURNVERBOSE)
 
output
-b*g^2/f^2/(f*g)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/(f*g)^(1/2))*n*ln( 
e*x+d)+b*g^2/f^2/(f*g)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/(f*g)^(1/2)) 
*ln((e*x+d)^n)-1/3*b*ln((e*x+d)^n)/f/x^3+b*ln((e*x+d)^n)/f^2*g/x-e*b*n/f^2 
*g/d*ln(e*x)+b*e*g*n*ln(e*x+d)/d/f^2-1/6*b*e*n/d/f/x^2+1/3*b*e^2*n/d^2/f/x 
+1/3*e^3*b*n/f/d^3*ln(e*x)-1/3*b*e^3*n*ln(e*x+d)/d^3/f+1/2*b*n*g^2/f^2*ln( 
e*x+d)/(-f*g)^(1/2)*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/f^2*ln(e*x+d)/(-f*g)^(1/2)*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/f^2/(-f*g)^(1/2)*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/f^2/(-f*g)^(1/2)*dilog((e 
*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e*(-f*g)^(1/2)-d*g))+(-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)*(-1/3/f/x^3+1/f^2*g/x+g^2/f^2/(f*g)^(1/2)*arct 
an(g*x/(f*g)^(1/2)))
 
3.3.65.5 Fricas [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{4}} \,d x } \]

input
integrate((a+b*log(c*(e*x+d)^n))/x^4/(g*x^2+f),x, algorithm="fricas")
 
output
integral((b*log((e*x + d)^n*c) + a)/(g*x^6 + f*x^4), x)
 
3.3.65.6 Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\text {Timed out} \]

input
integrate((a+b*ln(c*(e*x+d)**n))/x**4/(g*x**2+f),x)
 
output
Timed out
 
3.3.65.7 Maxima [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{4}} \,d x } \]

input
integrate((a+b*log(c*(e*x+d)^n))/x^4/(g*x^2+f),x, algorithm="maxima")
 
output
1/3*a*(3*g^2*arctan(g*x/sqrt(f*g))/(sqrt(f*g)*f^2) + (3*g*x^2 - f)/(f^2*x^ 
3)) + b*integrate((log((e*x + d)^n) + log(c))/(g*x^6 + f*x^4), x)
 
3.3.65.8 Giac [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{4}} \,d x } \]

input
integrate((a+b*log(c*(e*x+d)^n))/x^4/(g*x^2+f),x, algorithm="giac")
 
output
integrate((b*log((e*x + d)^n*c) + a)/((g*x^2 + f)*x^4), x)
 
3.3.65.9 Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^4\,\left (g\,x^2+f\right )} \,d x \]

input
int((a + b*log(c*(d + e*x)^n))/(x^4*(f + g*x^2)),x)
 
output
int((a + b*log(c*(d + e*x)^n))/(x^4*(f + g*x^2)), x)