\(\int \frac {a+b \log (c (d+e x)^n)}{(f+g x^2)^2} \, dx\) [273]

Optimal result

Integrand size = 24, antiderivative size = 503 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+g x^2\right )^2} \, dx=\frac {b e n \log (d+e x)}{4 f \left (e \sqrt {-f}+d \sqrt {g}\right ) \sqrt {g}}+\frac {b e n \log (d+e x)}{4 \left (e (-f)^{3/2}+d f \sqrt {g}\right ) \sqrt {g}}-\frac {a+b \log \left (c (d+e x)^n\right )}{4 f \sqrt {g} \left (\sqrt {-f}-\sqrt {g} x\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{4 f \sqrt {g} \left (\sqrt {-f}+\sqrt {g} x\right )}-\frac {b e n \log \left (\sqrt {-f}-\sqrt {g} x\right )}{4 f \left (e \sqrt {-f}+d \sqrt {g}\right ) \sqrt {g}}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{4 (-f)^{3/2} \sqrt {g}}-\frac {b e n \log \left (\sqrt {-f}+\sqrt {g} x\right )}{4 \left (e (-f)^{3/2}+d f \sqrt {g}\right ) \sqrt {g}}+\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{4 (-f)^{3/2} \sqrt {g}}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{4 (-f)^{3/2} \sqrt {g}}-\frac {b n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{4 (-f)^{3/2} \sqrt {g}} \] Output:

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

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 407, normalized size of antiderivative = 0.81 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+g x^2\right )^2} \, dx=\frac {1}{4} \left (\frac {a+b \log \left (c (d+e x)^n\right )}{f \left (\sqrt {-f} \sqrt {g}+g x\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{(-f)^{3/2} \sqrt {g}+f g x}+\frac {b e n \left (\log (d+e x)-\log \left (\sqrt {-f}-\sqrt {g} x\right )\right )}{e \sqrt {-f} f \sqrt {g}+d f g}+\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2} \sqrt {g}}+\frac {b e n \left (\log (d+e x)-\log \left (\sqrt {-f}+\sqrt {g} x\right )\right )}{e (-f)^{3/2} \sqrt {g}+d f g}+\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{3/2} \sqrt {g}}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{3/2} \sqrt {g}}+\frac {b f n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2} \sqrt {g}}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 503, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2856, 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.

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

\(\Big \downarrow \) 2856

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2856

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. 
)*(x_)^(r_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x) 
^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x] && I 
GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
 

Maple [C] (warning: unable to verify)

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

Time = 2.66 (sec) , antiderivative size = 1406, normalized size of antiderivative = 2.80

method	result	size
risch	\(\text {Expression too large to display}\)	\(1406\)

Input:

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

Output:

-1/2*b*e^2/f/(e^2*g*x^2+e^2*f)*x*n*ln(e*x+d)+1/2*b*e^2/f/(e^2*g*x^2+e^2*f) 
*x*ln((e*x+d)^n)-1/2*b/f/(g*f)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/(g*f 
)^(1/2))*n*ln(e*x+d)+1/2*b/f/(g*f)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/ 
(g*f)^(1/2))*ln((e*x+d)^n)-1/4*b*e*n/f/(d^2*g+e^2*f)*d*ln(g*(e*x+d)^2-2*(e 
*x+d)*d*g+d^2*g+f*e^2)-1/2*b*e^2*n/(d^2*g+e^2*f)/(g*f)^(1/2)*arctan(1/2*(2 
*g*(e*x+d)-2*d*g)/e/(g*f)^(1/2))+1/4*b*e^2*n*ln(e*x+d)/f/(d^2*g+e^2*f)/(e^ 
2*g*x^2+e^2*f)/(-g*f)^(1/2)*ln((e*(-g*f)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1 
/2)+d*g))*x^2*d^2*g^2+1/4*b*e^4*n*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f 
)/(-g*f)^(1/2)*ln((e*(-g*f)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1/2)+d*g))*x^2 
*g-1/4*b*e^2*n*ln(e*x+d)/f/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/(-g*f)^(1/2)*ln 
((e*(-g*f)^(1/2)+g*(e*x+d)-d*g)/(e*(-g*f)^(1/2)-d*g))*x^2*d^2*g^2-1/4*b*e^ 
4*n*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/(-g*f)^(1/2)*ln((e*(-g*f)^(1 
/2)+g*(e*x+d)-d*g)/(e*(-g*f)^(1/2)-d*g))*x^2*g+1/2*b*e^3*n*ln(e*x+d)/f/(d^ 
2*g+e^2*f)/(e^2*g*x^2+e^2*f)*x^2*d*g+1/4*b*e^2*n*ln(e*x+d)/(d^2*g+e^2*f)/( 
e^2*g*x^2+e^2*f)/(-g*f)^(1/2)*ln((e*(-g*f)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^ 
(1/2)+d*g))*d^2*g+1/4*b*e^4*n*ln(e*x+d)*f/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/ 
(-g*f)^(1/2)*ln((e*(-g*f)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1/2)+d*g))-1/4*b 
*e^2*n*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/(-g*f)^(1/2)*ln((e*(-g*f) 
^(1/2)+g*(e*x+d)-d*g)/(e*(-g*f)^(1/2)-d*g))*d^2*g-1/4*b*e^4*n*ln(e*x+d)*f/ 
(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/(-g*f)^(1/2)*ln((e*(-g*f)^(1/2)+g*(e*x+...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

1/2*a*(x/(f*g*x^2 + f^2) + arctan(g*x/sqrt(f*g))/(sqrt(f*g)*f)) + b*integr 
ate((log((e*x + d)^n) + log(c))/(g^2*x^4 + 2*f*g*x^2 + f^2), x)
 

Giac [F]

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

integrate((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)/(g*x^2 + f)^2, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+g x^2\right )^2} \, dx=\frac {\sqrt {g}\, \sqrt {f}\, \mathit {atan} \left (\frac {g x}{\sqrt {g}\, \sqrt {f}}\right ) a f +\sqrt {g}\, \sqrt {f}\, \mathit {atan} \left (\frac {g x}{\sqrt {g}\, \sqrt {f}}\right ) a g \,x^{2}+2 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{g^{2} x^{4}+2 f g \,x^{2}+f^{2}}d x \right ) b \,f^{3} g +2 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{g^{2} x^{4}+2 f g \,x^{2}+f^{2}}d x \right ) b \,f^{2} g^{2} x^{2}+a f g x}{2 f^{2} g \left (g \,x^{2}+f \right )} \] Input:

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

Output:

(sqrt(g)*sqrt(f)*atan((g*x)/(sqrt(g)*sqrt(f)))*a*f + sqrt(g)*sqrt(f)*atan( 
(g*x)/(sqrt(g)*sqrt(f)))*a*g*x**2 + 2*int(log((d + e*x)**n*c)/(f**2 + 2*f* 
g*x**2 + g**2*x**4),x)*b*f**3*g + 2*int(log((d + e*x)**n*c)/(f**2 + 2*f*g* 
x**2 + g**2*x**4),x)*b*f**2*g**2*x**2 + a*f*g*x)/(2*f**2*g*(f + g*x**2))