\(\int \frac {(a+b \log (c (d+e x)^n))^2}{x (f+g x^2)^2} \, dx\) [323]

Optimal result

Integrand size = 29, antiderivative size = 814 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{x \left (f+g x^2\right )^2} \, dx =\text {Too large to display} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.37 (sec) , antiderivative size = 1209, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{x \left (f+g x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((2*f*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2)/(f + g*x^2) + 4*Log 
[x]*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2 - 2*(a - b*n*Log[d + e 
*x] + b*Log[c*(d + e*x)^n])^2*Log[f + g*x^2] + 2*b*n*(a - b*n*Log[d + e*x] 
 + b*Log[c*(d + e*x)^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*S 
qrt[g])*(Sqrt[f] + I*Sqrt[g]*x)) + (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)) - 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*Sq 
rt[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*(Log[-((e*x)/d)]*Log[d + e*x] + PolyLog[2, 1 
 + (e*x)/d])) + b^2*n^2*(4*Log[-((e*x)/d)]*Log[d + e*x]^2 - 2*Log[d + e*x] 
^2*Log[1 - (Sqrt[g]*(d + e*x))/((-I)*e*Sqrt[f] + d*Sqrt[g])] - 2*Log[d + e 
*x]^2*Log[1 - (Sqrt[g]*(d + e*x))/(I*e*Sqrt[f] + d*Sqrt[g])] + (Sqrt[f]*(L 
og[d + e*x]*(I*Sqrt[g]*(d + e*x)*Log[d + e*x] + 2*e*(Sqrt[f] - I*Sqrt[g]*x 
)*Log[(e*(Sqrt[f] - I*Sqrt[g]*x))/(e*Sqrt[f] + I*d*Sqrt[g])]) + 2*e*(Sqrt[ 
f] - I*Sqrt[g]*x)*PolyLog[2, (I*Sqrt[g]*(d + e*x))/(e*Sqrt[f] + I*d*Sqrt[g 
])]))/((e*Sqrt[f] + I*d*Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x)) - 4*Log[d + e*x] 
*PolyLog[2, (Sqrt[g]*(d + e*x))/((-I)*e*Sqrt[f] + d*Sqrt[g])] - 4*Log[d...
 

Rubi [A] (verified)

Time = 2.83 (sec) , antiderivative size = 814, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {b^2 e \left (\sqrt {g} d+e \sqrt {-f}\right ) \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right ) n^2}{2 (-f)^{3/2} \left (g d^2+e^2 f\right )}+\frac {b^2 e \left (\sqrt {-f} \sqrt {g} d+e f\right ) \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right ) n^2}{2 f^2 \left (g d^2+e^2 f\right )}+\frac {b^2 \operatorname {PolyLog}\left (3,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right ) n^2}{f^2}+\frac {b^2 \operatorname {PolyLog}\left (3,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right ) n^2}{f^2}-\frac {2 b^2 \operatorname {PolyLog}\left (3,\frac {e x}{d}+1\right ) n^2}{f^2}+\frac {b e \left (\sqrt {-f} \sqrt {g} d+e f\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{\sqrt {g} d+e \sqrt {-f}}\right ) n}{2 f^2 \left (g d^2+e^2 f\right )}+\frac {b e \left (e f-d \sqrt {-f} \sqrt {g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {g} x+\sqrt {-f}\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) n}{2 f^2 \left (g d^2+e^2 f\right )}-\frac {b \left (a+b \log \left (c (d+e x)^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right ) n}{f^2}-\frac {b \left (a+b \log \left (c (d+e x)^n\right )\right ) \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right ) n}{f^2}+\frac {2 b \left (a+b \log \left (c (d+e x)^n\right )\right ) \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right ) n}{f^2}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{f^2}-\frac {e^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 f \left (g d^2+e^2 f\right )}+\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 f \left (g x^2+f\right )}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 f^2}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (\frac {e \left (\sqrt {g} x+\sqrt {-f}\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}\)

Input:

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

Output:

-1/2*(e^2*(a + b*Log[c*(d + e*x)^n])^2)/(f*(e^2*f + d^2*g)) + (a + b*Log[c 
*(d + e*x)^n])^2/(2*f*(f + g*x^2)) + (Log[-((e*x)/d)]*(a + b*Log[c*(d + e* 
x)^n])^2)/f^2 + (b*e*(e*f + d*Sqrt[-f]*Sqrt[g])*n*(a + b*Log[c*(d + e*x)^n 
])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*f^2*(e^2*f 
 + d^2*g)) - ((a + b*Log[c*(d + e*x)^n])^2*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/ 
(e*Sqrt[-f] + d*Sqrt[g])])/(2*f^2) + (b*e*(e*f - d*Sqrt[-f]*Sqrt[g])*n*(a 
+ b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqr 
t[g])])/(2*f^2*(e^2*f + d^2*g)) - ((a + b*Log[c*(d + e*x)^n])^2*Log[(e*(Sq 
rt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/(2*f^2) - (b^2*e*(e*Sqrt[- 
f] + d*Sqrt[g])*n^2*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[ 
g]))])/(2*(-f)^(3/2)*(e^2*f + d^2*g)) - (b*n*(a + b*Log[c*(d + e*x)^n])*Po 
lyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/f^2 + (b^2*e*(e 
*f + d*Sqrt[-f]*Sqrt[g])*n^2*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + 
d*Sqrt[g])])/(2*f^2*(e^2*f + d^2*g)) - (b*n*(a + b*Log[c*(d + e*x)^n])*Pol 
yLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/f^2 + (2*b*n*(a + b 
*Log[c*(d + e*x)^n])*PolyLog[2, 1 + (e*x)/d])/f^2 + (b^2*n^2*PolyLog[3, -( 
(Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/f^2 + (b^2*n^2*PolyLog[3, 
(Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/f^2 - (2*b^2*n^2*PolyLog[3, 
 1 + (e*x)/d])/f^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 [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b^2*log((e*x + d)^n*c)^2 + 2*a*b*log((e*x + d)^n*c) + a^2)/(g^2* 
x^5 + 2*f*g*x^3 + f^2*x), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

1/2*a^2*(1/(f*g*x^2 + f^2) - log(g*x^2 + f)/f^2 + 2*log(x)/f^2) + integrat 
e((b^2*log((e*x + d)^n)^2 + b^2*log(c)^2 + 2*a*b*log(c) + 2*(b^2*log(c) + 
a*b)*log((e*x + d)^n))/(g^2*x^5 + 2*f*g*x^3 + f^2*x), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{x \left (f+g x^2\right )^2} \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )^{2}}{g^{2} x^{5}+2 f g \,x^{3}+f^{2} x}d x \right ) b^{2} f^{3}+2 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )^{2}}{g^{2} x^{5}+2 f g \,x^{3}+f^{2} x}d x \right ) b^{2} f^{2} g \,x^{2}+4 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{g^{2} x^{5}+2 f g \,x^{3}+f^{2} x}d x \right ) a b \,f^{3}+4 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{g^{2} x^{5}+2 f g \,x^{3}+f^{2} x}d x \right ) a b \,f^{2} g \,x^{2}-\mathrm {log}\left (g \,x^{2}+f \right ) a^{2} f -\mathrm {log}\left (g \,x^{2}+f \right ) a^{2} g \,x^{2}+2 \,\mathrm {log}\left (x \right ) a^{2} f +2 \,\mathrm {log}\left (x \right ) a^{2} g \,x^{2}-a^{2} g \,x^{2}}{2 f^{2} \left (g \,x^{2}+f \right )} \] Input:

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

Output:

(2*int(log((d + e*x)**n*c)**2/(f**2*x + 2*f*g*x**3 + g**2*x**5),x)*b**2*f* 
*3 + 2*int(log((d + e*x)**n*c)**2/(f**2*x + 2*f*g*x**3 + g**2*x**5),x)*b** 
2*f**2*g*x**2 + 4*int(log((d + e*x)**n*c)/(f**2*x + 2*f*g*x**3 + g**2*x**5 
),x)*a*b*f**3 + 4*int(log((d + e*x)**n*c)/(f**2*x + 2*f*g*x**3 + g**2*x**5 
),x)*a*b*f**2*g*x**2 - log(f + g*x**2)*a**2*f - log(f + g*x**2)*a**2*g*x** 
2 + 2*log(x)*a**2*f + 2*log(x)*a**2*g*x**2 - a**2*g*x**2)/(2*f**2*(f + g*x 
**2))