\(\int \frac {(f+g x^2)^2 \log (c (d+e x^2)^p)}{x} \, dx\) [326]

Optimal result

Integrand size = 25, antiderivative size = 153 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x} \, dx=-f g p x^2+\frac {d g^2 p x^2}{4 e}-\frac {1}{8} g^2 p x^4-\frac {d^2 g^2 p \log \left (d+e x^2\right )}{4 e^2}+\frac {1}{4} g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )+\frac {f g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}+\frac {1}{2} f^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} f^2 p \operatorname {PolyLog}\left (2,1+\frac {e x^2}{d}\right ) \] Output:

-f*g*p*x^2+1/4*d*g^2*p*x^2/e-1/8*g^2*p*x^4-1/4*d^2*g^2*p*ln(e*x^2+d)/e^2+1 
/4*g^2*x^4*ln(c*(e*x^2+d)^p)+f*g*(e*x^2+d)*ln(c*(e*x^2+d)^p)/e+1/2*f^2*ln( 
-e*x^2/d)*ln(c*(e*x^2+d)^p)+1/2*f^2*p*polylog(2,1+e*x^2/d)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.79 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x} \, dx=\frac {-e g p x^2 \left (8 e f-2 d g+e g x^2\right )-2 d^2 g^2 p \log \left (d+e x^2\right )+2 e \left (g \left (4 d f+4 e f x^2+e g x^4\right )+2 e f^2 \log \left (-\frac {e x^2}{d}\right )\right ) \log \left (c \left (d+e x^2\right )^p\right )+4 e^2 f^2 p \operatorname {PolyLog}\left (2,1+\frac {e x^2}{d}\right )}{8 e^2} \] Input:

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

Output:

(-(e*g*p*x^2*(8*e*f - 2*d*g + e*g*x^2)) - 2*d^2*g^2*p*Log[d + e*x^2] + 2*e 
*(g*(4*d*f + 4*e*f*x^2 + e*g*x^4) + 2*e*f^2*Log[-((e*x^2)/d)])*Log[c*(d + 
e*x^2)^p] + 4*e^2*f^2*p*PolyLog[2, 1 + (e*x^2)/d])/(8*e^2)
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2925, 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.

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

\(\Big \downarrow \) 2925

\(\displaystyle \frac {1}{2} \int \frac {\left (g x^2+f\right )^2 \log \left (c \left (e x^2+d\right )^p\right )}{x^2}dx^2\)

\(\Big \downarrow \) 2863

\(\displaystyle \frac {1}{2} \int \left (\frac {\log \left (c \left (e x^2+d\right )^p\right ) f^2}{x^2}+2 g \log \left (c \left (e x^2+d\right )^p\right ) f+g^2 x^2 \log \left (c \left (e x^2+d\right )^p\right )\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (f^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {2 f g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}+\frac {1}{2} g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )-\frac {d^2 g^2 p \log \left (d+e x^2\right )}{2 e^2}+f^2 p \operatorname {PolyLog}\left (2,\frac {e x^2}{d}+1\right )+\frac {d g^2 p x^2}{2 e}-2 f g p x^2-\frac {1}{4} g^2 p x^4\right )\)

Input:

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

Output:

(-2*f*g*p*x^2 + (d*g^2*p*x^2)/(2*e) - (g^2*p*x^4)/4 - (d^2*g^2*p*Log[d + e 
*x^2])/(2*e^2) + (g^2*x^4*Log[c*(d + e*x^2)^p])/2 + (2*f*g*(d + e*x^2)*Log 
[c*(d + e*x^2)^p])/e + f^2*Log[-((e*x^2)/d)]*Log[c*(d + e*x^2)^p] + f^2*p* 
PolyLog[2, 1 + (e*x^2)/d])/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]
 

rule 2925

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Si 
mplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q, x], 
x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && Integer 
Q[r] && IntegerQ[s/n] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0 
] || IGtQ[q, 0])
 

Maple [A] (verified)

Time = 1.57 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.37

method	result	size
parts	\(\frac {g^{2} x^{4} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{4}+f g \,x^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )+\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) f^{2} \ln \left (x \right )-\frac {e p \left (g \left (\frac {\frac {1}{2} e g \,x^{4}-d g \,x^{2}+4 f \,x^{2} e}{2 e^{2}}+\frac {d \left (d g -4 e f \right ) \ln \left (e \,x^{2}+d \right )}{2 e^{3}}\right )+4 f^{2} \left (\frac {\ln \left (x \right ) \left (\ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )+\ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )\right )}{2 e}+\frac {\operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )+\operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}\right )\right )}{2}\)	\(209\)
risch	\(\frac {\ln \left (\left (e \,x^{2}+d \right )^{p}\right ) g^{2} x^{4}}{4}+\ln \left (\left (e \,x^{2}+d \right )^{p}\right ) g f \,x^{2}+\ln \left (\left (e \,x^{2}+d \right )^{p}\right ) f^{2} \ln \left (x \right )-\frac {g^{2} p \,x^{4}}{8}+\frac {d \,g^{2} p \,x^{2}}{4 e}-f g p \,x^{2}-\frac {d^{2} g^{2} p \ln \left (e \,x^{2}+d \right )}{4 e^{2}}+\frac {p g d \ln \left (e \,x^{2}+d \right ) f}{e}-p \,f^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )-p \,f^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )-p \,f^{2} \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )-p \,f^{2} \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {g^{2} x^{4}}{4}+f g \,x^{2}+f^{2} \ln \left (x \right )\right )\)	\(359\)

Input:

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

Output:

1/4*g^2*x^4*ln(c*(e*x^2+d)^p)+f*g*x^2*ln(c*(e*x^2+d)^p)+ln(c*(e*x^2+d)^p)* 
f^2*ln(x)-1/2*e*p*(g*(1/2/e^2*(1/2*e*g*x^4-d*g*x^2+4*f*x^2*e)+1/2*d*(d*g-4 
*e*f)/e^3*ln(e*x^2+d))+4*f^2*(1/2*ln(x)*(ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/ 
2))+ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2)))/e+1/2*(dilog((-e*x+(-d*e)^(1/2))/ 
(-d*e)^(1/2))+dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2)))/e))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate((g*x**2+f)**2*ln(c*(e*x**2+d)**p)/x,x)
 

Output:

Integral((f + g*x**2)**2*log(c*(d + e*x**2)**p)/x, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x} \, dx=\frac {8 \left (\int \frac {\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right )}{e \,x^{3}+d x}d x \right ) d \,e^{2} f^{2} p +2 {\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right )}^{2} e^{2} f^{2}-2 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{2} g^{2} p +8 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d e f g p +8 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{2} f g p \,x^{2}+2 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{2} g^{2} p \,x^{4}+2 d e \,g^{2} p^{2} x^{2}-8 e^{2} f g \,p^{2} x^{2}-e^{2} g^{2} p^{2} x^{4}}{8 e^{2} p} \] Input:

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

Output:

(8*int(log((d + e*x**2)**p*c)/(d*x + e*x**3),x)*d*e**2*f**2*p + 2*log((d + 
 e*x**2)**p*c)**2*e**2*f**2 - 2*log((d + e*x**2)**p*c)*d**2*g**2*p + 8*log 
((d + e*x**2)**p*c)*d*e*f*g*p + 8*log((d + e*x**2)**p*c)*e**2*f*g*p*x**2 + 
 2*log((d + e*x**2)**p*c)*e**2*g**2*p*x**4 + 2*d*e*g**2*p**2*x**2 - 8*e**2 
*f*g*p**2*x**2 - e**2*g**2*p**2*x**4)/(8*e**2*p)