\(\int x^2 (f+g x^2)^2 \log (c (d+e x^2)^p) \, dx\) [332]

Optimal result

Integrand size = 25, antiderivative size = 278 \[ \int x^2 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {2 d f^2 p x}{3 e}-\frac {4 d^2 f g p x}{5 e^2}+\frac {2 d^3 g^2 p x}{7 e^3}-\frac {2}{9} f^2 p x^3+\frac {4 d f g p x^3}{15 e}-\frac {2 d^2 g^2 p x^3}{21 e^2}-\frac {4}{25} f g p x^5+\frac {2 d g^2 p x^5}{35 e}-\frac {2}{49} g^2 p x^7-\frac {2 d^{3/2} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {4 d^{5/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^{7/2} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+\frac {1}{3} f^2 x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {2}{5} f g x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^2 x^7 \log \left (c \left (d+e x^2\right )^p\right ) \] Output:

2/3*d*f^2*p*x/e-4/5*d^2*f*g*p*x/e^2+2/7*d^3*g^2*p*x/e^3-2/9*f^2*p*x^3+4/15 
*d*f*g*p*x^3/e-2/21*d^2*g^2*p*x^3/e^2-4/25*f*g*p*x^5+2/35*d*g^2*p*x^5/e-2/ 
49*g^2*p*x^7-2/3*d^(3/2)*f^2*p*arctan(e^(1/2)*x/d^(1/2))/e^(3/2)+4/5*d^(5/ 
2)*f*g*p*arctan(e^(1/2)*x/d^(1/2))/e^(5/2)-2/7*d^(7/2)*g^2*p*arctan(e^(1/2 
)*x/d^(1/2))/e^(7/2)+1/3*f^2*x^3*ln(c*(e*x^2+d)^p)+2/5*f*g*x^5*ln(c*(e*x^2 
+d)^p)+1/7*g^2*x^7*ln(c*(e*x^2+d)^p)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.68 \[ \int x^2 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {-210 d^{3/2} \left (35 e^2 f^2-42 d e f g+15 d^2 g^2\right ) p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )+\sqrt {e} x \left (2 p \left (1575 d^3 g^2-105 d^2 e g \left (42 f+5 g x^2\right )+105 d e^2 \left (35 f^2+14 f g x^2+3 g^2 x^4\right )-e^3 x^2 \left (1225 f^2+882 f g x^2+225 g^2 x^4\right )\right )+105 e^3 x^2 \left (35 f^2+42 f g x^2+15 g^2 x^4\right ) \log \left (c \left (d+e x^2\right )^p\right )\right )}{11025 e^{7/2}} \] Input:

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

Output:

(-210*d^(3/2)*(35*e^2*f^2 - 42*d*e*f*g + 15*d^2*g^2)*p*ArcTan[(Sqrt[e]*x)/ 
Sqrt[d]] + Sqrt[e]*x*(2*p*(1575*d^3*g^2 - 105*d^2*e*g*(42*f + 5*g*x^2) + 1 
05*d*e^2*(35*f^2 + 14*f*g*x^2 + 3*g^2*x^4) - e^3*x^2*(1225*f^2 + 882*f*g*x 
^2 + 225*g^2*x^4)) + 105*e^3*x^2*(35*f^2 + 42*f*g*x^2 + 15*g^2*x^4)*Log[c* 
(d + e*x^2)^p]))/(11025*e^(7/2))
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2926, 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.

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

\(\Big \downarrow \) 2926

\(\displaystyle \int \left (f^2 x^2 \log \left (c \left (d+e x^2\right )^p\right )+2 f g x^4 \log \left (c \left (d+e x^2\right )^p\right )+g^2 x^6 \log \left (c \left (d+e x^2\right )^p\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 d^{3/2} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {4 d^{5/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^{7/2} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+\frac {1}{3} f^2 x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {2}{5} f g x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^2 x^7 \log \left (c \left (d+e x^2\right )^p\right )+\frac {2 d^3 g^2 p x}{7 e^3}-\frac {4 d^2 f g p x}{5 e^2}-\frac {2 d^2 g^2 p x^3}{21 e^2}+\frac {2 d f^2 p x}{3 e}+\frac {4 d f g p x^3}{15 e}+\frac {2 d g^2 p x^5}{35 e}-\frac {2}{9} f^2 p x^3-\frac {4}{25} f g p x^5-\frac {2}{49} g^2 p x^7\)

Input:

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

Output:

(2*d*f^2*p*x)/(3*e) - (4*d^2*f*g*p*x)/(5*e^2) + (2*d^3*g^2*p*x)/(7*e^3) - 
(2*f^2*p*x^3)/9 + (4*d*f*g*p*x^3)/(15*e) - (2*d^2*g^2*p*x^3)/(21*e^2) - (4 
*f*g*p*x^5)/25 + (2*d*g^2*p*x^5)/(35*e) - (2*g^2*p*x^7)/49 - (2*d^(3/2)*f^ 
2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*e^(3/2)) + (4*d^(5/2)*f*g*p*ArcTan[(Sq 
rt[e]*x)/Sqrt[d]])/(5*e^(5/2)) - (2*d^(7/2)*g^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[ 
d]])/(7*e^(7/2)) + (f^2*x^3*Log[c*(d + e*x^2)^p])/3 + (2*f*g*x^5*Log[c*(d 
+ e*x^2)^p])/5 + (g^2*x^7*Log[c*(d + e*x^2)^p])/7
 

Defintions of rubi rules used

rule 2009

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

rule 2926

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

Maple [A] (verified)

Time = 5.36 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.76

method	result	size
parts	\(\frac {g^{2} x^{7} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{7}+\frac {2 f g \,x^{5} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{5}+\frac {f^{2} x^{3} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{3}-\frac {2 e p \left (-\frac {-\frac {15}{7} e^{3} g^{2} x^{7}+3 d \,e^{2} g^{2} x^{5}-\frac {42}{5} e^{3} f g \,x^{5}-5 d^{2} e \,g^{2} x^{3}+14 d f g \,x^{3} e^{2}-\frac {35}{3} e^{3} f^{2} x^{3}+15 d^{3} x \,g^{2}-42 x \,d^{2} e f g +35 x d \,e^{2} f^{2}}{e^{4}}+\frac {d^{2} \left (15 d^{2} g^{2}-42 f g e d +35 f^{2} e^{2}\right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{e^{4} \sqrt {d e}}\right )}{105}\)	\(211\)
risch	\(\frac {2 \ln \left (c \right ) f g \,x^{5}}{5}-\frac {4 d^{2} f g p x}{5 e^{2}}+\frac {4 d f g p \,x^{3}}{15 e}-\frac {4 f g p \,x^{5}}{25}-\frac {i \pi \,f^{2} x^{3} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{6}-\frac {i \pi \,g^{2} x^{7} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{14}-\frac {2 \sqrt {-d e}\, p \,d^{2} \ln \left (-\sqrt {-d e}\, x -d \right ) f g}{5 e^{3}}-\frac {i \pi f g \,x^{5} \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 )}{5}+\frac {i \pi f g \,x^{5} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{5}-\frac {i \pi \,f^{2} x^{3} \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 )}{6}-\frac {\sqrt {-d e}\, p d \ln \left (\sqrt {-d e}\, x -d \right ) f^{2}}{3 e^{2}}+\frac {\sqrt {-d e}\, p d \ln \left (-\sqrt {-d e}\, x -d \right ) f^{2}}{3 e^{2}}+\left (\frac {1}{7} g^{2} x^{7}+\frac {2}{5} f g \,x^{5}+\frac {1}{3} f^{2} x^{3}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )+\frac {\sqrt {-d e}\, p \,d^{3} \ln \left (-\sqrt {-d e}\, x -d \right ) g^{2}}{7 e^{4}}-\frac {\sqrt {-d e}\, p \,d^{3} \ln \left (\sqrt {-d e}\, x -d \right ) g^{2}}{7 e^{4}}+\frac {2 \sqrt {-d e}\, p \,d^{2} \ln \left (\sqrt {-d e}\, x -d \right ) f g}{5 e^{3}}-\frac {i \pi \,g^{2} x^{7} \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 )}{14}+\frac {i \pi f g \,x^{5} \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}}{5}+\frac {\ln \left (c \right ) g^{2} x^{7}}{7}+\frac {\ln \left (c \right ) f^{2} x^{3}}{3}-\frac {2 f^{2} p \,x^{3}}{9}+\frac {2 d^{3} g^{2} p x}{7 e^{3}}-\frac {2 d^{2} g^{2} p \,x^{3}}{21 e^{2}}+\frac {2 d \,g^{2} p \,x^{5}}{35 e}-\frac {i \pi f g \,x^{5} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{5}+\frac {i \pi \,f^{2} x^{3} \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}}{6}+\frac {i \pi \,f^{2} x^{3} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{6}+\frac {i \pi \,g^{2} x^{7} \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}}{14}+\frac {i \pi \,g^{2} x^{7} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{14}-\frac {2 g^{2} p \,x^{7}}{49}+\frac {2 d \,f^{2} p x}{3 e}\)	\(761\)

Input:

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

Output:

1/7*g^2*x^7*ln(c*(e*x^2+d)^p)+2/5*f*g*x^5*ln(c*(e*x^2+d)^p)+1/3*f^2*x^3*ln 
(c*(e*x^2+d)^p)-2/105*e*p*(-1/e^4*(-15/7*e^3*g^2*x^7+3*d*e^2*g^2*x^5-42/5* 
e^3*f*g*x^5-5*d^2*e*g^2*x^3+14*d*f*g*x^3*e^2-35/3*e^3*f^2*x^3+15*d^3*x*g^2 
-42*x*d^2*e*f*g+35*x*d*e^2*f^2)+d^2*(15*d^2*g^2-42*d*e*f*g+35*e^2*f^2)/e^4 
/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 492, normalized size of antiderivative = 1.77 \[ \int x^2 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\left [-\frac {450 \, e^{3} g^{2} p x^{7} + 126 \, {\left (14 \, e^{3} f g - 5 \, d e^{2} g^{2}\right )} p x^{5} + 70 \, {\left (35 \, e^{3} f^{2} - 42 \, d e^{2} f g + 15 \, d^{2} e g^{2}\right )} p x^{3} - 105 \, {\left (35 \, d e^{2} f^{2} - 42 \, d^{2} e f g + 15 \, d^{3} g^{2}\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} - 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) - 210 \, {\left (35 \, d e^{2} f^{2} - 42 \, d^{2} e f g + 15 \, d^{3} g^{2}\right )} p x - 105 \, {\left (15 \, e^{3} g^{2} p x^{7} + 42 \, e^{3} f g p x^{5} + 35 \, e^{3} f^{2} p x^{3}\right )} \log \left (e x^{2} + d\right ) - 105 \, {\left (15 \, e^{3} g^{2} x^{7} + 42 \, e^{3} f g x^{5} + 35 \, e^{3} f^{2} x^{3}\right )} \log \left (c\right )}{11025 \, e^{3}}, -\frac {450 \, e^{3} g^{2} p x^{7} + 126 \, {\left (14 \, e^{3} f g - 5 \, d e^{2} g^{2}\right )} p x^{5} + 70 \, {\left (35 \, e^{3} f^{2} - 42 \, d e^{2} f g + 15 \, d^{2} e g^{2}\right )} p x^{3} + 210 \, {\left (35 \, d e^{2} f^{2} - 42 \, d^{2} e f g + 15 \, d^{3} g^{2}\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) - 210 \, {\left (35 \, d e^{2} f^{2} - 42 \, d^{2} e f g + 15 \, d^{3} g^{2}\right )} p x - 105 \, {\left (15 \, e^{3} g^{2} p x^{7} + 42 \, e^{3} f g p x^{5} + 35 \, e^{3} f^{2} p x^{3}\right )} \log \left (e x^{2} + d\right ) - 105 \, {\left (15 \, e^{3} g^{2} x^{7} + 42 \, e^{3} f g x^{5} + 35 \, e^{3} f^{2} x^{3}\right )} \log \left (c\right )}{11025 \, e^{3}}\right ] \] Input:

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

Output:

[-1/11025*(450*e^3*g^2*p*x^7 + 126*(14*e^3*f*g - 5*d*e^2*g^2)*p*x^5 + 70*( 
35*e^3*f^2 - 42*d*e^2*f*g + 15*d^2*e*g^2)*p*x^3 - 105*(35*d*e^2*f^2 - 42*d 
^2*e*f*g + 15*d^3*g^2)*p*sqrt(-d/e)*log((e*x^2 - 2*e*x*sqrt(-d/e) - d)/(e* 
x^2 + d)) - 210*(35*d*e^2*f^2 - 42*d^2*e*f*g + 15*d^3*g^2)*p*x - 105*(15*e 
^3*g^2*p*x^7 + 42*e^3*f*g*p*x^5 + 35*e^3*f^2*p*x^3)*log(e*x^2 + d) - 105*( 
15*e^3*g^2*x^7 + 42*e^3*f*g*x^5 + 35*e^3*f^2*x^3)*log(c))/e^3, -1/11025*(4 
50*e^3*g^2*p*x^7 + 126*(14*e^3*f*g - 5*d*e^2*g^2)*p*x^5 + 70*(35*e^3*f^2 - 
 42*d*e^2*f*g + 15*d^2*e*g^2)*p*x^3 + 210*(35*d*e^2*f^2 - 42*d^2*e*f*g + 1 
5*d^3*g^2)*p*sqrt(d/e)*arctan(e*x*sqrt(d/e)/d) - 210*(35*d*e^2*f^2 - 42*d^ 
2*e*f*g + 15*d^3*g^2)*p*x - 105*(15*e^3*g^2*p*x^7 + 42*e^3*f*g*p*x^5 + 35* 
e^3*f^2*p*x^3)*log(e*x^2 + d) - 105*(15*e^3*g^2*x^7 + 42*e^3*f*g*x^5 + 35* 
e^3*f^2*x^3)*log(c))/e^3]
 

Sympy [A] (verification not implemented)

Time = 125.21 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.01 \[ \int x^2 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\begin {cases} \left (\frac {f^{2} x^{3}}{3} + \frac {2 f g x^{5}}{5} + \frac {g^{2} x^{7}}{7}\right ) \log {\left (0^{p} c \right )} & \text {for}\: d = 0 \wedge e = 0 \\\left (\frac {f^{2} x^{3}}{3} + \frac {2 f g x^{5}}{5} + \frac {g^{2} x^{7}}{7}\right ) \log {\left (c d^{p} \right )} & \text {for}\: e = 0 \\- \frac {2 f^{2} p x^{3}}{9} + \frac {f^{2} x^{3} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{3} - \frac {4 f g p x^{5}}{25} + \frac {2 f g x^{5} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{5} - \frac {2 g^{2} p x^{7}}{49} + \frac {g^{2} x^{7} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{7} & \text {for}\: d = 0 \\- \frac {2 d^{4} g^{2} p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{7 e^{4} \sqrt {- \frac {d}{e}}} + \frac {d^{4} g^{2} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{7 e^{4} \sqrt {- \frac {d}{e}}} + \frac {4 d^{3} f g p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} - \frac {2 d^{3} f g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} + \frac {2 d^{3} g^{2} p x}{7 e^{3}} - \frac {2 d^{2} f^{2} p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} + \frac {d^{2} f^{2} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} - \frac {4 d^{2} f g p x}{5 e^{2}} - \frac {2 d^{2} g^{2} p x^{3}}{21 e^{2}} + \frac {2 d f^{2} p x}{3 e} + \frac {4 d f g p x^{3}}{15 e} + \frac {2 d g^{2} p x^{5}}{35 e} - \frac {2 f^{2} p x^{3}}{9} + \frac {f^{2} x^{3} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3} - \frac {4 f g p x^{5}}{25} + \frac {2 f g x^{5} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5} - \frac {2 g^{2} p x^{7}}{49} + \frac {g^{2} x^{7} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{7} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise(((f**2*x**3/3 + 2*f*g*x**5/5 + g**2*x**7/7)*log(0**p*c), Eq(d, 0 
) & Eq(e, 0)), ((f**2*x**3/3 + 2*f*g*x**5/5 + g**2*x**7/7)*log(c*d**p), Eq 
(e, 0)), (-2*f**2*p*x**3/9 + f**2*x**3*log(c*(e*x**2)**p)/3 - 4*f*g*p*x**5 
/25 + 2*f*g*x**5*log(c*(e*x**2)**p)/5 - 2*g**2*p*x**7/49 + g**2*x**7*log(c 
*(e*x**2)**p)/7, Eq(d, 0)), (-2*d**4*g**2*p*log(x - sqrt(-d/e))/(7*e**4*sq 
rt(-d/e)) + d**4*g**2*log(c*(d + e*x**2)**p)/(7*e**4*sqrt(-d/e)) + 4*d**3* 
f*g*p*log(x - sqrt(-d/e))/(5*e**3*sqrt(-d/e)) - 2*d**3*f*g*log(c*(d + e*x* 
*2)**p)/(5*e**3*sqrt(-d/e)) + 2*d**3*g**2*p*x/(7*e**3) - 2*d**2*f**2*p*log 
(x - sqrt(-d/e))/(3*e**2*sqrt(-d/e)) + d**2*f**2*log(c*(d + e*x**2)**p)/(3 
*e**2*sqrt(-d/e)) - 4*d**2*f*g*p*x/(5*e**2) - 2*d**2*g**2*p*x**3/(21*e**2) 
 + 2*d*f**2*p*x/(3*e) + 4*d*f*g*p*x**3/(15*e) + 2*d*g**2*p*x**5/(35*e) - 2 
*f**2*p*x**3/9 + f**2*x**3*log(c*(d + e*x**2)**p)/3 - 4*f*g*p*x**5/25 + 2* 
f*g*x**5*log(c*(d + e*x**2)**p)/5 - 2*g**2*p*x**7/49 + g**2*x**7*log(c*(d 
+ e*x**2)**p)/7, True))
 

Maxima [F(-2)]

Exception generated. \[ \int x^2 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.77 \[ \int x^2 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {1}{49} \, {\left (2 \, g^{2} p - 7 \, g^{2} \log \left (c\right )\right )} x^{7} - \frac {2 \, {\left (14 \, e f g p - 5 \, d g^{2} p - 35 \, e f g \log \left (c\right )\right )} x^{5}}{175 \, e} - \frac {{\left (70 \, e^{2} f^{2} p - 84 \, d e f g p + 30 \, d^{2} g^{2} p - 105 \, e^{2} f^{2} \log \left (c\right )\right )} x^{3}}{315 \, e^{2}} + \frac {1}{105} \, {\left (15 \, g^{2} p x^{7} + 42 \, f g p x^{5} + 35 \, f^{2} p x^{3}\right )} \log \left (e x^{2} + d\right ) + \frac {2 \, {\left (35 \, d e^{2} f^{2} p - 42 \, d^{2} e f g p + 15 \, d^{3} g^{2} p\right )} x}{105 \, e^{3}} - \frac {2 \, {\left (35 \, d^{2} e^{2} f^{2} p - 42 \, d^{3} e f g p + 15 \, d^{4} g^{2} p\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{105 \, \sqrt {d e} e^{3}} \] Input:

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

Output:

-1/49*(2*g^2*p - 7*g^2*log(c))*x^7 - 2/175*(14*e*f*g*p - 5*d*g^2*p - 35*e* 
f*g*log(c))*x^5/e - 1/315*(70*e^2*f^2*p - 84*d*e*f*g*p + 30*d^2*g^2*p - 10 
5*e^2*f^2*log(c))*x^3/e^2 + 1/105*(15*g^2*p*x^7 + 42*f*g*p*x^5 + 35*f^2*p* 
x^3)*log(e*x^2 + d) + 2/105*(35*d*e^2*f^2*p - 42*d^2*e*f*g*p + 15*d^3*g^2* 
p)*x/e^3 - 2/105*(35*d^2*e^2*f^2*p - 42*d^3*e*f*g*p + 15*d^4*g^2*p)*arctan 
(e*x/sqrt(d*e))/(sqrt(d*e)*e^3)
 

Mupad [B] (verification not implemented)

Time = 25.80 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.85 \[ \int x^2 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {f^2\,x^3}{3}+\frac {2\,f\,g\,x^5}{5}+\frac {g^2\,x^7}{7}\right )-x^3\,\left (\frac {2\,f^2\,p}{9}-\frac {d\,\left (\frac {4\,f\,g\,p}{5}-\frac {2\,d\,g^2\,p}{7\,e}\right )}{3\,e}\right )-x^5\,\left (\frac {4\,f\,g\,p}{25}-\frac {2\,d\,g^2\,p}{35\,e}\right )-\frac {2\,g^2\,p\,x^7}{49}+\frac {d\,x\,\left (\frac {2\,f^2\,p}{3}-\frac {d\,\left (\frac {4\,f\,g\,p}{5}-\frac {2\,d\,g^2\,p}{7\,e}\right )}{e}\right )}{e}-\frac {2\,d^{3/2}\,p\,\mathrm {atan}\left (\frac {d^{3/2}\,\sqrt {e}\,p\,x\,\left (15\,d^2\,g^2-42\,d\,e\,f\,g+35\,e^2\,f^2\right )}{15\,p\,d^4\,g^2-42\,p\,d^3\,e\,f\,g+35\,p\,d^2\,e^2\,f^2}\right )\,\left (15\,d^2\,g^2-42\,d\,e\,f\,g+35\,e^2\,f^2\right )}{105\,e^{7/2}} \] Input:

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

Output:

log(c*(d + e*x^2)^p)*((f^2*x^3)/3 + (g^2*x^7)/7 + (2*f*g*x^5)/5) - x^3*((2 
*f^2*p)/9 - (d*((4*f*g*p)/5 - (2*d*g^2*p)/(7*e)))/(3*e)) - x^5*((4*f*g*p)/ 
25 - (2*d*g^2*p)/(35*e)) - (2*g^2*p*x^7)/49 + (d*x*((2*f^2*p)/3 - (d*((4*f 
*g*p)/5 - (2*d*g^2*p)/(7*e)))/e))/e - (2*d^(3/2)*p*atan((d^(3/2)*e^(1/2)*p 
*x*(15*d^2*g^2 + 35*e^2*f^2 - 42*d*e*f*g))/(15*d^4*g^2*p + 35*d^2*e^2*f^2* 
p - 42*d^3*e*f*g*p))*(15*d^2*g^2 + 35*e^2*f^2 - 42*d*e*f*g))/(105*e^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.93 \[ \int x^2 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {-3150 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) d^{3} g^{2} p +8820 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) d^{2} e f g p -7350 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) d \,e^{2} f^{2} p +3675 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{4} f^{2} x^{3}+4410 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{4} f g \,x^{5}+1575 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{4} g^{2} x^{7}+3150 d^{3} e \,g^{2} p x -8820 d^{2} e^{2} f g p x -1050 d^{2} e^{2} g^{2} p \,x^{3}+7350 d \,e^{3} f^{2} p x +2940 d \,e^{3} f g p \,x^{3}+630 d \,e^{3} g^{2} p \,x^{5}-2450 e^{4} f^{2} p \,x^{3}-1764 e^{4} f g p \,x^{5}-450 e^{4} g^{2} p \,x^{7}}{11025 e^{4}} \] Input:

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

Output:

( - 3150*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*d**3*g**2*p + 8820* 
sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*d**2*e*f*g*p - 7350*sqrt(e)* 
sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*d*e**2*f**2*p + 3675*log((d + e*x**2 
)**p*c)*e**4*f**2*x**3 + 4410*log((d + e*x**2)**p*c)*e**4*f*g*x**5 + 1575* 
log((d + e*x**2)**p*c)*e**4*g**2*x**7 + 3150*d**3*e*g**2*p*x - 8820*d**2*e 
**2*f*g*p*x - 1050*d**2*e**2*g**2*p*x**3 + 7350*d*e**3*f**2*p*x + 2940*d*e 
**3*f*g*p*x**3 + 630*d*e**3*g**2*p*x**5 - 2450*e**4*f**2*p*x**3 - 1764*e** 
4*f*g*p*x**5 - 450*e**4*g**2*p*x**7)/(11025*e**4)