3.3.73 \(\int (f+g x^2)^2 \log ^2(c (d+e x^2)^p) \, dx\) [273]

3.3.73.1 Optimal result

Integrand size = 24, antiderivative size = 945 \[ \int \left (f+g x^2\right )^2 \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx=8 f^2 p^2 x-\frac {64 d f g p^2 x}{9 e}+\frac {184 d^2 g^2 p^2 x}{75 e^2}+\frac {16}{27} f g p^2 x^3-\frac {64 d g^2 p^2 x^3}{225 e}+\frac {8}{125} g^2 p^2 x^5-\frac {8 \sqrt {d} f^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {64 d^{3/2} f g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{9 e^{3/2}}-\frac {184 d^{5/2} g^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{75 e^{5/2}}+\frac {4 i \sqrt {d} f^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{\sqrt {e}}-\frac {8 i d^{3/2} f g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{3 e^{3/2}}+\frac {4 i d^{5/2} g^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{5 e^{5/2}}+\frac {8 \sqrt {d} f^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{\sqrt {e}}-\frac {16 d^{3/2} f g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{3 e^{3/2}}+\frac {8 d^{5/2} g^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{5 e^{5/2}}-4 f^2 p x \log \left (c \left (d+e x^2\right )^p\right )+\frac {8 d f g p x \log \left (c \left (d+e x^2\right )^p\right )}{3 e}-\frac {4 d^2 g^2 p x \log \left (c \left (d+e x^2\right )^p\right )}{5 e^2}-\frac {8}{9} f g p x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {4 d g^2 p x^3 \log \left (c \left (d+e x^2\right )^p\right )}{15 e}-\frac {4}{25} g^2 p x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {4 \sqrt {d} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{\sqrt {e}}-\frac {8 d^{3/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{3 e^{3/2}}+\frac {4 d^{5/2} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{5 e^{5/2}}+f^2 x \log ^2\left (c \left (d+e x^2\right )^p\right )+\frac {2}{3} f g x^3 \log ^2\left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^5 \log ^2\left (c \left (d+e x^2\right )^p\right )+\frac {4 i \sqrt {d} f^2 p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{\sqrt {e}}-\frac {8 i d^{3/2} f g p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{3 e^{3/2}}+\frac {4 i d^{5/2} g^2 p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )}{5 e^{5/2}} \]

output

8/3*d*f*g*p*x*ln(c*(e*x^2+d)^p)/e-8/3*d^(3/2)*f*g*p*arctan(x*e^(1/2)/d^(1/ 
2))*ln(c*(e*x^2+d)^p)/e^(3/2)-16/3*d^(3/2)*f*g*p^2*arctan(x*e^(1/2)/d^(1/2 
))*ln(2*d^(1/2)/(d^(1/2)+I*x*e^(1/2)))/e^(3/2)-8/3*I*d^(3/2)*f*g*p^2*arcta 
n(x*e^(1/2)/d^(1/2))^2/e^(3/2)-8/3*I*d^(3/2)*f*g*p^2*polylog(2,1-2*d^(1/2) 
/(d^(1/2)+I*x*e^(1/2)))/e^(3/2)+16/27*f*g*p^2*x^3-4*f^2*p*x*ln(c*(e*x^2+d) 
^p)-4/25*g^2*p*x^5*ln(c*(e*x^2+d)^p)+2/3*f*g*x^3*ln(c*(e*x^2+d)^p)^2+8*f^2 
*p^2*x+8/125*g^2*p^2*x^5+1/5*g^2*x^5*ln(c*(e*x^2+d)^p)^2-64/9*d*f*g*p^2*x/ 
e+8/5*d^(5/2)*g^2*p^2*arctan(x*e^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)+I*x* 
e^(1/2)))/e^(5/2)+4*f^2*p*arctan(x*e^(1/2)/d^(1/2))*ln(c*(e*x^2+d)^p)*d^(1 
/2)/e^(1/2)+8*f^2*p^2*arctan(x*e^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)+I*x* 
e^(1/2)))*d^(1/2)/e^(1/2)+64/9*d^(3/2)*f*g*p^2*arctan(x*e^(1/2)/d^(1/2))/e 
^(3/2)-4/5*d^2*g^2*p*x*ln(c*(e*x^2+d)^p)/e^2+4/15*d*g^2*p*x^3*ln(c*(e*x^2+ 
d)^p)/e+4/5*d^(5/2)*g^2*p*arctan(x*e^(1/2)/d^(1/2))*ln(c*(e*x^2+d)^p)/e^(5 
/2)+f^2*x*ln(c*(e*x^2+d)^p)^2+4/5*I*d^(5/2)*g^2*p^2*arctan(x*e^(1/2)/d^(1/ 
2))^2/e^(5/2)+4/5*I*d^(5/2)*g^2*p^2*polylog(2,1-2*d^(1/2)/(d^(1/2)+I*x*e^( 
1/2)))/e^(5/2)+4*I*f^2*p^2*arctan(x*e^(1/2)/d^(1/2))^2*d^(1/2)/e^(1/2)+4*I 
*f^2*p^2*polylog(2,1-2*d^(1/2)/(d^(1/2)+I*x*e^(1/2)))*d^(1/2)/e^(1/2)+184/ 
75*d^2*g^2*p^2*x/e^2-64/225*d*g^2*p^2*x^3/e-184/75*d^(5/2)*g^2*p^2*arctan( 
x*e^(1/2)/d^(1/2))/e^(5/2)-8/9*f*g*p*x^3*ln(c*(e*x^2+d)^p)-8*f^2*p^2*arcta 
n(x*e^(1/2)/d^(1/2))*d^(1/2)/e^(1/2)
 

3.3.73.2 Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 435, normalized size of antiderivative = 0.46 \[ \int \left (f+g x^2\right )^2 \log ^2\left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {900 i \sqrt {d} \left (15 e^2 f^2-10 d e f g+3 d^2 g^2\right ) p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2+60 \sqrt {d} p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (-2 \left (225 e^2 f^2-200 d e f g+69 d^2 g^2\right ) p+30 \left (15 e^2 f^2-10 d e f g+3 d^2 g^2\right ) p \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} x}\right )+15 \left (15 e^2 f^2-10 d e f g+3 d^2 g^2\right ) \log \left (c \left (d+e x^2\right )^p\right )\right )+\sqrt {e} x \left (8 p^2 \left (1035 d^2 g^2-120 d e g \left (25 f+g x^2\right )+e^2 \left (3375 f^2+250 f g x^2+27 g^2 x^4\right )\right )-60 p \left (45 d^2 g^2-15 d e g \left (10 f+g x^2\right )+e^2 \left (225 f^2+50 f g x^2+9 g^2 x^4\right )\right ) \log \left (c \left (d+e x^2\right )^p\right )+225 e^2 \left (15 f^2+10 f g x^2+3 g^2 x^4\right ) \log ^2\left (c \left (d+e x^2\right )^p\right )\right )+900 i \sqrt {d} \left (15 e^2 f^2-10 d e f g+3 d^2 g^2\right ) p^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {d}+\sqrt {e} x}{-i \sqrt {d}+\sqrt {e} x}\right )}{3375 e^{5/2}} \]

input

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

output

((900*I)*Sqrt[d]*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2)*p^2*ArcTan[(Sqrt[e] 
*x)/Sqrt[d]]^2 + 60*Sqrt[d]*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(-2*(225*e^2*f^2 
 - 200*d*e*f*g + 69*d^2*g^2)*p + 30*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2)* 
p*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*x)] + 15*(15*e^2*f^2 - 10*d*e*f*g + 
 3*d^2*g^2)*Log[c*(d + e*x^2)^p]) + Sqrt[e]*x*(8*p^2*(1035*d^2*g^2 - 120*d 
*e*g*(25*f + g*x^2) + e^2*(3375*f^2 + 250*f*g*x^2 + 27*g^2*x^4)) - 60*p*(4 
5*d^2*g^2 - 15*d*e*g*(10*f + g*x^2) + e^2*(225*f^2 + 50*f*g*x^2 + 9*g^2*x^ 
4))*Log[c*(d + e*x^2)^p] + 225*e^2*(15*f^2 + 10*f*g*x^2 + 3*g^2*x^4)*Log[c 
*(d + e*x^2)^p]^2) + (900*I)*Sqrt[d]*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2) 
*p^2*PolyLog[2, (I*Sqrt[d] + Sqrt[e]*x)/((-I)*Sqrt[d] + Sqrt[e]*x)])/(3375 
*e^(5/2))
 

3.3.73.3 Rubi [A] (verified)

Time = 1.38 (sec) , antiderivative size = 945, 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 = {2921, 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.

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

\(\Big \downarrow \) 2921

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {8}{125} g^2 p^2 x^5+\frac {1}{5} g^2 \log ^2\left (c \left (e x^2+d\right )^p\right ) x^5-\frac {4}{25} g^2 p \log \left (c \left (e x^2+d\right )^p\right ) x^5-\frac {64 d g^2 p^2 x^3}{225 e}+\frac {16}{27} f g p^2 x^3+\frac {2}{3} f g \log ^2\left (c \left (e x^2+d\right )^p\right ) x^3+\frac {4 d g^2 p \log \left (c \left (e x^2+d\right )^p\right ) x^3}{15 e}-\frac {8}{9} f g p \log \left (c \left (e x^2+d\right )^p\right ) x^3+8 f^2 p^2 x+\frac {184 d^2 g^2 p^2 x}{75 e^2}-\frac {64 d f g p^2 x}{9 e}+f^2 \log ^2\left (c \left (e x^2+d\right )^p\right ) x-4 f^2 p \log \left (c \left (e x^2+d\right )^p\right ) x-\frac {4 d^2 g^2 p \log \left (c \left (e x^2+d\right )^p\right ) x}{5 e^2}+\frac {8 d f g p \log \left (c \left (e x^2+d\right )^p\right ) x}{3 e}+\frac {4 i \sqrt {d} f^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{\sqrt {e}}+\frac {4 i d^{5/2} g^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{5 e^{5/2}}-\frac {8 i d^{3/2} f g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{3 e^{3/2}}-\frac {8 \sqrt {d} f^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {184 d^{5/2} g^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{75 e^{5/2}}+\frac {64 d^{3/2} f g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{9 e^{3/2}}+\frac {8 \sqrt {d} f^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{i \sqrt {e} x+\sqrt {d}}\right )}{\sqrt {e}}+\frac {8 d^{5/2} g^2 p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{i \sqrt {e} x+\sqrt {d}}\right )}{5 e^{5/2}}-\frac {16 d^{3/2} f g p^2 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{i \sqrt {e} x+\sqrt {d}}\right )}{3 e^{3/2}}+\frac {4 \sqrt {d} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (e x^2+d\right )^p\right )}{\sqrt {e}}+\frac {4 d^{5/2} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (e x^2+d\right )^p\right )}{5 e^{5/2}}-\frac {8 d^{3/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (c \left (e x^2+d\right )^p\right )}{3 e^{3/2}}+\frac {4 i \sqrt {d} f^2 p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{i \sqrt {e} x+\sqrt {d}}\right )}{\sqrt {e}}+\frac {4 i d^{5/2} g^2 p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{i \sqrt {e} x+\sqrt {d}}\right )}{5 e^{5/2}}-\frac {8 i d^{3/2} f g p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{i \sqrt {e} x+\sqrt {d}}\right )}{3 e^{3/2}}\)

input

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

output

8*f^2*p^2*x - (64*d*f*g*p^2*x)/(9*e) + (184*d^2*g^2*p^2*x)/(75*e^2) + (16* 
f*g*p^2*x^3)/27 - (64*d*g^2*p^2*x^3)/(225*e) + (8*g^2*p^2*x^5)/125 - (8*Sq 
rt[d]*f^2*p^2*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] + (64*d^(3/2)*f*g*p^2*A 
rcTan[(Sqrt[e]*x)/Sqrt[d]])/(9*e^(3/2)) - (184*d^(5/2)*g^2*p^2*ArcTan[(Sqr 
t[e]*x)/Sqrt[d]])/(75*e^(5/2)) + ((4*I)*Sqrt[d]*f^2*p^2*ArcTan[(Sqrt[e]*x) 
/Sqrt[d]]^2)/Sqrt[e] - (((8*I)/3)*d^(3/2)*f*g*p^2*ArcTan[(Sqrt[e]*x)/Sqrt[ 
d]]^2)/e^(3/2) + (((4*I)/5)*d^(5/2)*g^2*p^2*ArcTan[(Sqrt[e]*x)/Sqrt[d]]^2) 
/e^(5/2) + (8*Sqrt[d]*f^2*p^2*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*Log[(2*Sqrt[d])/ 
(Sqrt[d] + I*Sqrt[e]*x)])/Sqrt[e] - (16*d^(3/2)*f*g*p^2*ArcTan[(Sqrt[e]*x) 
/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*x)])/(3*e^(3/2)) + (8*d^(5/ 
2)*g^2*p^2*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e 
]*x)])/(5*e^(5/2)) - 4*f^2*p*x*Log[c*(d + e*x^2)^p] + (8*d*f*g*p*x*Log[c*( 
d + e*x^2)^p])/(3*e) - (4*d^2*g^2*p*x*Log[c*(d + e*x^2)^p])/(5*e^2) - (8*f 
*g*p*x^3*Log[c*(d + e*x^2)^p])/9 + (4*d*g^2*p*x^3*Log[c*(d + e*x^2)^p])/(1 
5*e) - (4*g^2*p*x^5*Log[c*(d + e*x^2)^p])/25 + (4*Sqrt[d]*f^2*p*ArcTan[(Sq 
rt[e]*x)/Sqrt[d]]*Log[c*(d + e*x^2)^p])/Sqrt[e] - (8*d^(3/2)*f*g*p*ArcTan[ 
(Sqrt[e]*x)/Sqrt[d]]*Log[c*(d + e*x^2)^p])/(3*e^(3/2)) + (4*d^(5/2)*g^2*p* 
ArcTan[(Sqrt[e]*x)/Sqrt[d]]*Log[c*(d + e*x^2)^p])/(5*e^(5/2)) + f^2*x*Log[ 
c*(d + e*x^2)^p]^2 + (2*f*g*x^3*Log[c*(d + e*x^2)^p]^2)/3 + (g^2*x^5*Log[c 
*(d + e*x^2)^p]^2)/5 + ((4*I)*Sqrt[d]*f^2*p^2*PolyLog[2, 1 - (2*Sqrt[d]...
 

3.3.73.3.1 Defintions of rubi rules used

rule 2009

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

rule 2921

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_) + 
 (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> With[{t = ExpandIntegrand[(a + b*Log[ 
c*(d + e*x^n)^p])^q, (f + g*x^s)^r, x]}, Int[t, x] /; SumQ[t]] /; FreeQ[{a, 
 b, c, d, e, f, g, n, p, q, r, s}, x] && IntegerQ[n] && IGtQ[q, 0] && Integ 
erQ[r] && IntegerQ[s] && (EqQ[q, 1] || (GtQ[r, 0] && GtQ[s, 1]) || (LtQ[s, 
0] && LtQ[r, 0]))
 

3.3.73.4 Maple [C] (warning: unable to verify)

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

Time = 2.22 (sec) , antiderivative size = 1077, normalized size of antiderivative = 1.14

method	result	size
risch	\(\text {Expression too large to display}\)	\(1077\)

input

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

output

-4/25*p*g^2*x^5*ln((e*x^2+d)^p)-4*p*f^2*x*ln((e*x^2+d)^p)-184/75*p^2/e^2*g 
^2*d^3/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))-4*p^2*d/(d*e)^(1/2)*arctan(x*e/ 
(d*e)^(1/2))*f^2*ln(e*x^2+d)+4/15*p/e*g^2*d*x^3*ln((e*x^2+d)^p)-4/5*p/e^2* 
g^2*d^2*x*ln((e*x^2+d)^p)+4*p*d/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*f^2*ln 
((e*x^2+d)^p)+2/3*ln((e*x^2+d)^p)^2*f*g*x^3+8/3*p/e*d*f*g*x*ln((e*x^2+d)^p 
)-8*p^2*d/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*f^2-8/9*p*f*g*x^3*ln((e*x^2+ 
d)^p)-8/3*p/e*d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*f*g*ln((e*x^2+d)^p)+ 
8/3*p^2/e*d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*f*g*ln(e*x^2+d)+1/5*ln(( 
e*x^2+d)^p)^2*g^2*x^5+ln((e*x^2+d)^p)^2*x*f^2-4/15*p^2*e*Sum(-1/2*(ln(x-_a 
lpha)*ln(e*x^2+d)-2*e*(1/4/_alpha/e*ln(x-_alpha)^2+1/2*_alpha/d*ln(x-_alph 
a)*ln(1/2*(x+_alpha)/_alpha)+1/2*_alpha/d*dilog(1/2*(x+_alpha)/_alpha)))*d 
*(3*d^2*g^2-10*d*e*f*g+15*e^2*f^2)/e^4/_alpha,_alpha=RootOf(_Z^2*e+d))+64/ 
9*p^2/e*d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*f*g-4/5*p^2/e^2*g^2*d^3/(d 
*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*ln(e*x^2+d)+(I*Pi*csgn(I*(e*x^2+d)^p)*cs 
gn(I*c*(e*x^2+d)^p)^2-I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn( 
I*c)-I*Pi*csgn(I*c*(e*x^2+d)^p)^3+I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+2 
*ln(c))*(1/5*ln((e*x^2+d)^p)*g^2*x^5+2/3*ln((e*x^2+d)^p)*f*g*x^3+ln((e*x^2 
+d)^p)*x*f^2-2/15*p*e*(1/e^3*(3/5*e^2*g^2*x^5-d*e*g^2*x^3+10/3*e^2*f*g*x^3 
+3*d^2*g^2*x-10*d*e*f*g*x+15*e^2*f^2*x)-d*(3*d^2*g^2-10*d*e*f*g+15*e^2*f^2 
)/e^3/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))))+4/5*p/e^2*g^2*d^3/(d*e)^(1/...
 

3.3.73.5 Fricas [F]

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

input

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

output

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

3.3.73.6 Sympy [F]

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

input

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

output

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

3.3.73.7 Maxima [F(-2)]

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

input

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

3.3.73.8 Giac [F]

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

input

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

output

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

3.3.73.9 Mupad [F(-1)]

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

input

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

output

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