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

Optimal. Leaf size=221 \[ f^2 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {2}{3} f g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac {4 d^{3/2} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^2 g^2 p x}{5 e^2}+\frac {2 \sqrt {d} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {4 d f g p x}{3 e}+\frac {2 d g^2 p x^3}{15 e}-2 f^2 p x-\frac {4}{9} f g p x^3-\frac {2}{25} g^2 p x^5 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2471, 2448, 321, 205, 2455, 302} \[ f^2 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {2}{3} f g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac {4 d^{3/2} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}-\frac {2 d^2 g^2 p x}{5 e^2}+\frac {2 d^{5/2} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}+\frac {2 \sqrt {d} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {4 d f g p x}{3 e}+\frac {2 d g^2 p x^3}{15 e}-2 f^2 p x-\frac {4}{9} f g p x^3-\frac {2}{25} g^2 p x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 2455

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[((f*x)^(m
+ 1)*(a + b*Log[c*(d + e*x^n)^p]))/(f*(m + 1)), x] - Dist[(b*e*n*p)/(f*(m + 1)), Int[(x^(n - 1)*(f*x)^(m + 1))
/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rule 2471

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]] /; Free
Q[{a, b, c, d, e, f, g, n, p, q, r, s}, x] && IntegerQ[n] && IGtQ[q, 0] && IntegerQ[r] && IntegerQ[s] && (EqQ[
q, 1] || (GtQ[r, 0] && GtQ[s, 1]) || (LtQ[s, 0] && LtQ[r, 0]))

Rubi steps

\begin {align*} \int \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx &=\int \left (f^2 \log \left (c \left (d+e x^2\right )^p\right )+2 f g x^2 \log \left (c \left (d+e x^2\right )^p\right )+g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )\right ) \, dx\\ &=f^2 \int \log \left (c \left (d+e x^2\right )^p\right ) \, dx+(2 f g) \int x^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx+g^2 \int x^4 \log \left (c \left (d+e x^2\right )^p\right ) \, dx\\ &=f^2 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {2}{3} f g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )-\left (2 e f^2 p\right ) \int \frac {x^2}{d+e x^2} \, dx-\frac {1}{3} (4 e f g p) \int \frac {x^4}{d+e x^2} \, dx-\frac {1}{5} \left (2 e g^2 p\right ) \int \frac {x^6}{d+e x^2} \, dx\\ &=-2 f^2 p x+f^2 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {2}{3} f g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )+\left (2 d f^2 p\right ) \int \frac {1}{d+e x^2} \, dx-\frac {1}{3} (4 e f g p) \int \left (-\frac {d}{e^2}+\frac {x^2}{e}+\frac {d^2}{e^2 \left (d+e x^2\right )}\right ) \, dx-\frac {1}{5} \left (2 e g^2 p\right ) \int \left (\frac {d^2}{e^3}-\frac {d x^2}{e^2}+\frac {x^4}{e}-\frac {d^3}{e^3 \left (d+e x^2\right )}\right ) \, dx\\ &=-2 f^2 p x+\frac {4 d f g p x}{3 e}-\frac {2 d^2 g^2 p x}{5 e^2}-\frac {4}{9} f g p x^3+\frac {2 d g^2 p x^3}{15 e}-\frac {2}{25} g^2 p x^5+\frac {2 \sqrt {d} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+f^2 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {2}{3} f g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac {\left (4 d^2 f g p\right ) \int \frac {1}{d+e x^2} \, dx}{3 e}+\frac {\left (2 d^3 g^2 p\right ) \int \frac {1}{d+e x^2} \, dx}{5 e^2}\\ &=-2 f^2 p x+\frac {4 d f g p x}{3 e}-\frac {2 d^2 g^2 p x}{5 e^2}-\frac {4}{9} f g p x^3+\frac {2 d g^2 p x^3}{15 e}-\frac {2}{25} g^2 p x^5+\frac {2 \sqrt {d} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {4 d^{3/2} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}+f^2 x \log \left (c \left (d+e x^2\right )^p\right )+\frac {2}{3} f g x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g^2 x^5 \log \left (c \left (d+e x^2\right )^p\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 151, normalized size = 0.68 \[ \frac {\sqrt {e} x \left (15 e^2 \left (15 f^2+10 f g x^2+3 g^2 x^4\right ) \log \left (c \left (d+e x^2\right )^p\right )-2 p \left (45 d^2 g^2-15 d e g \left (10 f+g x^2\right )+e^2 \left (225 f^2+50 f g x^2+9 g^2 x^4\right )\right )\right )+30 \sqrt {d} p \left (3 d^2 g^2-10 d e f g+15 e^2 f^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{225 e^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30*Sqrt[d]*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2)*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]] + Sqrt[e]*x*(-2*p*(45*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)) + 15*e^2*(15*f^2 + 10*f*g*x^2 + 3*g^2*x^4)
*Log[c*(d + e*x^2)^p]))/(225*e^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 404, normalized size = 1.83 \[ \left [-\frac {18 \, e^{2} g^{2} p x^{5} + 10 \, {\left (10 \, e^{2} f g - 3 \, d e g^{2}\right )} p x^{3} - 15 \, {\left (15 \, e^{2} f^{2} - 10 \, d e f g + 3 \, d^{2} g^{2}\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} + 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) + 30 \, {\left (15 \, e^{2} f^{2} - 10 \, d e f g + 3 \, d^{2} g^{2}\right )} p x - 15 \, {\left (3 \, e^{2} g^{2} p x^{5} + 10 \, e^{2} f g p x^{3} + 15 \, e^{2} f^{2} p x\right )} \log \left (e x^{2} + d\right ) - 15 \, {\left (3 \, e^{2} g^{2} x^{5} + 10 \, e^{2} f g x^{3} + 15 \, e^{2} f^{2} x\right )} \log \relax (c)}{225 \, e^{2}}, -\frac {18 \, e^{2} g^{2} p x^{5} + 10 \, {\left (10 \, e^{2} f g - 3 \, d e g^{2}\right )} p x^{3} - 30 \, {\left (15 \, e^{2} f^{2} - 10 \, d e f g + 3 \, d^{2} g^{2}\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) + 30 \, {\left (15 \, e^{2} f^{2} - 10 \, d e f g + 3 \, d^{2} g^{2}\right )} p x - 15 \, {\left (3 \, e^{2} g^{2} p x^{5} + 10 \, e^{2} f g p x^{3} + 15 \, e^{2} f^{2} p x\right )} \log \left (e x^{2} + d\right ) - 15 \, {\left (3 \, e^{2} g^{2} x^{5} + 10 \, e^{2} f g x^{3} + 15 \, e^{2} f^{2} x\right )} \log \relax (c)}{225 \, e^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/225*(18*e^2*g^2*p*x^5 + 10*(10*e^2*f*g - 3*d*e*g^2)*p*x^3 - 15*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2)*p*sqr
t(-d/e)*log((e*x^2 + 2*e*x*sqrt(-d/e) - d)/(e*x^2 + d)) + 30*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2)*p*x - 15*(3
*e^2*g^2*p*x^5 + 10*e^2*f*g*p*x^3 + 15*e^2*f^2*p*x)*log(e*x^2 + d) - 15*(3*e^2*g^2*x^5 + 10*e^2*f*g*x^3 + 15*e
^2*f^2*x)*log(c))/e^2, -1/225*(18*e^2*g^2*p*x^5 + 10*(10*e^2*f*g - 3*d*e*g^2)*p*x^3 - 30*(15*e^2*f^2 - 10*d*e*
f*g + 3*d^2*g^2)*p*sqrt(d/e)*arctan(e*x*sqrt(d/e)/d) + 30*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2)*p*x - 15*(3*e^
2*g^2*p*x^5 + 10*e^2*f*g*p*x^3 + 15*e^2*f^2*p*x)*log(e*x^2 + d) - 15*(3*e^2*g^2*x^5 + 10*e^2*f*g*x^3 + 15*e^2*
f^2*x)*log(c))/e^2]

________________________________________________________________________________________

giac [A]  time = 0.19, size = 201, normalized size = 0.91 \[ \frac {2 \, {\left (3 \, d^{3} g^{2} p - 10 \, d^{2} f g p e + 15 \, d f^{2} p e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {5}{2}\right )}}{15 \, \sqrt {d}} + \frac {1}{225} \, {\left (45 \, g^{2} p x^{5} e^{2} \log \left (x^{2} e + d\right ) - 18 \, g^{2} p x^{5} e^{2} + 45 \, g^{2} x^{5} e^{2} \log \relax (c) + 30 \, d g^{2} p x^{3} e + 150 \, f g p x^{3} e^{2} \log \left (x^{2} e + d\right ) - 100 \, f g p x^{3} e^{2} + 150 \, f g x^{3} e^{2} \log \relax (c) - 90 \, d^{2} g^{2} p x + 300 \, d f g p x e + 225 \, f^{2} p x e^{2} \log \left (x^{2} e + d\right ) - 450 \, f^{2} p x e^{2} + 225 \, f^{2} x e^{2} \log \relax (c)\right )} e^{\left (-2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*d^3*g^2*p - 10*d^2*f*g*p*e + 15*d*f^2*p*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-5/2)/sqrt(d) + 1/225*(45*g^
2*p*x^5*e^2*log(x^2*e + d) - 18*g^2*p*x^5*e^2 + 45*g^2*x^5*e^2*log(c) + 30*d*g^2*p*x^3*e + 150*f*g*p*x^3*e^2*l
og(x^2*e + d) - 100*f*g*p*x^3*e^2 + 150*f*g*x^3*e^2*log(c) - 90*d^2*g^2*p*x + 300*d*f*g*p*x*e + 225*f^2*p*x*e^
2*log(x^2*e + d) - 450*f^2*p*x*e^2 + 225*f^2*x*e^2*log(c))*e^(-2)

________________________________________________________________________________________

maple [C]  time = 0.50, size = 686, normalized size = 3.10 \[ \frac {g^{2} x^{5} \ln \relax (c )}{5}+f^{2} x \ln \relax (c )-2 f^{2} p x -\frac {2 g^{2} p \,x^{5}}{25}+\frac {2 f g \,x^{3} \ln \relax (c )}{3}+\frac {i \pi \,g^{2} x^{5} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{10}+\frac {i \pi \,g^{2} x^{5} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{10}-\frac {i \pi f g \,x^{3} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{3}+\frac {i \pi \,f^{2} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{2}+\frac {i \pi \,f^{2} x \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{2}-\frac {4 f g p \,x^{3}}{9}+\left (\frac {1}{5} g^{2} x^{5}+\frac {2}{3} f g \,x^{3}+f^{2} x \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )-\frac {2 d^{2} g^{2} p x}{5 e^{2}}+\frac {2 d \,g^{2} p \,x^{3}}{15 e}-\frac {i \pi f g \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}{3}-\frac {i \pi \,g^{2} x^{5} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{10}-\frac {i \pi \,f^{2} x \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{2}-\frac {\sqrt {-d e}\, d^{2} g^{2} p \ln \left (d +\sqrt {-d e}\, x \right )}{5 e^{3}}+\frac {\sqrt {-d e}\, d^{2} g^{2} p \ln \left (d -\sqrt {-d e}\, x \right )}{5 e^{3}}-\frac {\sqrt {-d e}\, f^{2} p \ln \left (d +\sqrt {-d e}\, x \right )}{e}+\frac {\sqrt {-d e}\, f^{2} p \ln \left (d -\sqrt {-d e}\, x \right )}{e}+\frac {2 \sqrt {-d e}\, d f g p \ln \left (d +\sqrt {-d e}\, x \right )}{3 e^{2}}-\frac {2 \sqrt {-d e}\, d f g p \ln \left (d -\sqrt {-d e}\, x \right )}{3 e^{2}}+\frac {4 d f g p x}{3 e}-\frac {i \pi \,g^{2} x^{5} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}{10}+\frac {i \pi f g \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{3}+\frac {i \pi f g \,x^{3} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{3}-\frac {i \pi \,f^{2} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*ln(c)*g^2*x^5+ln(c)*f^2*x-2*f^2*p*x-2/25*g^2*p*x^5+2/3*ln(c)*f*g*x^3-4/9*f*g*p*x^3+(1/5*g^2*x^5+2/3*f*g*x^
3+f^2*x)*ln((e*x^2+d)^p)-2/5*d^2*g^2*p*x/e^2+2/15*d*g^2*p*x^3/e-1/3*I*Pi*f*g*x^3*csgn(I*(e*x^2+d)^p)*csgn(I*c*
(e*x^2+d)^p)*csgn(I*c)-1/5/e^3*(-d*e)^(1/2)*p*ln((-d*e)^(1/2)*x+d)*g^2*d^2+1/5/e^3*(-d*e)^(1/2)*p*ln(-(-d*e)^(
1/2)*x+d)*g^2*d^2+1/10*I*Pi*g^2*x^5*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+1/10*I*Pi*g^2*x^5*csgn(I*(e*x^2+d)^p)*cs
gn(I*c*(e*x^2+d)^p)^2-1/3*I*Pi*f*g*x^3*csgn(I*c*(e*x^2+d)^p)^3+1/2*I*Pi*f^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*
x+1/2*I*Pi*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*x-1/e*(-d*e)^(1/2)*p*ln((-d*e)^(1/2)*x+d)*f^2+1/e*(
-d*e)^(1/2)*p*ln(-(-d*e)^(1/2)*x+d)*f^2-1/10*I*Pi*g^2*x^5*csgn(I*c*(e*x^2+d)^p)^3-1/2*I*Pi*f^2*csgn(I*c*(e*x^2
+d)^p)^3*x+2/3/e^2*(-d*e)^(1/2)*p*ln((-d*e)^(1/2)*x+d)*d*f*g-2/3/e^2*(-d*e)^(1/2)*p*ln(-(-d*e)^(1/2)*x+d)*d*f*
g-1/10*I*Pi*g^2*x^5*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+1/3*I*Pi*f*g*x^3*csgn(I*c*(e*x^2+d)^p)
^2*csgn(I*c)+1/3*I*Pi*f*g*x^3*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-1/2*I*Pi*f^2*csgn(I*(e*x^2+d)^p)*csg
n(I*c*(e*x^2+d)^p)*csgn(I*c)*x+4/3*d*f*g*p*x/e

________________________________________________________________________________________

maxima [A]  time = 1.01, size = 150, normalized size = 0.68 \[ \frac {2}{225} \, e p {\left (\frac {15 \, {\left (15 \, d e^{2} f^{2} - 10 \, d^{2} e f g + 3 \, d^{3} g^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e} e^{3}} - \frac {9 \, e^{2} g^{2} x^{5} + 5 \, {\left (10 \, e^{2} f g - 3 \, d e g^{2}\right )} x^{3} + 15 \, {\left (15 \, e^{2} f^{2} - 10 \, d e f g + 3 \, d^{2} g^{2}\right )} x}{e^{3}}\right )} + \frac {1}{15} \, {\left (3 \, g^{2} x^{5} + 10 \, f g x^{3} + 15 \, f^{2} x\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/225*e*p*(15*(15*d*e^2*f^2 - 10*d^2*e*f*g + 3*d^3*g^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*e^3) - (9*e^2*g^2*x^5
+ 5*(10*e^2*f*g - 3*d*e*g^2)*x^3 + 15*(15*e^2*f^2 - 10*d*e*f*g + 3*d^2*g^2)*x)/e^3) + 1/15*(3*g^2*x^5 + 10*f*
g*x^3 + 15*f^2*x)*log((e*x^2 + d)^p*c)

________________________________________________________________________________________

mupad [B]  time = 0.34, size = 193, normalized size = 0.87 \[ \ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (f^2\,x+\frac {2\,f\,g\,x^3}{3}+\frac {g^2\,x^5}{5}\right )-x\,\left (2\,f^2\,p-\frac {d\,\left (\frac {4\,f\,g\,p}{3}-\frac {2\,d\,g^2\,p}{5\,e}\right )}{e}\right )-x^3\,\left (\frac {4\,f\,g\,p}{9}-\frac {2\,d\,g^2\,p}{15\,e}\right )-\frac {2\,g^2\,p\,x^5}{25}+\frac {2\,\sqrt {d}\,p\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e}\,p\,x\,\left (3\,d^2\,g^2-10\,d\,e\,f\,g+15\,e^2\,f^2\right )}{3\,p\,d^3\,g^2-10\,p\,d^2\,e\,f\,g+15\,p\,d\,e^2\,f^2}\right )\,\left (3\,d^2\,g^2-10\,d\,e\,f\,g+15\,e^2\,f^2\right )}{15\,e^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*(d + e*x^2)^p)*(f^2*x + (g^2*x^5)/5 + (2*f*g*x^3)/3) - x*(2*f^2*p - (d*((4*f*g*p)/3 - (2*d*g^2*p)/(5*e))
)/e) - x^3*((4*f*g*p)/9 - (2*d*g^2*p)/(15*e)) - (2*g^2*p*x^5)/25 + (2*d^(1/2)*p*atan((d^(1/2)*e^(1/2)*p*x*(3*d
^2*g^2 + 15*e^2*f^2 - 10*d*e*f*g))/(3*d^3*g^2*p + 15*d*e^2*f^2*p - 10*d^2*e*f*g*p))*(3*d^2*g^2 + 15*e^2*f^2 -
10*d*e*f*g))/(15*e^(5/2))

________________________________________________________________________________________

sympy [A]  time = 89.77, size = 415, normalized size = 1.88 \[ \begin {cases} \frac {i d^{\frac {5}{2}} g^{2} p \log {\left (d + e x^{2} \right )}}{5 e^{3} \sqrt {\frac {1}{e}}} - \frac {2 i d^{\frac {5}{2}} g^{2} p \log {\left (- i \sqrt {d} \sqrt {\frac {1}{e}} + x \right )}}{5 e^{3} \sqrt {\frac {1}{e}}} - \frac {2 i d^{\frac {3}{2}} f g p \log {\left (d + e x^{2} \right )}}{3 e^{2} \sqrt {\frac {1}{e}}} + \frac {4 i d^{\frac {3}{2}} f g p \log {\left (- i \sqrt {d} \sqrt {\frac {1}{e}} + x \right )}}{3 e^{2} \sqrt {\frac {1}{e}}} + \frac {i \sqrt {d} f^{2} p \log {\left (d + e x^{2} \right )}}{e \sqrt {\frac {1}{e}}} - \frac {2 i \sqrt {d} f^{2} p \log {\left (- i \sqrt {d} \sqrt {\frac {1}{e}} + x \right )}}{e \sqrt {\frac {1}{e}}} - \frac {2 d^{2} g^{2} p x}{5 e^{2}} + \frac {4 d f g p x}{3 e} + \frac {2 d g^{2} p x^{3}}{15 e} + f^{2} p x \log {\left (d + e x^{2} \right )} - 2 f^{2} p x + f^{2} x \log {\relax (c )} + \frac {2 f g p x^{3} \log {\left (d + e x^{2} \right )}}{3} - \frac {4 f g p x^{3}}{9} + \frac {2 f g x^{3} \log {\relax (c )}}{3} + \frac {g^{2} p x^{5} \log {\left (d + e x^{2} \right )}}{5} - \frac {2 g^{2} p x^{5}}{25} + \frac {g^{2} x^{5} \log {\relax (c )}}{5} & \text {for}\: e \neq 0 \\\left (f^{2} x + \frac {2 f g x^{3}}{3} + \frac {g^{2} x^{5}}{5}\right ) \log {\left (c d^{p} \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*d**(5/2)*g**2*p*log(d + e*x**2)/(5*e**3*sqrt(1/e)) - 2*I*d**(5/2)*g**2*p*log(-I*sqrt(d)*sqrt(1/e)
+ x)/(5*e**3*sqrt(1/e)) - 2*I*d**(3/2)*f*g*p*log(d + e*x**2)/(3*e**2*sqrt(1/e)) + 4*I*d**(3/2)*f*g*p*log(-I*s
qrt(d)*sqrt(1/e) + x)/(3*e**2*sqrt(1/e)) + I*sqrt(d)*f**2*p*log(d + e*x**2)/(e*sqrt(1/e)) - 2*I*sqrt(d)*f**2*p
*log(-I*sqrt(d)*sqrt(1/e) + x)/(e*sqrt(1/e)) - 2*d**2*g**2*p*x/(5*e**2) + 4*d*f*g*p*x/(3*e) + 2*d*g**2*p*x**3/
(15*e) + f**2*p*x*log(d + e*x**2) - 2*f**2*p*x + f**2*x*log(c) + 2*f*g*p*x**3*log(d + e*x**2)/3 - 4*f*g*p*x**3
/9 + 2*f*g*x**3*log(c)/3 + g**2*p*x**5*log(d + e*x**2)/5 - 2*g**2*p*x**5/25 + g**2*x**5*log(c)/5, Ne(e, 0)), (
(f**2*x + 2*f*g*x**3/3 + g**2*x**5/5)*log(c*d**p), True))

________________________________________________________________________________________