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

Optimal. Leaf size=278 \[ \frac {2 d f^2 p x}{3 e}-\frac {4 d^2 f g p x}{5 e^2}+\frac {2 d^3 g^2 p x}{7 e^3}-\frac {2}{9} f^2 p x^3+\frac {4 d f g p x^3}{15 e}-\frac {2 d^2 g^2 p x^3}{21 e^2}-\frac {4}{25} f g p x^5+\frac {2 d g^2 p x^5}{35 e}-\frac {2}{49} g^2 p x^7-\frac {2 d^{3/2} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {4 d^{5/2} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^{7/2} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 e^{7/2}}+\frac {1}{3} f^2 x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {2}{5} f g x^5 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{7} g^2 x^7 \log \left (c \left (d+e x^2\right )^p\right ) \]

[Out]

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(x*e^(1/2)/d^(1/2))/e^(3/2)+4/5*d^(
5/2)*f*g*p*arctan(x*e^(1/2)/d^(1/2))/e^(5/2)-2/7*d^(7/2)*g^2*p*arctan(x*e^(1/2)/d^(1/2))/e^(7/2)+1/3*f^2*x^3*l
n(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)

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Rubi [A]
time = 0.15, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2526, 2505, 308, 211} \begin {gather*} -\frac {2 d^{3/2} f^2 p \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {4 d^{5/2} f g p \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^{7/2} g^2 p \text {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 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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)/(1
5*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[(Sqrt[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

Rule 211

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

Rule 308

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 2505

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 2526

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]

Rubi steps

\begin {align*} \int x^2 \left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx &=\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\\ &=f^2 \int x^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx+(2 f g) \int x^4 \log \left (c \left (d+e x^2\right )^p\right ) \, dx+g^2 \int x^6 \log \left (c \left (d+e x^2\right )^p\right ) \, dx\\ &=\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 {1}{3} \left (2 e f^2 p\right ) \int \frac {x^4}{d+e x^2} \, dx-\frac {1}{5} (4 e f g p) \int \frac {x^6}{d+e x^2} \, dx-\frac {1}{7} \left (2 e g^2 p\right ) \int \frac {x^8}{d+e x^2} \, dx\\ &=\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 {1}{3} \left (2 e f^2 p\right ) \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} (4 e f g p) \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-\frac {1}{7} \left (2 e g^2 p\right ) \int \left (-\frac {d^3}{e^4}+\frac {d^2 x^2}{e^3}-\frac {d x^4}{e^2}+\frac {x^6}{e}+\frac {d^4}{e^4 \left (d+e x^2\right )}\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 {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 {\left (2 d^2 f^2 p\right ) \int \frac {1}{d+e x^2} \, dx}{3 e}+\frac {\left (4 d^3 f g p\right ) \int \frac {1}{d+e x^2} \, dx}{5 e^2}-\frac {\left (2 d^4 g^2 p\right ) \int \frac {1}{d+e x^2} \, dx}{7 e^3}\\ &=\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 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {4 d^{5/2} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}-\frac {2 d^{7/2} g^2 p \tan ^{-1}\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 )\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 188, normalized size = 0.68 \begin {gather*} \frac {-210 d^{3/2} \left (35 e^2 f^2-42 d e f g+15 d^2 g^2\right ) p \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-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) + 105*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))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.68, size = 761, normalized size = 2.74

method	result	size
risch	\(-\frac {2 g^{2} p \,x^{7}}{49}-\frac {4 d^{2} f g p x}{5 e^{2}}+\frac {4 d f g p \,x^{3}}{15 e}+\frac {\ln \left (c \right ) g^{2} x^{7}}{7}-\frac {i \pi \,g^{2} x^{7} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{14}+\frac {i \pi \,g^{2} x^{7} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{14}-\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 \,g^{2} x^{7} \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}}{14}-\frac {i \pi \,g^{2} x^{7} \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 ) \mathrm {csgn}\left (i c \right )}{14}-\frac {4 f g p \,x^{5}}{25}+\frac {2 d \,f^{2} p x}{3 e}+\frac {\ln \left (c \right ) f^{2} x^{3}}{3}-\frac {i \pi f g \,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 ) \mathrm {csgn}\left (i c \right )}{5}-\frac {i \pi \,f^{2} x^{3} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{6}-\frac {i \pi f g \,x^{5} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{5}+\frac {i \pi \,f^{2} 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}}{6}+\frac {i \pi \,f^{2} x^{3} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{6}-\frac {\sqrt {-e d}\, p \,d^{3} \ln \left (\sqrt {-e d}\, x -d \right ) g^{2}}{7 e^{4}}+\frac {\sqrt {-e d}\, p \,d^{3} \ln \left (-\sqrt {-e d}\, x -d \right ) g^{2}}{7 e^{4}}+\frac {\sqrt {-e d}\, p d \ln \left (-\sqrt {-e d}\, x -d \right ) f^{2}}{3 e^{2}}-\frac {\sqrt {-e d}\, p d \ln \left (\sqrt {-e d}\, x -d \right ) f^{2}}{3 e^{2}}-\frac {i \pi \,f^{2} 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 ) \mathrm {csgn}\left (i c \right )}{6}+\frac {i \pi f g \,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}}{5}+\frac {i \pi f g \,x^{5} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{5}-\frac {2 \sqrt {-e d}\, p \,d^{2} \ln \left (-\sqrt {-e d}\, x -d \right ) f g}{5 e^{3}}+\frac {2 \sqrt {-e d}\, p \,d^{2} \ln \left (\sqrt {-e d}\, x -d \right ) f g}{5 e^{3}}+\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 {2 \ln \left (c \right ) f g \,x^{5}}{5}\)	\(761\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/49*g^2*p*x^7-4/5*d^2*f*g*p*x/e^2+4/15*d*f*g*p*x^3/e+1/7*ln(c)*g^2*x^7-2/9*f^2*p*x^3+2/7*d^3*g^2*p*x/e^3-2/2
1*d^2*g^2*p*x^3/e^2+2/35*d*g^2*p*x^5/e-4/25*f*g*p*x^5+2/3*d*f^2*p*x/e+1/3*ln(c)*f^2*x^3-1/5*I*Pi*f*g*x^5*csgn(
I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-1/14*I*Pi*g^2*x^7*csgn(I*c*(e*x^2+d)^p)^3-1/6*I*Pi*f^2*x^3*csgn
(I*c*(e*x^2+d)^p)^3-1/7/e^4*(-e*d)^(1/2)*p*d^3*ln((-e*d)^(1/2)*x-d)*g^2+1/7/e^4*(-e*d)^(1/2)*p*d^3*ln(-(-e*d)^
(1/2)*x-d)*g^2+1/3/e^2*(-e*d)^(1/2)*p*d*ln(-(-e*d)^(1/2)*x-d)*f^2-1/3/e^2*(-e*d)^(1/2)*p*d*ln((-e*d)^(1/2)*x-d
)*f^2-1/5*I*Pi*f*g*x^5*csgn(I*c*(e*x^2+d)^p)^3+1/14*I*Pi*g^2*x^7*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+1
/14*I*Pi*g^2*x^7*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+1/6*I*Pi*f^2*x^3*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^
2+1/6*I*Pi*f^2*x^3*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-2/5/e^3*(-e*d)^(1/2)*p*d^2*ln(-(-e*d)^(1/2)*x-d)*f*g+2/5/
e^3*(-e*d)^(1/2)*p*d^2*ln((-e*d)^(1/2)*x-d)*f*g-1/14*I*Pi*g^2*x^7*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*cs
gn(I*c)-1/6*I*Pi*f^2*x^3*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+1/5*I*Pi*f*g*x^5*csgn(I*(e*x^2+d)
^p)*csgn(I*c*(e*x^2+d)^p)^2+1/5*I*Pi*f*g*x^5*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+(1/7*g^2*x^7+2/5*f*g*x^5+1/3*f^
2*x^3)*ln((e*x^2+d)^p)+2/5*ln(c)*f*g*x^5

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Maxima [A]
time = 0.49, size = 182, normalized size = 0.65 \begin {gather*} -\frac {2}{11025} \, {\left (\frac {105 \, {\left (15 \, d^{4} g^{2} - 42 \, d^{3} f g e + 35 \, d^{2} f^{2} e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {9}{2}\right )}}{\sqrt {d}} + {\left (225 \, g^{2} x^{7} e^{3} - 63 \, {\left (5 \, d g^{2} e^{2} - 14 \, f g e^{3}\right )} x^{5} + 35 \, {\left (15 \, d^{2} g^{2} e - 42 \, d f g e^{2} + 35 \, f^{2} e^{3}\right )} x^{3} - 105 \, {\left (15 \, d^{3} g^{2} - 42 \, d^{2} f g e + 35 \, d f^{2} e^{2}\right )} x\right )} e^{\left (-4\right )}\right )} p e + \frac {1}{105} \, {\left (15 \, g^{2} x^{7} + 42 \, f g x^{5} + 35 \, f^{2} x^{3}\right )} \log \left ({\left (x^{2} e + d\right )}^{p} c\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/11025*(105*(15*d^4*g^2 - 42*d^3*f*g*e + 35*d^2*f^2*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-9/2)/sqrt(d) + (225*g
^2*x^7*e^3 - 63*(5*d*g^2*e^2 - 14*f*g*e^3)*x^5 + 35*(15*d^2*g^2*e - 42*d*f*g*e^2 + 35*f^2*e^3)*x^3 - 105*(15*d
^3*g^2 - 42*d^2*f*g*e + 35*d*f^2*e^2)*x)*e^(-4))*p*e + 1/105*(15*g^2*x^7 + 42*f*g*x^5 + 35*f^2*x^3)*log((x^2*e
+ d)^p*c)

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Fricas [A]
time = 0.40, size = 465, normalized size = 1.67 \begin {gather*} \left [\frac {1}{11025} \, {\left (3150 \, d^{3} g^{2} p x + 105 \, {\left (15 \, g^{2} p x^{7} + 42 \, f g p x^{5} + 35 \, f^{2} p x^{3}\right )} e^{3} \log \left (x^{2} e + d\right ) + 105 \, {\left (15 \, g^{2} x^{7} + 42 \, f g x^{5} + 35 \, f^{2} x^{3}\right )} e^{3} \log \left (c\right ) + 105 \, {\left (15 \, d^{3} g^{2} p - 42 \, d^{2} f g p e + 35 \, d f^{2} p e^{2}\right )} \sqrt {-d e^{\left (-1\right )}} \log \left (\frac {x^{2} e - 2 \, \sqrt {-d e^{\left (-1\right )}} x e - d}{x^{2} e + d}\right ) - 2 \, {\left (225 \, g^{2} p x^{7} + 882 \, f g p x^{5} + 1225 \, f^{2} p x^{3}\right )} e^{3} + 210 \, {\left (3 \, d g^{2} p x^{5} + 14 \, d f g p x^{3} + 35 \, d f^{2} p x\right )} e^{2} - 210 \, {\left (5 \, d^{2} g^{2} p x^{3} + 42 \, d^{2} f g p x\right )} e\right )} e^{\left (-3\right )}, \frac {1}{11025} \, {\left (3150 \, d^{3} g^{2} p x - 210 \, {\left (15 \, d^{3} g^{2} p - 42 \, d^{2} f g p e + 35 \, d f^{2} p e^{2}\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )} + 105 \, {\left (15 \, g^{2} p x^{7} + 42 \, f g p x^{5} + 35 \, f^{2} p x^{3}\right )} e^{3} \log \left (x^{2} e + d\right ) + 105 \, {\left (15 \, g^{2} x^{7} + 42 \, f g x^{5} + 35 \, f^{2} x^{3}\right )} e^{3} \log \left (c\right ) - 2 \, {\left (225 \, g^{2} p x^{7} + 882 \, f g p x^{5} + 1225 \, f^{2} p x^{3}\right )} e^{3} + 210 \, {\left (3 \, d g^{2} p x^{5} + 14 \, d f g p x^{3} + 35 \, d f^{2} p x\right )} e^{2} - 210 \, {\left (5 \, d^{2} g^{2} p x^{3} + 42 \, d^{2} f g p x\right )} e\right )} e^{\left (-3\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/11025*(3150*d^3*g^2*p*x + 105*(15*g^2*p*x^7 + 42*f*g*p*x^5 + 35*f^2*p*x^3)*e^3*log(x^2*e + d) + 105*(15*g^2
*x^7 + 42*f*g*x^5 + 35*f^2*x^3)*e^3*log(c) + 105*(15*d^3*g^2*p - 42*d^2*f*g*p*e + 35*d*f^2*p*e^2)*sqrt(-d*e^(-
1))*log((x^2*e - 2*sqrt(-d*e^(-1))*x*e - d)/(x^2*e + d)) - 2*(225*g^2*p*x^7 + 882*f*g*p*x^5 + 1225*f^2*p*x^3)*
e^3 + 210*(3*d*g^2*p*x^5 + 14*d*f*g*p*x^3 + 35*d*f^2*p*x)*e^2 - 210*(5*d^2*g^2*p*x^3 + 42*d^2*f*g*p*x)*e)*e^(-
3), 1/11025*(3150*d^3*g^2*p*x - 210*(15*d^3*g^2*p - 42*d^2*f*g*p*e + 35*d*f^2*p*e^2)*sqrt(d)*arctan(x*e^(1/2)/
sqrt(d))*e^(-1/2) + 105*(15*g^2*p*x^7 + 42*f*g*p*x^5 + 35*f^2*p*x^3)*e^3*log(x^2*e + d) + 105*(15*g^2*x^7 + 42
*f*g*x^5 + 35*f^2*x^3)*e^3*log(c) - 2*(225*g^2*p*x^7 + 882*f*g*p*x^5 + 1225*f^2*p*x^3)*e^3 + 210*(3*d*g^2*p*x^
5 + 14*d*f*g*p*x^3 + 35*d*f^2*p*x)*e^2 - 210*(5*d^2*g^2*p*x^3 + 42*d^2*f*g*p*x)*e)*e^(-3)]

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Sympy [A]
time = 132.53, size = 559, normalized size = 2.01 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*sqrt(-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))

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Giac [A]
time = 4.70, size = 246, normalized size = 0.88 \begin {gather*} -\frac {2 \, {\left (15 \, d^{4} g^{2} p - 42 \, d^{3} f g p e + 35 \, d^{2} f^{2} p e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {7}{2}\right )}}{105 \, \sqrt {d}} + \frac {1}{11025} \, {\left (1575 \, g^{2} p x^{7} e^{3} \log \left (x^{2} e + d\right ) - 450 \, g^{2} p x^{7} e^{3} + 1575 \, g^{2} x^{7} e^{3} \log \left (c\right ) + 630 \, d g^{2} p x^{5} e^{2} + 4410 \, f g p x^{5} e^{3} \log \left (x^{2} e + d\right ) - 1764 \, f g p x^{5} e^{3} - 1050 \, d^{2} g^{2} p x^{3} e + 4410 \, f g x^{5} e^{3} \log \left (c\right ) + 2940 \, d f g p x^{3} e^{2} + 3675 \, f^{2} p x^{3} e^{3} \log \left (x^{2} e + d\right ) + 3150 \, d^{3} g^{2} p x - 2450 \, f^{2} p x^{3} e^{3} - 8820 \, d^{2} f g p x e + 3675 \, f^{2} x^{3} e^{3} \log \left (c\right ) + 7350 \, d f^{2} p x e^{2}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*d^4*g^2*p - 42*d^3*f*g*p*e + 35*d^2*f^2*p*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-7/2)/sqrt(d) + 1/11025
*(1575*g^2*p*x^7*e^3*log(x^2*e + d) - 450*g^2*p*x^7*e^3 + 1575*g^2*x^7*e^3*log(c) + 630*d*g^2*p*x^5*e^2 + 4410
*f*g*p*x^5*e^3*log(x^2*e + d) - 1764*f*g*p*x^5*e^3 - 1050*d^2*g^2*p*x^3*e + 4410*f*g*x^5*e^3*log(c) + 2940*d*f
*g*p*x^3*e^2 + 3675*f^2*p*x^3*e^3*log(x^2*e + d) + 3150*d^3*g^2*p*x - 2450*f^2*p*x^3*e^3 - 8820*d^2*f*g*p*x*e
+ 3675*f^2*x^3*e^3*log(c) + 7350*d*f^2*p*x*e^2)*e^(-3)

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Mupad [B]
time = 0.36, size = 235, normalized size = 0.85 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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