3.335 \(\int \frac {(f+g x^2)^2 \log (c (d+e x^2)^p)}{x^4} \, dx\)

Optimal. Leaf size=169 \[ -\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{x}+g^2 x \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 e^{3/2} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {2 e f^2 p}{3 d x}+\frac {4 \sqrt {e} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}}+\frac {2 \sqrt {d} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-2 g^2 p x \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2476, 2448, 321, 205, 2455, 325} \[ -\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{x}+g^2 x \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 e^{3/2} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {2 e f^2 p}{3 d x}+\frac {4 \sqrt {e} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}}+\frac {2 \sqrt {d} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-2 g^2 p x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 2476

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

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Mathematica [C]  time = 0.14, size = 113, normalized size = 0.67 \[ -\frac {\left (f^2+6 f g x^2-3 g^2 x^4\right ) \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {2 e f^2 p \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {e x^2}{d}\right )}{3 d x}+\frac {2 g p (d g+2 e f) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}-2 g^2 p x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*g^2*p*x + (2*g*(2*e*f + d*g)*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*Sqrt[e]) - (2*e*f^2*p*Hypergeometric2F
1[-1/2, 1, 1/2, -((e*x^2)/d)])/(3*d*x) - ((f^2 + 6*f*g*x^2 - 3*g^2*x^4)*Log[c*(d + e*x^2)^p])/(3*x^3)

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fricas [A]  time = 0.98, size = 350, normalized size = 2.07 \[ \left [-\frac {6 \, d^{2} e g^{2} p x^{4} + 2 \, d e^{2} f^{2} p x^{2} - {\left (e^{2} f^{2} - 6 \, d e f g - 3 \, d^{2} g^{2}\right )} \sqrt {-d e} p x^{3} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) - {\left (3 \, d^{2} e g^{2} p x^{4} - 6 \, d^{2} e f g p x^{2} - d^{2} e f^{2} p\right )} \log \left (e x^{2} + d\right ) - {\left (3 \, d^{2} e g^{2} x^{4} - 6 \, d^{2} e f g x^{2} - d^{2} e f^{2}\right )} \log \relax (c)}{3 \, d^{2} e x^{3}}, -\frac {6 \, d^{2} e g^{2} p x^{4} + 2 \, d e^{2} f^{2} p x^{2} + 2 \, {\left (e^{2} f^{2} - 6 \, d e f g - 3 \, d^{2} g^{2}\right )} \sqrt {d e} p x^{3} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) - {\left (3 \, d^{2} e g^{2} p x^{4} - 6 \, d^{2} e f g p x^{2} - d^{2} e f^{2} p\right )} \log \left (e x^{2} + d\right ) - {\left (3 \, d^{2} e g^{2} x^{4} - 6 \, d^{2} e f g x^{2} - d^{2} e f^{2}\right )} \log \relax (c)}{3 \, d^{2} e x^{3}}\right ] \]

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^4,x, algorithm="fricas")

[Out]

[-1/3*(6*d^2*e*g^2*p*x^4 + 2*d*e^2*f^2*p*x^2 - (e^2*f^2 - 6*d*e*f*g - 3*d^2*g^2)*sqrt(-d*e)*p*x^3*log((e*x^2 -
2*sqrt(-d*e)*x - d)/(e*x^2 + d)) - (3*d^2*e*g^2*p*x^4 - 6*d^2*e*f*g*p*x^2 - d^2*e*f^2*p)*log(e*x^2 + d) - (3*
d^2*e*g^2*x^4 - 6*d^2*e*f*g*x^2 - d^2*e*f^2)*log(c))/(d^2*e*x^3), -1/3*(6*d^2*e*g^2*p*x^4 + 2*d*e^2*f^2*p*x^2
+ 2*(e^2*f^2 - 6*d*e*f*g - 3*d^2*g^2)*sqrt(d*e)*p*x^3*arctan(sqrt(d*e)*x/d) - (3*d^2*e*g^2*p*x^4 - 6*d^2*e*f*g
*p*x^2 - d^2*e*f^2*p)*log(e*x^2 + d) - (3*d^2*e*g^2*x^4 - 6*d^2*e*f*g*x^2 - d^2*e*f^2)*log(c))/(d^2*e*x^3)]

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giac [A]  time = 0.18, size = 154, normalized size = 0.91 \[ \frac {2 \, {\left (3 \, d^{2} g^{2} p + 6 \, d f g p e - f^{2} p e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{3 \, d^{\frac {3}{2}}} + \frac {3 \, d g^{2} p x^{4} \log \left (x^{2} e + d\right ) - 6 \, d g^{2} p x^{4} + 3 \, d g^{2} x^{4} \log \relax (c) - 6 \, d f g p x^{2} \log \left (x^{2} e + d\right ) - 2 \, f^{2} p x^{2} e - 6 \, d f g x^{2} \log \relax (c) - d f^{2} p \log \left (x^{2} e + d\right ) - d f^{2} \log \relax (c)}{3 \, d x^{3}} \]

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^4,x, algorithm="giac")

[Out]

2/3*(3*d^2*g^2*p + 6*d*f*g*p*e - f^2*p*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/d^(3/2) + 1/3*(3*d*g^2*p*x^4*lo
g(x^2*e + d) - 6*d*g^2*p*x^4 + 3*d*g^2*x^4*log(c) - 6*d*f*g*p*x^2*log(x^2*e + d) - 2*f^2*p*x^2*e - 6*d*f*g*x^2
*log(c) - d*f^2*p*log(x^2*e + d) - d*f^2*log(c))/(d*x^3)

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maple [C]  time = 0.62, size = 740, normalized size = 4.38 \[ -\frac {\left (-3 g^{2} x^{4}+6 f g \,x^{2}+f^{2}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{3 x^{3}}+\frac {-3 i \pi \,d^{2} e \,g^{2} x^{4} \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 i \pi \,d^{2} e \,g^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}+3 i \pi \,d^{2} e \,g^{2} x^{4} \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 i \pi \,d^{2} e \,g^{2} x^{4} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}+6 i \pi \,d^{2} e f g \,x^{2} \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 )-6 i \pi \,d^{2} e f g \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}-6 i \pi \,d^{2} e f g \,x^{2} \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 i \pi \,d^{2} e f g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}-12 d^{2} e \,g^{2} p \,x^{4}+6 d^{2} e \,g^{2} x^{4} \ln \relax (c )-6 \sqrt {-d e}\, d^{2} g^{2} p \,x^{3} \ln \left (-d -\sqrt {-d e}\, x \right )+6 \sqrt {-d e}\, d^{2} g^{2} p \,x^{3} \ln \left (d -\sqrt {-d e}\, x \right )-12 \sqrt {-d e}\, d e f g p \,x^{3} \ln \left (-d -\sqrt {-d e}\, x \right )+12 \sqrt {-d e}\, d e f g p \,x^{3} \ln \left (d -\sqrt {-d e}\, x \right )+2 \sqrt {-d e}\, e^{2} f^{2} p \,x^{3} \ln \left (-d -\sqrt {-d e}\, x \right )-2 \sqrt {-d e}\, e^{2} f^{2} p \,x^{3} \ln \left (d -\sqrt {-d e}\, x \right )+i \pi \,d^{2} e \,f^{2} \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 )-i \pi \,d^{2} e \,f^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}-i \pi \,d^{2} e \,f^{2} \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}+i \pi \,d^{2} e \,f^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}-12 d^{2} e f g \,x^{2} \ln \relax (c )-4 d \,e^{2} f^{2} p \,x^{2}-2 d^{2} e \,f^{2} \ln \relax (c )}{6 d^{2} e \,x^{3}} \]

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^4,x)

[Out]

-1/3*(-3*g^2*x^4+6*f*g*x^2+f^2)/x^3*ln((e*x^2+d)^p)+1/6*(-3*I*Pi*g^2*x^4*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d
)^p)*csgn(I*c)*e*d^2-3*I*Pi*g^2*x^4*csgn(I*c*(e*x^2+d)^p)^3*e*d^2+6*I*Pi*d^2*e*f*g*x^2*csgn(I*c*(e*x^2+d)^p)^3
+I*Pi*d^2*e*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+6*I*Pi*d^2*e*f*g*x^2*csgn(I*(e*x^2+d)^p)*c
sgn(I*c*(e*x^2+d)^p)*csgn(I*c)+I*Pi*d^2*e*f^2*csgn(I*c*(e*x^2+d)^p)^3-6*I*Pi*d^2*e*f*g*x^2*csgn(I*(e*x^2+d)^p)
*csgn(I*c*(e*x^2+d)^p)^2-6*I*Pi*d^2*e*f*g*x^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+6*ln(c)*g^2*x^4*e*d^2-I*Pi*d^2
*e*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+3*I*Pi*g^2*x^4*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*e*d^2-I*Pi
*d^2*e*f^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+3*I*Pi*g^2*x^4*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*e*d^2+
6*(-d*e)^(1/2)*p*ln(d-(-d*e)^(1/2)*x)*g^2*d^2*x^3+12*(-d*e)^(1/2)*p*ln(d-(-d*e)^(1/2)*x)*f*g*e*d*x^3-2*(-d*e)^
(1/2)*e^2*p*ln(d-(-d*e)^(1/2)*x)*f^2*x^3-6*(-d*e)^(1/2)*p*ln(-d-(-d*e)^(1/2)*x)*g^2*d^2*x^3-12*(-d*e)^(1/2)*p*
ln(-d-(-d*e)^(1/2)*x)*f*g*e*d*x^3+2*(-d*e)^(1/2)*e^2*p*ln(-d-(-d*e)^(1/2)*x)*f^2*x^3-12*d^2*e*g^2*p*x^4-12*ln(
c)*d^2*e*f*g*x^2-4*d*e^2*f^2*p*x^2-2*ln(c)*d^2*e*f^2)/e/d^2/x^3

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maxima [A]  time = 1.01, size = 105, normalized size = 0.62 \[ -\frac {2}{3} \, {\left (\frac {3 \, g^{2} x}{e} + \frac {f^{2}}{d x} + \frac {{\left (e^{2} f^{2} - 6 \, d e f g - 3 \, d^{2} g^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e} d e}\right )} e p + \frac {1}{3} \, {\left (3 \, g^{2} x - \frac {6 \, f g x^{2} + f^{2}}{x^{3}}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \]

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^4,x, algorithm="maxima")

[Out]

-2/3*(3*g^2*x/e + f^2/(d*x) + (e^2*f^2 - 6*d*e*f*g - 3*d^2*g^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d*e))*e*p + 1
/3*(3*g^2*x - (6*f*g*x^2 + f^2)/x^3)*log((e*x^2 + d)^p*c)

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mupad [B]  time = 0.39, size = 108, normalized size = 0.64 \[ \ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {8\,g^2\,x}{3}-\frac {\frac {f^2}{3}+2\,f\,g\,x^2+\frac {5\,g^2\,x^4}{3}}{x^3}\right )-2\,g^2\,p\,x+\frac {2\,p\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (3\,d^2\,g^2+6\,d\,e\,f\,g-e^2\,f^2\right )}{3\,d^{3/2}\,\sqrt {e}}-\frac {2\,e\,f^2\,p}{3\,d\,x} \]

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^4,x)

[Out]

log(c*(d + e*x^2)^p)*((8*g^2*x)/3 - (f^2/3 + (5*g^2*x^4)/3 + 2*f*g*x^2)/x^3) - 2*g^2*p*x + (2*p*atan((e^(1/2)*
x)/d^(1/2))*(3*d^2*g^2 - e^2*f^2 + 6*d*e*f*g))/(3*d^(3/2)*e^(1/2)) - (2*e*f^2*p)/(3*d*x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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**4,x)

[Out]

Timed out

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