3.336 \(\int \frac {(f+g x^2)^2 \log (c (d+e x^2)^p)}{x^6} \, dx\)

Optimal. Leaf size=200 \[ -\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x}+\frac {2 e^{5/2} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {4 e^{3/2} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}+\frac {2 e^2 f^2 p}{5 d^2 x}-\frac {2 e f^2 p}{15 d x^3}-\frac {4 e f g p}{3 d x}+\frac {2 \sqrt {e} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2476, 2455, 325, 205} \[ -\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x}+\frac {2 e^2 f^2 p}{5 d^2 x}+\frac {2 e^{5/2} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {4 e^{3/2} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {2 e f^2 p}{15 d x^3}-\frac {4 e f g p}{3 d x}+\frac {2 \sqrt {e} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x^2)^2*Log[c*(d + e*x^2)^p])/x^6,x]

[Out]

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

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

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Mathematica [C]  time = 0.07, size = 156, normalized size = 0.78 \[ -\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x}-\frac {2 e f^2 p \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\frac {e x^2}{d}\right )}{15 d x^3}-\frac {4 e f g p \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {e x^2}{d}\right )}{3 d x}+\frac {2 \sqrt {e} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x^2)^2*Log[c*(d + e*x^2)^p])/x^6,x]

[Out]

(2*Sqrt[e]*g^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[d] - (2*e*f^2*p*Hypergeometric2F1[-3/2, 1, -1/2, -((e*x^2)/
d)])/(15*d*x^3) - (4*e*f*g*p*Hypergeometric2F1[-1/2, 1, 1/2, -((e*x^2)/d)])/(3*d*x) - (f^2*Log[c*(d + e*x^2)^p
])/(5*x^5) - (2*f*g*Log[c*(d + e*x^2)^p])/(3*x^3) - (g^2*Log[c*(d + e*x^2)^p])/x

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fricas [A]  time = 0.56, size = 351, normalized size = 1.76 \[ \left [\frac {{\left (3 \, e^{2} f^{2} - 10 \, d e f g + 15 \, d^{2} g^{2}\right )} p x^{5} \sqrt {-\frac {e}{d}} \log \left (\frac {e x^{2} + 2 \, d x \sqrt {-\frac {e}{d}} - d}{e x^{2} + d}\right ) - 2 \, d e f^{2} p x^{2} + 2 \, {\left (3 \, e^{2} f^{2} - 10 \, d e f g\right )} p x^{4} - {\left (15 \, d^{2} g^{2} p x^{4} + 10 \, d^{2} f g p x^{2} + 3 \, d^{2} f^{2} p\right )} \log \left (e x^{2} + d\right ) - {\left (15 \, d^{2} g^{2} x^{4} + 10 \, d^{2} f g x^{2} + 3 \, d^{2} f^{2}\right )} \log \relax (c)}{15 \, d^{2} x^{5}}, \frac {2 \, {\left (3 \, e^{2} f^{2} - 10 \, d e f g + 15 \, d^{2} g^{2}\right )} p x^{5} \sqrt {\frac {e}{d}} \arctan \left (x \sqrt {\frac {e}{d}}\right ) - 2 \, d e f^{2} p x^{2} + 2 \, {\left (3 \, e^{2} f^{2} - 10 \, d e f g\right )} p x^{4} - {\left (15 \, d^{2} g^{2} p x^{4} + 10 \, d^{2} f g p x^{2} + 3 \, d^{2} f^{2} p\right )} \log \left (e x^{2} + d\right ) - {\left (15 \, d^{2} g^{2} x^{4} + 10 \, d^{2} f g x^{2} + 3 \, d^{2} f^{2}\right )} \log \relax (c)}{15 \, d^{2} x^{5}}\right ] \]

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^6,x, algorithm="fricas")

[Out]

[1/15*((3*e^2*f^2 - 10*d*e*f*g + 15*d^2*g^2)*p*x^5*sqrt(-e/d)*log((e*x^2 + 2*d*x*sqrt(-e/d) - d)/(e*x^2 + d))
- 2*d*e*f^2*p*x^2 + 2*(3*e^2*f^2 - 10*d*e*f*g)*p*x^4 - (15*d^2*g^2*p*x^4 + 10*d^2*f*g*p*x^2 + 3*d^2*f^2*p)*log
(e*x^2 + d) - (15*d^2*g^2*x^4 + 10*d^2*f*g*x^2 + 3*d^2*f^2)*log(c))/(d^2*x^5), 1/15*(2*(3*e^2*f^2 - 10*d*e*f*g
+ 15*d^2*g^2)*p*x^5*sqrt(e/d)*arctan(x*sqrt(e/d)) - 2*d*e*f^2*p*x^2 + 2*(3*e^2*f^2 - 10*d*e*f*g)*p*x^4 - (15*
d^2*g^2*p*x^4 + 10*d^2*f*g*p*x^2 + 3*d^2*f^2*p)*log(e*x^2 + d) - (15*d^2*g^2*x^4 + 10*d^2*f*g*x^2 + 3*d^2*f^2)
*log(c))/(d^2*x^5)]

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giac [A]  time = 0.19, size = 181, normalized size = 0.90 \[ \frac {2 \, {\left (15 \, d^{2} g^{2} p e - 10 \, d f g p e^{2} + 3 \, f^{2} p e^{3}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{15 \, d^{\frac {5}{2}}} - \frac {15 \, d^{2} g^{2} p x^{4} \log \left (x^{2} e + d\right ) + 20 \, d f g p x^{4} e + 15 \, d^{2} g^{2} x^{4} \log \relax (c) - 6 \, f^{2} p x^{4} e^{2} + 10 \, d^{2} f g p x^{2} \log \left (x^{2} e + d\right ) + 2 \, d f^{2} p x^{2} e + 10 \, d^{2} f g x^{2} \log \relax (c) + 3 \, d^{2} f^{2} p \log \left (x^{2} e + d\right ) + 3 \, d^{2} f^{2} \log \relax (c)}{15 \, d^{2} x^{5}} \]

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^6,x, algorithm="giac")

[Out]

2/15*(15*d^2*g^2*p*e - 10*d*f*g*p*e^2 + 3*f^2*p*e^3)*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/d^(5/2) - 1/15*(15*d^2
*g^2*p*x^4*log(x^2*e + d) + 20*d*f*g*p*x^4*e + 15*d^2*g^2*x^4*log(c) - 6*f^2*p*x^4*e^2 + 10*d^2*f*g*p*x^2*log(
x^2*e + d) + 2*d*f^2*p*x^2*e + 10*d^2*f*g*x^2*log(c) + 3*d^2*f^2*p*log(x^2*e + d) + 3*d^2*f^2*log(c))/(d^2*x^5
)

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maple [C]  time = 0.80, size = 753, normalized size = 3.76 \[ -\frac {\left (15 g^{2} x^{4}+10 f g \,x^{2}+3 f^{2}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{15 x^{5}}+\frac {-3 i \pi \,d^{2} 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}-15 i \pi \,d^{2} g^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}-15 i \pi \,d^{2} 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}+10 i \pi \,d^{2} 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 )+10 i \pi \,d^{2} f g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}+15 i \pi \,d^{2} 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 )-10 i \pi \,d^{2} f g \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}-10 i \pi \,d^{2} 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}-30 d^{2} g^{2} x^{4} \ln \relax (c )-40 d e f g p \,x^{4}+12 e^{2} f^{2} p \,x^{4}+2 d^{2} x^{5} \RootOf \left (225 d^{4} e \,g^{4} p^{2}-300 d^{3} e^{2} f \,g^{3} p^{2}+190 d^{2} e^{3} f^{2} g^{2} p^{2}-60 d \,e^{4} f^{3} g \,p^{2}+9 e^{5} f^{4} p^{2}+d^{5} \textit {\_Z}^{2}\right ) \ln \left (\left (-15 d^{5} g^{2} p +10 d^{4} e f g p -3 d^{3} e^{2} f^{2} p \right ) \RootOf \left (225 d^{4} e \,g^{4} p^{2}-300 d^{3} e^{2} f \,g^{3} p^{2}+190 d^{2} e^{3} f^{2} g^{2} p^{2}-60 d \,e^{4} f^{3} g \,p^{2}+9 e^{5} f^{4} p^{2}+d^{5} \textit {\_Z}^{2}\right )+\left (450 d^{4} e \,g^{4} p^{2}-600 d^{3} e^{2} f \,g^{3} p^{2}+380 d^{2} e^{3} f^{2} g^{2} p^{2}-120 d \,e^{4} f^{3} g \,p^{2}+18 e^{5} f^{4} p^{2}+3 \RootOf \left (225 d^{4} e \,g^{4} p^{2}-300 d^{3} e^{2} f \,g^{3} p^{2}+190 d^{2} e^{3} f^{2} g^{2} p^{2}-60 d \,e^{4} f^{3} g \,p^{2}+9 e^{5} f^{4} p^{2}+d^{5} \textit {\_Z}^{2}\right )^{2} d^{5}\right ) x \right )+15 i \pi \,d^{2} g^{2} x^{4} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}+3 i \pi \,d^{2} 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 )-3 i \pi \,d^{2} f^{2} \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} f^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}-20 d^{2} f g \,x^{2} \ln \relax (c )-4 d e \,f^{2} p \,x^{2}-6 d^{2} f^{2} \ln \relax (c )}{30 d^{2} x^{5}} \]

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^6,x)

[Out]

-1/15*(15*g^2*x^4+10*f*g*x^2+3*f^2)/x^5*ln((e*x^2+d)^p)+1/30*(-3*I*Pi*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+
d)^p)^2*d^2-15*I*Pi*d^2*g^2*x^4*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-15*I*Pi*d^2*g^2*x^4*csgn(I*(e*x^2+d)^p)*csgn
(I*c*(e*x^2+d)^p)^2+10*I*Pi*d^2*f*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+10*I*Pi*d^2*f*g*x^
2*csgn(I*c*(e*x^2+d)^p)^3+15*I*Pi*d^2*g^2*x^4*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-10*I*Pi*d^2*
f*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-10*I*Pi*d^2*f*g*x^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-30*l
n(c)*d^2*g^2*x^4+15*I*Pi*d^2*g^2*x^4*csgn(I*c*(e*x^2+d)^p)^3+3*I*Pi*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)
^p)*csgn(I*c)*d^2-3*I*Pi*f^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*d^2+3*I*Pi*f^2*csgn(I*c*(e*x^2+d)^p)^3*d^2-40*d
*e*f*g*p*x^4+12*e^2*f^2*p*x^4+2*sum(_R*ln((450*d^4*e*g^4*p^2-600*d^3*e^2*f*g^3*p^2+380*d^2*e^3*f^2*g^2*p^2-120
*d*e^4*f^3*g*p^2+18*e^5*f^4*p^2+3*_R^2*d^5)*x+(-15*d^5*g^2*p+10*d^4*e*f*g*p-3*d^3*e^2*f^2*p)*_R),_R=RootOf(225
*d^4*e*g^4*p^2-300*d^3*e^2*f*g^3*p^2+190*d^2*e^3*f^2*g^2*p^2-60*d*e^4*f^3*g*p^2+9*e^5*f^4*p^2+_Z^2*d^5))*d^2*x
^5-20*ln(c)*d^2*f*g*x^2-4*d*e*f^2*p*x^2-6*ln(c)*f^2*d^2)/d^2/x^5

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maxima [A]  time = 1.03, size = 116, normalized size = 0.58 \[ \frac {2}{15} \, e p {\left (\frac {{\left (3 \, e^{2} f^{2} - 10 \, d e f g + 15 \, d^{2} g^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e} d^{2}} - \frac {d f^{2} - {\left (3 \, e f^{2} - 10 \, d f g\right )} x^{2}}{d^{2} x^{3}}\right )} - \frac {{\left (15 \, g^{2} x^{4} + 10 \, f g x^{2} + 3 \, f^{2}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{15 \, x^{5}} \]

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^6,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.39, size = 115, normalized size = 0.58 \[ \frac {2\,\sqrt {e}\,p\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (15\,d^2\,g^2-10\,d\,e\,f\,g+3\,e^2\,f^2\right )}{15\,d^{5/2}}-\frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {f^2}{5}+\frac {2\,f\,g\,x^2}{3}+g^2\,x^4\right )}{x^5}-\frac {\frac {2\,e\,f^2\,p}{d}+\frac {2\,e\,f\,p\,x^2\,\left (10\,d\,g-3\,e\,f\right )}{d^2}}{15\,x^3} \]

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^6,x)

[Out]

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

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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**6,x)

[Out]

Timed out

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