\(\int \frac {(f+g x^2)^2 \log (c (d+e x^2)^p)}{x^8} \, dx\) [337]

Optimal result

Integrand size = 25, antiderivative size = 252 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=-\frac {2 e f^2 p}{35 d x^5}+\frac {2 e^2 f^2 p}{21 d^2 x^3}-\frac {4 e f g p}{15 d x^3}-\frac {2 e^3 f^2 p}{7 d^3 x}+\frac {4 e^2 f g p}{5 d^2 x}-\frac {2 e g^2 p}{3 d x}-\frac {2 e^{7/2} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 d^{7/2}}+\frac {4 e^{5/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {2 e^{3/2} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3} \]

[Out]

-2/35*e*f^2*p/d/x^5+2/21*e^2*f^2*p/d^2/x^3-4/15*e*f*g*p/d/x^3-2/7*e^3*f^2*p/d^3/x+4/5*e^2*f*g*p/d^2/x-2/3*e*g^
2*p/d/x-2/7*e^(7/2)*f^2*p*arctan(x*e^(1/2)/d^(1/2))/d^(7/2)+4/5*e^(5/2)*f*g*p*arctan(x*e^(1/2)/d^(1/2))/d^(5/2
)-2/3*e^(3/2)*g^2*p*arctan(x*e^(1/2)/d^(1/2))/d^(3/2)-1/7*f^2*ln(c*(e*x^2+d)^p)/x^7-2/5*f*g*ln(c*(e*x^2+d)^p)/
x^5-1/3*g^2*ln(c*(e*x^2+d)^p)/x^3

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2526, 2505, 331, 211} \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=-\frac {2 e^{7/2} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 d^{7/2}}+\frac {4 e^{5/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {2 e^{3/2} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {2 e^3 f^2 p}{7 d^3 x}+\frac {2 e^2 f^2 p}{21 d^2 x^3}+\frac {4 e^2 f g p}{5 d^2 x}-\frac {2 e f^2 p}{35 d x^5}-\frac {4 e f g p}{15 d x^3}-\frac {2 e g^2 p}{3 d x} \]

[In]

Int[((f + g*x^2)^2*Log[c*(d + e*x^2)^p])/x^8,x]

[Out]

(-2*e*f^2*p)/(35*d*x^5) + (2*e^2*f^2*p)/(21*d^2*x^3) - (4*e*f*g*p)/(15*d*x^3) - (2*e^3*f^2*p)/(7*d^3*x) + (4*e
^2*f*g*p)/(5*d^2*x) - (2*e*g^2*p)/(3*d*x) - (2*e^(7/2)*f^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(7*d^(7/2)) + (4*e^(
5/2)*f*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(5*d^(5/2)) - (2*e^(3/2)*g^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*d^(3/2)
) - (f^2*Log[c*(d + e*x^2)^p])/(7*x^7) - (2*f*g*Log[c*(d + e*x^2)^p])/(5*x^5) - (g^2*Log[c*(d + e*x^2)^p])/(3*
x^3)

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 331

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 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*} \text {integral}& = \int \left (\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8}+\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{x^6}+\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^4}\right ) \, dx \\ & = f^2 \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx+(2 f g) \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx+g^2 \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^4} \, dx \\ & = -\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}+\frac {1}{7} \left (2 e f^2 p\right ) \int \frac {1}{x^6 \left (d+e x^2\right )} \, dx+\frac {1}{5} (4 e f g p) \int \frac {1}{x^4 \left (d+e x^2\right )} \, dx+\frac {1}{3} \left (2 e g^2 p\right ) \int \frac {1}{x^2 \left (d+e x^2\right )} \, dx \\ & = -\frac {2 e f^2 p}{35 d x^5}-\frac {4 e f g p}{15 d x^3}-\frac {2 e g^2 p}{3 d x}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {\left (2 e^2 f^2 p\right ) \int \frac {1}{x^4 \left (d+e x^2\right )} \, dx}{7 d}-\frac {\left (4 e^2 f g p\right ) \int \frac {1}{x^2 \left (d+e x^2\right )} \, dx}{5 d}-\frac {\left (2 e^2 g^2 p\right ) \int \frac {1}{d+e x^2} \, dx}{3 d} \\ & = -\frac {2 e f^2 p}{35 d x^5}+\frac {2 e^2 f^2 p}{21 d^2 x^3}-\frac {4 e f g p}{15 d x^3}+\frac {4 e^2 f g p}{5 d^2 x}-\frac {2 e g^2 p}{3 d x}-\frac {2 e^{3/2} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}+\frac {\left (2 e^3 f^2 p\right ) \int \frac {1}{x^2 \left (d+e x^2\right )} \, dx}{7 d^2}+\frac {\left (4 e^3 f g p\right ) \int \frac {1}{d+e x^2} \, dx}{5 d^2} \\ & = -\frac {2 e f^2 p}{35 d x^5}+\frac {2 e^2 f^2 p}{21 d^2 x^3}-\frac {4 e f g p}{15 d x^3}-\frac {2 e^3 f^2 p}{7 d^3 x}+\frac {4 e^2 f g p}{5 d^2 x}-\frac {2 e g^2 p}{3 d x}+\frac {4 e^{5/2} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {2 e^{3/2} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {\left (2 e^4 f^2 p\right ) \int \frac {1}{d+e x^2} \, dx}{7 d^3} \\ & = -\frac {2 e f^2 p}{35 d x^5}+\frac {2 e^2 f^2 p}{21 d^2 x^3}-\frac {4 e f g p}{15 d x^3}-\frac {2 e^3 f^2 p}{7 d^3 x}+\frac {4 e^2 f g p}{5 d^2 x}-\frac {2 e g^2 p}{3 d x}-\frac {2 e^{7/2} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 d^{7/2}}+\frac {4 e^{5/2} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {2 e^{3/2} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.64 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=-\frac {2 e f^2 p \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\frac {e x^2}{d}\right )}{35 d x^5}-\frac {4 e f g p \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {e x^2}{d}\right )}{15 d x^3}-\frac {2 e g^2 p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {e x^2}{d}\right )}{3 d x}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3} \]

[In]

Integrate[((f + g*x^2)^2*Log[c*(d + e*x^2)^p])/x^8,x]

[Out]

(-2*e*f^2*p*Hypergeometric2F1[-5/2, 1, -3/2, -((e*x^2)/d)])/(35*d*x^5) - (4*e*f*g*p*Hypergeometric2F1[-3/2, 1,
-1/2, -((e*x^2)/d)])/(15*d*x^3) - (2*e*g^2*p*Hypergeometric2F1[-1/2, 1, 1/2, -((e*x^2)/d)])/(3*d*x) - (f^2*Lo
g[c*(d + e*x^2)^p])/(7*x^7) - (2*f*g*Log[c*(d + e*x^2)^p])/(5*x^5) - (g^2*Log[c*(d + e*x^2)^p])/(3*x^3)

Maple [A] (verified)

Time = 2.39 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.66

method	result	size
parts	\(-\frac {g^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{3 x^{3}}-\frac {2 f g \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{5 x^{5}}-\frac {f^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{7 x^{7}}-\frac {2 p e \left (-\frac {-35 g^{2} d^{2}+42 d e f g -15 e^{2} f^{2}}{d^{3} x}+\frac {3 f^{2}}{d \,x^{5}}+\frac {f \left (14 d g -5 e f \right )}{d^{2} x^{3}}+\frac {e \left (35 g^{2} d^{2}-42 d e f g +15 e^{2} f^{2}\right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{d^{3} \sqrt {d e}}\right )}{105}\)	\(167\)
risch	\(-\frac {\left (35 g^{2} x^{4}+42 f g \,x^{2}+15 f^{2}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{105 x^{7}}-\frac {30 \ln \left (c \right ) d^{4} f^{2}-70 \sqrt {-d e}\, p e \ln \left (-e x +\sqrt {-d e}\right ) g^{2} d^{2} x^{7}+70 \sqrt {-d e}\, p e \ln \left (-e x -\sqrt {-d e}\right ) g^{2} d^{2} x^{7}+84 \sqrt {-d e}\, p \,e^{2} \ln \left (-e x +\sqrt {-d e}\right ) f g d \,x^{7}-84 \sqrt {-d e}\, p \,e^{2} \ln \left (-e x -\sqrt {-d e}\right ) f g d \,x^{7}+30 \sqrt {-d e}\, p \,e^{3} \ln \left (-e x -\sqrt {-d e}\right ) f^{2} x^{7}-35 i \pi \,d^{4} g^{2} x^{4} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )+140 d^{3} e \,g^{2} p \,x^{6}+60 d \,e^{3} f^{2} p \,x^{6}-20 d^{2} e^{2} f^{2} p \,x^{4}+12 d^{3} e \,f^{2} p \,x^{2}-168 d^{2} e^{2} f g p \,x^{6}+56 d^{3} e f g p \,x^{4}+84 \ln \left (c \right ) d^{4} f g \,x^{2}-42 i \pi \,d^{4} f g \,x^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )+42 i \pi \,d^{4} f g \,x^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}+42 i \pi \,d^{4} f g \,x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+70 \ln \left (c \right ) d^{4} g^{2} x^{4}-30 \sqrt {-d e}\, p \,e^{3} \ln \left (-e x +\sqrt {-d e}\right ) f^{2} x^{7}-15 i \pi \,d^{4} f^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )+35 i \pi \,d^{4} g^{2} x^{4} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+35 i \pi \,d^{4} g^{2} x^{4} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}-42 i \pi \,d^{4} f g \,x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}-15 i \pi \,d^{4} f^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}-35 i \pi \,d^{4} g^{2} x^{4} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}+15 i \pi \,d^{4} f^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}+15 i \pi \,d^{4} f^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{210 d^{4} x^{7}}\)	\(784\)

[In]

int((g*x^2+f)^2*ln(c*(e*x^2+d)^p)/x^8,x,method=_RETURNVERBOSE)

[Out]

-1/3*g^2*ln(c*(e*x^2+d)^p)/x^3-2/5*f*g*ln(c*(e*x^2+d)^p)/x^5-1/7*f^2*ln(c*(e*x^2+d)^p)/x^7-2/105*p*e*(-1/d^3*(
-35*d^2*g^2+42*d*e*f*g-15*e^2*f^2)/x+3*f^2/d/x^5+f*(14*d*g-5*e*f)/d^2/x^3+e*(35*d^2*g^2-42*d*e*f*g+15*e^2*f^2)
/d^3/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.70 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=\left [\frac {{\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{7} \sqrt {-\frac {e}{d}} \log \left (\frac {e x^{2} - 2 \, d x \sqrt {-\frac {e}{d}} - d}{e x^{2} + d}\right ) - 6 \, d^{2} e f^{2} p x^{2} - 2 \, {\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{6} + 2 \, {\left (5 \, d e^{2} f^{2} - 14 \, d^{2} e f g\right )} p x^{4} - {\left (35 \, d^{3} g^{2} p x^{4} + 42 \, d^{3} f g p x^{2} + 15 \, d^{3} f^{2} p\right )} \log \left (e x^{2} + d\right ) - {\left (35 \, d^{3} g^{2} x^{4} + 42 \, d^{3} f g x^{2} + 15 \, d^{3} f^{2}\right )} \log \left (c\right )}{105 \, d^{3} x^{7}}, -\frac {2 \, {\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{7} \sqrt {\frac {e}{d}} \arctan \left (x \sqrt {\frac {e}{d}}\right ) + 6 \, d^{2} e f^{2} p x^{2} + 2 \, {\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{6} - 2 \, {\left (5 \, d e^{2} f^{2} - 14 \, d^{2} e f g\right )} p x^{4} + {\left (35 \, d^{3} g^{2} p x^{4} + 42 \, d^{3} f g p x^{2} + 15 \, d^{3} f^{2} p\right )} \log \left (e x^{2} + d\right ) + {\left (35 \, d^{3} g^{2} x^{4} + 42 \, d^{3} f g x^{2} + 15 \, d^{3} f^{2}\right )} \log \left (c\right )}{105 \, d^{3} x^{7}}\right ] \]

[In]

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^8,x, algorithm="fricas")

[Out]

[1/105*((15*e^3*f^2 - 42*d*e^2*f*g + 35*d^2*e*g^2)*p*x^7*sqrt(-e/d)*log((e*x^2 - 2*d*x*sqrt(-e/d) - d)/(e*x^2
+ d)) - 6*d^2*e*f^2*p*x^2 - 2*(15*e^3*f^2 - 42*d*e^2*f*g + 35*d^2*e*g^2)*p*x^6 + 2*(5*d*e^2*f^2 - 14*d^2*e*f*g
)*p*x^4 - (35*d^3*g^2*p*x^4 + 42*d^3*f*g*p*x^2 + 15*d^3*f^2*p)*log(e*x^2 + d) - (35*d^3*g^2*x^4 + 42*d^3*f*g*x
^2 + 15*d^3*f^2)*log(c))/(d^3*x^7), -1/105*(2*(15*e^3*f^2 - 42*d*e^2*f*g + 35*d^2*e*g^2)*p*x^7*sqrt(e/d)*arcta
n(x*sqrt(e/d)) + 6*d^2*e*f^2*p*x^2 + 2*(15*e^3*f^2 - 42*d*e^2*f*g + 35*d^2*e*g^2)*p*x^6 - 2*(5*d*e^2*f^2 - 14*
d^2*e*f*g)*p*x^4 + (35*d^3*g^2*p*x^4 + 42*d^3*f*g*p*x^2 + 15*d^3*f^2*p)*log(e*x^2 + d) + (35*d^3*g^2*x^4 + 42*
d^3*f*g*x^2 + 15*d^3*f^2)*log(c))/(d^3*x^7)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=\text {Timed out} \]

[In]

integrate((g*x**2+f)**2*ln(c*(e*x**2+d)**p)/x**8,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^8,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.82 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=-\frac {2 \, {\left (15 \, e^{4} f^{2} p - 42 \, d e^{3} f g p + 35 \, d^{2} e^{2} g^{2} p\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{105 \, \sqrt {d e} d^{3}} - \frac {{\left (35 \, g^{2} p x^{4} + 42 \, f g p x^{2} + 15 \, f^{2} p\right )} \log \left (e x^{2} + d\right )}{105 \, x^{7}} - \frac {30 \, e^{3} f^{2} p x^{6} - 84 \, d e^{2} f g p x^{6} + 70 \, d^{2} e g^{2} p x^{6} - 10 \, d e^{2} f^{2} p x^{4} + 28 \, d^{2} e f g p x^{4} + 35 \, d^{3} g^{2} x^{4} \log \left (c\right ) + 6 \, d^{2} e f^{2} p x^{2} + 42 \, d^{3} f g x^{2} \log \left (c\right ) + 15 \, d^{3} f^{2} \log \left (c\right )}{105 \, d^{3} x^{7}} \]

[In]

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^8,x, algorithm="giac")

[Out]

-2/105*(15*e^4*f^2*p - 42*d*e^3*f*g*p + 35*d^2*e^2*g^2*p)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^3) - 1/105*(35*g^
2*p*x^4 + 42*f*g*p*x^2 + 15*f^2*p)*log(e*x^2 + d)/x^7 - 1/105*(30*e^3*f^2*p*x^6 - 84*d*e^2*f*g*p*x^6 + 70*d^2*
e*g^2*p*x^6 - 10*d*e^2*f^2*p*x^4 + 28*d^2*e*f*g*p*x^4 + 35*d^3*g^2*x^4*log(c) + 6*d^2*e*f^2*p*x^2 + 42*d^3*f*g
*x^2*log(c) + 15*d^3*f^2*log(c))/(d^3*x^7)

Mupad [B] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.59 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=-\frac {\frac {6\,e\,f^2\,p}{d}+\frac {2\,e\,p\,x^4\,\left (35\,d^2\,g^2-42\,d\,e\,f\,g+15\,e^2\,f^2\right )}{d^3}+\frac {2\,e\,f\,p\,x^2\,\left (14\,d\,g-5\,e\,f\right )}{d^2}}{105\,x^5}-\frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {f^2}{7}+\frac {2\,f\,g\,x^2}{5}+\frac {g^2\,x^4}{3}\right )}{x^7}-\frac {2\,e^{3/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (35\,d^2\,g^2-42\,d\,e\,f\,g+15\,e^2\,f^2\right )}{105\,d^{7/2}} \]

[In]

int((log(c*(d + e*x^2)^p)*(f + g*x^2)^2)/x^8,x)

[Out]

- ((6*e*f^2*p)/d + (2*e*p*x^4*(35*d^2*g^2 + 15*e^2*f^2 - 42*d*e*f*g))/d^3 + (2*e*f*p*x^2*(14*d*g - 5*e*f))/d^2
)/(105*x^5) - (log(c*(d + e*x^2)^p)*(f^2/7 + (g^2*x^4)/3 + (2*f*g*x^2)/5))/x^7 - (2*e^(3/2)*p*atan((e^(1/2)*x)
/d^(1/2))*(35*d^2*g^2 + 15*e^2*f^2 - 42*d*e*f*g))/(105*d^(7/2))