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

Optimal. Leaf size=252 \[ -\frac {2 e f^2 p}{35 d x^5}+\frac {2 e^2 f^2 p}{21 d^2 x^3}-\frac {4 e f g p}{15 d x^3}-\frac {2 e^3 f^2 p}{7 d^3 x}+\frac {4 e^2 f g p}{5 d^2 x}-\frac {2 e g^2 p}{3 d x}-\frac {2 e^{7/2} f^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 d^{7/2}}+\frac {4 e^{5/2} f g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {2 e^{3/2} g^2 p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3} \]

[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

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Rubi [A]
time = 0.14, antiderivative size = 252, 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, 331, 211} \begin {gather*} -\frac {2 e^{7/2} f^2 p \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 d^{7/2}}+\frac {4 e^{5/2} f g p \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {2 e^{3/2} g^2 p \text {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} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx &=\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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 161, normalized size = 0.64 \begin {gather*} -\frac {2 e f^2 p \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-\frac {e x^2}{d}\right )}{35 d x^5}-\frac {4 e f g p \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\frac {e x^2}{d}\right )}{15 d x^3}-\frac {2 e g^2 p \, _2F_1\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} \end {gather*}

Antiderivative was successfully verified.

[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)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.43, size = 784, normalized size = 3.11

method	result	size
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 {-35 i \pi \,d^{4} 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 ) \mathrm {csgn}\left (i c \right )+42 i \pi \,d^{4} 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}+42 i \pi \,d^{4} f g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )+70 \ln \left (c \right ) d^{4} g^{2} x^{4}+30 \ln \left (c \right ) d^{4} f^{2}-15 i \pi \,d^{4} f^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}+84 \ln \left (c \right ) d^{4} f g \,x^{2}+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}-42 i \pi \,d^{4} 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 ) \mathrm {csgn}\left (i c \right )+35 i \pi \,d^{4} g^{2} x^{4} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )-42 i \pi \,d^{4} f g \,x^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}-15 i \pi \,d^{4} 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 ) \mathrm {csgn}\left (i c \right )+35 i \pi \,d^{4} 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}-35 i \pi \,d^{4} g^{2} x^{4} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}+15 i \pi \,d^{4} 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^{4} f^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )-84 \sqrt {-e d}\, p \,e^{2} \ln \left (-e x -\sqrt {-e d}\right ) f g d \,x^{7}+84 \sqrt {-e d}\, p \,e^{2} \ln \left (-e x +\sqrt {-e d}\right ) f g d \,x^{7}+30 \sqrt {-e d}\, p \,e^{3} \ln \left (-e x -\sqrt {-e d}\right ) f^{2} x^{7}-30 \sqrt {-e d}\, p \,e^{3} \ln \left (-e x +\sqrt {-e d}\right ) f^{2} x^{7}+70 \sqrt {-e d}\, p e \ln \left (-e x -\sqrt {-e d}\right ) g^{2} d^{2} x^{7}-70 \sqrt {-e d}\, p e \ln \left (-e x +\sqrt {-e d}\right ) g^{2} d^{2} x^{7}}{210 d^{4} x^{7}}\)	\(784\)

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^8,x,method=_RETURNVERBOSE)

[Out]

-1/105*(35*g^2*x^4+42*f*g*x^2+15*f^2)/x^7*ln((e*x^2+d)^p)-1/210*(70*ln(c)*d^4*g^2*x^4+30*ln(c)*d^4*f^2+84*ln(c
)*d^4*f*g*x^2+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+42*I*Pi*d^4*f*g*x^2*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-84*(-e*d)^(1/2)*p*e^2*ln(-e*x-(-
e*d)^(1/2))*f*g*d*x^7+84*(-e*d)^(1/2)*p*e^2*ln(-e*x+(-e*d)^(1/2))*f*g*d*x^7-35*I*Pi*d^4*g^2*x^4*csgn(I*(e*x^2+
d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+42*I*Pi*d^4*f*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+30*(-e*d
)^(1/2)*p*e^3*ln(-e*x-(-e*d)^(1/2))*f^2*x^7-30*(-e*d)^(1/2)*p*e^3*ln(-e*x+(-e*d)^(1/2))*f^2*x^7-35*I*Pi*d^4*g^
2*x^4*csgn(I*c*(e*x^2+d)^p)^3+15*I*Pi*d^4*f^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+15*I*Pi*d^4*f^2*csgn
(I*c*(e*x^2+d)^p)^2*csgn(I*c)-42*I*Pi*d^4*f*g*x^2*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-15*I*Pi*
d^4*f^2*csgn(I*c*(e*x^2+d)^p)^3+35*I*Pi*d^4*g^2*x^4*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2+35*I*Pi*d^4*g^
2*x^4*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)-42*I*Pi*d^4*f*g*x^2*csgn(I*c*(e*x^2+d)^p)^3-15*I*Pi*d^4*f^2*csgn(I*(e*
x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)+70*(-e*d)^(1/2)*p*e*ln(-e*x-(-e*d)^(1/2))*g^2*d^2*x^7-70*(-e*d)^(1/2
)*p*e*ln(-e*x+(-e*d)^(1/2))*g^2*d^2*x^7)/d^4/x^7

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Maxima [A]
time = 0.56, size = 148, normalized size = 0.59 \begin {gather*} -\frac {2}{105} \, p {\left (\frac {{\left (35 \, d^{2} g^{2} e - 42 \, d f g e^{2} + 15 \, f^{2} e^{3}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{d^{\frac {7}{2}}} + \frac {{\left (35 \, d^{2} g^{2} - 42 \, d f g e + 15 \, f^{2} e^{2}\right )} x^{4} + 3 \, d^{2} f^{2} + {\left (14 \, d^{2} f g - 5 \, d f^{2} e\right )} x^{2}}{d^{3} x^{5}}\right )} e - \frac {{\left (35 \, g^{2} x^{4} + 42 \, f g x^{2} + 15 \, f^{2}\right )} \log \left ({\left (x^{2} e + d\right )}^{p} c\right )}{105 \, x^{7}} \end {gather*}

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^8,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 456, normalized size = 1.81 \begin {gather*} \left [-\frac {30 \, f^{2} p x^{6} e^{3} - {\left (35 \, d^{2} g^{2} p x^{7} e - 42 \, d f g p x^{7} e^{2} + 15 \, f^{2} p x^{7} e^{3}\right )} \sqrt {-\frac {e}{d}} \log \left (\frac {x^{2} e - 2 \, d x \sqrt {-\frac {e}{d}} - d}{x^{2} e + d}\right ) - 2 \, {\left (42 \, d f g p x^{6} + 5 \, d f^{2} p x^{4}\right )} e^{2} + 2 \, {\left (35 \, d^{2} g^{2} p x^{6} + 14 \, d^{2} f g p x^{4} + 3 \, d^{2} f^{2} p x^{2}\right )} e + {\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 (x^{2} e + 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 {30 \, f^{2} p x^{6} e^{3} + \frac {2 \, {\left (35 \, d^{2} g^{2} p x^{7} e - 42 \, d f g p x^{7} e^{2} + 15 \, f^{2} p x^{7} e^{3}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}}}{\sqrt {d}} - 2 \, {\left (42 \, d f g p x^{6} + 5 \, d f^{2} p x^{4}\right )} e^{2} + 2 \, {\left (35 \, d^{2} g^{2} p x^{6} + 14 \, d^{2} f g p x^{4} + 3 \, d^{2} f^{2} p x^{2}\right )} e + {\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 (x^{2} e + 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 ] \end {gather*}

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^8,x, algorithm="fricas")

[Out]

[-1/105*(30*f^2*p*x^6*e^3 - (35*d^2*g^2*p*x^7*e - 42*d*f*g*p*x^7*e^2 + 15*f^2*p*x^7*e^3)*sqrt(-e/d)*log((x^2*e
- 2*d*x*sqrt(-e/d) - d)/(x^2*e + d)) - 2*(42*d*f*g*p*x^6 + 5*d*f^2*p*x^4)*e^2 + 2*(35*d^2*g^2*p*x^6 + 14*d^2*
f*g*p*x^4 + 3*d^2*f^2*p*x^2)*e + (35*d^3*g^2*p*x^4 + 42*d^3*f*g*p*x^2 + 15*d^3*f^2*p)*log(x^2*e + 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*(30*f^2*p*x^6*e^3 + 2*(35*d^2*g^2*p*x^7*e -
42*d*f*g*p*x^7*e^2 + 15*f^2*p*x^7*e^3)*arctan(x*e^(1/2)/sqrt(d))*e^(1/2)/sqrt(d) - 2*(42*d*f*g*p*x^6 + 5*d*f^2
*p*x^4)*e^2 + 2*(35*d^2*g^2*p*x^6 + 14*d^2*f*g*p*x^4 + 3*d^2*f^2*p*x^2)*e + (35*d^3*g^2*p*x^4 + 42*d^3*f*g*p*x
^2 + 15*d^3*f^2*p)*log(x^2*e + 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)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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**8,x)

[Out]

Timed out

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Giac [A]
time = 6.53, size = 222, normalized size = 0.88 \begin {gather*} -\frac {2 \, {\left (35 \, d^{2} g^{2} p e^{2} - 42 \, d f g p e^{3} + 15 \, f^{2} p e^{4}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{105 \, d^{\frac {7}{2}}} - \frac {70 \, d^{2} g^{2} p x^{6} e - 84 \, d f g p x^{6} e^{2} + 35 \, d^{3} g^{2} p x^{4} \log \left (x^{2} e + d\right ) + 30 \, f^{2} p x^{6} e^{3} + 28 \, d^{2} f g p x^{4} e + 35 \, d^{3} g^{2} x^{4} \log \left (c\right ) - 10 \, d f^{2} p x^{4} e^{2} + 42 \, d^{3} f g p x^{2} \log \left (x^{2} e + d\right ) + 6 \, d^{2} f^{2} p x^{2} e + 42 \, d^{3} f g x^{2} \log \left (c\right ) + 15 \, d^{3} f^{2} p \log \left (x^{2} e + d\right ) + 15 \, d^{3} f^{2} \log \left (c\right )}{105 \, d^{3} x^{7}} \end {gather*}

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^8,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.43, size = 149, normalized size = 0.59 \begin {gather*} -\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}} \end {gather*}

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^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))

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