\(\int \frac {(f+g x^2)^2 \log (c (d+e x^2)^p)}{x^6} \, dx\) [336]

Optimal result

Integrand size = 25, antiderivative size = 200 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx=-\frac {2 e f^2 p}{15 d x^3}+\frac {2 e^2 f^2 p}{5 d^2 x}-\frac {4 e f g p}{3 d x}+\frac {2 e^{5/2} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {4 e^{3/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}+\frac {2 \sqrt {e} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x} \]

[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)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2526, 2505, 331, 211} \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx=\frac {2 e^{5/2} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {4 e^{3/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}+\frac {2 \sqrt {e} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x}+\frac {2 e^2 f^2 p}{5 d^2 x}-\frac {2 e f^2 p}{15 d x^3}-\frac {4 e f g p}{3 d x} \]

[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 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^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*}

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.78 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx=\frac {2 \sqrt {e} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}}-\frac {2 e f^2 p \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {e x^2}{d}\right )}{15 d x^3}-\frac {4 e f g p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {e x^2}{d}\right )}{3 d x}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x} \]

[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

Maple [A] (verified)

Time = 1.90 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.67

method	result	size
parts	\(-\frac {g^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{x}-\frac {2 f g \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{3 x^{3}}-\frac {f^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{5 x^{5}}-\frac {2 p e \left (\frac {f^{2}}{d \,x^{3}}+\frac {f \left (10 d g -3 e f \right )}{d^{2} x}+\frac {\left (-15 g^{2} d^{2}+10 d e f g -3 e^{2} f^{2}\right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{d^{2} \sqrt {d e}}\right )}{15}\)	\(134\)
risch	\(-\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 {-10 i \pi \,d^{3} f g \,x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}-15 i \pi \,d^{3} g^{2} x^{4} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}-3 i \pi \,d^{3} 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 )+3 i \pi \,d^{3} 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}-30 \sqrt {-d e}\, p \ln \left (-e x -\sqrt {-d e}\right ) g^{2} d^{2} x^{5}+20 \sqrt {-d e}\, p \ln \left (-e x -\sqrt {-d e}\right ) e f g d \,x^{5}-6 \sqrt {-d e}\, p \ln \left (-e x -\sqrt {-d e}\right ) e^{2} f^{2} x^{5}+30 \sqrt {-d e}\, p \ln \left (-e x +\sqrt {-d e}\right ) g^{2} d^{2} x^{5}-20 \sqrt {-d e}\, p \ln \left (-e x +\sqrt {-d e}\right ) e f g d \,x^{5}+6 \sqrt {-d e}\, p \ln \left (-e x +\sqrt {-d e}\right ) e^{2} f^{2} x^{5}-10 i \pi \,d^{3} 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 )+10 i \pi \,d^{3} 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}-15 i \pi \,d^{3} 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 )+3 i \pi \,d^{3} f^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+15 i \pi \,d^{3} g^{2} x^{4} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+10 i \pi \,d^{3} f g \,x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-3 i \pi \,d^{3} f^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}+15 i \pi \,d^{3} 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}+30 \ln \left (c \right ) d^{3} g^{2} x^{4}+40 d^{2} e f g p \,x^{4}-12 d \,e^{2} f^{2} p \,x^{4}+20 \ln \left (c \right ) d^{3} f g \,x^{2}+4 d^{2} e \,f^{2} p \,x^{2}+6 \ln \left (c \right ) d^{3} f^{2}}{30 d^{3} x^{5}}\)	\(736\)

[In]

int((g*x^2+f)^2*ln(c*(e*x^2+d)^p)/x^6,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.76 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx=\left [\frac {{\left (3 \, e^{2} f^{2} - 10 \, d e f g + 15 \, d^{2} g^{2}\right )} p x^{5} \sqrt {-\frac {e}{d}} \log \left (\frac {e x^{2} + 2 \, d x \sqrt {-\frac {e}{d}} - d}{e x^{2} + d}\right ) - 2 \, d e f^{2} p x^{2} + 2 \, {\left (3 \, e^{2} f^{2} - 10 \, d e f g\right )} p x^{4} - {\left (15 \, d^{2} g^{2} p x^{4} + 10 \, d^{2} f g p x^{2} + 3 \, d^{2} f^{2} p\right )} \log \left (e x^{2} + d\right ) - {\left (15 \, d^{2} g^{2} x^{4} + 10 \, d^{2} f g x^{2} + 3 \, d^{2} f^{2}\right )} \log \left (c\right )}{15 \, d^{2} x^{5}}, \frac {2 \, {\left (3 \, e^{2} f^{2} - 10 \, d e f g + 15 \, d^{2} g^{2}\right )} p x^{5} \sqrt {\frac {e}{d}} \arctan \left (x \sqrt {\frac {e}{d}}\right ) - 2 \, d e f^{2} p x^{2} + 2 \, {\left (3 \, e^{2} f^{2} - 10 \, d e f g\right )} p x^{4} - {\left (15 \, d^{2} g^{2} p x^{4} + 10 \, d^{2} f g p x^{2} + 3 \, d^{2} f^{2} p\right )} \log \left (e x^{2} + d\right ) - {\left (15 \, d^{2} g^{2} x^{4} + 10 \, d^{2} f g x^{2} + 3 \, d^{2} f^{2}\right )} \log \left (c\right )}{15 \, d^{2} x^{5}}\right ] \]

[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)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1603 vs. \(2 (199) = 398\).

Time = 156.42 (sec) , antiderivative size = 1603, normalized size of antiderivative = 8.02 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx=\text {Too large to display} \]

[In]

integrate((g*x**2+f)**2*ln(c*(e*x**2+d)**p)/x**6,x)

[Out]

Piecewise(((-f**2/(5*x**5) - 2*f*g/(3*x**3) - g**2/x)*log(0**p*c), Eq(d, 0) & Eq(e, 0)), ((-f**2/(5*x**5) - 2*
f*g/(3*x**3) - g**2/x)*log(c*d**p), Eq(e, 0)), (-2*f**2*p/(25*x**5) - f**2*log(c*(e*x**2)**p)/(5*x**5) - 4*f*g
*p/(9*x**3) - 2*f*g*log(c*(e*x**2)**p)/(3*x**3) - 2*g**2*p/x - g**2*log(c*(e*x**2)**p)/x, Eq(d, 0)), ((-f**2/(
5*x**5) - 2*f*g/(3*x**3) - g**2/x)*log(0**p*c), Eq(d, -e*x**2)), (-3*d**3*f**2*sqrt(-d/e)*log(c*(d + e*x**2)**
p)/(15*d**3*x**5*sqrt(-d/e) + 15*d**2*e*x**7*sqrt(-d/e)) - 10*d**3*f*g*x**2*sqrt(-d/e)*log(c*(d + e*x**2)**p)/
(15*d**3*x**5*sqrt(-d/e) + 15*d**2*e*x**7*sqrt(-d/e)) + 30*d**3*g**2*p*x**5*log(x - sqrt(-d/e))/(15*d**3*x**5*
sqrt(-d/e) + 15*d**2*e*x**7*sqrt(-d/e)) - 15*d**3*g**2*x**5*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e) +
15*d**2*e*x**7*sqrt(-d/e)) - 15*d**3*g**2*x**4*sqrt(-d/e)*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e) + 15
*d**2*e*x**7*sqrt(-d/e)) - 2*d**2*f**2*p*x**2*sqrt(-d/e)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e))
- 3*d**2*f**2*x**2*sqrt(-d/e)*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) -
20*d**2*f*g*p*x**5*log(x - sqrt(-d/e))/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 20*d**2*f*g*p*x
**4*sqrt(-d/e)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) + 10*d**2*f*g*x**5*log(c*(d + e*x**2)**p)
/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 10*d**2*f*g*x**4*sqrt(-d/e)*log(c*(d + e*x**2)**p)/(1
5*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) + 30*d**2*g**2*p*x**7*log(x - sqrt(-d/e))/(15*d**3*x**5*sq
rt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 15*d**2*g**2*x**7*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e)/e +
15*d**2*x**7*sqrt(-d/e)) - 15*d**2*g**2*x**6*sqrt(-d/e)*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e)/e + 15
*d**2*x**7*sqrt(-d/e)) + 6*d*e*f**2*p*x**5*log(x - sqrt(-d/e))/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(
-d/e)) + 4*d*e*f**2*p*x**4*sqrt(-d/e)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 3*d*e*f**2*x**5*
log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 20*d*e*f*g*p*x**7*log(x - sqrt(
-d/e))/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 20*d*e*f*g*p*x**6*sqrt(-d/e)/(15*d**3*x**5*sqrt
(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) + 10*d*e*f*g*x**7*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d
**2*x**7*sqrt(-d/e)) + 6*e**2*f**2*p*x**7*log(x - sqrt(-d/e))/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-
d/e)) + 6*e**2*f**2*p*x**6*sqrt(-d/e)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)) - 3*e**2*f**2*x**7
*log(c*(d + e*x**2)**p)/(15*d**3*x**5*sqrt(-d/e)/e + 15*d**2*x**7*sqrt(-d/e)), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^6,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 = 0.32 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.82 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx=\frac {2 \, {\left (3 \, e^{3} f^{2} p - 10 \, d e^{2} f g p + 15 \, d^{2} e g^{2} p\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{15 \, \sqrt {d e} d^{2}} - \frac {{\left (15 \, g^{2} p x^{4} + 10 \, f g p x^{2} + 3 \, f^{2} p\right )} \log \left (e x^{2} + d\right )}{15 \, x^{5}} + \frac {6 \, e^{2} f^{2} p x^{4} - 20 \, d e f g p x^{4} - 15 \, d^{2} g^{2} x^{4} \log \left (c\right ) - 2 \, d e f^{2} p x^{2} - 10 \, d^{2} f g x^{2} \log \left (c\right ) - 3 \, d^{2} f^{2} \log \left (c\right )}{15 \, d^{2} x^{5}} \]

[In]

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^6,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.61 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.58 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^6} \, dx=\frac {2\,\sqrt {e}\,p\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (15\,d^2\,g^2-10\,d\,e\,f\,g+3\,e^2\,f^2\right )}{15\,d^{5/2}}-\frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {f^2}{5}+\frac {2\,f\,g\,x^2}{3}+g^2\,x^4\right )}{x^5}-\frac {\frac {2\,e\,f^2\,p}{d}+\frac {2\,e\,f\,p\,x^2\,\left (10\,d\,g-3\,e\,f\right )}{d^2}}{15\,x^3} \]

[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)