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

Optimal. Leaf size=651 \[ \frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{5/2}}+\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 x}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{3 f x^3}-\frac {2 e^{3/2} p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2} f}+\frac {i g^{3/2} p \text {Li}_2\left (\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}-\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}+1\right )}{2 f^{5/2}}+\frac {i g^{3/2} p \text {Li}_2\left (1-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{2 f^{5/2}}-\frac {g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (-\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}\right )}\right )}{f^{5/2}}-\frac {g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}\right )}\right )}{f^{5/2}}-\frac {2 \sqrt {e} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} f^2}-\frac {2 e p}{3 d f x}-\frac {i g^{3/2} p \text {Li}_2\left (1-\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{f^{5/2}}+\frac {2 g^{3/2} p \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right ) \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{f^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.65, antiderivative size = 651, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {2476, 2455, 325, 205, 2470, 12, 4928, 4856, 2402, 2315, 2447} \[ \frac {i g^{3/2} p \text {PolyLog}\left (2,1+\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (-\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}\right )}\right )}{2 f^{5/2}}+\frac {i g^{3/2} p \text {PolyLog}\left (2,1-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}\right )}\right )}{2 f^{5/2}}-\frac {i g^{3/2} p \text {PolyLog}\left (2,1-\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{f^{5/2}}+\frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{5/2}}+\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 x}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{3 f x^3}-\frac {2 e^{3/2} p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2} f}-\frac {g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (-\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}\right )}\right )}{f^{5/2}}-\frac {g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}\right )}\right )}{f^{5/2}}-\frac {2 \sqrt {e} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} f^2}-\frac {2 e p}{3 d f x}+\frac {2 g^{3/2} p \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right ) \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{f^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*p)/(3*d*f*x) - (2*e^(3/2)*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(3*d^(3/2)*f) - (2*Sqrt[e]*g*p*ArcTan[(Sqrt[e]*
x)/Sqrt[d]])/(Sqrt[d]*f^2) + (2*g^(3/2)*p*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[(2*Sqrt[f])/(Sqrt[f] - I*Sqrt[g]*x)]
)/f^(5/2) - (g^(3/2)*p*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[(-2*Sqrt[f]*Sqrt[g]*(Sqrt[-d] - Sqrt[e]*x))/((I*Sqrt[e]
*Sqrt[f] - Sqrt[-d]*Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/f^(5/2) - (g^(3/2)*p*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[(
2*Sqrt[f]*Sqrt[g]*(Sqrt[-d] + Sqrt[e]*x))/((I*Sqrt[e]*Sqrt[f] + Sqrt[-d]*Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/f
^(5/2) - Log[c*(d + e*x^2)^p]/(3*f*x^3) + (g*Log[c*(d + e*x^2)^p])/(f^2*x) + (g^(3/2)*ArcTan[(Sqrt[g]*x)/Sqrt[
f]]*Log[c*(d + e*x^2)^p])/f^(5/2) - (I*g^(3/2)*p*PolyLog[2, 1 - (2*Sqrt[f])/(Sqrt[f] - I*Sqrt[g]*x)])/f^(5/2)
+ ((I/2)*g^(3/2)*p*PolyLog[2, 1 + (2*Sqrt[f]*Sqrt[g]*(Sqrt[-d] - Sqrt[e]*x))/((I*Sqrt[e]*Sqrt[f] - Sqrt[-d]*Sq
rt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/f^(5/2) + ((I/2)*g^(3/2)*p*PolyLog[2, 1 - (2*Sqrt[f]*Sqrt[g]*(Sqrt[-d] + Sqr
t[e]*x))/((I*Sqrt[e]*Sqrt[f] + Sqrt[-d]*Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/f^(5/2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, 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 2470

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_) + (g_.)*(x_)^2), x_Symbol] :> With[{u = In
tHide[1/(f + g*x^2), x]}, Simp[u*(a + b*Log[c*(d + e*x^n)^p]), x] - Dist[b*e*n*p, Int[(u*x^(n - 1))/(d + e*x^n
), x], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[n]

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]

Rule 4856

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])*Log[2/(1 -
I*c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d
+ e*x))/((c*d + I*e)*(1 - I*c*x))]/(1 + c^2*x^2), x], x] + Simp[((a + b*ArcTan[c*x])*Log[(2*c*(d + e*x))/((c*d
+ I*e)*(1 - I*c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 4928

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[a
+ b*ArcTan[c*x], x^m/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[m] &&  !(EqQ[m, 1] && NeQ[a,
0])

Rubi steps

\begin {align*} \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^4 \left (f+g x^2\right )} \, dx &=\int \left (\frac {\log \left (c \left (d+e x^2\right )^p\right )}{f x^4}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 x^2}+\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{f^2 \left (f+g x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^4} \, dx}{f}-\frac {g \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^2} \, dx}{f^2}+\frac {g^2 \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{f+g x^2} \, dx}{f^2}\\ &=-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{3 f x^3}+\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 x}+\frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{5/2}}+\frac {(2 e p) \int \frac {1}{x^2 \left (d+e x^2\right )} \, dx}{3 f}-\frac {(2 e g p) \int \frac {1}{d+e x^2} \, dx}{f^2}-\frac {\left (2 e g^2 p\right ) \int \frac {x \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{\sqrt {f} \sqrt {g} \left (d+e x^2\right )} \, dx}{f^2}\\ &=-\frac {2 e p}{3 d f x}-\frac {2 \sqrt {e} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} f^2}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{3 f x^3}+\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 x}+\frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{5/2}}-\frac {\left (2 e^2 p\right ) \int \frac {1}{d+e x^2} \, dx}{3 d f}-\frac {\left (2 e g^{3/2} p\right ) \int \frac {x \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{d+e x^2} \, dx}{f^{5/2}}\\ &=-\frac {2 e p}{3 d f x}-\frac {2 e^{3/2} p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2} f}-\frac {2 \sqrt {e} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} f^2}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{3 f x^3}+\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 x}+\frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{5/2}}-\frac {\left (2 e g^{3/2} p\right ) \int \left (-\frac {\tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {e} \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {e} \left (\sqrt {-d}+\sqrt {e} x\right )}\right ) \, dx}{f^{5/2}}\\ &=-\frac {2 e p}{3 d f x}-\frac {2 e^{3/2} p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2} f}-\frac {2 \sqrt {e} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} f^2}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{3 f x^3}+\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 x}+\frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{5/2}}+\frac {\left (\sqrt {e} g^{3/2} p\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{\sqrt {-d}-\sqrt {e} x} \, dx}{f^{5/2}}-\frac {\left (\sqrt {e} g^{3/2} p\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{\sqrt {-d}+\sqrt {e} x} \, dx}{f^{5/2}}\\ &=-\frac {2 e p}{3 d f x}-\frac {2 e^{3/2} p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2} f}-\frac {2 \sqrt {e} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} f^2}+\frac {2 g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{f^{5/2}}-\frac {g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}-\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{f^{5/2}}-\frac {g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{f^{5/2}}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{3 f x^3}+\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 x}+\frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{5/2}}-2 \frac {\left (g^2 p\right ) \int \frac {\log \left (\frac {2}{1-\frac {i \sqrt {g} x}{\sqrt {f}}}\right )}{1+\frac {g x^2}{f}} \, dx}{f^3}+\frac {\left (g^2 p\right ) \int \frac {\log \left (\frac {2 \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\sqrt {f} \left (-i \sqrt {e}+\frac {\sqrt {-d} \sqrt {g}}{\sqrt {f}}\right ) \left (1-\frac {i \sqrt {g} x}{\sqrt {f}}\right )}\right )}{1+\frac {g x^2}{f}} \, dx}{f^3}+\frac {\left (g^2 p\right ) \int \frac {\log \left (\frac {2 \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\sqrt {f} \left (i \sqrt {e}+\frac {\sqrt {-d} \sqrt {g}}{\sqrt {f}}\right ) \left (1-\frac {i \sqrt {g} x}{\sqrt {f}}\right )}\right )}{1+\frac {g x^2}{f}} \, dx}{f^3}\\ &=-\frac {2 e p}{3 d f x}-\frac {2 e^{3/2} p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2} f}-\frac {2 \sqrt {e} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} f^2}+\frac {2 g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{f^{5/2}}-\frac {g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}-\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{f^{5/2}}-\frac {g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{f^{5/2}}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{3 f x^3}+\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 x}+\frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{5/2}}+\frac {i g^{3/2} p \text {Li}_2\left (1+\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}-\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{2 f^{5/2}}+\frac {i g^{3/2} p \text {Li}_2\left (1-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{2 f^{5/2}}-2 \frac {\left (i g^{3/2} p\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-\frac {i \sqrt {g} x}{\sqrt {f}}}\right )}{f^{5/2}}\\ &=-\frac {2 e p}{3 d f x}-\frac {2 e^{3/2} p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2} f}-\frac {2 \sqrt {e} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} f^2}+\frac {2 g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{f^{5/2}}-\frac {g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}-\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{f^{5/2}}-\frac {g^{3/2} p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{f^{5/2}}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{3 f x^3}+\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 x}+\frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{5/2}}-\frac {i g^{3/2} p \text {Li}_2\left (1-\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{f^{5/2}}+\frac {i g^{3/2} p \text {Li}_2\left (1+\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}-\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{2 f^{5/2}}+\frac {i g^{3/2} p \text {Li}_2\left (1-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{2 f^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 754, normalized size = 1.16 \[ \frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{5/2}}+\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 x}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{3 f x^3}-\frac {2 e g^{3/2} p \left (\frac {i \left (\frac {\text {Li}_2\left (\frac {\sqrt {e} \left (\sqrt {f}-i \sqrt {g} x\right )}{\sqrt {e} \sqrt {f}-i \sqrt {-d} \sqrt {g}}\right )}{\sqrt {e}}+\frac {\log \left (1-\frac {i \sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {\sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}}\right )}{\sqrt {e}}\right )}{4 \sqrt {e}}+\frac {i \left (\frac {\text {Li}_2\left (\frac {\sqrt {e} \left (\sqrt {f}-i \sqrt {g} x\right )}{\sqrt {e} \sqrt {f}+i \sqrt {-d} \sqrt {g}}\right )}{\sqrt {e}}+\frac {\log \left (1-\frac {i \sqrt {g} x}{\sqrt {f}}\right ) \log \left (-\frac {\sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{-\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}}\right )}{\sqrt {e}}\right )}{4 \sqrt {e}}-\frac {i \left (\frac {\text {Li}_2\left (\frac {\sqrt {e} \left (i \sqrt {g} x+\sqrt {f}\right )}{\sqrt {e} \sqrt {f}-i \sqrt {-d} \sqrt {g}}\right )}{\sqrt {e}}+\frac {\log \left (1+\frac {i \sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {\sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}}\right )}{\sqrt {e}}\right )}{4 \sqrt {e}}-\frac {i \left (\frac {\text {Li}_2\left (\frac {\sqrt {e} \left (i \sqrt {g} x+\sqrt {f}\right )}{\sqrt {e} \sqrt {f}+i \sqrt {-d} \sqrt {g}}\right )}{\sqrt {e}}+\frac {\log \left (1+\frac {i \sqrt {g} x}{\sqrt {f}}\right ) \log \left (-\frac {\sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{-\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}}\right )}{\sqrt {e}}\right )}{4 \sqrt {e}}\right )}{f^{5/2}}-\frac {2 \sqrt {e} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} f^2}-\frac {2 e p \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {e x^2}{d}\right )}{3 d f x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Sqrt[e]*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*f^2) - (2*e*p*Hypergeometric2F1[-1/2, 1, 1/2, -((e*x^2)/
d)])/(3*d*f*x) - Log[c*(d + e*x^2)^p]/(3*f*x^3) + (g*Log[c*(d + e*x^2)^p])/(f^2*x) + (g^(3/2)*ArcTan[(Sqrt[g]*
x)/Sqrt[f]]*Log[c*(d + e*x^2)^p])/f^(5/2) - (2*e*g^(3/2)*p*(((I/4)*((Log[(Sqrt[g]*(Sqrt[-d] - Sqrt[e]*x))/(I*S
qrt[e]*Sqrt[f] + Sqrt[-d]*Sqrt[g])]*Log[1 - (I*Sqrt[g]*x)/Sqrt[f]])/Sqrt[e] + PolyLog[2, (Sqrt[e]*(Sqrt[f] - I
*Sqrt[g]*x))/(Sqrt[e]*Sqrt[f] - I*Sqrt[-d]*Sqrt[g])]/Sqrt[e]))/Sqrt[e] + ((I/4)*((Log[-((Sqrt[g]*(Sqrt[-d] + S
qrt[e]*x))/(I*Sqrt[e]*Sqrt[f] - Sqrt[-d]*Sqrt[g]))]*Log[1 - (I*Sqrt[g]*x)/Sqrt[f]])/Sqrt[e] + PolyLog[2, (Sqrt
[e]*(Sqrt[f] - I*Sqrt[g]*x))/(Sqrt[e]*Sqrt[f] + I*Sqrt[-d]*Sqrt[g])]/Sqrt[e]))/Sqrt[e] - ((I/4)*((Log[(Sqrt[g]
*(Sqrt[-d] + Sqrt[e]*x))/(I*Sqrt[e]*Sqrt[f] + Sqrt[-d]*Sqrt[g])]*Log[1 + (I*Sqrt[g]*x)/Sqrt[f]])/Sqrt[e] + Pol
yLog[2, (Sqrt[e]*(Sqrt[f] + I*Sqrt[g]*x))/(Sqrt[e]*Sqrt[f] - I*Sqrt[-d]*Sqrt[g])]/Sqrt[e]))/Sqrt[e] - ((I/4)*(
(Log[-((Sqrt[g]*(Sqrt[-d] - Sqrt[e]*x))/(I*Sqrt[e]*Sqrt[f] - Sqrt[-d]*Sqrt[g]))]*Log[1 + (I*Sqrt[g]*x)/Sqrt[f]
])/Sqrt[e] + PolyLog[2, (Sqrt[e]*(Sqrt[f] + I*Sqrt[g]*x))/(Sqrt[e]*Sqrt[f] + I*Sqrt[-d]*Sqrt[g])]/Sqrt[e]))/Sq
rt[e]))/f^(5/2)

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{g x^{6} + f x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(e*x^2+d)^p)/x^4/(g*x^2+f),x, algorithm="fricas")

[Out]

integral(log((e*x^2 + d)^p*c)/(g*x^6 + f*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(e*x^2+d)^p)/x^4/(g*x^2+f),x, algorithm="giac")

[Out]

integrate(log((e*x^2 + d)^p*c)/((g*x^2 + f)*x^4), x)

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maple [C]  time = 0.59, size = 1005, normalized size = 1.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(e*x^2+d)^p)/x^4/(g*x^2+f),x)

[Out]

(-p*ln(e*x^2+d)+ln((e*x^2+d)^p))*g^2/f^2/(f*g)^(1/2)*arctan(1/(f*g)^(1/2)*g*x)-1/3*(-p*ln(e*x^2+d)+ln((e*x^2+d
)^p))/f/x^3+(-p*ln(e*x^2+d)+ln((e*x^2+d)^p))*g/f^2/x+p*g/f^2/x*ln(e*x^2+d)-2*p*g/f^2*e/(d*e)^(1/2)*arctan(1/(d
*e)^(1/2)*e*x)+p*Sum(1/2*(ln(-_alpha+x)*ln(e*x^2+d)-2*e*(1/2*ln(-_alpha+x)*(ln((RootOf(_Z^2*e*g+2*_Z*_alpha*e*
g+d*g-e*f,index=1)-x+_alpha)/RootOf(_Z^2*e*g+2*_Z*_alpha*e*g+d*g-e*f,index=1))+ln((RootOf(_Z^2*e*g+2*_Z*_alpha
*e*g+d*g-e*f,index=2)-x+_alpha)/RootOf(_Z^2*e*g+2*_Z*_alpha*e*g+d*g-e*f,index=2)))/e+1/2*(dilog((RootOf(_Z^2*e
*g+2*_Z*_alpha*e*g+d*g-e*f,index=1)-x+_alpha)/RootOf(_Z^2*e*g+2*_Z*_alpha*e*g+d*g-e*f,index=1))+dilog((RootOf(
_Z^2*e*g+2*_Z*_alpha*e*g+d*g-e*f,index=2)-x+_alpha)/RootOf(_Z^2*e*g+2*_Z*_alpha*e*g+d*g-e*f,index=2)))/e))*g/f
^2/_alpha,_alpha=RootOf(_Z^2*g+f))-1/3*p/f/x^3*ln(e*x^2+d)-2/3*e*p/d/f/x-2/3*p/f*e^2/d/(d*e)^(1/2)*arctan(1/(d
*e)^(1/2)*e*x)-1/6*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2/f/x^3+1/6*I*Pi*csgn(I*c*(e*x^2+d)^p)^3/f/x
^3-1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*g/f^2/x+1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(
I*c)*g^2/f^2/(f*g)^(1/2)*arctan(1/(f*g)^(1/2)*g*x)+1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*g/f^2/x+1/2*I*Pi
*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*g^2/f^2/(f*g)^(1/2)*arctan(1/(f*g)^(1/2)*g*x)+1/2*I*Pi*csgn(I*(e*
x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*g/f^2/x-1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*g/f^2/x-1/2*I*Pi*csgn(I*c*(e*x^2+d)
^p)^3*g^2/f^2/(f*g)^(1/2)*arctan(1/(f*g)^(1/2)*g*x)+1/6*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*
c)/f/x^3-1/6*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)/f/x^3-1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*c
sgn(I*c)*g^2/f^2/(f*g)^(1/2)*arctan(1/(f*g)^(1/2)*g*x)+ln(c)*g^2/f^2/(f*g)^(1/2)*arctan(1/(f*g)^(1/2)*g*x)-1/3
*ln(c)/f/x^3+ln(c)*g/f^2/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(e*x^2+d)^p)/x^4/(g*x^2+f),x, algorithm="maxima")

[Out]

integrate(log((e*x^2 + d)^p*c)/((g*x^2 + f)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{x^4\,\left (g\,x^2+f\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(d + e*x^2)^p)/(x^4*(f + g*x^2)),x)

[Out]

int(log(c*(d + e*x^2)^p)/(x^4*(f + g*x^2)), 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(ln(c*(e*x**2+d)**p)/x**4/(g*x**2+f),x)

[Out]

Timed out

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