3.352 \(\int \frac{\log (c (d+e x^2)^p)}{x^3 (f+g x^2)^2} \, dx\)

Optimal. Leaf size=251 \[ \frac{g p \text{PolyLog}\left (2,-\frac{g \left (d+e x^2\right )}{e f-d g}\right )}{f^3}-\frac{g p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )}{f^3}-\frac{g \log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^3}-\frac{g \log \left (c \left (d+e x^2\right )^p\right )}{2 f^2 \left (f+g x^2\right )}+\frac{g \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )}{f^3}-\frac{\log \left (c \left (d+e x^2\right )^p\right )}{2 f^2 x^2}+\frac{e g p \log \left (d+e x^2\right )}{2 f^2 (e f-d g)}-\frac{e g p \log \left (f+g x^2\right )}{2 f^2 (e f-d g)}-\frac{e p \log \left (d+e x^2\right )}{2 d f^2}+\frac{e p \log (x)}{d f^2} \]

[Out]

(e*p*Log[x])/(d*f^2) - (e*p*Log[d + e*x^2])/(2*d*f^2) + (e*g*p*Log[d + e*x^2])/(2*f^2*(e*f - d*g)) - Log[c*(d
+ e*x^2)^p]/(2*f^2*x^2) - (g*Log[c*(d + e*x^2)^p])/(2*f^2*(f + g*x^2)) - (g*Log[-((e*x^2)/d)]*Log[c*(d + e*x^2
)^p])/f^3 - (e*g*p*Log[f + g*x^2])/(2*f^2*(e*f - d*g)) + (g*Log[c*(d + e*x^2)^p]*Log[(e*(f + g*x^2))/(e*f - d*
g)])/f^3 + (g*p*PolyLog[2, -((g*(d + e*x^2))/(e*f - d*g))])/f^3 - (g*p*PolyLog[2, 1 + (e*x^2)/d])/f^3

________________________________________________________________________________________

Rubi [A]  time = 0.340879, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {2475, 44, 2416, 2395, 36, 29, 31, 2394, 2315, 2393, 2391} \[ \frac{g p \text{PolyLog}\left (2,-\frac{g \left (d+e x^2\right )}{e f-d g}\right )}{f^3}-\frac{g p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )}{f^3}-\frac{g \log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^3}-\frac{g \log \left (c \left (d+e x^2\right )^p\right )}{2 f^2 \left (f+g x^2\right )}+\frac{g \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )}{f^3}-\frac{\log \left (c \left (d+e x^2\right )^p\right )}{2 f^2 x^2}+\frac{e g p \log \left (d+e x^2\right )}{2 f^2 (e f-d g)}-\frac{e g p \log \left (f+g x^2\right )}{2 f^2 (e f-d g)}-\frac{e p \log \left (d+e x^2\right )}{2 d f^2}+\frac{e p \log (x)}{d f^2} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(d + e*x^2)^p]/(x^3*(f + g*x^2)^2),x]

[Out]

(e*p*Log[x])/(d*f^2) - (e*p*Log[d + e*x^2])/(2*d*f^2) + (e*g*p*Log[d + e*x^2])/(2*f^2*(e*f - d*g)) - Log[c*(d
+ e*x^2)^p]/(2*f^2*x^2) - (g*Log[c*(d + e*x^2)^p])/(2*f^2*(f + g*x^2)) - (g*Log[-((e*x^2)/d)]*Log[c*(d + e*x^2
)^p])/f^3 - (e*g*p*Log[f + g*x^2])/(2*f^2*(e*f - d*g)) + (g*Log[c*(d + e*x^2)^p]*Log[(e*(f + g*x^2))/(e*f - d*
g)])/f^3 + (g*p*PolyLog[2, -((g*(d + e*x^2))/(e*f - d*g))])/f^3 - (g*p*PolyLog[2, 1 + (e*x^2)/d])/f^3

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 36

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

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 2393

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

Rule 2391

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

Rubi steps

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

Mathematica [A]  time = 0.174308, size = 208, normalized size = 0.83 \[ \frac{2 g \left (p \text{PolyLog}\left (2,\frac{g \left (d+e x^2\right )}{d g-e f}\right )+\log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )\right )-2 g \left (p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )+\log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )\right )-\frac{f g \log \left (c \left (d+e x^2\right )^p\right )}{f+g x^2}-\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{x^2}+\frac{e f g p \left (\log \left (d+e x^2\right )-\log \left (f+g x^2\right )\right )}{e f-d g}+\frac{e f p \left (2 \log (x)-\log \left (d+e x^2\right )\right )}{d}}{2 f^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(d + e*x^2)^p]/(x^3*(f + g*x^2)^2),x]

[Out]

((e*f*p*(2*Log[x] - Log[d + e*x^2]))/d - (f*Log[c*(d + e*x^2)^p])/x^2 - (f*g*Log[c*(d + e*x^2)^p])/(f + g*x^2)
+ (e*f*g*p*(Log[d + e*x^2] - Log[f + g*x^2]))/(e*f - d*g) + 2*g*(Log[c*(d + e*x^2)^p]*Log[(e*(f + g*x^2))/(e*
f - d*g)] + p*PolyLog[2, (g*(d + e*x^2))/(-(e*f) + d*g)]) - 2*g*(Log[-((e*x^2)/d)]*Log[c*(d + e*x^2)^p] + p*Po
lyLog[2, 1 + (e*x^2)/d]))/(2*f^3)

________________________________________________________________________________________

Maple [C]  time = 0.724, size = 1216, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(e*x^2+d)^p)/x^3/(g*x^2+f)^2,x)

[Out]

-I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*g/f^3*ln(x)-1/4*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*g/f^2
/(g*x^2+f)-1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*g/f^3*ln(g*x^2+f)-1/2*ln((e*x^2+d)^p)/f^2/x^2-1/4*I*Pi*csgn(I*(e*x
^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2/f^2/x^2+e*p*ln(x)/d/f^2+2*p*g/f^3*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+2*p
*g/f^3*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-p*g/f^3*sum(ln(x-_alpha)*ln(g*x^2+f)-ln(x-_alpha)*(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)))-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,inde
x=2)),_alpha=RootOf(_Z^2*e+d))+ln(c)*g/f^3*ln(g*x^2+f)-1/2*ln(c)*g/f^2/(g*x^2+f)-2*ln(c)*g/f^3*ln(x)+ln((e*x^2
+d)^p)*g/f^3*ln(g*x^2+f)-1/2*ln((e*x^2+d)^p)*g/f^2/(g*x^2+f)-2*ln((e*x^2+d)^p)*g/f^3*ln(x)-1/4*I*Pi*csgn(I*c*(
e*x^2+d)^p)^2*csgn(I*c)/f^2/x^2-1/2*ln(c)/f^2/x^2+1/4*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)
*g/f^2/(g*x^2+f)+I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*g/f^3*ln(x)-1/2*I*Pi*csgn(I*(e*x^2+d
)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*g/f^3*ln(g*x^2+f)-p*e/f^2/(d*g-e*f)*ln(e*x^2+d)*g+1/2*p*e^2/f/d/(d*g-e*f)
*ln(e*x^2+d)+1/2*p*e/f^2*g/(d*g-e*f)*ln(g*x^2+f)+1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*g/f^3*ln
(g*x^2+f)+1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*g/f^2/(g*x^2+f)+I*Pi*csgn(I*c*(e*x^2+d)^p)^3*g/f^3*ln(x)+2*p*g/f^3*
ln(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+2*p*g/f^3*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/4*I*Pi*csgn(I
*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)/f^2/x^2+1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*g/f^3*ln(g*x^
2+f)-I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*g/f^3*ln(x)-1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*g
/f^2/(g*x^2+f)+1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^3/f^2/x^2

________________________________________________________________________________________

Maxima [A]  time = 1.26513, size = 398, normalized size = 1.59 \begin{align*} -\frac{1}{2} \,{\left (f{\left (\frac{e \log \left (e x^{2} + d\right )}{d e f^{3} - d^{2} f^{2} g} - \frac{g \log \left (g x^{2} + f\right )}{e f^{4} - d f^{3} g} - \frac{\log \left (x^{2}\right )}{d f^{3}}\right )} - 2 \, g{\left (\frac{\log \left (e x^{2} + d\right )}{e f^{3} - d f^{2} g} - \frac{\log \left (g x^{2} + f\right )}{e f^{3} - d f^{2} g}\right )} - \frac{2 \,{\left (2 \, \log \left (\frac{e x^{2}}{d} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{e x^{2}}{d}\right )\right )} g}{e f^{3}} + \frac{2 \,{\left (\log \left (g x^{2} + f\right ) \log \left (-\frac{e g x^{2} + e f}{e f - d g} + 1\right ) +{\rm Li}_2\left (\frac{e g x^{2} + e f}{e f - d g}\right )\right )} g}{e f^{3}}\right )} e p - \frac{1}{2} \,{\left (\frac{2 \, g x^{2} + f}{f^{2} g x^{4} + f^{3} x^{2}} - \frac{2 \, g \log \left (g x^{2} + f\right )}{f^{3}} + \frac{2 \, g \log \left (x^{2}\right )}{f^{3}}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(e*x^2+d)^p)/x^3/(g*x^2+f)^2,x, algorithm="maxima")

[Out]

-1/2*(f*(e*log(e*x^2 + d)/(d*e*f^3 - d^2*f^2*g) - g*log(g*x^2 + f)/(e*f^4 - d*f^3*g) - log(x^2)/(d*f^3)) - 2*g
*(log(e*x^2 + d)/(e*f^3 - d*f^2*g) - log(g*x^2 + f)/(e*f^3 - d*f^2*g)) - 2*(2*log(e*x^2/d + 1)*log(x) + dilog(
-e*x^2/d))*g/(e*f^3) + 2*(log(g*x^2 + f)*log(-(e*g*x^2 + e*f)/(e*f - d*g) + 1) + dilog((e*g*x^2 + e*f)/(e*f -
d*g)))*g/(e*f^3))*e*p - 1/2*((2*g*x^2 + f)/(f^2*g*x^4 + f^3*x^2) - 2*g*log(g*x^2 + f)/f^3 + 2*g*log(x^2)/f^3)*
log((e*x^2 + d)^p*c)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{g^{2} x^{7} + 2 \, f g x^{5} + f^{2} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(e*x^2+d)^p)/x^3/(g*x^2+f)^2,x, algorithm="fricas")

[Out]

integral(log((e*x^2 + d)^p*c)/(g^2*x^7 + 2*f*g*x^5 + f^2*x^3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(e*x**2+d)**p)/x**3/(g*x**2+f)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )}^{2} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(e*x^2+d)^p)/x^3/(g*x^2+f)^2,x, algorithm="giac")

[Out]

integrate(log((e*x^2 + d)^p*c)/((g*x^2 + f)^2*x^3), x)