3.4.52 \(\int \frac {\log (c (d+e x^2)^p)}{x^3 (f+g x^2)^2} \, dx\) [352]

Optimal. Leaf size=251 \[ \frac {e p \log (x)}{d f^2}-\frac {e p \log \left (d+e x^2\right )}{2 d f^2}+\frac {e g p \log \left (d+e x^2\right )}{2 f^2 (e f-d g)}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{2 f^2 x^2}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{2 f^2 \left (f+g x^2\right )}-\frac {g \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^3}-\frac {e g p \log \left (f+g x^2\right )}{2 f^2 (e f-d g)}+\frac {g \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )}{f^3}+\frac {g p \text {Li}_2\left (-\frac {g \left (d+e x^2\right )}{e f-d g}\right )}{f^3}-\frac {g p \text {Li}_2\left (1+\frac {e x^2}{d}\right )}{f^3} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {2525, 46, 2463, 2442, 36, 29, 31, 2441, 2352, 2440, 2438} \begin {gather*} \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 ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\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 {\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} \end {gather*}

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 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 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 46

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] && Lt
Q[m + n + 2, 0])

Rule 2352

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

Rule 2438

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]

Rule 2440

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 2441

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 2442

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 2463

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 2525

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])

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} \text {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} \text {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 {\text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x^2} \, dx,x,x^2\right )}{2 f^2}-\frac {g \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^2\right )}{f^3}+\frac {g^2 \text {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 \text {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) \text {Subst}\left (\int \frac {1}{x (d+e x)} \, dx,x,x^2\right )}{2 f^2}+\frac {(e g p) \text {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) \text {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) \text {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) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d f^2}-\frac {\left (e^2 p\right ) \text {Subst}\left (\int \frac {1}{d+e x} \, dx,x,x^2\right )}{2 d f^2}-\frac {(g p) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.12, size = 208, normalized size = 0.83 \begin {gather*} \frac {\frac {e f p \left (2 \log (x)-\log \left (d+e x^2\right )\right )}{d}-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{x^2}-\frac {f g \log \left (c \left (d+e x^2\right )^p\right )}{f+g 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}+2 g \left (\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 )+p \text {Li}_2\left (\frac {g \left (d+e x^2\right )}{-e f+d g}\right )\right )-2 g \left (\log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+p \text {Li}_2\left (1+\frac {e x^2}{d}\right )\right )}{2 f^3} \end {gather*}

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)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.45, size = 1216, normalized size = 4.84

method	result	size
risch	\(\text {Expression too large to display}\)	\(1216\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.64, size = 310, normalized size = 1.24 \begin {gather*} -\frac {1}{2} \, {\left (f {\left (\frac {g \log \left (g x^{2} + f\right )}{d f^{3} g - f^{4} e} - \frac {e \log \left (x^{2} e + d\right )}{d^{2} f^{2} g - d f^{3} e} - \frac {\log \left (x^{2}\right )}{d f^{3}}\right )} - 2 \, g {\left (\frac {\log \left (g x^{2} + f\right )}{d f^{2} g - f^{3} e} - \frac {\log \left (x^{2} e + d\right )}{d f^{2} g - f^{3} e}\right )} - \frac {2 \, {\left (2 \, \log \left (\frac {x^{2} e}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {x^{2} e}{d}\right )\right )} g e^{\left (-1\right )}}{f^{3}} + \frac {2 \, {\left (\log \left (g x^{2} + f\right ) \log \left (\frac {g x^{2} e + f e}{d g - f e} + 1\right ) + {\rm Li}_2\left (-\frac {g x^{2} e + f e}{d g - f e}\right )\right )} g e^{\left (-1\right )}}{f^{3}}\right )} p e - \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 (x^{2} e + d\right )}^{p} c\right ) \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{x^3\,{\left (g\,x^2+f\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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