3.3.64 \(\int \frac {\log (c (d+\frac {e}{x})^p)}{f+g x^2} \, dx\) [264]

Optimal. Leaf size=360 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+\frac {e}{x}\right )^p\right )}{\sqrt {f} \sqrt {g}}+\frac {p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{\sqrt {f} \sqrt {g}}-\frac {p \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f} \sqrt {g} (e+d x)}{\left (i d \sqrt {f}+e \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{\sqrt {f} \sqrt {g}}+\frac {i p \text {Li}_2\left (-\frac {i \sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g}}-\frac {i p \text {Li}_2\left (\frac {i \sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g}}-\frac {i p \text {Li}_2\left (1-\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{2 \sqrt {f} \sqrt {g}}+\frac {i p \text {Li}_2\left (1-\frac {2 \sqrt {f} \sqrt {g} (e+d x)}{\left (i d \sqrt {f}+e \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{2 \sqrt {f} \sqrt {g}} \]

[Out]

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

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Rubi [A]
time = 0.30, antiderivative size = 360, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {211, 2520, 12, 266, 6820, 4996, 4940, 2438, 4966, 2449, 2352, 2497} \begin {gather*} \frac {i p \text {PolyLog}\left (2,1-\frac {2 \sqrt {f} \sqrt {g} (d x+e)}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (e \sqrt {g}+i d \sqrt {f}\right )}\right )}{2 \sqrt {f} \sqrt {g}}+\frac {i p \text {PolyLog}\left (2,-\frac {i \sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g}}-\frac {i p \text {PolyLog}\left (2,\frac {i \sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g}}-\frac {i p \text {PolyLog}\left (2,1-\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{2 \sqrt {f} \sqrt {g}}+\frac {\text {ArcTan}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+\frac {e}{x}\right )^p\right )}{\sqrt {f} \sqrt {g}}-\frac {p \text {ArcTan}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f} \sqrt {g} (d x+e)}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (e \sqrt {g}+i d \sqrt {f}\right )}\right )}{\sqrt {f} \sqrt {g}}+\frac {p \text {ArcTan}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{\sqrt {f} \sqrt {g}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[c*(d + e/x)^p]/(f + g*x^2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 266

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

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 2449

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 2497

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 2520

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 4940

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

Rule 4966

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 4996

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]
time = 0.16, size = 373, normalized size = 1.04 \begin {gather*} \frac {\log \left (c \left (d+\frac {e}{x}\right )^p\right ) \log \left (\sqrt {-f}-\sqrt {g} x\right )+p \log \left (\frac {\sqrt {g} x}{\sqrt {-f}}\right ) \log \left (\sqrt {-f}-\sqrt {g} x\right )-p \log \left (\frac {\sqrt {g} (e+d x)}{d \sqrt {-f}+e \sqrt {g}}\right ) \log \left (\sqrt {-f}-\sqrt {g} x\right )-\log \left (c \left (d+\frac {e}{x}\right )^p\right ) \log \left (\sqrt {-f}+\sqrt {g} x\right )-p \log \left (\frac {f \sqrt {g} x}{(-f)^{3/2}}\right ) \log \left (\sqrt {-f}+\sqrt {g} x\right )+p \log \left (-\frac {\sqrt {g} (e+d x)}{d \sqrt {-f}-e \sqrt {g}}\right ) \log \left (\sqrt {-f}+\sqrt {g} x\right )-p \text {Li}_2\left (\frac {d \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {-f}+e \sqrt {g}}\right )+p \text {Li}_2\left (\frac {d \left (\sqrt {-f}+\sqrt {g} x\right )}{d \sqrt {-f}-e \sqrt {g}}\right )-p \text {Li}_2\left (1+\frac {\sqrt {g} x}{\sqrt {-f}}\right )+p \text {Li}_2\left (1+\frac {f \sqrt {g} x}{(-f)^{3/2}}\right )}{2 \sqrt {-f} \sqrt {g}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.25, size = 0, normalized size = 0.00 \[\int \frac {\ln \left (c \left (d +\frac {e}{x}\right )^{p}\right )}{g \,x^{2}+f}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.55, size = 381, normalized size = 1.06 \begin {gather*} \frac {{\left (4 \, \arctan \left (\frac {g x}{\sqrt {f g}}\right ) e^{\left (-1\right )} \log \left (d + \frac {e}{x}\right ) - {\left ({\left (\pi - 2 \, \arctan \left (\frac {{\left (d^{2} x + d e\right )} \sqrt {f} \sqrt {g}}{d^{2} f + g e^{2}}, \frac {d g x e + g e^{2}}{d^{2} f + g e^{2}}\right )\right )} \log \left (g x^{2} + f\right ) - 4 \, \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) + 2 \, \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {d^{2} g x^{2} + 2 \, d g x e + g e^{2}}{d^{2} f + g e^{2}}\right ) + 2 i \, {\rm Li}_2\left (\frac {i \, \sqrt {g} x + \sqrt {f}}{\sqrt {f}}\right ) - 2 i \, {\rm Li}_2\left (-\frac {i \, \sqrt {g} x - \sqrt {f}}{\sqrt {f}}\right ) + 2 i \, {\rm Li}_2\left (\frac {d g x e + d^{2} f - {\left (i \, d^{2} x - i \, d e\right )} \sqrt {f} \sqrt {g}}{d^{2} f + 2 i \, d \sqrt {f} \sqrt {g} e - g e^{2}}\right ) - 2 i \, {\rm Li}_2\left (\frac {d g x e + d^{2} f + {\left (i \, d^{2} x - i \, d e\right )} \sqrt {f} \sqrt {g}}{d^{2} f - 2 i \, d \sqrt {f} \sqrt {g} e - g e^{2}}\right )\right )} e^{\left (-1\right )}\right )} p e}{4 \, \sqrt {f g}} - \frac {p \arctan \left (\frac {g x}{\sqrt {f g}}\right ) \log \left (d + \frac {e}{x}\right )}{\sqrt {f g}} + \frac {\arctan \left (\frac {g x}{\sqrt {f g}}\right ) \log \left (c {\left (d + \frac {e}{x}\right )}^{p}\right )}{\sqrt {f g}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*arctan(g*x/sqrt(f*g))*e^(-1)*log(d + e/x) - ((pi - 2*arctan2((d^2*x + d*e)*sqrt(f)*sqrt(g)/(d^2*f + g*e
^2), (d*g*x*e + g*e^2)/(d^2*f + g*e^2)))*log(g*x^2 + f) - 4*arctan(sqrt(g)*x/sqrt(f))*log(sqrt(g)*x/sqrt(f)) +
2*arctan(sqrt(g)*x/sqrt(f))*log((d^2*g*x^2 + 2*d*g*x*e + g*e^2)/(d^2*f + g*e^2)) + 2*I*dilog((I*sqrt(g)*x + s
qrt(f))/sqrt(f)) - 2*I*dilog(-(I*sqrt(g)*x - sqrt(f))/sqrt(f)) + 2*I*dilog((d*g*x*e + d^2*f - (I*d^2*x - I*d*e
)*sqrt(f)*sqrt(g))/(d^2*f + 2*I*d*sqrt(f)*sqrt(g)*e - g*e^2)) - 2*I*dilog((d*g*x*e + d^2*f + (I*d^2*x - I*d*e)
*sqrt(f)*sqrt(g))/(d^2*f - 2*I*d*sqrt(f)*sqrt(g)*e - g*e^2)))*e^(-1))*p*e/sqrt(f*g) - p*arctan(g*x/sqrt(f*g))*
log(d + e/x)/sqrt(f*g) + arctan(g*x/sqrt(f*g))*log(c*(d + e/x)^p)/sqrt(f*g)

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\log {\left (c \left (d + \frac {e}{x}\right )^{p} \right )}}{f + g x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(c*(d + e/x)**p)/(f + g*x**2), x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(d + e/x)^p)/(f + g*x^2),x)

[Out]

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

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