3.3.59 \(\int \frac {a+b \log (c (d+e x)^n)}{x (f+g x^2)} \, dx\) [259]

Optimal. Leaf size=245 \[ \frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f}-\frac {b n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f}-\frac {b n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f}+\frac {b n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f} \]

[Out]

ln(-e*x/d)*(a+b*ln(c*(e*x+d)^n))/f-1/2*(a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1/2)-x*g^(1/2))/(e*(-f)^(1/2)+d*g^(1/
2)))/f-1/2*(a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1/2)+x*g^(1/2))/(e*(-f)^(1/2)-d*g^(1/2)))/f+b*n*polylog(2,1+e*x/d
)/f-1/2*b*n*polylog(2,-(e*x+d)*g^(1/2)/(e*(-f)^(1/2)-d*g^(1/2)))/f-1/2*b*n*polylog(2,(e*x+d)*g^(1/2)/(e*(-f)^(
1/2)+d*g^(1/2)))/f

________________________________________________________________________________________

Rubi [A]
time = 0.23, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {272, 36, 29, 31, 2463, 2441, 2352, 266, 2440, 2438} \begin {gather*} -\frac {b n \text {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f}-\frac {b n \text {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{d \sqrt {g}+e \sqrt {-f}}\right )}{2 f}+\frac {b n \text {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f}-\frac {\log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f}-\frac {\log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])/(x*(f + g*x^2)),x]

[Out]

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

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 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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.06, size = 224, normalized size = 0.91 \begin {gather*} -\frac {-2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )+\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )+b n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+b n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )-2 b n \text {Li}_2\left (1+\frac {e x}{d}\right )}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(-2*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]) + (a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x)
)/(e*Sqrt[-f] + d*Sqrt[g])] + (a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g
])] + b*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))] + b*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*
Sqrt[-f] + d*Sqrt[g])] - 2*b*n*PolyLog[2, 1 + (e*x)/d])/f

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.33, size = 604, normalized size = 2.47

method	result	size
risch	\(-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g \,x^{2}+f \right )}{2 f}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (x \right )}{f}-\frac {b n \dilog \left (\frac {e x +d}{d}\right )}{f}-\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f}+\frac {b n \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{2 f}-\frac {b n \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 f}-\frac {b n \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 f}-\frac {b n \dilog \left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 f}-\frac {b n \dilog \left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 f}-\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} \ln \left (x \right )}{2 f}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (x \right )}{2 f}+\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (x \right )}{2 f}+\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} \ln \left (g \,x^{2}+f \right )}{4 f}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (g \,x^{2}+f \right )}{4 f}-\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (g \,x^{2}+f \right )}{4 f}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) \ln \left (g \,x^{2}+f \right )}{4 f}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) \ln \left (x \right )}{2 f}-\frac {b \ln \left (c \right ) \ln \left (g \,x^{2}+f \right )}{2 f}+\frac {b \ln \left (c \right ) \ln \left (x \right )}{f}-\frac {a \ln \left (g \,x^{2}+f \right )}{2 f}+\frac {a \ln \left (x \right )}{f}\)	\(604\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)^n))/x/(g*x^2+f),x,method=_RETURNVERBOSE)

[Out]

-1/2*b*ln((e*x+d)^n)/f*ln(g*x^2+f)+b*ln((e*x+d)^n)/f*ln(x)-b*n/f*dilog((e*x+d)/d)-b*n/f*ln(x)*ln((e*x+d)/d)+1/
2*b*n/f*ln(e*x+d)*ln(g*x^2+f)-1/2*b*n/f*ln(e*x+d)*ln((e*(-f*g)^(1/2)-g*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+d*g))-1/2*
b*n/f*ln(e*x+d)*ln((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e*(-f*g)^(1/2)-d*g))-1/2*b*n/f*dilog((e*(-f*g)^(1/2)-g*(e*x
+d)+d*g)/(e*(-f*g)^(1/2)+d*g))-1/2*b*n/f*dilog((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e*(-f*g)^(1/2)-d*g))-1/2*I*b*Pi
*csgn(I*c*(e*x+d)^n)^3/f*ln(x)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/f*ln(x)+1/2*I*b*Pi*csgn(I*(e*x+d)^n)
*csgn(I*c*(e*x+d)^n)^2/f*ln(x)+1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/f*ln(g*x^2+f)-1/4*I*b*Pi*csgn(I*c)*csgn(I*c*(e
*x+d)^n)^2/f*ln(g*x^2+f)-1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/f*ln(g*x^2+f)+1/4*I*b*Pi*csgn(I*c)
*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/f*ln(g*x^2+f)-1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n
)/f*ln(x)-1/2*b*ln(c)/f*ln(g*x^2+f)+b*ln(c)/f*ln(x)-1/2*a/f*ln(g*x^2+f)+a/f*ln(x)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(log(g*x^2 + f)/f - 2*log(x)/f) + b*integrate((log((x*e + d)^n) + log(c))/(g*x^3 + f*x), x)

________________________________________________________________________________________

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((a+b*log(c*(e*x+d)^n))/x/(g*x^2+f),x, algorithm="fricas")

[Out]

integral((b*log((x*e + d)^n*c) + a)/(g*x^3 + f*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(e*x+d)**n))/x/(g*x**2+f),x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))/(x*(f + g*x**2)), x)

________________________________________________________________________________________

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((a+b*log(c*(e*x+d)^n))/x/(g*x^2+f),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________