3.4.45 \(\int \frac {\log (c (a+b x)^n)}{d x+e x^2} \, dx\) [345]

Optimal. Leaf size=97 \[ \frac {\log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^n\right )}{d}-\frac {\log \left (c (a+b x)^n\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}-\frac {n \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{d}+\frac {n \text {Li}_2\left (1+\frac {b x}{a}\right )}{d} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1607, 36, 29, 31, 2463, 2441, 2352, 2440, 2438} \begin {gather*} -\frac {n \text {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d}+\frac {n \text {PolyLog}\left (2,\frac {b x}{a}+1\right )}{d}-\frac {\log \left (c (a+b x)^n\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}+\frac {\log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^n\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

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Mathematica [A]
time = 0.02, size = 98, normalized size = 1.01 \begin {gather*} \frac {\log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^n\right )}{d}-\frac {\log \left (c (a+b x)^n\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}+\frac {n \text {Li}_2\left (\frac {a+b x}{a}\right )}{d}-\frac {n \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.38, size = 420, normalized size = 4.33

method	result	size
risch	\(-\frac {\ln \left (e x +d \right ) \ln \left (\left (b x +a \right )^{n}\right )}{d}+\frac {\ln \left (\left (b x +a \right )^{n}\right ) \ln \left (x \right )}{d}-\frac {n \dilog \left (\frac {b x +a}{a}\right )}{d}-\frac {n \ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{d}+\frac {n \dilog \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d}+\frac {n \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2} \mathrm {csgn}\left (i c \right ) \ln \left (e x +d \right )}{2 d}-\frac {i \pi \,\mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \ln \left (x \right )}{2 d}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{3} \ln \left (x \right )}{2 d}+\frac {i \pi \,\mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \ln \left (e x +d \right )}{2 d}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{3} \ln \left (e x +d \right )}{2 d}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2} \mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \ln \left (e x +d \right )}{2 d}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2} \mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \ln \left (x \right )}{2 d}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2} \mathrm {csgn}\left (i c \right ) \ln \left (x \right )}{2 d}-\frac {\ln \left (c \right ) \ln \left (e x +d \right )}{d}+\frac {\ln \left (c \right ) \ln \left (x \right )}{d}\)	\(420\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 129, normalized size = 1.33 \begin {gather*} -b n {\left (\frac {\log \left (\frac {b x}{a} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x}{a}\right )}{b d} - \frac {\log \left (x e + d\right ) \log \left (-\frac {b x e + b d}{b d - a e} + 1\right ) + {\rm Li}_2\left (\frac {b x e + b d}{b d - a e}\right )}{b d}\right )} - {\left (\frac {\log \left (x e + d\right )}{d} - \frac {\log \left (x\right )}{d}\right )} \log \left ({\left (b x + a\right )}^{n} c\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*n*((log(b*x/a + 1)*log(x) + dilog(-b*x/a))/(b*d) - (log(x*e + d)*log(-(b*x*e + b*d)/(b*d - a*e) + 1) + dilo
g((b*x*e + b*d)/(b*d - a*e)))/(b*d)) - (log(x*e + d)/d - log(x)/d)*log((b*x + a)^n*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*(b*x+a)^n)/(e*x^2+d*x),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\ln \left (c\,{\left (a+b\,x\right )}^n\right )}{e\,x^2+d\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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