3.65 \(\int \frac {\log (d (b x+c x^2)^n)}{x} \, dx\)

Optimal. Leaf size=53 \[ \log (x) \log \left (d \left (b x+c x^2\right )^n\right )-n \text {Li}_2\left (-\frac {c x}{b}\right )-n \log (x) \log \left (\frac {c x}{b}+1\right )-\frac {1}{2} n \log ^2(x) \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2524, 1593, 2357, 2301, 2317, 2391} \[ -n \text {PolyLog}\left (2,-\frac {c x}{b}\right )+\log (x) \log \left (d \left (b x+c x^2\right )^n\right )-n \log (x) \log \left (\frac {c x}{b}+1\right )-\frac {1}{2} n \log ^2(x) \]

Antiderivative was successfully verified.

[In]

Int[Log[d*(b*x + c*x^2)^n]/x,x]

[Out]

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

Rule 1593

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 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]

Rule 2524

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[d + e*x]*(a + b
*Log[c*RFx^p])^n)/e, x] - Dist[(b*n*p)/e, Int[(Log[d + e*x]*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x
] /; FreeQ[{a, b, c, d, e, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 50, normalized size = 0.94 \[ \log (x) \log \left (d (x (b+c x))^n\right )-n \left (\text {Li}_2\left (-\frac {c x}{b}\right )+\log (x) \log \left (\frac {b+c x}{b}\right )+\frac {\log ^2(x)}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[d*(b*x + c*x^2)^n]/x,x]

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (c x^{2} + b x\right )}^{n} d\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log((c*x^2 + b*x)^n*d)/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (c x^{2} + b x\right )}^{n} d\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (d \left (c \,x^{2}+b x \right )^{n}\right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(d*(c*x^2+b*x)^n)/x,x)

[Out]

int(ln(d*(c*x^2+b*x)^n)/x,x)

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maxima [A]  time = 0.45, size = 80, normalized size = 1.51 \[ -n \log \left (c x^{2} + b x\right ) \log \relax (x) + \frac {1}{2} \, {\left (2 \, \log \left (c x^{2} + b x\right ) \log \relax (x) - 2 \, \log \left (\frac {c x}{b} + 1\right ) \log \relax (x) - \log \relax (x)^{2} - 2 \, {\rm Li}_2\left (-\frac {c x}{b}\right )\right )} n + \log \left ({\left (c x^{2} + b x\right )}^{n} d\right ) \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-n*log(c*x^2 + b*x)*log(x) + 1/2*(2*log(c*x^2 + b*x)*log(x) - 2*log(c*x/b + 1)*log(x) - log(x)^2 - 2*dilog(-c*
x/b))*n + log((c*x^2 + b*x)^n*d)*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\ln \left (d\,{\left (c\,x^2+b\,x\right )}^n\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(d*(b*x + c*x^2)^n)/x,x)

[Out]

int(log(d*(b*x + c*x^2)^n)/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (d \left (b x + c x^{2}\right )^{n} \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(d*(b*x + c*x**2)**n)/x, x)

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