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

Optimal. Leaf size=47 \[ -\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x}+\frac {c n \log (x)}{b}-\frac {c n \log (b+c x)}{b}-\frac {n}{x} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2525, 77} \[ -\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x}+\frac {c n \log (x)}{b}-\frac {c n \log (b+c x)}{b}-\frac {n}{x} \]

Antiderivative was successfully verified.

[In]

Int[Log[d*(b*x + c*x^2)^n]/x^2,x]

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2525

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 45, normalized size = 0.96 \[ -\frac {\log \left (d (x (b+c x))^n\right )}{x}+\frac {c n \log (x)}{b}-\frac {c n \log (b+c x)}{b}-\frac {n}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 46, normalized size = 0.98 \[ -\frac {c n x \log \left (c x + b\right ) - c n x \log \relax (x) + b n \log \left (c x^{2} + b x\right ) + b n + b \log \relax (d)}{b x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 47, normalized size = 1.00 \[ -\frac {c n \log \left (c x + b\right )}{b} + \frac {c n \log \relax (x)}{b} - \frac {n \log \left (c x^{2} + b x\right )}{x} - \frac {n + \log \relax (d)}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (d \left (c \,x^{2}+b x \right )^{n}\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 46, normalized size = 0.98 \[ -n {\left (\frac {c \log \left (c x + b\right )}{b} - \frac {c \log \relax (x)}{b} + \frac {1}{x}\right )} - \frac {\log \left ({\left (c x^{2} + b x\right )}^{n} d\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.80, size = 43, normalized size = 0.91 \[ -\frac {\ln \left (d\,{\left (c\,x^2+b\,x\right )}^n\right )}{x}-\frac {n}{x}-\frac {2\,c\,n\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(d*(b*x + c*x^2)^n)/x^2,x)

[Out]

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

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sympy [A]  time = 3.41, size = 76, normalized size = 1.62 \[ \begin {cases} - \frac {n \log {\left (b x + c x^{2} \right )}}{x} - \frac {n}{x} - \frac {\log {\relax (d )}}{x} - \frac {2 c n \log {\left (b + c x \right )}}{b} + \frac {c n \log {\left (b x + c x^{2} \right )}}{b} & \text {for}\: b \neq 0 \\- \frac {n \log {\relax (c )}}{x} - \frac {2 n \log {\relax (x )}}{x} - \frac {2 n}{x} - \frac {\log {\relax (d )}}{x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-n*log(b*x + c*x**2)/x - n/x - log(d)/x - 2*c*n*log(b + c*x)/b + c*n*log(b*x + c*x**2)/b, Ne(b, 0))
, (-n*log(c)/x - 2*n*log(x)/x - 2*n/x - log(d)/x, True))

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