3.67 \(\int \frac {\log (d (b x+c x^2)^n)}{x^3} \, dx\)

Optimal. Leaf size=72 \[ -\frac {c^2 n \log (x)}{2 b^2}+\frac {c^2 n \log (b+c x)}{2 b^2}-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{2 x^2}-\frac {c n}{2 b x}-\frac {n}{4 x^2} \]

[Out]

-1/4*n/x^2-1/2*c*n/b/x-1/2*c^2*n*ln(x)/b^2+1/2*c^2*n*ln(c*x+b)/b^2-1/2*ln(d*(c*x^2+b*x)^n)/x^2

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Rubi [A]  time = 0.05, antiderivative size = 72, 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} \[ -\frac {c^2 n \log (x)}{2 b^2}+\frac {c^2 n \log (b+c x)}{2 b^2}-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{2 x^2}-\frac {c n}{2 b x}-\frac {n}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Log[d*(b*x + c*x^2)^n]/x^3,x]

[Out]

-n/(4*x^2) - (c*n)/(2*b*x) - (c^2*n*Log[x])/(2*b^2) + (c^2*n*Log[b + c*x])/(2*b^2) - Log[d*(b*x + c*x^2)^n]/(2
*x^2)

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^3} \, dx &=-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{2 x^2}+\frac {1}{2} n \int \frac {b+2 c x}{x^3 (b+c x)} \, dx\\ &=-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{2 x^2}+\frac {1}{2} n \int \left (\frac {1}{x^3}+\frac {c}{b x^2}-\frac {c^2}{b^2 x}+\frac {c^3}{b^2 (b+c x)}\right ) \, dx\\ &=-\frac {n}{4 x^2}-\frac {c n}{2 b x}-\frac {c^2 n \log (x)}{2 b^2}+\frac {c^2 n \log (b+c x)}{2 b^2}-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{2 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 65, normalized size = 0.90 \[ \frac {1}{2} n \left (-\frac {c^2 \log (x)}{b^2}+\frac {c^2 \log (b+c x)}{b^2}-\frac {c}{b x}-\frac {1}{2 x^2}\right )-\frac {\log \left (d (x (b+c x))^n\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[d*(b*x + c*x^2)^n]/x^3,x]

[Out]

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

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fricas [A]  time = 0.48, size = 70, normalized size = 0.97 \[ \frac {2 \, c^{2} n x^{2} \log \left (c x + b\right ) - 2 \, c^{2} n x^{2} \log \relax (x) - 2 \, b c n x - 2 \, b^{2} n \log \left (c x^{2} + b x\right ) - b^{2} n - 2 \, b^{2} \log \relax (d)}{4 \, b^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x)^n)/x^3,x, algorithm="fricas")

[Out]

1/4*(2*c^2*n*x^2*log(c*x + b) - 2*c^2*n*x^2*log(x) - 2*b*c*n*x - 2*b^2*n*log(c*x^2 + b*x) - b^2*n - 2*b^2*log(
d))/(b^2*x^2)

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giac [A]  time = 0.17, size = 65, normalized size = 0.90 \[ \frac {c^{2} n \log \left (c x + b\right )}{2 \, b^{2}} - \frac {c^{2} n \log \relax (x)}{2 \, b^{2}} - \frac {n \log \left (c x^{2} + b x\right )}{2 \, x^{2}} - \frac {2 \, c n x + b n + 2 \, b \log \relax (d)}{4 \, b x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x)^n)/x^3,x, algorithm="giac")

[Out]

1/2*c^2*n*log(c*x + b)/b^2 - 1/2*c^2*n*log(x)/b^2 - 1/2*n*log(c*x^2 + b*x)/x^2 - 1/4*(2*c*n*x + b*n + 2*b*log(
d))/(b*x^2)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (d \left (c \,x^{2}+b x \right )^{n}\right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(d*(c*x^2+b*x)^n)/x^3,x)

[Out]

int(ln(d*(c*x^2+b*x)^n)/x^3,x)

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maxima [A]  time = 0.70, size = 62, normalized size = 0.86 \[ \frac {1}{4} \, n {\left (\frac {2 \, c^{2} \log \left (c x + b\right )}{b^{2}} - \frac {2 \, c^{2} \log \relax (x)}{b^{2}} - \frac {2 \, c x + b}{b x^{2}}\right )} - \frac {\log \left ({\left (c x^{2} + b x\right )}^{n} d\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x)^n)/x^3,x, algorithm="maxima")

[Out]

1/4*n*(2*c^2*log(c*x + b)/b^2 - 2*c^2*log(x)/b^2 - (2*c*x + b)/(b*x^2)) - 1/2*log((c*x^2 + b*x)^n*d)/x^2

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mupad [B]  time = 0.48, size = 54, normalized size = 0.75 \[ \frac {c^2\,n\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )}{b^2}-\frac {\frac {n}{2}+\frac {c\,n\,x}{b}}{2\,x^2}-\frac {\ln \left (d\,{\left (c\,x^2+b\,x\right )}^n\right )}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(d*(b*x + c*x^2)^n)/x^3,x)

[Out]

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

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sympy [A]  time = 6.95, size = 110, normalized size = 1.53 \[ \begin {cases} - \frac {n \log {\left (b x + c x^{2} \right )}}{2 x^{2}} - \frac {n}{4 x^{2}} - \frac {\log {\relax (d )}}{2 x^{2}} - \frac {c n}{2 b x} + \frac {c^{2} n \log {\left (b + c x \right )}}{b^{2}} - \frac {c^{2} n \log {\left (b x + c x^{2} \right )}}{2 b^{2}} & \text {for}\: b \neq 0 \\- \frac {n \log {\relax (c )}}{2 x^{2}} - \frac {n \log {\relax (x )}}{x^{2}} - \frac {n}{2 x^{2}} - \frac {\log {\relax (d )}}{2 x^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(d*(c*x**2+b*x)**n)/x**3,x)

[Out]

Piecewise((-n*log(b*x + c*x**2)/(2*x**2) - n/(4*x**2) - log(d)/(2*x**2) - c*n/(2*b*x) + c**2*n*log(b + c*x)/b*
*2 - c**2*n*log(b*x + c*x**2)/(2*b**2), Ne(b, 0)), (-n*log(c)/(2*x**2) - n*log(x)/x**2 - n/(2*x**2) - log(d)/(
2*x**2), True))

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