\(\int \frac {\log (d (b x+c x^2)^n)}{x^3} \, dx\) [67]

Optimal result

Integrand size = 18, antiderivative size = 72 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^3} \, dx=-\frac {n}{4 x^2}-\frac {c n}{2 b x}-\frac {c^2 n \log (x)}{2 b^2}+\frac {c^2 n \log (b+c x)}{2 b^2}-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{2 x^2} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2605, 78} \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^3} \, dx=-\frac {c^2 n \log (x)}{2 b^2}+\frac {c^2 n \log (b+c x)}{2 b^2}-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{2 x^2}-\frac {c n}{2 b x}-\frac {n}{4 x^2} \]

[In]

Int[Log[d*(b*x + c*x^2)^n]/x^3,x]

[Out]

-1/4*n/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 78

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 2605

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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.90 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^3} \, dx=\frac {1}{2} n \left (-\frac {1}{2 x^2}-\frac {c}{b x}-\frac {c^2 \log (x)}{b^2}+\frac {c^2 \log (b+c x)}{b^2}\right )-\frac {\log \left (d (x (b+c x))^n\right )}{2 x^2} \]

[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)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86

method	result	size
parts	\(-\frac {\ln \left (d \left (c \,x^{2}+b x \right )^{n}\right )}{2 x^{2}}+\frac {n \left (-\frac {1}{2 x^{2}}-\frac {c}{b x}-\frac {c^{2} \ln \left (x \right )}{b^{2}}+\frac {c^{2} \ln \left (x c +b \right )}{b^{2}}\right )}{2}\)	\(62\)
parallelrisch	\(-\frac {2 \ln \left (x \right ) x^{2} c^{2} n -2 \ln \left (x c +b \right ) x^{2} c^{2} n -2 x^{2} c^{2} n +2 x b c n +2 \ln \left (d \left (x \left (x c +b \right )\right )^{n}\right ) b^{2}+b^{2} n}{4 x^{2} b^{2}}\)	\(73\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^3} \, dx=\frac {2 \, c^{2} n x^{2} \log \left (c x + b\right ) - 2 \, c^{2} n x^{2} \log \left (x\right ) - 2 \, b c n x - 2 \, b^{2} n \log \left (c x^{2} + b x\right ) - b^{2} n - 2 \, b^{2} \log \left (d\right )}{4 \, b^{2} x^{2}} \]

[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)

Sympy [A] (verification not implemented)

Time = 1.90 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.31 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^3} \, dx=\begin {cases} - \frac {n}{4 x^{2}} - \frac {\log {\left (d \left (b x + c x^{2}\right )^{n} \right )}}{2 x^{2}} - \frac {c n}{2 b x} + \frac {c^{2} n \log {\left (b + c x \right )}}{b^{2}} - \frac {c^{2} \log {\left (d \left (b x + c x^{2}\right )^{n} \right )}}{2 b^{2}} & \text {for}\: b \neq 0 \\- \frac {n}{2 x^{2}} - \frac {\log {\left (d \left (c x^{2}\right )^{n} \right )}}{2 x^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^3} \, dx=\frac {1}{4} \, n {\left (\frac {2 \, c^{2} \log \left (c x + b\right )}{b^{2}} - \frac {2 \, c^{2} \log \left (x\right )}{b^{2}} - \frac {2 \, c x + b}{b x^{2}}\right )} - \frac {\log \left ({\left (c x^{2} + b x\right )}^{n} d\right )}{2 \, x^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.90 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^3} \, dx=\frac {c^{2} n \log \left (c x + b\right )}{2 \, b^{2}} - \frac {c^{2} n \log \left (x\right )}{2 \, b^{2}} - \frac {n \log \left (c x^{2} + b x\right )}{2 \, x^{2}} - \frac {2 \, c n x + b n + 2 \, b \log \left (d\right )}{4 \, b x^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.75 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^3} \, dx=\frac {c^2\,n\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )}{b^2}-\frac {\frac {n}{2}+\frac {c\,n\,x}{b}}{2\,x^2}-\frac {\ln \left (d\,{\left (c\,x^2+b\,x\right )}^n\right )}{2\,x^2} \]

[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)