\(\int \frac {\log (d (b x+c x^2)^n)}{x^4} \, dx\) [68]

Optimal result

Integrand size = 18, antiderivative size = 86 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^4} \, dx=-\frac {n}{9 x^3}-\frac {c n}{6 b x^2}+\frac {c^2 n}{3 b^2 x}+\frac {c^3 n \log (x)}{3 b^3}-\frac {c^3 n \log (b+c x)}{3 b^3}-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{3 x^3} \]

[Out]

-1/9*n/x^3-1/6*c*n/b/x^2+1/3*c^2*n/b^2/x+1/3*c^3*n*ln(x)/b^3-1/3*c^3*n*ln(c*x+b)/b^3-1/3*ln(d*(c*x^2+b*x)^n)/x
^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 86, 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} \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^4} \, dx=\frac {c^3 n \log (x)}{3 b^3}-\frac {c^3 n \log (b+c x)}{3 b^3}+\frac {c^2 n}{3 b^2 x}-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{3 x^3}-\frac {c n}{6 b x^2}-\frac {n}{9 x^3} \]

[In]

Int[Log[d*(b*x + c*x^2)^n]/x^4,x]

[Out]

-1/9*n/x^3 - (c*n)/(6*b*x^2) + (c^2*n)/(3*b^2*x) + (c^3*n*Log[x])/(3*b^3) - (c^3*n*Log[b + c*x])/(3*b^3) - Log
[d*(b*x + c*x^2)^n]/(3*x^3)

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 )}{3 x^3}+\frac {1}{3} n \int \frac {b+2 c x}{x^4 (b+c x)} \, dx \\ & = -\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{3 x^3}+\frac {1}{3} n \int \left (\frac {1}{x^4}+\frac {c}{b x^3}-\frac {c^2}{b^2 x^2}+\frac {c^3}{b^3 x}-\frac {c^4}{b^3 (b+c x)}\right ) \, dx \\ & = -\frac {n}{9 x^3}-\frac {c n}{6 b x^2}+\frac {c^2 n}{3 b^2 x}+\frac {c^3 n \log (x)}{3 b^3}-\frac {c^3 n \log (b+c x)}{3 b^3}-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^4} \, dx=\frac {1}{3} n \left (-\frac {1}{3 x^3}-\frac {c}{2 b x^2}+\frac {c^2}{b^2 x}+\frac {c^3 \log (x)}{b^3}-\frac {c^3 \log (b+c x)}{b^3}\right )-\frac {\log \left (d (x (b+c x))^n\right )}{3 x^3} \]

[In]

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

[Out]

(n*(-1/3*1/x^3 - c/(2*b*x^2) + c^2/(b^2*x) + (c^3*Log[x])/b^3 - (c^3*Log[b + c*x])/b^3))/3 - Log[d*(x*(b + c*x
))^n]/(3*x^3)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84

method	result	size
parts	\(-\frac {\ln \left (d \left (c \,x^{2}+b x \right )^{n}\right )}{3 x^{3}}+\frac {n \left (-\frac {1}{3 x^{3}}-\frac {c}{2 b \,x^{2}}+\frac {c^{3} \ln \left (x \right )}{b^{3}}+\frac {c^{2}}{b^{2} x}-\frac {c^{3} \ln \left (x c +b \right )}{b^{3}}\right )}{3}\)	\(72\)
parallelrisch	\(-\frac {-6 \ln \left (x \right ) x^{3} c^{3} n +6 \ln \left (x c +b \right ) x^{3} c^{3} n +6 x^{3} c^{3} n -6 x^{2} b \,c^{2} n +3 x \,b^{2} c n +6 \ln \left (d \left (x \left (x c +b \right )\right )^{n}\right ) b^{3}+2 b^{3} n}{18 x^{3} b^{3}}\)	\(86\)

[In]

int(ln(d*(c*x^2+b*x)^n)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*ln(d*(c*x^2+b*x)^n)/x^3+1/3*n*(-1/3/x^3-1/2*c/b/x^2+c^3/b^3*ln(x)+c^2/b^2/x-c^3/b^3*ln(c*x+b))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^4} \, dx=-\frac {6 \, c^{3} n x^{3} \log \left (c x + b\right ) - 6 \, c^{3} n x^{3} \log \left (x\right ) - 6 \, b c^{2} n x^{2} + 3 \, b^{2} c n x + 6 \, b^{3} n \log \left (c x^{2} + b x\right ) + 2 \, b^{3} n + 6 \, b^{3} \log \left (d\right )}{18 \, b^{3} x^{3}} \]

[In]

integrate(log(d*(c*x^2+b*x)^n)/x^4,x, algorithm="fricas")

[Out]

-1/18*(6*c^3*n*x^3*log(c*x + b) - 6*c^3*n*x^3*log(x) - 6*b*c^2*n*x^2 + 3*b^2*c*n*x + 6*b^3*n*log(c*x^2 + b*x)
+ 2*b^3*n + 6*b^3*log(d))/(b^3*x^3)

Sympy [A] (verification not implemented)

Time = 3.85 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^4} \, dx=\begin {cases} - \frac {n}{9 x^{3}} - \frac {\log {\left (d \left (b x + c x^{2}\right )^{n} \right )}}{3 x^{3}} - \frac {c n}{6 b x^{2}} + \frac {c^{2} n}{3 b^{2} x} - \frac {2 c^{3} n \log {\left (b + c x \right )}}{3 b^{3}} + \frac {c^{3} \log {\left (d \left (b x + c x^{2}\right )^{n} \right )}}{3 b^{3}} & \text {for}\: b \neq 0 \\- \frac {2 n}{9 x^{3}} - \frac {\log {\left (d \left (c x^{2}\right )^{n} \right )}}{3 x^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(ln(d*(c*x**2+b*x)**n)/x**4,x)

[Out]

Piecewise((-n/(9*x**3) - log(d*(b*x + c*x**2)**n)/(3*x**3) - c*n/(6*b*x**2) + c**2*n/(3*b**2*x) - 2*c**3*n*log
(b + c*x)/(3*b**3) + c**3*log(d*(b*x + c*x**2)**n)/(3*b**3), Ne(b, 0)), (-2*n/(9*x**3) - log(d*(c*x**2)**n)/(3
*x**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^4} \, dx=-\frac {1}{18} \, n {\left (\frac {6 \, c^{3} \log \left (c x + b\right )}{b^{3}} - \frac {6 \, c^{3} \log \left (x\right )}{b^{3}} - \frac {6 \, c^{2} x^{2} - 3 \, b c x - 2 \, b^{2}}{b^{2} x^{3}}\right )} - \frac {\log \left ({\left (c x^{2} + b x\right )}^{n} d\right )}{3 \, x^{3}} \]

[In]

integrate(log(d*(c*x^2+b*x)^n)/x^4,x, algorithm="maxima")

[Out]

-1/18*n*(6*c^3*log(c*x + b)/b^3 - 6*c^3*log(x)/b^3 - (6*c^2*x^2 - 3*b*c*x - 2*b^2)/(b^2*x^3)) - 1/3*log((c*x^2
+ b*x)^n*d)/x^3

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^4} \, dx=-\frac {c^{3} n \log \left (c x + b\right )}{3 \, b^{3}} + \frac {c^{3} n \log \left (x\right )}{3 \, b^{3}} - \frac {n \log \left (c x^{2} + b x\right )}{3 \, x^{3}} + \frac {6 \, c^{2} n x^{2} - 3 \, b c n x - 2 \, b^{2} n - 6 \, b^{2} \log \left (d\right )}{18 \, b^{2} x^{3}} \]

[In]

integrate(log(d*(c*x^2+b*x)^n)/x^4,x, algorithm="giac")

[Out]

-1/3*c^3*n*log(c*x + b)/b^3 + 1/3*c^3*n*log(x)/b^3 - 1/3*n*log(c*x^2 + b*x)/x^3 + 1/18*(6*c^2*n*x^2 - 3*b*c*n*
x - 2*b^2*n - 6*b^2*log(d))/(b^2*x^3)

Mupad [B] (verification not implemented)

Time = 1.68 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^4} \, dx=-\frac {\ln \left (d\,{\left (c\,x^2+b\,x\right )}^n\right )}{3\,x^3}-\frac {\frac {n}{3}-\frac {c^2\,n\,x^2}{b^2}+\frac {c\,n\,x}{2\,b}}{3\,x^3}-\frac {2\,c^3\,n\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )}{3\,b^3} \]

[In]

int(log(d*(b*x + c*x^2)^n)/x^4,x)

[Out]

- log(d*(b*x + c*x^2)^n)/(3*x^3) - (n/3 - (c^2*n*x^2)/b^2 + (c*n*x)/(2*b))/(3*x^3) - (2*c^3*n*atanh((2*c*x)/b
+ 1))/(3*b^3)