\(\int \frac {\log (d (b x+c x^2)^n)}{x^5} \, dx\) [69]

Optimal result

Integrand size = 18, antiderivative size = 100 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^5} \, dx=-\frac {n}{16 x^4}-\frac {c n}{12 b x^3}+\frac {c^2 n}{8 b^2 x^2}-\frac {c^3 n}{4 b^3 x}-\frac {c^4 n \log (x)}{4 b^4}+\frac {c^4 n \log (b+c x)}{4 b^4}-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{4 x^4} \]

[Out]

-1/16*n/x^4-1/12*c*n/b/x^3+1/8*c^2*n/b^2/x^2-1/4*c^3*n/b^3/x-1/4*c^4*n*ln(x)/b^4+1/4*c^4*n*ln(c*x+b)/b^4-1/4*l
n(d*(c*x^2+b*x)^n)/x^4

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 100, 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} \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^5} \, dx=-\frac {c^4 n \log (x)}{4 b^4}+\frac {c^4 n \log (b+c x)}{4 b^4}-\frac {c^3 n}{4 b^3 x}+\frac {c^2 n}{8 b^2 x^2}-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{4 x^4}-\frac {c n}{12 b x^3}-\frac {n}{16 x^4} \]

[In]

Int[Log[d*(b*x + c*x^2)^n]/x^5,x]

[Out]

-1/16*n/x^4 - (c*n)/(12*b*x^3) + (c^2*n)/(8*b^2*x^2) - (c^3*n)/(4*b^3*x) - (c^4*n*Log[x])/(4*b^4) + (c^4*n*Log
[b + c*x])/(4*b^4) - Log[d*(b*x + c*x^2)^n]/(4*x^4)

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 )}{4 x^4}+\frac {1}{4} n \int \frac {b+2 c x}{x^5 (b+c x)} \, dx \\ & = -\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{4 x^4}+\frac {1}{4} n \int \left (\frac {1}{x^5}+\frac {c}{b x^4}-\frac {c^2}{b^2 x^3}+\frac {c^3}{b^3 x^2}-\frac {c^4}{b^4 x}+\frac {c^5}{b^4 (b+c x)}\right ) \, dx \\ & = -\frac {n}{16 x^4}-\frac {c n}{12 b x^3}+\frac {c^2 n}{8 b^2 x^2}-\frac {c^3 n}{4 b^3 x}-\frac {c^4 n \log (x)}{4 b^4}+\frac {c^4 n \log (b+c x)}{4 b^4}-\frac {\log \left (d \left (b x+c x^2\right )^n\right )}{4 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^5} \, dx=-\frac {b n \left (3 b^3+4 b^2 c x-6 b c^2 x^2+12 c^3 x^3\right )+12 c^4 n x^4 \log (x)-12 c^4 n x^4 \log (b+c x)+12 b^4 \log \left (d (x (b+c x))^n\right )}{48 b^4 x^4} \]

[In]

Integrate[Log[d*(b*x + c*x^2)^n]/x^5,x]

[Out]

-1/48*(b*n*(3*b^3 + 4*b^2*c*x - 6*b*c^2*x^2 + 12*c^3*x^3) + 12*c^4*n*x^4*Log[x] - 12*c^4*n*x^4*Log[b + c*x] +
12*b^4*Log[d*(x*(b + c*x))^n])/(b^4*x^4)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.84

method	result	size
parts	\(-\frac {\ln \left (d \left (c \,x^{2}+b x \right )^{n}\right )}{4 x^{4}}+\frac {n \left (-\frac {1}{4 x^{4}}-\frac {c}{3 b \,x^{3}}-\frac {c^{3}}{b^{3} x}+\frac {c^{2}}{2 b^{2} x^{2}}-\frac {c^{4} \ln \left (x \right )}{b^{4}}+\frac {c^{4} \ln \left (x c +b \right )}{b^{4}}\right )}{4}\)	\(84\)
parallelrisch	\(-\frac {12 \ln \left (x \right ) x^{4} c^{4} n -12 \ln \left (x c +b \right ) x^{4} c^{4} n -12 x^{4} c^{4} n +12 x^{3} b \,c^{3} n -6 x^{2} b^{2} c^{2} n +4 x \,b^{3} c n +12 \ln \left (d \left (x \left (x c +b \right )\right )^{n}\right ) b^{4}+3 b^{4} n}{48 x^{4} b^{4}}\)	\(98\)

[In]

int(ln(d*(c*x^2+b*x)^n)/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4*ln(d*(c*x^2+b*x)^n)/x^4+1/4*n*(-1/4/x^4-1/3*c/b/x^3-c^3/b^3/x+1/2*c^2/b^2/x^2-c^4/b^4*ln(x)+c^4/b^4*ln(c*
x+b))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^5} \, dx=\frac {12 \, c^{4} n x^{4} \log \left (c x + b\right ) - 12 \, c^{4} n x^{4} \log \left (x\right ) - 12 \, b c^{3} n x^{3} + 6 \, b^{2} c^{2} n x^{2} - 4 \, b^{3} c n x - 12 \, b^{4} n \log \left (c x^{2} + b x\right ) - 3 \, b^{4} n - 12 \, b^{4} \log \left (d\right )}{48 \, b^{4} x^{4}} \]

[In]

integrate(log(d*(c*x^2+b*x)^n)/x^5,x, algorithm="fricas")

[Out]

1/48*(12*c^4*n*x^4*log(c*x + b) - 12*c^4*n*x^4*log(x) - 12*b*c^3*n*x^3 + 6*b^2*c^2*n*x^2 - 4*b^3*c*n*x - 12*b^
4*n*log(c*x^2 + b*x) - 3*b^4*n - 12*b^4*log(d))/(b^4*x^4)

Sympy [A] (verification not implemented)

Time = 7.52 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.22 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^5} \, dx=\begin {cases} - \frac {n}{16 x^{4}} - \frac {\log {\left (d \left (b x + c x^{2}\right )^{n} \right )}}{4 x^{4}} - \frac {c n}{12 b x^{3}} + \frac {c^{2} n}{8 b^{2} x^{2}} - \frac {c^{3} n}{4 b^{3} x} + \frac {c^{4} n \log {\left (b + c x \right )}}{2 b^{4}} - \frac {c^{4} \log {\left (d \left (b x + c x^{2}\right )^{n} \right )}}{4 b^{4}} & \text {for}\: b \neq 0 \\- \frac {n}{8 x^{4}} - \frac {\log {\left (d \left (c x^{2}\right )^{n} \right )}}{4 x^{4}} & \text {otherwise} \end {cases} \]

[In]

integrate(ln(d*(c*x**2+b*x)**n)/x**5,x)

[Out]

Piecewise((-n/(16*x**4) - log(d*(b*x + c*x**2)**n)/(4*x**4) - c*n/(12*b*x**3) + c**2*n/(8*b**2*x**2) - c**3*n/
(4*b**3*x) + c**4*n*log(b + c*x)/(2*b**4) - c**4*log(d*(b*x + c*x**2)**n)/(4*b**4), Ne(b, 0)), (-n/(8*x**4) -
log(d*(c*x**2)**n)/(4*x**4), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^5} \, dx=\frac {1}{48} \, n {\left (\frac {12 \, c^{4} \log \left (c x + b\right )}{b^{4}} - \frac {12 \, c^{4} \log \left (x\right )}{b^{4}} - \frac {12 \, c^{3} x^{3} - 6 \, b c^{2} x^{2} + 4 \, b^{2} c x + 3 \, b^{3}}{b^{3} x^{4}}\right )} - \frac {\log \left ({\left (c x^{2} + b x\right )}^{n} d\right )}{4 \, x^{4}} \]

[In]

integrate(log(d*(c*x^2+b*x)^n)/x^5,x, algorithm="maxima")

[Out]

1/48*n*(12*c^4*log(c*x + b)/b^4 - 12*c^4*log(x)/b^4 - (12*c^3*x^3 - 6*b*c^2*x^2 + 4*b^2*c*x + 3*b^3)/(b^3*x^4)
) - 1/4*log((c*x^2 + b*x)^n*d)/x^4

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.92 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^5} \, dx=\frac {c^{4} n \log \left (c x + b\right )}{4 \, b^{4}} - \frac {c^{4} n \log \left (x\right )}{4 \, b^{4}} - \frac {n \log \left (c x^{2} + b x\right )}{4 \, x^{4}} - \frac {12 \, c^{3} n x^{3} - 6 \, b c^{2} n x^{2} + 4 \, b^{2} c n x + 3 \, b^{3} n + 12 \, b^{3} \log \left (d\right )}{48 \, b^{3} x^{4}} \]

[In]

integrate(log(d*(c*x^2+b*x)^n)/x^5,x, algorithm="giac")

[Out]

1/4*c^4*n*log(c*x + b)/b^4 - 1/4*c^4*n*log(x)/b^4 - 1/4*n*log(c*x^2 + b*x)/x^4 - 1/48*(12*c^3*n*x^3 - 6*b*c^2*
n*x^2 + 4*b^2*c*n*x + 3*b^3*n + 12*b^3*log(d))/(b^3*x^4)

Mupad [B] (verification not implemented)

Time = 1.76 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79 \[ \int \frac {\log \left (d \left (b x+c x^2\right )^n\right )}{x^5} \, dx=\frac {c^4\,n\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )}{2\,b^4}-\frac {\ln \left (d\,{\left (c\,x^2+b\,x\right )}^n\right )}{4\,x^4}-\frac {\frac {n}{4}-\frac {c^2\,n\,x^2}{2\,b^2}+\frac {c^3\,n\,x^3}{b^3}+\frac {c\,n\,x}{3\,b}}{4\,x^4} \]

[In]

int(log(d*(b*x + c*x^2)^n)/x^5,x)

[Out]

(c^4*n*atanh((2*c*x)/b + 1))/(2*b^4) - log(d*(b*x + c*x^2)^n)/(4*x^4) - (n/4 - (c^2*n*x^2)/(2*b^2) + (c^3*n*x^
3)/b^3 + (c*n*x)/(3*b))/(4*x^4)