\(\int x^4 \log (d (b x+c x^2)^n) \, dx\) [60]

Optimal result

Integrand size = 18, antiderivative size = 99 \[ \int x^4 \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=-\frac {b^4 n x}{5 c^4}+\frac {b^3 n x^2}{10 c^3}-\frac {b^2 n x^3}{15 c^2}+\frac {b n x^4}{20 c}-\frac {2 n x^5}{25}+\frac {b^5 n \log (b+c x)}{5 c^5}+\frac {1}{5} x^5 \log \left (d \left (b x+c x^2\right )^n\right ) \]

[Out]

-1/5*b^4*n*x/c^4+1/10*b^3*n*x^2/c^3-1/15*b^2*n*x^3/c^2+1/20*b*n*x^4/c-2/25*n*x^5+1/5*b^5*n*ln(c*x+b)/c^5+1/5*x
^5*ln(d*(c*x^2+b*x)^n)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 99, 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} \[ \int x^4 \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\frac {b^5 n \log (b+c x)}{5 c^5}-\frac {b^4 n x}{5 c^4}+\frac {b^3 n x^2}{10 c^3}-\frac {b^2 n x^3}{15 c^2}+\frac {1}{5} x^5 \log \left (d \left (b x+c x^2\right )^n\right )+\frac {b n x^4}{20 c}-\frac {2 n x^5}{25} \]

[In]

Int[x^4*Log[d*(b*x + c*x^2)^n],x]

[Out]

-1/5*(b^4*n*x)/c^4 + (b^3*n*x^2)/(10*c^3) - (b^2*n*x^3)/(15*c^2) + (b*n*x^4)/(20*c) - (2*n*x^5)/25 + (b^5*n*Lo
g[b + c*x])/(5*c^5) + (x^5*Log[d*(b*x + c*x^2)^n])/5

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 {1}{5} x^5 \log \left (d \left (b x+c x^2\right )^n\right )-\frac {1}{5} n \int \frac {x^4 (b+2 c x)}{b+c x} \, dx \\ & = \frac {1}{5} x^5 \log \left (d \left (b x+c x^2\right )^n\right )-\frac {1}{5} n \int \left (\frac {b^4}{c^4}-\frac {b^3 x}{c^3}+\frac {b^2 x^2}{c^2}-\frac {b x^3}{c}+2 x^4-\frac {b^5}{c^4 (b+c x)}\right ) \, dx \\ & = -\frac {b^4 n x}{5 c^4}+\frac {b^3 n x^2}{10 c^3}-\frac {b^2 n x^3}{15 c^2}+\frac {b n x^4}{20 c}-\frac {2 n x^5}{25}+\frac {b^5 n \log (b+c x)}{5 c^5}+\frac {1}{5} x^5 \log \left (d \left (b x+c x^2\right )^n\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.86 \[ \int x^4 \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\frac {c n x \left (-60 b^4+30 b^3 c x-20 b^2 c^2 x^2+15 b c^3 x^3-24 c^4 x^4\right )+60 b^5 n \log (b+c x)+60 c^5 x^5 \log \left (d (x (b+c x))^n\right )}{300 c^5} \]

[In]

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

[Out]

(c*n*x*(-60*b^4 + 30*b^3*c*x - 20*b^2*c^2*x^2 + 15*b*c^3*x^3 - 24*c^4*x^4) + 60*b^5*n*Log[b + c*x] + 60*c^5*x^
5*Log[d*(x*(b + c*x))^n])/(300*c^5)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.87

method	result	size
parts	\(\frac {x^{5} \ln \left (d \left (c \,x^{2}+b x \right )^{n}\right )}{5}-\frac {n \left (\frac {\frac {2}{5} c^{4} x^{5}-\frac {1}{4} b \,x^{4} c^{3}+\frac {1}{3} b^{2} c^{2} x^{3}-\frac {1}{2} b^{3} c \,x^{2}+b^{4} x}{c^{4}}-\frac {b^{5} \ln \left (x c +b \right )}{c^{5}}\right )}{5}\)	\(86\)
parallelrisch	\(-\frac {-60 x^{5} \ln \left (d \left (x \left (x c +b \right )\right )^{n}\right ) c^{5} n +24 x^{5} c^{5} n^{2}-15 x^{4} b \,c^{4} n^{2}+20 x^{3} b^{2} c^{3} n^{2}-30 x^{2} b^{3} c^{2} n^{2}+60 \ln \left (x \right ) b^{5} n^{2}+60 x \,b^{4} c \,n^{2}-60 \ln \left (d \left (x \left (x c +b \right )\right )^{n}\right ) b^{5} n -60 b^{5} n^{2}}{300 c^{5} n}\)	\(128\)

[In]

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

[Out]

1/5*x^5*ln(d*(c*x^2+b*x)^n)-1/5*n*(1/c^4*(2/5*c^4*x^5-1/4*b*x^4*c^3+1/3*b^2*c^2*x^3-1/2*b^3*c*x^2+b^4*x)-b^5/c
^5*ln(c*x+b))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.99 \[ \int x^4 \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\frac {60 \, c^{5} n x^{5} \log \left (c x^{2} + b x\right ) - 24 \, c^{5} n x^{5} + 60 \, c^{5} x^{5} \log \left (d\right ) + 15 \, b c^{4} n x^{4} - 20 \, b^{2} c^{3} n x^{3} + 30 \, b^{3} c^{2} n x^{2} - 60 \, b^{4} c n x + 60 \, b^{5} n \log \left (c x + b\right )}{300 \, c^{5}} \]

[In]

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

[Out]

1/300*(60*c^5*n*x^5*log(c*x^2 + b*x) - 24*c^5*n*x^5 + 60*c^5*x^5*log(d) + 15*b*c^4*n*x^4 - 20*b^2*c^3*n*x^3 +
30*b^3*c^2*n*x^2 - 60*b^4*c*n*x + 60*b^5*n*log(c*x + b))/c^5

Sympy [A] (verification not implemented)

Time = 4.45 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.13 \[ \int x^4 \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\begin {cases} \frac {b^{5} n \log {\left (b + c x \right )}}{5 c^{5}} - \frac {b^{4} n x}{5 c^{4}} + \frac {b^{3} n x^{2}}{10 c^{3}} - \frac {b^{2} n x^{3}}{15 c^{2}} + \frac {b n x^{4}}{20 c} - \frac {2 n x^{5}}{25} + \frac {x^{5} \log {\left (d \left (b x + c x^{2}\right )^{n} \right )}}{5} & \text {for}\: c \neq 0 \\- \frac {n x^{5}}{25} + \frac {x^{5} \log {\left (d \left (b x\right )^{n} \right )}}{5} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((b**5*n*log(b + c*x)/(5*c**5) - b**4*n*x/(5*c**4) + b**3*n*x**2/(10*c**3) - b**2*n*x**3/(15*c**2) +
b*n*x**4/(20*c) - 2*n*x**5/25 + x**5*log(d*(b*x + c*x**2)**n)/5, Ne(c, 0)), (-n*x**5/25 + x**5*log(d*(b*x)**n)
/5, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.88 \[ \int x^4 \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\frac {1}{5} \, x^{5} \log \left ({\left (c x^{2} + b x\right )}^{n} d\right ) + \frac {1}{300} \, n {\left (\frac {60 \, b^{5} \log \left (c x + b\right )}{c^{5}} - \frac {24 \, c^{4} x^{5} - 15 \, b c^{3} x^{4} + 20 \, b^{2} c^{2} x^{3} - 30 \, b^{3} c x^{2} + 60 \, b^{4} x}{c^{4}}\right )} \]

[In]

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

[Out]

1/5*x^5*log((c*x^2 + b*x)^n*d) + 1/300*n*(60*b^5*log(c*x + b)/c^5 - (24*c^4*x^5 - 15*b*c^3*x^4 + 20*b^2*c^2*x^
3 - 30*b^3*c*x^2 + 60*b^4*x)/c^4)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.90 \[ \int x^4 \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\frac {1}{5} \, n x^{5} \log \left (c x^{2} + b x\right ) - \frac {1}{25} \, {\left (2 \, n - 5 \, \log \left (d\right )\right )} x^{5} + \frac {b n x^{4}}{20 \, c} - \frac {b^{2} n x^{3}}{15 \, c^{2}} + \frac {b^{3} n x^{2}}{10 \, c^{3}} - \frac {b^{4} n x}{5 \, c^{4}} + \frac {b^{5} n \log \left (c x + b\right )}{5 \, c^{5}} \]

[In]

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

[Out]

1/5*n*x^5*log(c*x^2 + b*x) - 1/25*(2*n - 5*log(d))*x^5 + 1/20*b*n*x^4/c - 1/15*b^2*n*x^3/c^2 + 1/10*b^3*n*x^2/
c^3 - 1/5*b^4*n*x/c^4 + 1/5*b^5*n*log(c*x + b)/c^5

Mupad [B] (verification not implemented)

Time = 1.57 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.86 \[ \int x^4 \log \left (d \left (b x+c x^2\right )^n\right ) \, dx=\frac {x^5\,\ln \left (d\,{\left (c\,x^2+b\,x\right )}^n\right )}{5}-\frac {2\,n\,x^5}{25}-\frac {b^2\,n\,x^3}{15\,c^2}+\frac {b^3\,n\,x^2}{10\,c^3}+\frac {b^5\,n\,\ln \left (b+c\,x\right )}{5\,c^5}+\frac {b\,n\,x^4}{20\,c}-\frac {b^4\,n\,x}{5\,c^4} \]

[In]

int(x^4*log(d*(b*x + c*x^2)^n),x)

[Out]

(x^5*log(d*(b*x + c*x^2)^n))/5 - (2*n*x^5)/25 - (b^2*n*x^3)/(15*c^2) + (b^3*n*x^2)/(10*c^3) + (b^5*n*log(b + c
*x))/(5*c^5) + (b*n*x^4)/(20*c) - (b^4*n*x)/(5*c^4)