3.63 \(\int x \log (d (b x+c x^2)^n) \, dx\)

Optimal. Leaf size=57 \[ -\frac {b^2 n \log (b+c x)}{2 c^2}+\frac {1}{2} x^2 \log \left (d \left (b x+c x^2\right )^n\right )+\frac {b n x}{2 c}-\frac {n x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2525, 77} \[ -\frac {b^2 n \log (b+c x)}{2 c^2}+\frac {1}{2} x^2 \log \left (d \left (b x+c x^2\right )^n\right )+\frac {b n x}{2 c}-\frac {n x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[x*Log[d*(b*x + c*x^2)^n],x]

[Out]

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.86 \[ \frac {1}{2} x^2 \log \left (d (x (b+c x))^n\right )-\frac {1}{2} n \left (\frac {b^2 \log (b+c x)}{c^2}-\frac {b x}{c}+x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Log[d*(b*x + c*x^2)^n],x]

[Out]

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

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fricas [A]  time = 0.43, size = 59, normalized size = 1.04 \[ \frac {c^{2} n x^{2} \log \left (c x^{2} + b x\right ) - c^{2} n x^{2} + c^{2} x^{2} \log \relax (d) + b c n x - b^{2} n \log \left (c x + b\right )}{2 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 51, normalized size = 0.89 \[ \frac {1}{2} \, n x^{2} \log \left (c x^{2} + b x\right ) - \frac {1}{2} \, {\left (n - \log \relax (d)\right )} x^{2} + \frac {b n x}{2 \, c} - \frac {b^{2} n \log \left (c x + b\right )}{2 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int x \ln \left (d \left (c \,x^{2}+b x \right )^{n}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 51, normalized size = 0.89 \[ \frac {1}{2} \, x^{2} \log \left ({\left (c x^{2} + b x\right )}^{n} d\right ) - \frac {1}{2} \, n {\left (\frac {b^{2} \log \left (c x + b\right )}{c^{2}} + \frac {c x^{2} - b x}{c}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.37, size = 49, normalized size = 0.86 \[ \frac {x^2\,\ln \left (d\,{\left (c\,x^2+b\,x\right )}^n\right )}{2}-\frac {n\,x^2}{2}+\frac {b\,n\,x}{2\,c}-\frac {b^2\,n\,\ln \left (b+c\,x\right )}{2\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(d*(b*x + c*x^2)^n),x)

[Out]

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

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sympy [A]  time = 2.69, size = 92, normalized size = 1.61 \[ \begin {cases} - \frac {b^{2} n \log {\left (b + c x \right )}}{2 c^{2}} + \frac {b n x}{2 c} + \frac {n x^{2} \log {\left (b x + c x^{2} \right )}}{2} - \frac {n x^{2}}{2} + \frac {x^{2} \log {\relax (d )}}{2} & \text {for}\: c \neq 0 \\\frac {n x^{2} \log {\relax (b )}}{2} + \frac {n x^{2} \log {\relax (x )}}{2} - \frac {n x^{2}}{4} + \frac {x^{2} \log {\relax (d )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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