3.177 \(\int x (a+b \log (c x))^p \, dx\)

Optimal. Leaf size=63 \[ \frac{2^{-p-1} e^{-\frac{2 a}{b}} (a+b \log (c x))^p \left (-\frac{a+b \log (c x)}{b}\right )^{-p} \text{Gamma}\left (p+1,-\frac{2 (a+b \log (c x))}{b}\right )}{c^2} \]

[Out]

(2^(-1 - p)*Gamma[1 + p, (-2*(a + b*Log[c*x]))/b]*(a + b*Log[c*x])^p)/(c^2*E^((2*a)/b)*(-((a + b*Log[c*x])/b))
^p)

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Rubi [A]  time = 0.0474052, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2309, 2181} \[ \frac{2^{-p-1} e^{-\frac{2 a}{b}} (a+b \log (c x))^p \left (-\frac{a+b \log (c x)}{b}\right )^{-p} \text{Gamma}\left (p+1,-\frac{2 (a+b \log (c x))}{b}\right )}{c^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Log[c*x])^p,x]

[Out]

(2^(-1 - p)*Gamma[1 + p, (-2*(a + b*Log[c*x]))/b]*(a + b*Log[c*x])^p)/(c^2*E^((2*a)/b)*(-((a + b*Log[c*x])/b))
^p)

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

\begin{align*} \int x (a+b \log (c x))^p \, dx &=\frac{\operatorname{Subst}\left (\int e^{2 x} (a+b x)^p \, dx,x,\log (c x)\right )}{c^2}\\ &=\frac{2^{-1-p} e^{-\frac{2 a}{b}} \Gamma \left (1+p,-\frac{2 (a+b \log (c x))}{b}\right ) (a+b \log (c x))^p \left (-\frac{a+b \log (c x)}{b}\right )^{-p}}{c^2}\\ \end{align*}

Mathematica [A]  time = 0.0312426, size = 63, normalized size = 1. \[ \frac{2^{-p-1} e^{-\frac{2 a}{b}} (a+b \log (c x))^p \left (-\frac{a+b \log (c x)}{b}\right )^{-p} \text{Gamma}\left (p+1,-\frac{2 (a+b \log (c x))}{b}\right )}{c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*Log[c*x])^p,x]

[Out]

(2^(-1 - p)*Gamma[1 + p, (-2*(a + b*Log[c*x]))/b]*(a + b*Log[c*x])^p)/(c^2*E^((2*a)/b)*(-((a + b*Log[c*x])/b))
^p)

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Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b\ln \left ( cx \right ) \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*ln(c*x))^p,x)

[Out]

int(x*(a+b*ln(c*x))^p,x)

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Maxima [A]  time = 1.26148, size = 59, normalized size = 0.94 \begin{align*} -\frac{{\left (b \log \left (c x\right ) + a\right )}^{p + 1} e^{\left (-\frac{2 \, a}{b}\right )} E_{-p}\left (-\frac{2 \,{\left (b \log \left (c x\right ) + a\right )}}{b}\right )}{b c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x))^p,x, algorithm="maxima")

[Out]

-(b*log(c*x) + a)^(p + 1)*e^(-2*a/b)*exp_integral_e(-p, -2*(b*log(c*x) + a)/b)/(b*c^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \log \left (c x\right ) + a\right )}^{p} x, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x))^p,x, algorithm="fricas")

[Out]

integral((b*log(c*x) + a)^p*x, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (a + b \log{\left (c x \right )}\right )^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*ln(c*x))**p,x)

[Out]

Integral(x*(a + b*log(c*x))**p, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \log \left (c x\right ) + a\right )}^{p} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x))^p,x, algorithm="giac")

[Out]

integrate((b*log(c*x) + a)^p*x, x)