3.180 \(\int \frac{(a+b \log (c x))^p}{x^2} \, dx\)

Optimal. Leaf size=52 \[ -c e^{a/b} (a+b \log (c x))^p \left (\frac{a+b \log (c x)}{b}\right )^{-p} \text{Gamma}\left (p+1,\frac{a+b \log (c x)}{b}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0493041, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2309, 2181} \[ -c e^{a/b} (a+b \log (c x))^p \left (\frac{a+b \log (c x)}{b}\right )^{-p} \text{Gamma}\left (p+1,\frac{a+b \log (c x)}{b}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x])^p/x^2,x]

[Out]

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

Mathematica [A]  time = 0.0329271, size = 48, normalized size = 0.92 \[ -c e^{a/b} \left (\frac{a}{b}+\log (c x)\right )^{-p} (a+b \log (c x))^p \text{Gamma}\left (p+1,\frac{a}{b}+\log (c x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x])^p/x^2,x]

[Out]

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

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Maple [F]  time = 0.054, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\ln \left ( cx \right ) \right ) ^{p}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x))^p/x^2,x)

[Out]

int((a+b*ln(c*x))^p/x^2,x)

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Maxima [A]  time = 1.25913, size = 54, normalized size = 1.04 \begin{align*} -\frac{{\left (b \log \left (c x\right ) + a\right )}^{p + 1} c e^{\frac{a}{b}} E_{-p}\left (\frac{b \log \left (c x\right ) + a}{b}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x))^p/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b \log \left (c x\right ) + a\right )}^{p}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x))^p/x^2,x, algorithm="fricas")

[Out]

integral((b*log(c*x) + a)^p/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \log{\left (c x \right )}\right )^{p}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x))**p/x**2,x)

[Out]

Integral((a + b*log(c*x))**p/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x\right ) + a\right )}^{p}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x))^p/x^2,x, algorithm="giac")

[Out]

integrate((b*log(c*x) + a)^p/x^2, x)