### 3.285 $$\int e^{a+b x} \log (x) \, dx$$

Optimal. Leaf size=26 $\frac{\log (x) e^{a+b x}}{b}-\frac{e^a \text{Ei}(b x)}{b}$

[Out]

-((E^a*ExpIntegralEi[b*x])/b) + (E^(a + b*x)*Log[x])/b

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Rubi [A]  time = 0.0354785, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 4, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.4, Rules used = {2194, 2554, 12, 2178} $\frac{\log (x) e^{a+b x}}{b}-\frac{e^a \text{Ei}(b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(a + b*x)*Log[x],x]

[Out]

-((E^a*ExpIntegralEi[b*x])/b) + (E^(a + b*x)*Log[x])/b

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !\$UseGamma === True

Rubi steps

\begin{align*} \int e^{a+b x} \log (x) \, dx &=\frac{e^{a+b x} \log (x)}{b}-\int \frac{e^{a+b x}}{b x} \, dx\\ &=\frac{e^{a+b x} \log (x)}{b}-\frac{\int \frac{e^{a+b x}}{x} \, dx}{b}\\ &=-\frac{e^a \text{Ei}(b x)}{b}+\frac{e^{a+b x} \log (x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0148314, size = 22, normalized size = 0.85 $\frac{e^a \left (e^{b x} \log (x)-\text{Ei}(b x)\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(a + b*x)*Log[x],x]

[Out]

(E^a*(-ExpIntegralEi[b*x] + E^(b*x)*Log[x]))/b

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Maple [A]  time = 0.016, size = 26, normalized size = 1. \begin{align*}{\frac{{{\rm e}^{bx+a}}\ln \left ( x \right ) }{b}}+{\frac{{{\rm e}^{a}}{\it Ei} \left ( 1,-bx \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*ln(x),x)

[Out]

exp(b*x+a)*ln(x)/b+1/b*exp(a)*Ei(1,-b*x)

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Maxima [A]  time = 1.13359, size = 32, normalized size = 1.23 \begin{align*} -\frac{{\rm Ei}\left (b x\right ) e^{a}}{b} + \frac{e^{\left (b x + a\right )} \log \left (x\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*log(x),x, algorithm="maxima")

[Out]

-Ei(b*x)*e^a/b + e^(b*x + a)*log(x)/b

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Fricas [A]  time = 2.14951, size = 53, normalized size = 2.04 \begin{align*} -\frac{{\rm Ei}\left (b x\right ) e^{a} - e^{\left (b x + a\right )} \log \left (x\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*log(x),x, algorithm="fricas")

[Out]

-(Ei(b*x)*e^a - e^(b*x + a)*log(x))/b

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Sympy [A]  time = 6.91723, size = 26, normalized size = 1. \begin{align*} \left (\begin{cases} x & \text{for}\: b = 0 \\\frac{e^{b x}}{b} & \text{otherwise} \end{cases}\right ) e^{a} \log{\left (x \right )} - \left (\begin{cases} x & \text{for}\: b = 0 \\\frac{\operatorname{Ei}{\left (b x \right )}}{b} & \text{otherwise} \end{cases}\right ) e^{a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*ln(x),x)

[Out]

Piecewise((x, Eq(b, 0)), (exp(b*x)/b, True))*exp(a)*log(x) - Piecewise((x, Eq(b, 0)), (Ei(b*x)/b, True))*exp(a
)

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Giac [A]  time = 1.30655, size = 32, normalized size = 1.23 \begin{align*} -\frac{{\rm Ei}\left (b x\right ) e^{a}}{b} + \frac{e^{\left (b x + a\right )} \log \left (x\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*log(x),x, algorithm="giac")

[Out]

-Ei(b*x)*e^a/b + e^(b*x + a)*log(x)/b