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]

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

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Rubi [A]  time = 0.04, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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

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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.42, size = 23, normalized size = 0.88 \[ -\frac {{\rm Ei}\left (b x\right ) e^{a} - e^{\left (b x + a\right )} \log \relax (x)}{b} \]

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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giac [A]  time = 0.17, size = 24, normalized size = 0.92 \[ -\frac {{\rm Ei}\left (b x\right ) e^{a}}{b} + \frac {e^{\left (b x + a\right )} \log \relax (x)}{b} \]

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

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maple [A]  time = 0.62, size = 26, normalized size = 1.00 \[ \frac {\Ei \left (1, -b x \right ) {\mathrm e}^{a}}{b}+\frac {{\mathrm e}^{b x +a} \ln \relax (x )}{b} \]

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 = 0.86, size = 24, normalized size = 0.92 \[ -\frac {{\rm Ei}\left (b x\right ) e^{a}}{b} + \frac {e^{\left (b x + a\right )} \log \relax (x)}{b} \]

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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mupad [B]  time = 0.37, size = 20, normalized size = 0.77 \[ -\frac {{\mathrm {e}}^a\,\left (\mathrm {ei}\left (b\,x\right )-{\mathrm {e}}^{b\,x}\,\ln \relax (x)\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)*log(x),x)

[Out]

-(exp(a)*(ei(b*x) - exp(b*x)*log(x)))/b

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sympy [A]  time = 7.38, size = 26, normalized size = 1.00 \[ \left (\begin {cases} x & \text {for}\: b = 0 \\\frac {e^{b x}}{b} & \text {otherwise} \end {cases}\right ) e^{a} \log {\relax (x )} - \left (\begin {cases} x & \text {for}\: b = 0 \\\frac {\operatorname {Ei}{\left (b x \right )}}{b} & \text {otherwise} \end {cases}\right ) e^{a} \]

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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