3.23.20 \(\int \frac {e^x (4+(-4+4 x) \log (5 x))}{x^2} \, dx\)

Optimal. Leaf size=12 \[ \frac {4 e^x \log (5 x)}{x} \]

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Rubi [A]  time = 0.23, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {6742, 2177, 2178, 2197, 2554} \begin {gather*} \frac {4 e^x \log (5 x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(4 + (-4 + 4*x)*Log[5*x]))/x^2,x]

[Out]

(4*E^x*Log[5*x])/x

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, 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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {4 e^x}{x^2}+\frac {4 e^x (-1+x) \log (5 x)}{x^2}\right ) \, dx\\ &=4 \int \frac {e^x}{x^2} \, dx+4 \int \frac {e^x (-1+x) \log (5 x)}{x^2} \, dx\\ &=-\frac {4 e^x}{x}+\frac {4 e^x \log (5 x)}{x}-4 \int \frac {e^x}{x^2} \, dx+4 \int \frac {e^x}{x} \, dx\\ &=4 \text {Ei}(x)+\frac {4 e^x \log (5 x)}{x}-4 \int \frac {e^x}{x} \, dx\\ &=\frac {4 e^x \log (5 x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 12, normalized size = 1.00 \begin {gather*} \frac {4 e^x \log (5 x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(4 + (-4 + 4*x)*Log[5*x]))/x^2,x]

[Out]

(4*E^x*Log[5*x])/x

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fricas [A]  time = 0.66, size = 11, normalized size = 0.92 \begin {gather*} \frac {4 \, e^{x} \log \left (5 \, x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-4)*log(5*x)+4)/x^2/exp(-x),x, algorithm="fricas")

[Out]

4*e^x*log(5*x)/x

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giac [A]  time = 0.16, size = 11, normalized size = 0.92 \begin {gather*} \frac {4 \, e^{x} \log \left (5 \, x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-4)*log(5*x)+4)/x^2/exp(-x),x, algorithm="giac")

[Out]

4*e^x*log(5*x)/x

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maple [A]  time = 0.06, size = 12, normalized size = 1.00

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{x} \ln \left (5 x \right )}{x}\)	\(12\)
norman	\(\frac {4 \,{\mathrm e}^{x} \ln \left (5 x \right )}{x}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x-4)*ln(5*x)+4)/x^2/exp(-x),x,method=_RETURNVERBOSE)

[Out]

4/x*exp(x)*ln(5*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {4 \, e^{x} \log \relax (x)}{x} + 4 \, \Gamma \left (-1, -x\right ) + 4 \, \int \frac {{\left (x \log \relax (5) - \log \relax (5) - 1\right )} e^{x}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-4)*log(5*x)+4)/x^2/exp(-x),x, algorithm="maxima")

[Out]

4*e^x*log(x)/x + 4*gamma(-1, -x) + 4*integrate((x*log(5) - log(5) - 1)*e^x/x^2, x)

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mupad [B]  time = 1.29, size = 11, normalized size = 0.92 \begin {gather*} \frac {4\,\ln \left (5\,x\right )\,{\mathrm {e}}^x}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(log(5*x)*(4*x - 4) + 4))/x^2,x)

[Out]

(4*log(5*x)*exp(x))/x

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sympy [A]  time = 0.25, size = 10, normalized size = 0.83 \begin {gather*} \frac {4 e^{x} \log {\left (5 x \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-4)*ln(5*x)+4)/x**2/exp(-x),x)

[Out]

4*exp(x)*log(5*x)/x

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