3.93.40 \(\int \frac {1}{18} (e^{\frac {5+27 x}{9 x}} (5-18 x) \log (5)-54 x \log (5)) \, dx\)

Optimal. Leaf size=28 \[ \left (3+e^{3+\frac {5}{9 x}}\right ) x \left (2 x-\frac {1}{2} x (4+\log (5))\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 30, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {12, 2225, 2226, 2206, 2210, 2214} \begin {gather*} -\frac {1}{2} e^{\frac {5}{9 x}+3} x^2 \log (5)-\frac {3}{2} x^2 \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((5 + 27*x)/(9*x))*(5 - 18*x)*Log[5] - 54*x*Log[5])/18,x]

[Out]

(-3*x^2*Log[5])/2 - (E^(3 + 5/(9*x))*x^2*Log[5])/2

Rule 12

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

Rule 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 2225

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && BinomialQ[v, x] &&  !(LinearMatchQ[u, x] && BinomialMatchQ[v, x])

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{18} \int \left (e^{\frac {5+27 x}{9 x}} (5-18 x) \log (5)-54 x \log (5)\right ) \, dx\\ &=-\frac {3}{2} x^2 \log (5)+\frac {1}{18} \log (5) \int e^{\frac {5+27 x}{9 x}} (5-18 x) \, dx\\ &=-\frac {3}{2} x^2 \log (5)+\frac {1}{18} \log (5) \int e^{3+\frac {5}{9 x}} (5-18 x) \, dx\\ &=-\frac {3}{2} x^2 \log (5)+\frac {1}{18} \log (5) \int \left (5 e^{3+\frac {5}{9 x}}-18 e^{3+\frac {5}{9 x}} x\right ) \, dx\\ &=-\frac {3}{2} x^2 \log (5)+\frac {1}{18} (5 \log (5)) \int e^{3+\frac {5}{9 x}} \, dx-\log (5) \int e^{3+\frac {5}{9 x}} x \, dx\\ &=\frac {5}{18} e^{3+\frac {5}{9 x}} x \log (5)-\frac {3}{2} x^2 \log (5)-\frac {1}{2} e^{3+\frac {5}{9 x}} x^2 \log (5)+\frac {1}{162} (25 \log (5)) \int \frac {e^{3+\frac {5}{9 x}}}{x} \, dx-\frac {1}{18} (5 \log (5)) \int e^{3+\frac {5}{9 x}} \, dx\\ &=-\frac {3}{2} x^2 \log (5)-\frac {1}{2} e^{3+\frac {5}{9 x}} x^2 \log (5)-\frac {25}{162} e^3 \text {Ei}\left (\frac {5}{9 x}\right ) \log (5)-\frac {1}{162} (25 \log (5)) \int \frac {e^{3+\frac {5}{9 x}}}{x} \, dx\\ &=-\frac {3}{2} x^2 \log (5)-\frac {1}{2} e^{3+\frac {5}{9 x}} x^2 \log (5)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 28, normalized size = 1.00 \begin {gather*} -\frac {1}{18} \left (27 x^2+9 e^{3+\frac {5}{9 x}} x^2\right ) \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((5 + 27*x)/(9*x))*(5 - 18*x)*Log[5] - 54*x*Log[5])/18,x]

[Out]

-1/18*((27*x^2 + 9*E^(3 + 5/(9*x))*x^2)*Log[5])

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fricas [A]  time = 0.51, size = 26, normalized size = 0.93 \begin {gather*} -\frac {1}{2} \, x^{2} e^{\left (\frac {27 \, x + 5}{9 \, x}\right )} \log \relax (5) - \frac {3}{2} \, x^{2} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(-18*x+5)*log(5)*exp(1/9*(27*x+5)/x)-3*x*log(5),x, algorithm="fricas")

[Out]

-1/2*x^2*e^(1/9*(27*x + 5)/x)*log(5) - 3/2*x^2*log(5)

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giac [B]  time = 0.25, size = 48, normalized size = 1.71 \begin {gather*} -\frac {3}{2} \, x^{2} \log \relax (5) - \frac {25 \, e^{\left (\frac {27 \, x + 5}{9 \, x}\right )} \log \relax (5)}{2 \, {\left (\frac {{\left (27 \, x + 5\right )}^{2}}{x^{2}} - \frac {54 \, {\left (27 \, x + 5\right )}}{x} + 729\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(-18*x+5)*log(5)*exp(1/9*(27*x+5)/x)-3*x*log(5),x, algorithm="giac")

[Out]

-3/2*x^2*log(5) - 25/2*e^(1/9*(27*x + 5)/x)*log(5)/((27*x + 5)^2/x^2 - 54*(27*x + 5)/x + 729)

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maple [A]  time = 0.20, size = 24, normalized size = 0.86

method	result	size
derivativedivides	\(-\frac {3 x^{2} \ln \relax (5)}{2}-\frac {x^{2} \ln \relax (5) {\mathrm e}^{\frac {5}{9 x}+3}}{2}\)	\(24\)
default	\(-\frac {3 x^{2} \ln \relax (5)}{2}-\frac {x^{2} \ln \relax (5) {\mathrm e}^{\frac {5}{9 x}+3}}{2}\)	\(24\)
norman	\(-\frac {3 x^{2} \ln \relax (5)}{2}-\frac {x^{2} \ln \relax (5) {\mathrm e}^{\frac {27 x +5}{9 x}}}{2}\)	\(27\)
risch	\(-\frac {3 x^{2} \ln \relax (5)}{2}-\frac {x^{2} \ln \relax (5) {\mathrm e}^{\frac {27 x +5}{9 x}}}{2}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/18*(-18*x+5)*ln(5)*exp(1/9*(27*x+5)/x)-3*x*ln(5),x,method=_RETURNVERBOSE)

[Out]

-3/2*x^2*ln(5)-1/2*x^2*ln(5)*exp(5/9/x+3)

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maxima [C]  time = 0.39, size = 34, normalized size = 1.21 \begin {gather*} -\frac {3}{2} \, x^{2} \log \relax (5) - \frac {25}{162} \, {\left (e^{3} \Gamma \left (-1, -\frac {5}{9 \, x}\right ) + 2 \, e^{3} \Gamma \left (-2, -\frac {5}{9 \, x}\right )\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(-18*x+5)*log(5)*exp(1/9*(27*x+5)/x)-3*x*log(5),x, algorithm="maxima")

[Out]

-3/2*x^2*log(5) - 25/162*(e^3*gamma(-1, -5/9/x) + 2*e^3*gamma(-2, -5/9/x))*log(5)

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mupad [B]  time = 6.97, size = 17, normalized size = 0.61 \begin {gather*} -\frac {x^2\,\ln \relax (5)\,\left ({\mathrm {e}}^{\frac {5}{9\,x}+3}+3\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(- 3*x*log(5) - (exp((3*x + 5/9)/x)*log(5)*(18*x - 5))/18,x)

[Out]

-(x^2*log(5)*(exp(5/(9*x) + 3) + 3))/2

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sympy [A]  time = 0.20, size = 29, normalized size = 1.04 \begin {gather*} - \frac {x^{2} e^{\frac {3 x + \frac {5}{9}}{x}} \log {\relax (5 )}}{2} - \frac {3 x^{2} \log {\relax (5 )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(-18*x+5)*ln(5)*exp(1/9*(27*x+5)/x)-3*x*ln(5),x)

[Out]

-x**2*exp((3*x + 5/9)/x)*log(5)/2 - 3*x**2*log(5)/2

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