3.95.88 \(\int \frac {175+25 e^{-25+15 x}+375 e^{-25+15 x} x \log (x)}{x} \, dx\)

Optimal. Leaf size=15 \[ 25 \left (7+e^{5 (-5+3 x)}\right ) \log (x) \]

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Rubi [A]  time = 0.05, antiderivative size = 16, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {14, 2288} \begin {gather*} 25 e^{15 x-25} \log (x)+175 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(175 + 25*E^(-25 + 15*x) + 375*E^(-25 + 15*x)*x*Log[x])/x,x]

[Out]

175*Log[x] + 25*E^(-25 + 15*x)*Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {175}{x}+\frac {25 e^{-25+15 x} (1+15 x \log (x))}{x}\right ) \, dx\\ &=175 \log (x)+25 \int \frac {e^{-25+15 x} (1+15 x \log (x))}{x} \, dx\\ &=175 \log (x)+25 e^{-25+15 x} \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.40 \begin {gather*} \frac {25 \left (7 e^{25} \log (x)+e^{15 x} \log (x)\right )}{e^{25}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(175 + 25*E^(-25 + 15*x) + 375*E^(-25 + 15*x)*x*Log[x])/x,x]

[Out]

(25*(7*E^25*Log[x] + E^(15*x)*Log[x]))/E^25

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fricas [A]  time = 0.62, size = 12, normalized size = 0.80 \begin {gather*} 25 \, {\left (e^{\left (15 \, x - 25\right )} + 7\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((375*x*exp(15*x-25)*log(x)+25*exp(15*x-25)+175)/x,x, algorithm="fricas")

[Out]

25*(e^(15*x - 25) + 7)*log(x)

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giac [A]  time = 0.16, size = 18, normalized size = 1.20 \begin {gather*} 25 \, {\left (7 \, e^{25} \log \relax (x) + e^{\left (15 \, x\right )} \log \relax (x)\right )} e^{\left (-25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((375*x*exp(15*x-25)*log(x)+25*exp(15*x-25)+175)/x,x, algorithm="giac")

[Out]

25*(7*e^25*log(x) + e^(15*x)*log(x))*e^(-25)

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maple [A]  time = 0.06, size = 16, normalized size = 1.07




method result size



default \(25 \,{\mathrm e}^{15 x -25} \ln \relax (x )+175 \ln \relax (x )\) \(16\)
norman \(25 \,{\mathrm e}^{15 x -25} \ln \relax (x )+175 \ln \relax (x )\) \(16\)
risch \(25 \,{\mathrm e}^{15 x -25} \ln \relax (x )+175 \ln \relax (x )\) \(16\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((375*x*exp(15*x-25)*ln(x)+25*exp(15*x-25)+175)/x,x,method=_RETURNVERBOSE)

[Out]

25*exp(15*x-25)*ln(x)+175*ln(x)

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maxima [A]  time = 0.38, size = 15, normalized size = 1.00 \begin {gather*} 25 \, e^{\left (15 \, x - 25\right )} \log \relax (x) + 175 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((375*x*exp(15*x-25)*log(x)+25*exp(15*x-25)+175)/x,x, algorithm="maxima")

[Out]

25*e^(15*x - 25)*log(x) + 175*log(x)

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mupad [B]  time = 7.66, size = 15, normalized size = 1.00 \begin {gather*} 25\,{\mathrm {e}}^{-25}\,\ln \relax (x)\,\left ({\mathrm {e}}^{15\,x}+7\,{\mathrm {e}}^{25}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*exp(15*x - 25) + 375*x*exp(15*x - 25)*log(x) + 175)/x,x)

[Out]

25*exp(-25)*log(x)*(exp(15*x) + 7*exp(25))

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sympy [A]  time = 0.27, size = 15, normalized size = 1.00 \begin {gather*} 25 e^{15 x - 25} \log {\relax (x )} + 175 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((375*x*exp(15*x-25)*ln(x)+25*exp(15*x-25)+175)/x,x)

[Out]

25*exp(15*x - 25)*log(x) + 175*log(x)

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