3.14.92 \(\int \frac {e^{\frac {1}{4} (15-4 e^x+20 x-5 \log (3))} (-e^x x \log (4)+(-1+5 x) \log (4))}{x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {e^{-e^x+5 \left (1+x+\frac {1}{4} (-1-\log (3))\right )} \log (4)}{x} \]

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Rubi [A]  time = 0.76, antiderivative size = 49, normalized size of antiderivative = 1.69, number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {2274, 12, 6741, 2288} \begin {gather*} \frac {e^{\frac {1}{4} \left (20 x-4 e^x+15\right )} \left (5 x-e^x x\right ) \log (4)}{3 \sqrt [4]{3} \left (5-e^x\right ) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((15 - 4*E^x + 20*x - 5*Log[3])/4)*(-(E^x*x*Log[4]) + (-1 + 5*x)*Log[4]))/x^2,x]

[Out]

(E^((15 - 4*E^x + 20*x)/4)*(5*x - E^x*x)*Log[4])/(3*3^(1/4)*(5 - E^x)*x^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 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

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]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{3 \sqrt [4]{3} x^2} \, dx\\ &=\frac {\int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{x^2} \, dx}{3 \sqrt [4]{3}}\\ &=\frac {\int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x\right )} \left (-1+5 x-e^x x\right ) \log (4)}{x^2} \, dx}{3 \sqrt [4]{3}}\\ &=\frac {\log (4) \int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x\right )} \left (-1+5 x-e^x x\right )}{x^2} \, dx}{3 \sqrt [4]{3}}\\ &=\frac {e^{\frac {1}{4} \left (15-4 e^x+20 x\right )} \left (5 x-e^x x\right ) \log (4)}{3 \sqrt [4]{3} \left (5-e^x\right ) x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 28, normalized size = 0.97 \begin {gather*} \frac {e^{\frac {15}{4}-e^x+5 x} \log (4)}{3 \sqrt [4]{3} x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((15 - 4*E^x + 20*x - 5*Log[3])/4)*(-(E^x*x*Log[4]) + (-1 + 5*x)*Log[4]))/x^2,x]

[Out]

(E^(15/4 - E^x + 5*x)*Log[4])/(3*3^(1/4)*x)

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fricas [A]  time = 0.86, size = 21, normalized size = 0.72 \begin {gather*} \frac {2 \, e^{\left (5 \, x - e^{x} - \frac {5}{4} \, \log \relax (3) + \frac {15}{4}\right )} \log \relax (2)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(2)*exp(x)+2*(5*x-1)*log(2))/x^2/exp(exp(x)+5/4*log(3)-5*x-15/4),x, algorithm="fricas")

[Out]

2*e^(5*x - e^x - 5/4*log(3) + 15/4)*log(2)/x

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(2)*exp(x)+2*(5*x-1)*log(2))/x^2/exp(exp(x)+5/4*log(3)-5*x-15/4),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Polynomial exponent overflow. Error: Bad Argument Value

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maple [A]  time = 0.06, size = 21, normalized size = 0.72

method	result	size
risch	\(\frac {2 \ln \relax (2) 3^{\frac {3}{4}} {\mathrm e}^{-{\mathrm e}^{x}+\frac {15}{4}+5 x}}{9 x}\)	\(21\)
norman	\(\frac {2 \ln \relax (2) {\mathrm e}^{-{\mathrm e}^{x}-\frac {5 \ln \relax (3)}{4}+5 x +\frac {15}{4}}}{x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(2)*exp(x)+2*(5*x-1)*ln(2))/x^2/exp(exp(x)+5/4*ln(3)-5*x-15/4),x,method=_RETURNVERBOSE)

[Out]

2/9*ln(2)/x*3^(3/4)*exp(-exp(x)+15/4+5*x)

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maxima [A]  time = 0.63, size = 20, normalized size = 0.69 \begin {gather*} \frac {2 \cdot 3^{\frac {3}{4}} e^{\left (5 \, x - e^{x} + \frac {15}{4}\right )} \log \relax (2)}{9 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(2)*exp(x)+2*(5*x-1)*log(2))/x^2/exp(exp(x)+5/4*log(3)-5*x-15/4),x, algorithm="maxima")

[Out]

2/9*3^(3/4)*e^(5*x - e^x + 15/4)*log(2)/x

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mupad [B]  time = 0.10, size = 21, normalized size = 0.72 \begin {gather*} \frac {2\,3^{3/4}\,{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{15/4}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,\ln \relax (2)}{9\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5*x - (5*log(3))/4 - exp(x) + 15/4)*(2*log(2)*(5*x - 1) - 2*x*exp(x)*log(2)))/x^2,x)

[Out]

(2*3^(3/4)*exp(5*x)*exp(15/4)*exp(-exp(x))*log(2))/(9*x)

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sympy [A]  time = 0.26, size = 24, normalized size = 0.83 \begin {gather*} \frac {2 \cdot 3^{\frac {3}{4}} e^{5 x - e^{x} + \frac {15}{4}} \log {\relax (2 )}}{9 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(2)*exp(x)+2*(5*x-1)*ln(2))/x**2/exp(exp(x)+5/4*ln(3)-5*x-15/4),x)

[Out]

2*3**(3/4)*exp(5*x - exp(x) + 15/4)*log(2)/(9*x)

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