3.23.2 \(\int \frac {-x^2+e^{\frac {x^3+x \log (3)-e^x \log (4)}{x^2}} (x^3-x \log (3)+e^x (2-x) \log (4))}{x^3} \, dx\)

Optimal. Leaf size=26 \[ e^{x+\frac {\log (3)-\frac {e^x \log (4)}{x}}{x}}-\log (x) \]

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Rubi [A]  time = 0.27, antiderivative size = 24, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 2, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {14, 2288} \begin {gather*} 3^{\frac {1}{x}} 4^{-\frac {e^x}{x^2}} e^x-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3^x^(-1)*E^x)/4^(E^x/x^2) - 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 {1}{x}-\frac {3^{\frac {1}{x}} 4^{-\frac {e^x}{x^2}} e^x \left (-x^3+x \log (3)+e^x x \log (4)-e^x \log (16)\right )}{x^3}\right ) \, dx\\ &=-\log (x)-\int \frac {3^{\frac {1}{x}} 4^{-\frac {e^x}{x^2}} e^x \left (-x^3+x \log (3)+e^x x \log (4)-e^x \log (16)\right )}{x^3} \, dx\\ &=3^{\frac {1}{x}} 4^{-\frac {e^x}{x^2}} e^x-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 24, normalized size = 0.92 \begin {gather*} 3^{\frac {1}{x}} 4^{-\frac {e^x}{x^2}} e^x-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3^x^(-1)*E^x)/4^(E^x/x^2) - Log[x]

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fricas [A]  time = 0.86, size = 24, normalized size = 0.92 \begin {gather*} e^{\left (\frac {x^{3} + x \log \relax (3) - 2 \, e^{x} \log \relax (2)}{x^{2}}\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.66, size = 23, normalized size = 0.88 \begin {gather*} e^{\left (x + \frac {\log \relax (3)}{x} - \frac {2 \, e^{x} \log \relax (2)}{x^{2}}\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 22, normalized size = 0.85

method	result	size
risch	\(\left (\frac {1}{4}\right )^{\frac {{\mathrm e}^{x}}{x^{2}}} 3^{\frac {1}{x}} {\mathrm e}^{x}-\ln \relax (x )\)	\(22\)
norman	\({\mathrm e}^{\frac {-2 \,{\mathrm e}^{x} \ln \relax (2)+x \ln \relax (3)+x^{3}}{x^{2}}}-\ln \relax (x )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*(2-x)*ln(2)*exp(x)-x*ln(3)+x^3)*exp((-2*exp(x)*ln(2)+x*ln(3)+x^3)/x^2)-x^2)/x^3,x,method=_RETURNVERBOS
E)

[Out]

(1/4)^(exp(x)/x^2)*3^(1/x)*exp(x)-ln(x)

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maxima [A]  time = 0.60, size = 23, normalized size = 0.88 \begin {gather*} e^{\left (x + \frac {\log \relax (3)}{x} - \frac {2 \, e^{x} \log \relax (2)}{x^{2}}\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.42, size = 24, normalized size = 0.92 \begin {gather*} \frac {3^{1/x}\,{\mathrm {e}}^x}{2^{\frac {2\,{\mathrm {e}}^x}{x^2}}}-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x*log(3) - 2*exp(x)*log(2) + x^3)/x^2)*(x*log(3) - x^3 + 2*exp(x)*log(2)*(x - 2)) + x^2)/x^3,x)

[Out]

(3^(1/x)*exp(x))/2^((2*exp(x))/x^2) - log(x)

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sympy [A]  time = 0.29, size = 24, normalized size = 0.92 \begin {gather*} e^{\frac {x^{3} + x \log {\relax (3 )} - 2 e^{x} \log {\relax (2 )}}{x^{2}}} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2-x)*ln(2)*exp(x)-x*ln(3)+x**3)*exp((-2*exp(x)*ln(2)+x*ln(3)+x**3)/x**2)-x**2)/x**3,x)

[Out]

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

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