3.14.53 \(\int \frac {2 x^3+e^{.\frac {3}{2}/x} (3+2 x) \log (2) \log ^2(\log (4))}{2 x^3} \, dx\)

Optimal. Leaf size=27 \[ -2+x+\frac {\left (-e^{\left .\frac {3}{2}\right /x}+x\right ) \log (2) \log ^2(\log (4))}{x} \]

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Rubi [A]  time = 0.09, antiderivative size = 23, normalized size of antiderivative = 0.85, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 14, 2288} \begin {gather*} x-\frac {e^{\left .\frac {3}{2}\right /x} \log (2) \log ^2(\log (4))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\frac {1}{2} \int \frac {2 x^3+e^{\left .\frac {3}{2}\right /x} (3+2 x) \log (2) \log ^2(\log (4))}{x^3} \, dx\\ &=\frac {1}{2} \int \left (2+\frac {e^{\left .\frac {3}{2}\right /x} (3+2 x) \log (2) \log ^2(\log (4))}{x^3}\right ) \, dx\\ &=x+\frac {1}{2} \left (\log (2) \log ^2(\log (4))\right ) \int \frac {e^{\left .\frac {3}{2}\right /x} (3+2 x)}{x^3} \, dx\\ &=x-\frac {e^{\left .\frac {3}{2}\right /x} \log (2) \log ^2(\log (4))}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.85 \begin {gather*} x-\frac {e^{\left .\frac {3}{2}\right /x} \log (2) \log ^2(\log (4))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.65, size = 27, normalized size = 1.00 \begin {gather*} -\frac {e^{\left (\frac {3}{2 \, x}\right )} \log \relax (2) \log \left (2 \, \log \relax (2)\right )^{2} - x^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.33, size = 54, normalized size = 2.00 \begin {gather*} -{\left (\frac {e^{\left (\frac {3}{2 \, x}\right )} \log \relax (2)^{3}}{x^{2}} + \frac {2 \, e^{\left (\frac {3}{2 \, x}\right )} \log \relax (2)^{2} \log \left (\log \relax (2)\right )}{x^{2}} + \frac {e^{\left (\frac {3}{2 \, x}\right )} \log \relax (2) \log \left (\log \relax (2)\right )^{2}}{x^{2}} - 1\right )} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 33, normalized size = 1.22

method	result	size
risch	\(x -\frac {\ln \relax (2) \left (\ln \relax (2)^{2}+2 \ln \relax (2) \ln \left (\ln \relax (2)\right )+\ln \left (\ln \relax (2)\right )^{2}\right ) {\mathrm e}^{\frac {3}{2 x}}}{x}\)	\(33\)
norman	\(\frac {x^{3}+\left (-\ln \relax (2)^{3}-2 \ln \relax (2)^{2} \ln \left (\ln \relax (2)\right )-\ln \relax (2) \ln \left (\ln \relax (2)\right )^{2}\right ) x \,{\mathrm e}^{\frac {3}{2 x}}}{x^{2}}\)	\(42\)
derivativedivides	\(x -\frac {\ln \relax (2)^{3} {\mathrm e}^{\frac {3}{2 x}}}{x}-\frac {\ln \relax (2) {\mathrm e}^{\frac {3}{2 x}} \ln \left (\ln \relax (2)\right )^{2}}{x}-\frac {2 \ln \relax (2)^{2} {\mathrm e}^{\frac {3}{2 x}} \ln \left (\ln \relax (2)\right )}{x}\)	\(54\)
default	\(x -\frac {\ln \relax (2)^{3} {\mathrm e}^{\frac {3}{2 x}}}{x}-\frac {\ln \relax (2) {\mathrm e}^{\frac {3}{2 x}} \ln \left (\ln \relax (2)\right )^{2}}{x}-\frac {2 \ln \relax (2)^{2} {\mathrm e}^{\frac {3}{2 x}} \ln \left (\ln \relax (2)\right )}{x}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.83, size = 37, normalized size = 1.37 \begin {gather*} -\frac {2}{3} \, e^{\left (\frac {3}{2 \, x}\right )} \log \relax (2) \log \left (2 \, \log \relax (2)\right )^{2} + \frac {2}{3} \, \Gamma \left (2, -\frac {3}{2 \, x}\right ) \log \relax (2) \log \left (2 \, \log \relax (2)\right )^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.95, size = 20, normalized size = 0.74 \begin {gather*} x-\frac {{\mathrm {e}}^{\frac {3}{2\,x}}\,\ln \relax (2)\,{\ln \left (\ln \relax (4)\right )}^2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.15, size = 37, normalized size = 1.37 \begin {gather*} x + \frac {\left (- \log {\relax (2 )}^{3} - \log {\relax (2 )} \log {\left (\log {\relax (2 )} \right )}^{2} - 2 \log {\relax (2 )}^{2} \log {\left (\log {\relax (2 )} \right )}\right ) e^{\frac {3}{2 x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x+3)*ln(2)*exp(3/2/x)*ln(2*ln(2))**2+2*x**3)/x**3,x)

[Out]

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

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