3.25.90 \(\int \frac {2+(x^2+2 e^x x^2) \log (\log (3))}{2 x^2 \log (\log (3))} \, dx\) [2490]

Optimal. Leaf size=23 \[ 7+e^3+e^x+\frac {x}{2}-\frac {1}{x \log (\log (3))} \]

[Out]

1/2*x+exp(3)-1/x/ln(ln(3))+exp(x)+7

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Rubi [A]
time = 0.01, antiderivative size = 19, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 14, 2225} \begin {gather*} \frac {x}{2}+e^x-\frac {1}{x \log (\log (3))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x + x/2 - 1/(x*Log[Log[3]])

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 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 19, normalized size = 0.83 \begin {gather*} e^x+\frac {x}{2}-\frac {1}{x \log (\log (3))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x + x/2 - 1/(x*Log[Log[3]])

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Maple [A]
time = 0.14, size = 26, normalized size = 1.13

method	result	size
risch	\(-\frac {1}{x \ln \left (\ln \left (3\right )\right )}+{\mathrm e}^{x}+\frac {x}{2}\)	\(17\)
norman	\(\frac {{\mathrm e}^{x} x +\frac {x^{2}}{2}-\frac {1}{\ln \left (\ln \left (3\right )\right )}}{x}\)	\(22\)
default	\(\frac {-\frac {2}{x}+2 \,{\mathrm e}^{x} \ln \left (\ln \left (3\right )\right )+\ln \left (\ln \left (3\right )\right ) x}{2 \ln \left (\ln \left (3\right )\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 25, normalized size = 1.09 \begin {gather*} \frac {x \log \left (\log \left (3\right )\right ) + 2 \, e^{x} \log \left (\log \left (3\right )\right ) - \frac {2}{x}}{2 \, \log \left (\log \left (3\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 25, normalized size = 1.09 \begin {gather*} \frac {{\left (x^{2} + 2 \, x e^{x}\right )} \log \left (\log \left (3\right )\right ) - 2}{2 \, x \log \left (\log \left (3\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.06, size = 19, normalized size = 0.83 \begin {gather*} \frac {x \log {\left (\log {\left (3 \right )} \right )} - \frac {2}{x}}{2 \log {\left (\log {\left (3 \right )} \right )}} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.40, size = 27, normalized size = 1.17 \begin {gather*} \frac {x^{2} \log \left (\log \left (3\right )\right ) + 2 \, x e^{x} \log \left (\log \left (3\right )\right ) - 2}{2 \, x \log \left (\log \left (3\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.36, size = 16, normalized size = 0.70 \begin {gather*} \frac {x}{2}+{\mathrm {e}}^x-\frac {1}{x\,\ln \left (\ln \left (3\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(log(3))*(2*x^2*exp(x) + x^2))/2 + 1)/(x^2*log(log(3))),x)

[Out]

x/2 + exp(x) - 1/(x*log(log(3)))

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