3.102.26 \(\int \frac {2 e^x+e^x x \log (x^2) \log (\log (x^2))}{2 x \log (x^2)} \, dx\)

Optimal. Leaf size=12 \[ \frac {1}{2} e^x \log \left (\log \left (x^2\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 6688, 2288} \begin {gather*} \frac {1}{2} e^x \log \left (\log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^x*Log[Log[x^2]])/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 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 6688

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 12, normalized size = 1.00 \begin {gather*} \frac {1}{2} e^x \log \left (\log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^x*Log[Log[x^2]])/2

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 9, normalized size = 0.75 \begin {gather*} \frac {1}{2} \, e^{x} \log \left (\log \left (x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^x*log(log(x^2))

________________________________________________________________________________________

giac [A]  time = 0.15, size = 9, normalized size = 0.75 \begin {gather*} \frac {1}{2} \, e^{x} \log \left (\log \left (x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^x*log(log(x^2))

________________________________________________________________________________________

maple [C]  time = 0.08, size = 39, normalized size = 3.25

method	result	size
risch	\(\frac {{\mathrm e}^{x} \ln \left (2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )}{2}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*exp(x)*ln(2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2)

________________________________________________________________________________________

maxima [A]  time = 0.47, size = 14, normalized size = 1.17 \begin {gather*} \frac {1}{2} \, e^{x} \log \relax (2) + \frac {1}{2} \, e^{x} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^x*log(2) + 1/2*e^x*log(log(x))

________________________________________________________________________________________

mupad [B]  time = 7.90, size = 9, normalized size = 0.75 \begin {gather*} \frac {{\mathrm {e}}^x\,\ln \left (\ln \left (x^2\right )\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(x)*log(log(x^2)))/2

________________________________________________________________________________________

sympy [A]  time = 3.94, size = 10, normalized size = 0.83 \begin {gather*} \frac {e^{x} \log {\left (\log {\left (x^{2} \right )} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)*log(log(x**2))/2

________________________________________________________________________________________