3.89.93 \(\int \frac {4 e^5 x^2 \log ^2(x)+e^{\frac {3}{e^5 x^2 \log (x)}} \log (2+e^2) (6+12 \log (x)-4 e^5 x^2 \log ^2(x))+e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2(2+e^2) (-3-6 \log (x)+e^5 x^2 \log ^2(x))}{2 e^5 x \log ^2(2+e^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=32 \[ \left (-\frac {1}{2} e^{\frac {3}{e^5 x^2 \log (x)}} x+\frac {x}{\log \left (2+e^2\right )}\right )^2 \]

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Rubi [B]  time = 0.94, antiderivative size = 137, normalized size of antiderivative = 4.28, number of steps used = 5, number of rules used = 3, integrand size = 114, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {12, 6742, 2288} \begin {gather*} \frac {x^2}{\log ^2\left (2+e^2\right )}+\frac {e^{\frac {6}{e^5 x^2 \log (x)}-5} (2 \log (x)+1)}{4 x \left (\frac {1}{e^5 x^3 \log ^2(x)}+\frac {2}{e^5 x^3 \log (x)}\right ) \log ^2(x)}-\frac {e^{\frac {3}{e^5 x^2 \log (x)}-5} (2 \log (x)+1)}{x \log \left (2+e^2\right ) \left (\frac {1}{e^5 x^3 \log ^2(x)}+\frac {2}{e^5 x^3 \log (x)}\right ) \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^5*x^2*Log[x]^2 + E^(3/(E^5*x^2*Log[x]))*Log[2 + E^2]*(6 + 12*Log[x] - 4*E^5*x^2*Log[x]^2) + E^(6/(E^5
*x^2*Log[x]))*Log[2 + E^2]^2*(-3 - 6*Log[x] + E^5*x^2*Log[x]^2))/(2*E^5*x*Log[2 + E^2]^2*Log[x]^2),x]

[Out]

x^2/Log[2 + E^2]^2 + (E^(-5 + 6/(E^5*x^2*Log[x]))*(1 + 2*Log[x]))/(4*x*(1/(E^5*x^3*Log[x]^2) + 2/(E^5*x^3*Log[
x]))*Log[x]^2) - (E^(-5 + 3/(E^5*x^2*Log[x]))*(1 + 2*Log[x]))/(x*Log[2 + E^2]*(1/(E^5*x^3*Log[x]^2) + 2/(E^5*x
^3*Log[x]))*Log[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 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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {4 e^5 x^2 \log ^2(x)+e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right ) \left (6+12 \log (x)-4 e^5 x^2 \log ^2(x)\right )+e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2\left (2+e^2\right ) \left (-3-6 \log (x)+e^5 x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx}{2 e^5 \log ^2\left (2+e^2\right )}\\ &=\frac {\int \left (4 e^5 x+\frac {e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2\left (2+e^2\right ) \left (-3-6 \log (x)+e^5 x^2 \log ^2(x)\right )}{x \log ^2(x)}-\frac {2 e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right ) \left (-3-6 \log (x)+2 e^5 x^2 \log ^2(x)\right )}{x \log ^2(x)}\right ) \, dx}{2 e^5 \log ^2\left (2+e^2\right )}\\ &=\frac {x^2}{\log ^2\left (2+e^2\right )}+\frac {\int \frac {e^{\frac {6}{e^5 x^2 \log (x)}} \left (-3-6 \log (x)+e^5 x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx}{2 e^5}-\frac {\int \frac {e^{\frac {3}{e^5 x^2 \log (x)}} \left (-3-6 \log (x)+2 e^5 x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx}{e^5 \log \left (2+e^2\right )}\\ &=\frac {x^2}{\log ^2\left (2+e^2\right )}+\frac {e^{-5+\frac {6}{e^5 x^2 \log (x)}} (1+2 \log (x))}{4 x \left (\frac {1}{e^5 x^3 \log ^2(x)}+\frac {2}{e^5 x^3 \log (x)}\right ) \log ^2(x)}-\frac {e^{-5+\frac {3}{e^5 x^2 \log (x)}} (1+2 \log (x))}{x \log \left (2+e^2\right ) \left (\frac {1}{e^5 x^3 \log ^2(x)}+\frac {2}{e^5 x^3 \log (x)}\right ) \log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 40, normalized size = 1.25 \begin {gather*} \frac {x^2 \left (-2+e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right )\right )^2}{4 \log ^2\left (2+e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^5*x^2*Log[x]^2 + E^(3/(E^5*x^2*Log[x]))*Log[2 + E^2]*(6 + 12*Log[x] - 4*E^5*x^2*Log[x]^2) + E^(
6/(E^5*x^2*Log[x]))*Log[2 + E^2]^2*(-3 - 6*Log[x] + E^5*x^2*Log[x]^2))/(2*E^5*x*Log[2 + E^2]^2*Log[x]^2),x]

[Out]

(x^2*(-2 + E^(3/(E^5*x^2*Log[x]))*Log[2 + E^2])^2)/(4*Log[2 + E^2]^2)

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fricas [B]  time = 0.52, size = 60, normalized size = 1.88 \begin {gather*} \frac {x^{2} e^{\left (\frac {6 \, e^{\left (-5\right )}}{x^{2} \log \relax (x)}\right )} \log \left (e^{2} + 2\right )^{2} - 4 \, x^{2} e^{\left (\frac {3 \, e^{\left (-5\right )}}{x^{2} \log \relax (x)}\right )} \log \left (e^{2} + 2\right ) + 4 \, x^{2}}{4 \, \log \left (e^{2} + 2\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x^2*exp(5)*log(x)^2-6*log(x)-3)*log(exp(2)+2)^2*exp(3/x^2/exp(5)/log(x))^2+(-4*x^2*exp(5)*log(
x)^2+12*log(x)+6)*log(exp(2)+2)*exp(3/x^2/exp(5)/log(x))+4*x^2*exp(5)*log(x)^2)/x/exp(5)/log(x)^2/log(exp(2)+2
)^2,x, algorithm="fricas")

[Out]

1/4*(x^2*e^(6*e^(-5)/(x^2*log(x)))*log(e^2 + 2)^2 - 4*x^2*e^(3*e^(-5)/(x^2*log(x)))*log(e^2 + 2) + 4*x^2)/log(
e^2 + 2)^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (4 \, x^{2} e^{5} \log \relax (x)^{2} + {\left (x^{2} e^{5} \log \relax (x)^{2} - 6 \, \log \relax (x) - 3\right )} e^{\left (\frac {6 \, e^{\left (-5\right )}}{x^{2} \log \relax (x)}\right )} \log \left (e^{2} + 2\right )^{2} - 2 \, {\left (2 \, x^{2} e^{5} \log \relax (x)^{2} - 6 \, \log \relax (x) - 3\right )} e^{\left (\frac {3 \, e^{\left (-5\right )}}{x^{2} \log \relax (x)}\right )} \log \left (e^{2} + 2\right )\right )} e^{\left (-5\right )}}{2 \, x \log \relax (x)^{2} \log \left (e^{2} + 2\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x^2*exp(5)*log(x)^2-6*log(x)-3)*log(exp(2)+2)^2*exp(3/x^2/exp(5)/log(x))^2+(-4*x^2*exp(5)*log(
x)^2+12*log(x)+6)*log(exp(2)+2)*exp(3/x^2/exp(5)/log(x))+4*x^2*exp(5)*log(x)^2)/x/exp(5)/log(x)^2/log(exp(2)+2
)^2,x, algorithm="giac")

[Out]

integrate(1/2*(4*x^2*e^5*log(x)^2 + (x^2*e^5*log(x)^2 - 6*log(x) - 3)*e^(6*e^(-5)/(x^2*log(x)))*log(e^2 + 2)^2
- 2*(2*x^2*e^5*log(x)^2 - 6*log(x) - 3)*e^(3*e^(-5)/(x^2*log(x)))*log(e^2 + 2))*e^(-5)/(x*log(x)^2*log(e^2 +
2)^2), x)

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maple [A]  time = 0.28, size = 54, normalized size = 1.69

method	result	size
risch	\(\frac {x^{2}}{\ln \left ({\mathrm e}^{2}+2\right )^{2}}+\frac {x^{2} {\mathrm e}^{\frac {6 \,{\mathrm e}^{-5}}{x^{2} \ln \relax (x )}}}{4}-\frac {x^{2} {\mathrm e}^{\frac {3 \,{\mathrm e}^{-5}}{x^{2} \ln \relax (x )}}}{\ln \left ({\mathrm e}^{2}+2\right )}\)	\(54\)
default	\(\frac {{\mathrm e}^{-5} \left (-2 \,{\mathrm e}^{5} \ln \left ({\mathrm e}^{2}+2\right ) x^{2} {\mathrm e}^{\frac {3 \,{\mathrm e}^{-5}}{x^{2} \ln \relax (x )}}+\frac {{\mathrm e}^{5} \ln \left ({\mathrm e}^{2}+2\right )^{2} x^{2} {\mathrm e}^{\frac {6 \,{\mathrm e}^{-5}}{x^{2} \ln \relax (x )}}}{2}+2 x^{2} {\mathrm e}^{5}\right )}{2 \ln \left ({\mathrm e}^{2}+2\right )^{2}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((x^2*exp(5)*ln(x)^2-6*ln(x)-3)*ln(exp(2)+2)^2*exp(3/x^2/exp(5)/ln(x))^2+(-4*x^2*exp(5)*ln(x)^2+12*ln(
x)+6)*ln(exp(2)+2)*exp(3/x^2/exp(5)/ln(x))+4*x^2*exp(5)*ln(x)^2)/x/exp(5)/ln(x)^2/ln(exp(2)+2)^2,x,method=_RET
URNVERBOSE)

[Out]

1/ln(exp(2)+2)^2*x^2+1/4*x^2*exp(6/x^2*exp(-5)/ln(x))-1/ln(exp(2)+2)*x^2*exp(3/x^2*exp(-5)/ln(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x^2*exp(5)*log(x)^2-6*log(x)-3)*log(exp(2)+2)^2*exp(3/x^2/exp(5)/log(x))^2+(-4*x^2*exp(5)*log(
x)^2+12*log(x)+6)*log(exp(2)+2)*exp(3/x^2/exp(5)/log(x))+4*x^2*exp(5)*log(x)^2)/x/exp(5)/log(x)^2/log(exp(2)+2
)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 5.89, size = 34, normalized size = 1.06 \begin {gather*} \frac {x^2\,{\left (\ln \left ({\mathrm {e}}^2+2\right )\,{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{-5}}{x^2\,\ln \relax (x)}}-2\right )}^2}{4\,{\ln \left ({\mathrm {e}}^2+2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-5)*((log(exp(2) + 2)*exp((3*exp(-5))/(x^2*log(x)))*(12*log(x) - 4*x^2*exp(5)*log(x)^2 + 6))/2 - (log
(exp(2) + 2)^2*exp((6*exp(-5))/(x^2*log(x)))*(6*log(x) - x^2*exp(5)*log(x)^2 + 3))/2 + 2*x^2*exp(5)*log(x)^2))
/(x*log(exp(2) + 2)^2*log(x)^2),x)

[Out]

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

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sympy [B]  time = 0.57, size = 61, normalized size = 1.91 \begin {gather*} \frac {x^{2}}{\log {\left (2 + e^{2} \right )}^{2}} + \frac {x^{2} e^{\frac {6}{x^{2} e^{5} \log {\relax (x )}}} \log {\left (2 + e^{2} \right )} - 4 x^{2} e^{\frac {3}{x^{2} e^{5} \log {\relax (x )}}}}{4 \log {\left (2 + e^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x**2*exp(5)*ln(x)**2-6*ln(x)-3)*ln(exp(2)+2)**2*exp(3/x**2/exp(5)/ln(x))**2+(-4*x**2*exp(5)*ln
(x)**2+12*ln(x)+6)*ln(exp(2)+2)*exp(3/x**2/exp(5)/ln(x))+4*x**2*exp(5)*ln(x)**2)/x/exp(5)/ln(x)**2/ln(exp(2)+2
)**2,x)

[Out]

x**2/log(2 + exp(2))**2 + (x**2*exp(6*exp(-5)/(x**2*log(x)))*log(2 + exp(2)) - 4*x**2*exp(3*exp(-5)/(x**2*log(
x))))/(4*log(2 + exp(2)))

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