3.88.100 \(\int \frac {-4 \log (5) \log ^2(\frac {16}{x^2})+e^{\frac {x}{\log (\frac {16}{x^2})}} (8 x-8 x^2+(4 x-4 x^2) \log (\frac {16}{x^2})-4 \log ^2(\frac {16}{x^2}))}{3 x^2 \log ^2(\frac {16}{x^2})} \, dx\)

Optimal. Leaf size=27 \[ \frac {(4-4 x) \left (e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}}+\log (5)\right )}{3 x} \]

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Rubi [B]  time = 0.80, antiderivative size = 86, normalized size of antiderivative = 3.19, number of steps used = 4, number of rules used = 3, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {12, 6742, 2288} \begin {gather*} \frac {4 e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (-2 x^2+x^2 \left (-\log \left (\frac {16}{x^2}\right )\right )+x \log \left (\frac {16}{x^2}\right )+2 x\right )}{3 x^2 \left (\frac {2}{\log ^2\left (\frac {16}{x^2}\right )}+\frac {1}{\log \left (\frac {16}{x^2}\right )}\right ) \log ^2\left (\frac {16}{x^2}\right )}+\frac {4 \log (5)}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*Log[5]*Log[16/x^2]^2 + E^(x/Log[16/x^2])*(8*x - 8*x^2 + (4*x - 4*x^2)*Log[16/x^2] - 4*Log[16/x^2]^2))/
(3*x^2*Log[16/x^2]^2),x]

[Out]

(4*Log[5])/(3*x) + (4*E^(x/Log[16/x^2])*(2*x - 2*x^2 + x*Log[16/x^2] - x^2*Log[16/x^2]))/(3*x^2*(2/Log[16/x^2]
^2 + Log[16/x^2]^(-1))*Log[16/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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {-4 \log (5) \log ^2\left (\frac {16}{x^2}\right )+e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (8 x-8 x^2+\left (4 x-4 x^2\right ) \log \left (\frac {16}{x^2}\right )-4 \log ^2\left (\frac {16}{x^2}\right )\right )}{x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx\\ &=\frac {1}{3} \int \left (-\frac {4 \log (5)}{x^2}-\frac {4 e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (-2 x+2 x^2-x \log \left (\frac {16}{x^2}\right )+x^2 \log \left (\frac {16}{x^2}\right )+\log ^2\left (\frac {16}{x^2}\right )\right )}{x^2 \log ^2\left (\frac {16}{x^2}\right )}\right ) \, dx\\ &=\frac {4 \log (5)}{3 x}-\frac {4}{3} \int \frac {e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (-2 x+2 x^2-x \log \left (\frac {16}{x^2}\right )+x^2 \log \left (\frac {16}{x^2}\right )+\log ^2\left (\frac {16}{x^2}\right )\right )}{x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx\\ &=\frac {4 \log (5)}{3 x}+\frac {4 e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (2 x-2 x^2+x \log \left (\frac {16}{x^2}\right )-x^2 \log \left (\frac {16}{x^2}\right )\right )}{3 x^2 \left (\frac {2}{\log ^2\left (\frac {16}{x^2}\right )}+\frac {1}{\log \left (\frac {16}{x^2}\right )}\right ) \log ^2\left (\frac {16}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 28, normalized size = 1.04 \begin {gather*} -\frac {4 \left (e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} (-1+x)-\log (5)\right )}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*Log[5]*Log[16/x^2]^2 + E^(x/Log[16/x^2])*(8*x - 8*x^2 + (4*x - 4*x^2)*Log[16/x^2] - 4*Log[16/x^2
]^2))/(3*x^2*Log[16/x^2]^2),x]

[Out]

(-4*(E^(x/Log[16/x^2])*(-1 + x) - Log[5]))/(3*x)

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fricas [A]  time = 0.79, size = 25, normalized size = 0.93 \begin {gather*} -\frac {4 \, {\left ({\left (x - 1\right )} e^{\left (\frac {x}{\log \left (\frac {16}{x^{2}}\right )}\right )} - \log \relax (5)\right )}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*log(16/x^2)^2+(-4*x^2+4*x)*log(16/x^2)-8*x^2+8*x)*exp(x/log(16/x^2))-4*log(5)*log(16/x^2)^2
)/x^2/log(16/x^2)^2,x, algorithm="fricas")

[Out]

-4/3*((x - 1)*e^(x/log(16/x^2)) - log(5))/x

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giac [A]  time = 0.45, size = 36, normalized size = 1.33 \begin {gather*} -\frac {4 \, {\left (x e^{\left (\frac {x}{\log \left (\frac {16}{x^{2}}\right )}\right )} - e^{\left (\frac {x}{\log \left (\frac {16}{x^{2}}\right )}\right )} - \log \relax (5)\right )}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*log(16/x^2)^2+(-4*x^2+4*x)*log(16/x^2)-8*x^2+8*x)*exp(x/log(16/x^2))-4*log(5)*log(16/x^2)^2
)/x^2/log(16/x^2)^2,x, algorithm="giac")

[Out]

-4/3*(x*e^(x/log(16/x^2)) - e^(x/log(16/x^2)) - log(5))/x

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maple [A]  time = 0.12, size = 28, normalized size = 1.04

method	result	size
risch	\(\frac {4 \ln \relax (5)}{3 x}-\frac {4 \left (x -1\right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}}{3 x}\)	\(28\)
norman	\(\frac {\frac {4 \ln \relax (5) \ln \left (\frac {16}{x^{2}}\right )}{3}+\frac {4 \ln \left (\frac {16}{x^{2}}\right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}}{3}-\frac {4 \ln \left (\frac {16}{x^{2}}\right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}} x}{3}}{x \ln \left (\frac {16}{x^{2}}\right )}\)	\(63\)
default	\(-\frac {4 \left (x \left (\ln \left (\frac {16}{x^{2}}\right )+2 \ln \relax (x )\right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}-\left (\ln \left (\frac {16}{x^{2}}\right )+2 \ln \relax (x )\right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}+2 \ln \relax (x ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}-2 x \ln \relax (x ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}\right )}{3 x \ln \left (\frac {16}{x^{2}}\right )}+\frac {4 \ln \relax (5)}{3 x}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-4*ln(16/x^2)^2+(-4*x^2+4*x)*ln(16/x^2)-8*x^2+8*x)*exp(x/ln(16/x^2))-4*ln(5)*ln(16/x^2)^2)/x^2/ln(16
/x^2)^2,x,method=_RETURNVERBOSE)

[Out]

4/3*ln(5)/x-4/3*(x-1)/x*exp(x/ln(16/x^2))

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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(1/3*((-4*log(16/x^2)^2+(-4*x^2+4*x)*log(16/x^2)-8*x^2+8*x)*exp(x/log(16/x^2))-4*log(5)*log(16/x^2)^2
)/x^2/log(16/x^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 = 5.55, size = 36, normalized size = 1.33 \begin {gather*} \frac {\frac {4\,{\mathrm {e}}^{\frac {x}{\ln \left (\frac {16}{x^2}\right )}}}{3}+\ln \left ({625}^{1/3}\right )}{x}-\frac {4\,{\mathrm {e}}^{\frac {x}{\ln \left (\frac {16}{x^2}\right )}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((4*log(5)*log(16/x^2)^2)/3 - (exp(x/log(16/x^2))*(8*x - 4*log(16/x^2)^2 + log(16/x^2)*(4*x - 4*x^2) - 8*
x^2))/3)/(x^2*log(16/x^2)^2),x)

[Out]

((4*exp(x/log(16/x^2)))/3 + log(625^(1/3)))/x - (4*exp(x/log(16/x^2)))/3

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sympy [A]  time = 0.36, size = 26, normalized size = 0.96 \begin {gather*} \frac {\left (4 - 4 x\right ) e^{\frac {x}{\log {\left (\frac {16}{x^{2}} \right )}}}}{3 x} + \frac {4 \log {\relax (5 )}}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*ln(16/x**2)**2+(-4*x**2+4*x)*ln(16/x**2)-8*x**2+8*x)*exp(x/ln(16/x**2))-4*ln(5)*ln(16/x**2)
**2)/x**2/ln(16/x**2)**2,x)

[Out]

(4 - 4*x)*exp(x/log(16/x**2))/(3*x) + 4*log(5)/(3*x)

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