3.12.61 \(\int \frac {x^2+e^{\frac {2 (x \log (5)+e \log (5) \log (x^2))}{e x}} (-4 \log (5)+2 \log (5) \log (x^2))}{x^2} \, dx\)

Optimal. Leaf size=21 \[ -5^{\frac {2 \left (\frac {x}{e}+\log \left (x^2\right )\right )}{x}}+x \]

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Rubi [A]  time = 0.22, antiderivative size = 22, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 2, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {14, 6706} \begin {gather*} x-5^{\frac {2 \left (e \log \left (x^2\right )+x\right )}{e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-5^((2*(x + E*Log[x^2]))/(E*x)) + 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 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 21, normalized size = 1.00 \begin {gather*} -5^{\frac {2}{e}+\frac {2 \log \left (x^2\right )}{x}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-5^(2/E + (2*Log[x^2])/x) + x

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fricas [A]  time = 0.83, size = 26, normalized size = 1.24 \begin {gather*} x - e^{\left (\frac {2 \, {\left (e \log \relax (5) \log \left (x^{2}\right ) + x \log \relax (5)\right )} e^{\left (-1\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(5)*log(x^2)-4*log(5))*exp((exp(1)*log(5)*log(x^2)+x*log(5))/x/exp(1))^2+x^2)/x^2,x, algorith
m="fricas")

[Out]

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

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giac [A]  time = 1.01, size = 26, normalized size = 1.24 \begin {gather*} x - e^{\left (\frac {2 \, {\left (e \log \relax (5) \log \left (x^{2}\right ) + x \log \relax (5)\right )} e^{\left (-1\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(5)*log(x^2)-4*log(5))*exp((exp(1)*log(5)*log(x^2)+x*log(5))/x/exp(1))^2+x^2)/x^2,x, algorith
m="giac")

[Out]

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

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maple [A]  time = 0.19, size = 22, normalized size = 1.05

method	result	size
risch	\(x -5^{\frac {2 \ln \left (x^{2}\right )+2 \,{\mathrm e}^{-1} x}{x}}\)	\(22\)
default	\(x -{\mathrm e}^{\frac {2 \left ({\mathrm e} \ln \relax (5) \ln \left (x^{2}\right )+x \ln \relax (5)\right ) {\mathrm e}^{-1}}{x}}\)	\(30\)
norman	\(\frac {x^{2}-x \,{\mathrm e}^{\frac {2 \left ({\mathrm e} \ln \relax (5) \ln \left (x^{2}\right )+x \ln \relax (5)\right ) {\mathrm e}^{-1}}{x}}}{x}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.70, size = 20, normalized size = 0.95 \begin {gather*} -5^{2 \, e^{\left (-1\right )}} e^{\left (\frac {4 \, \log \relax (5) \log \relax (x)}{x}\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(5)*log(x^2)-4*log(5))*exp((exp(1)*log(5)*log(x^2)+x*log(5))/x/exp(1))^2+x^2)/x^2,x, algorith
m="maxima")

[Out]

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

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mupad [B]  time = 0.83, size = 21, normalized size = 1.00 \begin {gather*} x-5^{2\,{\mathrm {e}}^{-1}}\,{\left (x^2\right )}^{\frac {2\,\ln \relax (5)}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*exp(-1)*(x*log(5) + log(x^2)*exp(1)*log(5)))/x)*(4*log(5) - 2*log(x^2)*log(5)) - x^2)/x^2,x)

[Out]

x - 5^(2*exp(-1))*(x^2)^((2*log(5))/x)

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sympy [A]  time = 0.33, size = 27, normalized size = 1.29 \begin {gather*} x - e^{\frac {2 \left (x \log {\relax (5 )} + e \log {\relax (5 )} \log {\left (x^{2} \right )}\right )}{e x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - exp(2*(x*log(5) + E*log(5)*log(x**2))*exp(-1)/x)

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