3.71.36 \(\int \frac {-12+10 e^{16}+10 x^2}{(-6 x+5 e^{16} x-5 x^3) \log ^2(\frac {6-5 e^{16}+5 x^2}{5 x})} \, dx\)

Optimal. Leaf size=20 \[ \frac {2}{\log \left (\frac {\frac {6}{5}-e^{16}}{x}+x\right )} \]

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Rubi [A]  time = 0.17, antiderivative size = 24, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 3, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {6, 1593, 6686} \begin {gather*} \frac {2}{\log \left (\frac {5 x^2-5 e^{16}+6}{5 x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-12 + 10*E^16 + 10*x^2)/((-6*x + 5*E^16*x - 5*x^3)*Log[(6 - 5*E^16 + 5*x^2)/(5*x)]^2),x]

[Out]

2/Log[(6 - 5*E^16 + 5*x^2)/(5*x)]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-12+10 e^{16}+10 x^2}{\left (\left (-6+5 e^{16}\right ) x-5 x^3\right ) \log ^2\left (\frac {6-5 e^{16}+5 x^2}{5 x}\right )} \, dx\\ &=\int \frac {-12+10 e^{16}+10 x^2}{x \left (-6+5 e^{16}-5 x^2\right ) \log ^2\left (\frac {6-5 e^{16}+5 x^2}{5 x}\right )} \, dx\\ &=\frac {2}{\log \left (\frac {6-5 e^{16}+5 x^2}{5 x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 1.00 \begin {gather*} \frac {2}{\log \left (\frac {\frac {6}{5}-e^{16}}{x}+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12 + 10*E^16 + 10*x^2)/((-6*x + 5*E^16*x - 5*x^3)*Log[(6 - 5*E^16 + 5*x^2)/(5*x)]^2),x]

[Out]

2/Log[(6/5 - E^16)/x + x]

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fricas [A]  time = 0.53, size = 21, normalized size = 1.05 \begin {gather*} \frac {2}{\log \left (\frac {5 \, x^{2} - 5 \, e^{16} + 6}{5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(16)+10*x^2-12)/(5*x*exp(16)-5*x^3-6*x)/log(1/5*(-5*exp(16)+5*x^2+6)/x)^2,x, algorithm="frica
s")

[Out]

2/log(1/5*(5*x^2 - 5*e^16 + 6)/x)

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giac [A]  time = 0.22, size = 21, normalized size = 1.05 \begin {gather*} \frac {2}{\log \left (\frac {5 \, x^{2} - 5 \, e^{16} + 6}{5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(16)+10*x^2-12)/(5*x*exp(16)-5*x^3-6*x)/log(1/5*(-5*exp(16)+5*x^2+6)/x)^2,x, algorithm="giac"
)

[Out]

2/log(1/5*(5*x^2 - 5*e^16 + 6)/x)

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maple [A]  time = 0.12, size = 22, normalized size = 1.10

method	result	size
norman	\(\frac {2}{\ln \left (\frac {-5 \,{\mathrm e}^{16}+5 x^{2}+6}{5 x}\right )}\)	\(22\)
risch	\(\frac {2}{\ln \left (\frac {-5 \,{\mathrm e}^{16}+5 x^{2}+6}{5 x}\right )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(16)+10*x^2-12)/(5*x*exp(16)-5*x^3-6*x)/ln(1/5*(-5*exp(16)+5*x^2+6)/x)^2,x,method=_RETURNVERBOSE)

[Out]

2/ln(1/5*(-5*exp(16)+5*x^2+6)/x)

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maxima [A]  time = 0.52, size = 23, normalized size = 1.15 \begin {gather*} -\frac {2}{\log \relax (5) - \log \left (5 \, x^{2} - 5 \, e^{16} + 6\right ) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(16)+10*x^2-12)/(5*x*exp(16)-5*x^3-6*x)/log(1/5*(-5*exp(16)+5*x^2+6)/x)^2,x, algorithm="maxim
a")

[Out]

-2/(log(5) - log(5*x^2 - 5*e^16 + 6) + log(x))

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mupad [B]  time = 4.45, size = 18, normalized size = 0.90 \begin {gather*} \frac {2}{\ln \left (\frac {x^2-{\mathrm {e}}^{16}+\frac {6}{5}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*exp(16) + 10*x^2 - 12)/(log((x^2 - exp(16) + 6/5)/x)^2*(6*x - 5*x*exp(16) + 5*x^3)),x)

[Out]

2/log((x^2 - exp(16) + 6/5)/x)

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sympy [A]  time = 0.17, size = 14, normalized size = 0.70 \begin {gather*} \frac {2}{\log {\left (\frac {x^{2} - e^{16} + \frac {6}{5}}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(16)+10*x**2-12)/(5*x*exp(16)-5*x**3-6*x)/ln(1/5*(-5*exp(16)+5*x**2+6)/x)**2,x)

[Out]

2/log((x**2 - exp(16) + 6/5)/x)

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