3.93.46 \(\int \frac {-15 e^5-20 x}{5 e^5 x+4 x^2} \, dx\)

Optimal. Leaf size=19 \[ \log \left (\frac {108}{x^3 \left (-x+5 \left (e^5+x\right )\right )^2}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {631} \begin {gather*} -3 \log (x)-2 \log \left (4 x+5 e^5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*E^5 - 20*x)/(5*E^5*x + 4*x^2),x]

[Out]

-3*Log[x] - 2*Log[5*E^5 + 4*x]

Rule 631

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)
*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0]
|| EqQ[a, 0])

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 23, normalized size = 1.21 \begin {gather*} -5 \left (\frac {3 \log (x)}{5}+\frac {2}{5} \log \left (5 e^5+4 x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*E^5 - 20*x)/(5*E^5*x + 4*x^2),x]

[Out]

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

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fricas [A]  time = 0.96, size = 16, normalized size = 0.84 \begin {gather*} -2 \, \log \left (4 \, x + 5 \, e^{5}\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*exp(5)-20*x)/(5*x*exp(5)+4*x^2),x, algorithm="fricas")

[Out]

-2*log(4*x + 5*e^5) - 3*log(x)

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giac [A]  time = 0.13, size = 18, normalized size = 0.95 \begin {gather*} -2 \, \log \left ({\left | 4 \, x + 5 \, e^{5} \right |}\right ) - 3 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*exp(5)-20*x)/(5*x*exp(5)+4*x^2),x, algorithm="giac")

[Out]

-2*log(abs(4*x + 5*e^5)) - 3*log(abs(x))

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maple [A]  time = 0.54, size = 17, normalized size = 0.89

method	result	size
default	\(-3 \ln \relax (x )-2 \ln \left (5 \,{\mathrm e}^{5}+4 x \right )\)	\(17\)
norman	\(-3 \ln \relax (x )-2 \ln \left (5 \,{\mathrm e}^{5}+4 x \right )\)	\(17\)
risch	\(-3 \ln \relax (x )-2 \ln \left (5 \,{\mathrm e}^{5}+4 x \right )\)	\(17\)
meijerg	\(-3 \ln \relax (x )+3 \ln \relax (5)-6 \ln \relax (2)+15-2 \ln \left (1+\frac {4 x \,{\mathrm e}^{-5}}{5}\right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-15*exp(5)-20*x)/(5*x*exp(5)+4*x^2),x,method=_RETURNVERBOSE)

[Out]

-3*ln(x)-2*ln(5*exp(5)+4*x)

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maxima [A]  time = 0.35, size = 16, normalized size = 0.84 \begin {gather*} -2 \, \log \left (4 \, x + 5 \, e^{5}\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*exp(5)-20*x)/(5*x*exp(5)+4*x^2),x, algorithm="maxima")

[Out]

-2*log(4*x + 5*e^5) - 3*log(x)

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mupad [B]  time = 0.17, size = 14, normalized size = 0.74 \begin {gather*} -2\,\ln \left (x+\frac {5\,{\mathrm {e}}^5}{4}\right )-3\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x + 15*exp(5))/(5*x*exp(5) + 4*x^2),x)

[Out]

- 2*log(x + (5*exp(5))/4) - 3*log(x)

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sympy [A]  time = 0.21, size = 17, normalized size = 0.89 \begin {gather*} - 3 \log {\relax (x )} - 2 \log {\left (x + \frac {5 e^{5}}{4} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*exp(5)-20*x)/(5*x*exp(5)+4*x**2),x)

[Out]

-3*log(x) - 2*log(x + 5*exp(5)/4)

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