3.38.13 \(\int \frac {-2 e^3-5 x^3}{20 x^3-5 x^4+e^3 (x+10 x^3)} \, dx\)

Optimal. Leaf size=21 \[ \log \left (4-x+\frac {2 e^3 \left (\frac {1}{10}+x^2\right )}{x^2}\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 25, normalized size of antiderivative = 1.19, number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2074, 1587} \begin {gather*} \log \left (-5 x^3+10 \left (2+e^3\right ) x^2+e^3\right )-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^3 - 5*x^3)/(20*x^3 - 5*x^4 + E^3*(x + 10*x^3)),x]

[Out]

-2*Log[x] + Log[E^3 + 10*(2 + E^3)*x^2 - 5*x^3]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2}{x}+\frac {5 \left (8+4 e^3-3 x\right ) x}{e^3+10 \left (2+e^3\right ) x^2-5 x^3}\right ) \, dx\\ &=-2 \log (x)+5 \int \frac {\left (8+4 e^3-3 x\right ) x}{e^3+10 \left (2+e^3\right ) x^2-5 x^3} \, dx\\ &=-2 \log (x)+\log \left (e^3+10 \left (2+e^3\right ) x^2-5 x^3\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 28, normalized size = 1.33 \begin {gather*} -2 \log (x)+\log \left (e^3+20 x^2+10 e^3 x^2-5 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^3 - 5*x^3)/(20*x^3 - 5*x^4 + E^3*(x + 10*x^3)),x]

[Out]

-2*Log[x] + Log[E^3 + 20*x^2 + 10*E^3*x^2 - 5*x^3]

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fricas [A]  time = 0.51, size = 28, normalized size = 1.33 \begin {gather*} \log \left (5 \, x^{3} - 20 \, x^{2} - {\left (10 \, x^{2} + 1\right )} e^{3}\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(3)-5*x^3)/((10*x^3+x)*exp(3)-5*x^4+20*x^3),x, algorithm="fricas")

[Out]

log(5*x^3 - 20*x^2 - (10*x^2 + 1)*e^3) - 2*log(x)

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giac [A]  time = 0.12, size = 30, normalized size = 1.43 \begin {gather*} \log \left ({\left | 5 \, x^{3} - 10 \, x^{2} e^{3} - 20 \, x^{2} - e^{3} \right |}\right ) - 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(3)-5*x^3)/((10*x^3+x)*exp(3)-5*x^4+20*x^3),x, algorithm="giac")

[Out]

log(abs(5*x^3 - 10*x^2*e^3 - 20*x^2 - e^3)) - 2*log(abs(x))

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maple [A]  time = 0.06, size = 27, normalized size = 1.29

method	result	size
norman	\(-2 \ln \relax (x )+\ln \left (10 x^{2} {\mathrm e}^{3}-5 x^{3}+20 x^{2}+{\mathrm e}^{3}\right )\)	\(27\)
default	\(-2 \ln \relax (x )+\ln \left (-10 x^{2} {\mathrm e}^{3}+5 x^{3}-20 x^{2}-{\mathrm e}^{3}\right )\)	\(29\)
risch	\(-2 \ln \left (-x \right )+\ln \left (5 x^{3}+\left (-10 \,{\mathrm e}^{3}-20\right ) x^{2}-{\mathrm e}^{3}\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*exp(3)-5*x^3)/((10*x^3+x)*exp(3)-5*x^4+20*x^3),x,method=_RETURNVERBOSE)

[Out]

-2*ln(x)+ln(10*x^2*exp(3)-5*x^3+20*x^2+exp(3))

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maxima [A]  time = 0.37, size = 25, normalized size = 1.19 \begin {gather*} \log \left (5 \, x^{3} - 10 \, x^{2} {\left (e^{3} + 2\right )} - e^{3}\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(3)-5*x^3)/((10*x^3+x)*exp(3)-5*x^4+20*x^3),x, algorithm="maxima")

[Out]

log(5*x^3 - 10*x^2*(e^3 + 2) - e^3) - 2*log(x)

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mupad [B]  time = 0.20, size = 26, normalized size = 1.24 \begin {gather*} \ln \left (x^3-2\,x^2\,{\mathrm {e}}^3-4\,x^2-\frac {{\mathrm {e}}^3}{5}\right )-2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(3) + 5*x^3)/(exp(3)*(x + 10*x^3) + 20*x^3 - 5*x^4),x)

[Out]

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

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sympy [A]  time = 1.06, size = 26, normalized size = 1.24 \begin {gather*} - 2 \log {\relax (x )} + \log {\left (x^{3} + x^{2} \left (- 2 e^{3} - 4\right ) - \frac {e^{3}}{5} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(3)-5*x**3)/((10*x**3+x)*exp(3)-5*x**4+20*x**3),x)

[Out]

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

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