3.29.32 \(\int \frac {-40-6 x^3}{20 x+11 x^3+e^{10} x^3-6 x^4+x^3 \log (4)} \, dx\)

Optimal. Leaf size=23 \[ \log \left (1+e^{10}+5 \left (2+\frac {4}{x^2}-x\right )-x+\log (4)\right ) \]

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Rubi [A]  time = 0.34, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {6, 1594, 6742, 1587} \begin {gather*} \log \left (-6 x^3+x^2 \left (11+e^{10}+\log (4)\right )+20\right )-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-40 - 6*x^3)/(20*x + 11*x^3 + E^10*x^3 - 6*x^4 + x^3*Log[4]),x]

[Out]

-2*Log[x] + Log[20 - 6*x^3 + x^2*(11 + E^10 + Log[4])]

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 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 1594

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-40-6 x^3}{20 x+\left (11+e^{10}\right ) x^3-6 x^4+x^3 \log (4)} \, dx\\ &=\int \frac {-40-6 x^3}{20 x-6 x^4+x^3 \left (11+e^{10}+\log (4)\right )} \, dx\\ &=\int \frac {-40-6 x^3}{x \left (20-6 x^3+x^2 \left (11+e^{10}+\log (4)\right )\right )} \, dx\\ &=\int \left (-\frac {2}{x}+\frac {2 x \left (11+e^{10}-9 x+\log (4)\right )}{20-6 x^3+x^2 \left (11+e^{10}+\log (4)\right )}\right ) \, dx\\ &=-2 \log (x)+2 \int \frac {x \left (11+e^{10}-9 x+\log (4)\right )}{20-6 x^3+x^2 \left (11+e^{10}+\log (4)\right )} \, dx\\ &=-2 \log (x)+\log \left (20-6 x^3+x^2 \left (11+e^{10}+\log (4)\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 1.35 \begin {gather*} -2 \log (x)+\log \left (20+11 x^2+e^{10} x^2-6 x^3+x^2 \log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40 - 6*x^3)/(20*x + 11*x^3 + E^10*x^3 - 6*x^4 + x^3*Log[4]),x]

[Out]

-2*Log[x] + Log[20 + 11*x^2 + E^10*x^2 - 6*x^3 + x^2*Log[4]]

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fricas [A]  time = 0.67, size = 32, normalized size = 1.39 \begin {gather*} \log \left (6 \, x^{3} - x^{2} e^{10} - 2 \, x^{2} \log \relax (2) - 11 \, x^{2} - 20\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x^3-40)/(2*x^3*log(2)+x^3*exp(10)-6*x^4+11*x^3+20*x),x, algorithm="fricas")

[Out]

log(6*x^3 - x^2*e^10 - 2*x^2*log(2) - 11*x^2 - 20) - 2*log(x)

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giac [A]  time = 0.27, size = 34, normalized size = 1.48 \begin {gather*} \log \left ({\left | 6 \, x^{3} - x^{2} e^{10} - 2 \, x^{2} \log \relax (2) - 11 \, x^{2} - 20 \right |}\right ) - 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x^3-40)/(2*x^3*log(2)+x^3*exp(10)-6*x^4+11*x^3+20*x),x, algorithm="giac")

[Out]

log(abs(6*x^3 - x^2*e^10 - 2*x^2*log(2) - 11*x^2 - 20)) - 2*log(abs(x))

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maple [A]  time = 0.06, size = 30, normalized size = 1.30




method result size



risch \(-2 \ln \left (-x \right )+\ln \left (-20+6 x^{3}+\left (-2 \ln \relax (2)-{\mathrm e}^{10}-11\right ) x^{2}\right )\) \(30\)
norman \(-2 \ln \relax (x )+\ln \left (2 x^{2} \ln \relax (2)+x^{2} {\mathrm e}^{10}-6 x^{3}+11 x^{2}+20\right )\) \(32\)
default \(-2 \ln \relax (x )+\ln \left (-2 x^{2} \ln \relax (2)-x^{2} {\mathrm e}^{10}+6 x^{3}-11 x^{2}-20\right )\) \(33\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-6*x^3-40)/(2*x^3*ln(2)+x^3*exp(10)-6*x^4+11*x^3+20*x),x,method=_RETURNVERBOSE)

[Out]

-2*ln(-x)+ln(-20+6*x^3+(-2*ln(2)-exp(10)-11)*x^2)

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maxima [A]  time = 0.39, size = 26, normalized size = 1.13 \begin {gather*} \log \left (6 \, x^{3} - x^{2} {\left (e^{10} + 2 \, \log \relax (2) + 11\right )} - 20\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x^3-40)/(2*x^3*log(2)+x^3*exp(10)-6*x^4+11*x^3+20*x),x, algorithm="maxima")

[Out]

log(6*x^3 - x^2*(e^10 + 2*log(2) + 11) - 20) - 2*log(x)

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mupad [B]  time = 0.22, size = 32, normalized size = 1.39 \begin {gather*} \ln \left (3\,x^2\,{\mathrm {e}}^{10}+6\,x^2\,\ln \relax (2)+33\,x^2-18\,x^3+60\right )-2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*x^3 + 40)/(20*x + x^3*exp(10) + 2*x^3*log(2) + 11*x^3 - 6*x^4),x)

[Out]

log(3*x^2*exp(10) + 6*x^2*log(2) + 33*x^2 - 18*x^3 + 60) - 2*log(x)

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sympy [A]  time = 3.15, size = 31, normalized size = 1.35 \begin {gather*} - 2 \log {\relax (x )} + \log {\left (x^{3} + x^{2} \left (- \frac {e^{10}}{6} - \frac {11}{6} - \frac {\log {\relax (2 )}}{3}\right ) - \frac {10}{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x**3-40)/(2*x**3*ln(2)+x**3*exp(10)-6*x**4+11*x**3+20*x),x)

[Out]

-2*log(x) + log(x**3 + x**2*(-exp(10)/6 - 11/6 - log(2)/3) - 10/3)

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