3.42.76 \(\int \frac {20-23 x^2+2 x^4+e^{12} (-8 x+2 x^3)}{-4 x+x^3} \, dx\)

Optimal. Leaf size=24 \[ \left (e^{12}+x\right )^2+5 \left (4-\log \left (x \left (8-2 x^2\right )\right )\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {1593, 1802, 260} \begin {gather*} x^2-5 \log \left (4-x^2\right )+2 e^{12} x-5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 - 23*x^2 + 2*x^4 + E^12*(-8*x + 2*x^3))/(-4*x + x^3),x]

[Out]

2*E^12*x + x^2 - 5*Log[x] - 5*Log[4 - x^2]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {20-23 x^2+2 x^4+e^{12} \left (-8 x+2 x^3\right )}{x \left (-4+x^2\right )} \, dx\\ &=\int \left (2 e^{12}-\frac {5}{x}+2 x-\frac {10 x}{-4+x^2}\right ) \, dx\\ &=2 e^{12} x+x^2-5 \log (x)-10 \int \frac {x}{-4+x^2} \, dx\\ &=2 e^{12} x+x^2-5 \log (x)-5 \log \left (4-x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.00 \begin {gather*} 2 e^{12} x+x^2-5 \log (x)-5 \log \left (4-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 - 23*x^2 + 2*x^4 + E^12*(-8*x + 2*x^3))/(-4*x + x^3),x]

[Out]

2*E^12*x + x^2 - 5*Log[x] - 5*Log[4 - x^2]

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fricas [A]  time = 1.51, size = 19, normalized size = 0.79 \begin {gather*} x^{2} + 2 \, x e^{12} - 5 \, \log \left (x^{3} - 4 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-8*x)*exp(12)+2*x^4-23*x^2+20)/(x^3-4*x),x, algorithm="fricas")

[Out]

x^2 + 2*x*e^12 - 5*log(x^3 - 4*x)

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giac [A]  time = 0.21, size = 28, normalized size = 1.17 \begin {gather*} x^{2} + 2 \, x e^{12} - 5 \, \log \left ({\left | x + 2 \right |}\right ) - 5 \, \log \left ({\left | x - 2 \right |}\right ) - 5 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-8*x)*exp(12)+2*x^4-23*x^2+20)/(x^3-4*x),x, algorithm="giac")

[Out]

x^2 + 2*x*e^12 - 5*log(abs(x + 2)) - 5*log(abs(x - 2)) - 5*log(abs(x))

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maple [A]  time = 0.09, size = 20, normalized size = 0.83

method	result	size
risch	\(2 x \,{\mathrm e}^{12}+x^{2}-5 \ln \left (x^{3}-4 x \right )\)	\(20\)
default	\(x^{2}+2 x \,{\mathrm e}^{12}-5 \ln \left (2+x \right )-5 \ln \left (x -2\right )-5 \ln \relax (x )\)	\(26\)
norman	\(x^{2}+2 x \,{\mathrm e}^{12}-5 \ln \left (2+x \right )-5 \ln \left (x -2\right )-5 \ln \relax (x )\)	\(26\)
meijerg	\(-5 \ln \relax (x )+5 \ln \relax (2)-\frac {5 i \pi }{2}-5 \ln \left (1-\frac {x^{2}}{4}\right )-2 i {\mathrm e}^{12} \left (i x -2 i \arctanh \left (\frac {x}{2}\right )\right )+4 \,{\mathrm e}^{12} \arctanh \left (\frac {x}{2}\right )+x^{2}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3-8*x)*exp(12)+2*x^4-23*x^2+20)/(x^3-4*x),x,method=_RETURNVERBOSE)

[Out]

2*x*exp(12)+x^2-5*ln(x^3-4*x)

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maxima [A]  time = 0.36, size = 25, normalized size = 1.04 \begin {gather*} x^{2} + 2 \, x e^{12} - 5 \, \log \left (x + 2\right ) - 5 \, \log \left (x - 2\right ) - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-8*x)*exp(12)+2*x^4-23*x^2+20)/(x^3-4*x),x, algorithm="maxima")

[Out]

x^2 + 2*x*e^12 - 5*log(x + 2) - 5*log(x - 2) - 5*log(x)

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mupad [B]  time = 0.08, size = 19, normalized size = 0.79 \begin {gather*} 2\,x\,{\mathrm {e}}^{12}-5\,\ln \left (x^3-4\,x\right )+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(12)*(8*x - 2*x^3) + 23*x^2 - 2*x^4 - 20)/(4*x - x^3),x)

[Out]

2*x*exp(12) - 5*log(x^3 - 4*x) + x^2

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sympy [A]  time = 0.11, size = 19, normalized size = 0.79 \begin {gather*} x^{2} + 2 x e^{12} - 5 \log {\left (x^{3} - 4 x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3-8*x)*exp(12)+2*x**4-23*x**2+20)/(x**3-4*x),x)

[Out]

x**2 + 2*x*exp(12) - 5*log(x**3 - 4*x)

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