3.23.6 \(\int \frac {-60-24 x+23 x^2+10 x^3+x^4+e^{e^2} (250 x^2+100 x^3+10 x^4)}{50 x^2+20 x^3+2 x^4} \, dx\)

Optimal. Leaf size=30 \[ 5 e^{e^2} x+\frac {1}{x (5+x)}+\frac {2+x^2}{2 x} \]

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Rubi [A]  time = 0.09, antiderivative size = 31, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {1594, 27, 12, 1620} \begin {gather*} \frac {1}{2} \left (1+10 e^{e^2}\right ) x-\frac {1}{5 (x+5)}+\frac {6}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-60 - 24*x + 23*x^2 + 10*x^3 + x^4 + E^E^2*(250*x^2 + 100*x^3 + 10*x^4))/(50*x^2 + 20*x^3 + 2*x^4),x]

[Out]

6/(5*x) + ((1 + 10*E^E^2)*x)/2 - 1/(5*(5 + x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-60-24 x+23 x^2+10 x^3+x^4+e^{e^2} \left (250 x^2+100 x^3+10 x^4\right )}{x^2 \left (50+20 x+2 x^2\right )} \, dx\\ &=\int \frac {-60-24 x+23 x^2+10 x^3+x^4+e^{e^2} \left (250 x^2+100 x^3+10 x^4\right )}{2 x^2 (5+x)^2} \, dx\\ &=\frac {1}{2} \int \frac {-60-24 x+23 x^2+10 x^3+x^4+e^{e^2} \left (250 x^2+100 x^3+10 x^4\right )}{x^2 (5+x)^2} \, dx\\ &=\frac {1}{2} \int \left (1+10 e^{e^2}-\frac {12}{5 x^2}+\frac {2}{5 (5+x)^2}\right ) \, dx\\ &=\frac {6}{5 x}+\frac {1}{2} \left (1+10 e^{e^2}\right ) x-\frac {1}{5 (5+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 32, normalized size = 1.07 \begin {gather*} \frac {1}{2} \left (\frac {12}{5 x}+\left (1+10 e^{e^2}\right ) x-\frac {2}{5 (5+x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-60 - 24*x + 23*x^2 + 10*x^3 + x^4 + E^E^2*(250*x^2 + 100*x^3 + 10*x^4))/(50*x^2 + 20*x^3 + 2*x^4),
x]

[Out]

(12/(5*x) + (1 + 10*E^E^2)*x - 2/(5*(5 + x)))/2

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fricas [A]  time = 0.59, size = 38, normalized size = 1.27 \begin {gather*} \frac {x^{3} + 5 \, x^{2} + 10 \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (e^{2}\right )} + 2 \, x + 12}{2 \, {\left (x^{2} + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^4+100*x^3+250*x^2)*exp(exp(2))+x^4+10*x^3+23*x^2-24*x-60)/(2*x^4+20*x^3+50*x^2),x, algorithm=
"fricas")

[Out]

1/2*(x^3 + 5*x^2 + 10*(x^3 + 5*x^2)*e^(e^2) + 2*x + 12)/(x^2 + 5*x)

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giac [A]  time = 0.37, size = 23, normalized size = 0.77 \begin {gather*} 5 \, x e^{\left (e^{2}\right )} + \frac {1}{2} \, x + \frac {x + 6}{x^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^4+100*x^3+250*x^2)*exp(exp(2))+x^4+10*x^3+23*x^2-24*x-60)/(2*x^4+20*x^3+50*x^2),x, algorithm=
"giac")

[Out]

5*x*e^(e^2) + 1/2*x + (x + 6)/(x^2 + 5*x)

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maple [A]  time = 0.05, size = 23, normalized size = 0.77

method	result	size
default	\(5 x \,{\mathrm e}^{{\mathrm e}^{2}}+\frac {x}{2}-\frac {1}{5 \left (5+x \right )}+\frac {6}{5 x}\)	\(23\)
risch	\(5 x \,{\mathrm e}^{{\mathrm e}^{2}}+\frac {x}{2}+\frac {x +6}{\left (5+x \right ) x}\)	\(23\)
norman	\(\frac {6+\left (5 \,{\mathrm e}^{{\mathrm e}^{2}}+\frac {1}{2}\right ) x^{3}+\left (-\frac {23}{2}-125 \,{\mathrm e}^{{\mathrm e}^{2}}\right ) x}{\left (5+x \right ) x}\)	\(32\)
gosper	\(\frac {10 x^{3} {\mathrm e}^{{\mathrm e}^{2}}+x^{3}-250 x \,{\mathrm e}^{{\mathrm e}^{2}}-23 x +12}{2 x \left (5+x \right )}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^4+100*x^3+250*x^2)*exp(exp(2))+x^4+10*x^3+23*x^2-24*x-60)/(2*x^4+20*x^3+50*x^2),x,method=_RETURNVER
BOSE)

[Out]

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

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maxima [A]  time = 0.55, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{2} \, x {\left (10 \, e^{\left (e^{2}\right )} + 1\right )} + \frac {x + 6}{x^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^4+100*x^3+250*x^2)*exp(exp(2))+x^4+10*x^3+23*x^2-24*x-60)/(2*x^4+20*x^3+50*x^2),x, algorithm=
"maxima")

[Out]

1/2*x*(10*e^(e^2) + 1) + (x + 6)/(x^2 + 5*x)

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mupad [B]  time = 0.09, size = 22, normalized size = 0.73 \begin {gather*} x\,\left (5\,{\mathrm {e}}^{{\mathrm {e}}^2}+\frac {1}{2}\right )+\frac {x+6}{x\,\left (x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(2))*(250*x^2 + 100*x^3 + 10*x^4) - 24*x + 23*x^2 + 10*x^3 + x^4 - 60)/(50*x^2 + 20*x^3 + 2*x^4),x
)

[Out]

x*(5*exp(exp(2)) + 1/2) + (x + 6)/(x*(x + 5))

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sympy [A]  time = 0.16, size = 20, normalized size = 0.67 \begin {gather*} x \left (\frac {1}{2} + 5 e^{e^{2}}\right ) + \frac {x + 6}{x^{2} + 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**4+100*x**3+250*x**2)*exp(exp(2))+x**4+10*x**3+23*x**2-24*x-60)/(2*x**4+20*x**3+50*x**2),x)

[Out]

x*(1/2 + 5*exp(exp(2))) + (x + 6)/(x**2 + 5*x)

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