3.26.78 \(\int \frac {-50 e^4-2 x^2+8 x^3+e^2 (-20 x+60 x^2)}{75 e^4+30 e^2 x+3 x^2} \, dx\)

Optimal. Leaf size=23 \[ 3-\frac {2 x}{3}+\frac {4 x^3}{3 \left (5 e^2+x\right )} \]

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Rubi [A]  time = 0.05, antiderivative size = 36, normalized size of antiderivative = 1.57, number of steps used = 4, number of rules used = 3, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {27, 12, 1850} \begin {gather*} \frac {4 x^2}{3}-\frac {2}{3} \left (1+10 e^2\right ) x-\frac {500 e^6}{3 \left (x+5 e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50*E^4 - 2*x^2 + 8*x^3 + E^2*(-20*x + 60*x^2))/(75*E^4 + 30*E^2*x + 3*x^2),x]

[Out]

(-2*(1 + 10*E^2)*x)/3 + (4*x^2)/3 - (500*E^6)/(3*(5*E^2 + 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 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-50 e^4-2 x^2+8 x^3+e^2 \left (-20 x+60 x^2\right )}{3 \left (5 e^2+x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {-50 e^4-2 x^2+8 x^3+e^2 \left (-20 x+60 x^2\right )}{\left (5 e^2+x\right )^2} \, dx\\ &=\frac {1}{3} \int \left (-2 \left (1+10 e^2\right )+8 x+\frac {500 e^6}{\left (5 e^2+x\right )^2}\right ) \, dx\\ &=-\frac {2}{3} \left (1+10 e^2\right ) x+\frac {4 x^2}{3}-\frac {500 e^6}{3 \left (5 e^2+x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 40, normalized size = 1.74 \begin {gather*} -\frac {2}{3} \left (100 e^4+x-2 x^2+\frac {250 e^6}{5 e^2+x}+5 e^2 (1+2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50*E^4 - 2*x^2 + 8*x^3 + E^2*(-20*x + 60*x^2))/(75*E^4 + 30*E^2*x + 3*x^2),x]

[Out]

(-2*(100*E^4 + x - 2*x^2 + (250*E^6)/(5*E^2 + x) + 5*E^2*(1 + 2*x)))/3

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fricas [A]  time = 0.48, size = 35, normalized size = 1.52 \begin {gather*} \frac {2 \, {\left (2 \, x^{3} - x^{2} - 50 \, x e^{4} - 5 \, x e^{2} - 250 \, e^{6}\right )}}{3 \, {\left (x + 5 \, e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(2)^2+(60*x^2-20*x)*exp(2)+8*x^3-2*x^2)/(75*exp(2)^2+30*exp(2)*x+3*x^2),x, algorithm="fricas
")

[Out]

2/3*(2*x^3 - x^2 - 50*x*e^4 - 5*x*e^2 - 250*e^6)/(x + 5*e^2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(2)^2+(60*x^2-20*x)*exp(2)+8*x^3-2*x^2)/(75*exp(2)^2+30*exp(2)*x+3*x^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 2/3*(2*sageVARx^2-10*sageVARx*exp(2)-sag
eVARx+(50*exp(2)^2-50*exp(4))*ln(sageVARx^2+10*sageVARx*exp(2)+25*exp(4))+(-500*exp(2)^3+750*exp(2)*exp(4))*1/
10/sqrt(exp(2)^2-exp(

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maple [A]  time = 0.46, size = 27, normalized size = 1.17

method	result	size
norman	\(\frac {-\frac {2 x^{2}}{3}+\frac {4 x^{3}}{3}+\frac {50 \,{\mathrm e}^{4}}{3}}{5 \,{\mathrm e}^{2}+x}\)	\(27\)
risch	\(-\frac {20 \,{\mathrm e}^{2} x}{3}+\frac {4 x^{2}}{3}-\frac {2 x}{3}-\frac {100 \,{\mathrm e}^{6}}{3 \left ({\mathrm e}^{2}+\frac {x}{5}\right )}\)	\(27\)
gosper	\(\frac {-\frac {2 x^{2}}{3}+\frac {4 x^{3}}{3}+\frac {50 \,{\mathrm e}^{4}}{3}}{5 \,{\mathrm e}^{2}+x}\)	\(28\)
meijerg	\(-\frac {2 x}{3 \left (1+\frac {x \,{\mathrm e}^{-2}}{5}\right )}+\frac {5 \left (60 \,{\mathrm e}^{2}-2\right ) {\mathrm e}^{2} \left (\frac {x \,{\mathrm e}^{-2} \left (\frac {3 x \,{\mathrm e}^{-2}}{5}+6\right )}{15+3 x \,{\mathrm e}^{-2}}-2 \ln \left (1+\frac {x \,{\mathrm e}^{-2}}{5}\right )\right )}{3}-\frac {20 \,{\mathrm e}^{2} \left (-\frac {x \,{\mathrm e}^{-2}}{5 \left (1+\frac {x \,{\mathrm e}^{-2}}{5}\right )}+\ln \left (1+\frac {x \,{\mathrm e}^{-2}}{5}\right )\right )}{3}+\frac {200 \,{\mathrm e}^{4} \left (-\frac {x \,{\mathrm e}^{-2} \left (-\frac {2 x^{2} {\mathrm e}^{-4}}{25}+\frac {6 x \,{\mathrm e}^{-2}}{5}+12\right )}{20 \left (1+\frac {x \,{\mathrm e}^{-2}}{5}\right )}+3 \ln \left (1+\frac {x \,{\mathrm e}^{-2}}{5}\right )\right )}{3}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-50*exp(2)^2+(60*x^2-20*x)*exp(2)+8*x^3-2*x^2)/(75*exp(2)^2+30*exp(2)*x+3*x^2),x,method=_RETURNVERBOSE)

[Out]

(-2/3*x^2+4/3*x^3+50/3*exp(2)^2)/(5*exp(2)+x)

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maxima [A]  time = 0.40, size = 27, normalized size = 1.17 \begin {gather*} \frac {4}{3} \, x^{2} - \frac {2}{3} \, x {\left (10 \, e^{2} + 1\right )} - \frac {500 \, e^{6}}{3 \, {\left (x + 5 \, e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(2)^2+(60*x^2-20*x)*exp(2)+8*x^3-2*x^2)/(75*exp(2)^2+30*exp(2)*x+3*x^2),x, algorithm="maxima
")

[Out]

4/3*x^2 - 2/3*x*(10*e^2 + 1) - 500/3*e^6/(x + 5*e^2)

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mupad [B]  time = 1.58, size = 29, normalized size = 1.26 \begin {gather*} \frac {4\,x^2}{3}-\frac {500\,{\mathrm {e}}^6}{3\,x+15\,{\mathrm {e}}^2}-x\,\left (\frac {20\,{\mathrm {e}}^2}{3}+\frac {2}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(50*exp(4) + exp(2)*(20*x - 60*x^2) + 2*x^2 - 8*x^3)/(75*exp(4) + 30*x*exp(2) + 3*x^2),x)

[Out]

(4*x^2)/3 - (500*exp(6))/(3*x + 15*exp(2)) - x*((20*exp(2))/3 + 2/3)

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sympy [A]  time = 0.14, size = 32, normalized size = 1.39 \begin {gather*} \frac {4 x^{2}}{3} + x \left (- \frac {20 e^{2}}{3} - \frac {2}{3}\right ) - \frac {500 e^{6}}{3 x + 15 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(2)**2+(60*x**2-20*x)*exp(2)+8*x**3-2*x**2)/(75*exp(2)**2+30*exp(2)*x+3*x**2),x)

[Out]

4*x**2/3 + x*(-20*exp(2)/3 - 2/3) - 500*exp(6)/(3*x + 15*exp(2))

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