3.85.77 \(\int \frac {-400+16 x^2+e^{25} (100-4 x^2)}{243 (15625+17500 x+6150 x^2+700 x^3+25 x^4)} \, dx\)

Optimal. Leaf size=23 \[ \frac {\left (-4+e^{25}\right ) x}{6075 \left (x+\frac {1}{4} (5+x)^2\right )} \]

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Rubi [A]  time = 0.05, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 1680, 1814, 8} \begin {gather*} \frac {4 \left (4-e^{25}\right ) x}{6075 \left (24-(x+7)^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-400 + 16*x^2 + E^25*(100 - 4*x^2))/(243*(15625 + 17500*x + 6150*x^2 + 700*x^3 + 25*x^4)),x]

[Out]

(4*(4 - E^25)*x)/(6075*(24 - (7 + x)^2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{243} \int \frac {-400+16 x^2+e^{25} \left (100-4 x^2\right )}{15625+17500 x+6150 x^2+700 x^3+25 x^4} \, dx\\ &=\frac {1}{243} \operatorname {Subst}\left (\int \frac {4 \left (4-e^{25}\right ) \left (24-14 x+x^2\right )}{25 \left (24-x^2\right )^2} \, dx,x,7+x\right )\\ &=\frac {\left (4 \left (4-e^{25}\right )\right ) \operatorname {Subst}\left (\int \frac {24-14 x+x^2}{\left (24-x^2\right )^2} \, dx,x,7+x\right )}{6075}\\ &=\frac {4 \left (4-e^{25}\right ) x}{6075 \left (24-(7+x)^2\right )}+\frac {\left (-4+e^{25}\right ) \operatorname {Subst}(\int 0 \, dx,x,7+x)}{72900}\\ &=\frac {4 \left (4-e^{25}\right ) x}{6075 \left (24-(7+x)^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.87 \begin {gather*} \frac {4 \left (-4+e^{25}\right ) x}{6075 \left (25+14 x+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-400 + 16*x^2 + E^25*(100 - 4*x^2))/(243*(15625 + 17500*x + 6150*x^2 + 700*x^3 + 25*x^4)),x]

[Out]

(4*(-4 + E^25)*x)/(6075*(25 + 14*x + x^2))

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fricas [A]  time = 0.76, size = 20, normalized size = 0.87 \begin {gather*} \frac {4 \, {\left (x e^{25} - 4 \, x\right )}}{6075 \, {\left (x^{2} + 14 \, x + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/243*((-4*x^2+100)*exp(25)+16*x^2-400)/(25*x^4+700*x^3+6150*x^2+17500*x+15625),x, algorithm="fricas
")

[Out]

4/6075*(x*e^25 - 4*x)/(x^2 + 14*x + 25)

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giac [A]  time = 0.20, size = 20, normalized size = 0.87 \begin {gather*} \frac {4 \, {\left (x e^{25} - 4 \, x\right )}}{6075 \, {\left (x^{2} + 14 \, x + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/243*((-4*x^2+100)*exp(25)+16*x^2-400)/(25*x^4+700*x^3+6150*x^2+17500*x+15625),x, algorithm="giac")

[Out]

4/6075*(x*e^25 - 4*x)/(x^2 + 14*x + 25)

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maple [A]  time = 0.04, size = 18, normalized size = 0.78

method	result	size
gosper	\(\frac {4 x \left ({\mathrm e}^{25}-4\right )}{6075 \left (x^{2}+14 x +25\right )}\)	\(18\)
norman	\(\frac {\left (-\frac {16}{6075}+\frac {4 \,{\mathrm e}^{25}}{6075}\right ) x}{x^{2}+14 x +25}\)	\(19\)
default	\(\frac {\left (4 \,{\mathrm e}^{25}-16\right ) x}{6075 x^{2}+85050 x +151875}\)	\(20\)
risch	\(\frac {x \left (\frac {4 \,{\mathrm e}^{25}}{25}-\frac {16}{25}\right )}{243 x^{2}+3402 x +6075}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/243*((-4*x^2+100)*exp(25)+16*x^2-400)/(25*x^4+700*x^3+6150*x^2+17500*x+15625),x,method=_RETURNVERBOSE)

[Out]

4/6075*x*(exp(25)-4)/(x^2+14*x+25)

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maxima [A]  time = 0.36, size = 17, normalized size = 0.74 \begin {gather*} \frac {4 \, x {\left (e^{25} - 4\right )}}{6075 \, {\left (x^{2} + 14 \, x + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/243*((-4*x^2+100)*exp(25)+16*x^2-400)/(25*x^4+700*x^3+6150*x^2+17500*x+15625),x, algorithm="maxima
")

[Out]

4/6075*x*(e^25 - 4)/(x^2 + 14*x + 25)

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mupad [B]  time = 0.08, size = 19, normalized size = 0.83 \begin {gather*} \frac {4\,x\,\left ({\mathrm {e}}^{25}-4\right )}{6075\,\left (x^2+14\,x+25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(25)*(4*x^2 - 100))/243 - (16*x^2)/243 + 400/243)/(17500*x + 6150*x^2 + 700*x^3 + 25*x^4 + 15625),x)

[Out]

(4*x*(exp(25) - 4))/(6075*(14*x + x^2 + 25))

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sympy [A]  time = 0.32, size = 17, normalized size = 0.74 \begin {gather*} \frac {x \left (-16 + 4 e^{25}\right )}{6075 x^{2} + 85050 x + 151875} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/243*((-4*x**2+100)*exp(25)+16*x**2-400)/(25*x**4+700*x**3+6150*x**2+17500*x+15625),x)

[Out]

x*(-16 + 4*exp(25))/(6075*x**2 + 85050*x + 151875)

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