3.2.28 \(\int \frac {497-139 x^2+10 x^4+e^x (245+245 x-70 x^2-70 x^3+5 x^4+5 x^5)}{49-14 x^2+x^4} \, dx\)

Optimal. Leaf size=26 \[ \frac {5}{4} \left (4+4 \left (2+e^x\right ) x\right )+\frac {x}{7-x^2} \]

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Rubi [A]  time = 0.15, antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {28, 6742, 2176, 2194, 1157, 21, 8} \begin {gather*} \frac {x}{7-x^2}+10 x-5 e^x+5 e^x (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(497 - 139*x^2 + 10*x^4 + E^x*(245 + 245*x - 70*x^2 - 70*x^3 + 5*x^4 + 5*x^5))/(49 - 14*x^2 + x^4),x]

[Out]

-5*E^x + 10*x + 5*E^x*(1 + x) + x/(7 - x^2)

Rule 8

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

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 {497-139 x^2+10 x^4+e^x \left (245+245 x-70 x^2-70 x^3+5 x^4+5 x^5\right )}{\left (-7+x^2\right )^2} \, dx\\ &=\int \left (5 e^x (1+x)+\frac {497-139 x^2+10 x^4}{\left (-7+x^2\right )^2}\right ) \, dx\\ &=5 \int e^x (1+x) \, dx+\int \frac {497-139 x^2+10 x^4}{\left (-7+x^2\right )^2} \, dx\\ &=5 e^x (1+x)+\frac {x}{7-x^2}+\frac {1}{14} \int \frac {-980+140 x^2}{-7+x^2} \, dx-5 \int e^x \, dx\\ &=-5 e^x+5 e^x (1+x)+\frac {x}{7-x^2}+10 \int 1 \, dx\\ &=-5 e^x+10 x+5 e^x (1+x)+\frac {x}{7-x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 18, normalized size = 0.69 \begin {gather*} x \left (10+5 e^x+\frac {1}{7-x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(497 - 139*x^2 + 10*x^4 + E^x*(245 + 245*x - 70*x^2 - 70*x^3 + 5*x^4 + 5*x^5))/(49 - 14*x^2 + x^4),x
]

[Out]

x*(10 + 5*E^x + (7 - x^2)^(-1))

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fricas [A]  time = 1.12, size = 28, normalized size = 1.08 \begin {gather*} \frac {10 \, x^{3} + 5 \, {\left (x^{3} - 7 \, x\right )} e^{x} - 71 \, x}{x^{2} - 7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^5+5*x^4-70*x^3-70*x^2+245*x+245)*exp(x)+10*x^4-139*x^2+497)/(x^4-14*x^2+49),x, algorithm="fric
as")

[Out]

(10*x^3 + 5*(x^3 - 7*x)*e^x - 71*x)/(x^2 - 7)

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giac [A]  time = 0.34, size = 29, normalized size = 1.12 \begin {gather*} \frac {5 \, x^{3} e^{x} + 10 \, x^{3} - 35 \, x e^{x} - 71 \, x}{x^{2} - 7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^5+5*x^4-70*x^3-70*x^2+245*x+245)*exp(x)+10*x^4-139*x^2+497)/(x^4-14*x^2+49),x, algorithm="giac
")

[Out]

(5*x^3*e^x + 10*x^3 - 35*x*e^x - 71*x)/(x^2 - 7)

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maple [A]  time = 0.09, size = 20, normalized size = 0.77

method	result	size
risch	\(-\frac {x}{x^{2}-7}+10 x +5 \,{\mathrm e}^{x} x\)	\(20\)
default	\(-\frac {x}{x^{2}-7}+10 x +5 \left (x -1\right ) {\mathrm e}^{x}+5 \,{\mathrm e}^{x}\)	\(26\)
norman	\(\frac {-71 x +10 x^{3}-35 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x} x^{3}}{x^{2}-7}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^5+5*x^4-70*x^3-70*x^2+245*x+245)*exp(x)+10*x^4-139*x^2+497)/(x^4-14*x^2+49),x,method=_RETURNVERBOSE)

[Out]

-x/(x^2-7)+10*x+5*exp(x)*x

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maxima [A]  time = 0.66, size = 19, normalized size = 0.73 \begin {gather*} 5 \, x e^{x} + 10 \, x - \frac {x}{x^{2} - 7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^5+5*x^4-70*x^3-70*x^2+245*x+245)*exp(x)+10*x^4-139*x^2+497)/(x^4-14*x^2+49),x, algorithm="maxi
ma")

[Out]

5*x*e^x + 10*x - x/(x^2 - 7)

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mupad [B]  time = 0.36, size = 19, normalized size = 0.73 \begin {gather*} 10\,x-\frac {x}{x^2-7}+5\,x\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(245*x - 70*x^2 - 70*x^3 + 5*x^4 + 5*x^5 + 245) - 139*x^2 + 10*x^4 + 497)/(x^4 - 14*x^2 + 49),x)

[Out]

10*x - x/(x^2 - 7) + 5*x*exp(x)

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sympy [A]  time = 0.15, size = 15, normalized size = 0.58 \begin {gather*} 5 x e^{x} + 10 x - \frac {x}{x^{2} - 7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**5+5*x**4-70*x**3-70*x**2+245*x+245)*exp(x)+10*x**4-139*x**2+497)/(x**4-14*x**2+49),x)

[Out]

5*x*exp(x) + 10*x - x/(x**2 - 7)

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