3.7.8 \(\int \frac {630-172 x+65 x^2+5 x^4+e^x (405+90 x^2+5 x^4)}{405+90 x^2+5 x^4} \, dx\)

Optimal. Leaf size=31 \[ e^x+x-\frac {\frac {4}{5}+(3-x)^2+x+2 x^2}{9+x^2} \]

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Rubi [A]  time = 0.16, antiderivative size = 46, normalized size of antiderivative = 1.48, number of steps used = 12, number of rules used = 8, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {28, 6742, 2194, 199, 203, 261, 288, 321} \begin {gather*} \frac {x}{2 \left (x^2+9\right )}+\frac {86}{5 \left (x^2+9\right )}-\frac {x^3}{2 \left (x^2+9\right )}+\frac {3 x}{2}+e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(630 - 172*x + 65*x^2 + 5*x^4 + E^x*(405 + 90*x^2 + 5*x^4))/(405 + 90*x^2 + 5*x^4),x]

[Out]

E^x + (3*x)/2 + 86/(5*(9 + x^2)) + x/(2*(9 + x^2)) - x^3/(2*(9 + x^2))

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 261

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

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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} &=5 \int \frac {630-172 x+65 x^2+5 x^4+e^x \left (405+90 x^2+5 x^4\right )}{\left (45+5 x^2\right )^2} \, dx\\ &=5 \int \left (\frac {e^x}{5}+\frac {126}{5 \left (9+x^2\right )^2}-\frac {172 x}{25 \left (9+x^2\right )^2}+\frac {13 x^2}{5 \left (9+x^2\right )^2}+\frac {x^4}{5 \left (9+x^2\right )^2}\right ) \, dx\\ &=13 \int \frac {x^2}{\left (9+x^2\right )^2} \, dx-\frac {172}{5} \int \frac {x}{\left (9+x^2\right )^2} \, dx+126 \int \frac {1}{\left (9+x^2\right )^2} \, dx+\int e^x \, dx+\int \frac {x^4}{\left (9+x^2\right )^2} \, dx\\ &=e^x+\frac {86}{5 \left (9+x^2\right )}+\frac {x}{2 \left (9+x^2\right )}-\frac {x^3}{2 \left (9+x^2\right )}+\frac {3}{2} \int \frac {x^2}{9+x^2} \, dx+\frac {13}{2} \int \frac {1}{9+x^2} \, dx+7 \int \frac {1}{9+x^2} \, dx\\ &=e^x+\frac {3 x}{2}+\frac {86}{5 \left (9+x^2\right )}+\frac {x}{2 \left (9+x^2\right )}-\frac {x^3}{2 \left (9+x^2\right )}+\frac {9}{2} \tan ^{-1}\left (\frac {x}{3}\right )-\frac {27}{2} \int \frac {1}{9+x^2} \, dx\\ &=e^x+\frac {3 x}{2}+\frac {86}{5 \left (9+x^2\right )}+\frac {x}{2 \left (9+x^2\right )}-\frac {x^3}{2 \left (9+x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 31, normalized size = 1.00 \begin {gather*} \frac {86+70 x+5 x^3+5 e^x \left (9+x^2\right )}{5 \left (9+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(630 - 172*x + 65*x^2 + 5*x^4 + E^x*(405 + 90*x^2 + 5*x^4))/(405 + 90*x^2 + 5*x^4),x]

[Out]

(86 + 70*x + 5*x^3 + 5*E^x*(9 + x^2))/(5*(9 + x^2))

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fricas [A]  time = 0.82, size = 28, normalized size = 0.90 \begin {gather*} \frac {5 \, x^{3} + 5 \, {\left (x^{2} + 9\right )} e^{x} + 70 \, x + 86}{5 \, {\left (x^{2} + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^4+90*x^2+405)*exp(x)+5*x^4+65*x^2-172*x+630)/(5*x^4+90*x^2+405),x, algorithm="fricas")

[Out]

1/5*(5*x^3 + 5*(x^2 + 9)*e^x + 70*x + 86)/(x^2 + 9)

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giac [A]  time = 0.49, size = 30, normalized size = 0.97 \begin {gather*} \frac {5 \, x^{3} + 5 \, x^{2} e^{x} + 70 \, x + 45 \, e^{x} + 86}{5 \, {\left (x^{2} + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^4+90*x^2+405)*exp(x)+5*x^4+65*x^2-172*x+630)/(5*x^4+90*x^2+405),x, algorithm="giac")

[Out]

1/5*(5*x^3 + 5*x^2*e^x + 70*x + 45*e^x + 86)/(x^2 + 9)

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maple [A]  time = 0.13, size = 18, normalized size = 0.58

method	result	size
risch	\(x +\frac {5 x +\frac {86}{5}}{x^{2}+9}+{\mathrm e}^{x}\)	\(18\)
default	\(\frac {5 x}{x^{2}+9}+\frac {86}{5 \left (x^{2}+9\right )}+x +{\mathrm e}^{x}\)	\(24\)
norman	\(\frac {x^{3}+{\mathrm e}^{x} x^{2}+14 x +9 \,{\mathrm e}^{x}+\frac {86}{5}}{x^{2}+9}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^4+90*x^2+405)*exp(x)+5*x^4+65*x^2-172*x+630)/(5*x^4+90*x^2+405),x,method=_RETURNVERBOSE)

[Out]

x+(5*x+86/5)/(x^2+9)+exp(x)

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maxima [A]  time = 0.57, size = 23, normalized size = 0.74 \begin {gather*} x + \frac {5 \, x}{x^{2} + 9} + \frac {86}{5 \, {\left (x^{2} + 9\right )}} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^4+90*x^2+405)*exp(x)+5*x^4+65*x^2-172*x+630)/(5*x^4+90*x^2+405),x, algorithm="maxima")

[Out]

x + 5*x/(x^2 + 9) + 86/5/(x^2 + 9) + e^x

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mupad [B]  time = 0.16, size = 19, normalized size = 0.61 \begin {gather*} x+{\mathrm {e}}^x+\frac {25\,x+86}{5\,x^2+45} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(90*x^2 + 5*x^4 + 405) - 172*x + 65*x^2 + 5*x^4 + 630)/(90*x^2 + 5*x^4 + 405),x)

[Out]

x + exp(x) + (25*x + 86)/(5*x^2 + 45)

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sympy [A]  time = 0.15, size = 15, normalized size = 0.48 \begin {gather*} x + \frac {25 x + 86}{5 x^{2} + 45} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**4+90*x**2+405)*exp(x)+5*x**4+65*x**2-172*x+630)/(5*x**4+90*x**2+405),x)

[Out]

x + (25*x + 86)/(5*x**2 + 45) + exp(x)

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