3.64.59 \(\int \frac {45-4 x-30 x^2+5 x^4+e^{2+x} (18-12 x^2+2 x^4)}{18-12 x^2+2 x^4} \, dx\)

Optimal. Leaf size=19 \[ -7+e^{2+x}+\frac {5 x}{2}+\frac {1}{-3+x^2} \]

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Rubi [B]  time = 0.18, antiderivative size = 52, normalized size of antiderivative = 2.74, number of steps used = 12, number of rules used = 8, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {28, 6742, 2194, 199, 207, 261, 288, 321} \begin {gather*} -\frac {15 x}{4 \left (3-x^2\right )}-\frac {1}{3-x^2}+\frac {5 x^3}{4 \left (3-x^2\right )}+\frac {15 x}{4}+e^{x+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(45 - 4*x - 30*x^2 + 5*x^4 + E^(2 + x)*(18 - 12*x^2 + 2*x^4))/(18 - 12*x^2 + 2*x^4),x]

[Out]

E^(2 + x) + (15*x)/4 - (3 - x^2)^(-1) - (15*x)/(4*(3 - x^2)) + (5*x^3)/(4*(3 - 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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[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} &=2 \int \frac {45-4 x-30 x^2+5 x^4+e^{2+x} \left (18-12 x^2+2 x^4\right )}{\left (-6+2 x^2\right )^2} \, dx\\ &=2 \int \left (\frac {e^{2+x}}{2}+\frac {45}{4 \left (-3+x^2\right )^2}-\frac {x}{\left (-3+x^2\right )^2}-\frac {15 x^2}{2 \left (-3+x^2\right )^2}+\frac {5 x^4}{4 \left (-3+x^2\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {x}{\left (-3+x^2\right )^2} \, dx\right )+\frac {5}{2} \int \frac {x^4}{\left (-3+x^2\right )^2} \, dx-15 \int \frac {x^2}{\left (-3+x^2\right )^2} \, dx+\frac {45}{2} \int \frac {1}{\left (-3+x^2\right )^2} \, dx+\int e^{2+x} \, dx\\ &=e^{2+x}-\frac {1}{3-x^2}-\frac {15 x}{4 \left (3-x^2\right )}+\frac {5 x^3}{4 \left (3-x^2\right )}-\frac {15}{4} \int \frac {1}{-3+x^2} \, dx+\frac {15}{4} \int \frac {x^2}{-3+x^2} \, dx-\frac {15}{2} \int \frac {1}{-3+x^2} \, dx\\ &=e^{2+x}+\frac {15 x}{4}-\frac {1}{3-x^2}-\frac {15 x}{4 \left (3-x^2\right )}+\frac {5 x^3}{4 \left (3-x^2\right )}+\frac {15}{4} \sqrt {3} \tanh ^{-1}\left (\frac {x}{\sqrt {3}}\right )+\frac {45}{4} \int \frac {1}{-3+x^2} \, dx\\ &=e^{2+x}+\frac {15 x}{4}-\frac {1}{3-x^2}-\frac {15 x}{4 \left (3-x^2\right )}+\frac {5 x^3}{4 \left (3-x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 33, normalized size = 1.74 \begin {gather*} \frac {2-15 x+5 x^3+2 e^{2+x} \left (-3+x^2\right )}{2 \left (-3+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(45 - 4*x - 30*x^2 + 5*x^4 + E^(2 + x)*(18 - 12*x^2 + 2*x^4))/(18 - 12*x^2 + 2*x^4),x]

[Out]

(2 - 15*x + 5*x^3 + 2*E^(2 + x)*(-3 + x^2))/(2*(-3 + x^2))

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fricas [A]  time = 0.70, size = 30, normalized size = 1.58 \begin {gather*} \frac {5 \, x^{3} + 2 \, {\left (x^{2} - 3\right )} e^{\left (x + 2\right )} - 15 \, x + 2}{2 \, {\left (x^{2} - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4-12*x^2+18)*exp(2+x)+5*x^4-30*x^2-4*x+45)/(2*x^4-12*x^2+18),x, algorithm="fricas")

[Out]

1/2*(5*x^3 + 2*(x^2 - 3)*e^(x + 2) - 15*x + 2)/(x^2 - 3)

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giac [B]  time = 0.15, size = 34, normalized size = 1.79 \begin {gather*} \frac {5 \, x^{3} + 2 \, x^{2} e^{\left (x + 2\right )} - 15 \, x - 6 \, e^{\left (x + 2\right )} + 2}{2 \, {\left (x^{2} - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4-12*x^2+18)*exp(2+x)+5*x^4-30*x^2-4*x+45)/(2*x^4-12*x^2+18),x, algorithm="giac")

[Out]

1/2*(5*x^3 + 2*x^2*e^(x + 2) - 15*x - 6*e^(x + 2) + 2)/(x^2 - 3)

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maple [A]  time = 0.18, size = 16, normalized size = 0.84

method	result	size
risch	\(\frac {5 x}{2}+\frac {1}{x^{2}-3}+{\mathrm e}^{2+x}\)	\(16\)
norman	\(\frac {x^{2} {\mathrm e}^{2+x}-\frac {15 x}{2}+\frac {5 x^{3}}{2}-3 \,{\mathrm e}^{2+x}+1}{x^{2}-3}\)	\(33\)
derivativedivides	\(-\frac {13 x}{12 \left (\left (2+x \right )^{2}-7-4 x \right )}+\frac {\frac {22 x}{3}+11}{\left (2+x \right )^{2}-7-4 x}+\frac {-90-\frac {105 x}{2}}{\left (2+x \right )^{2}-7-4 x}-\frac {20 \left (-\frac {15}{2}-\frac {13 x}{3}\right )}{\left (2+x \right )^{2}-7-4 x}+5+\frac {5 x}{2}+\frac {-70-\frac {485 x}{12}}{\left (2+x \right )^{2}-7-4 x}-\frac {{\mathrm e}^{2+x} x}{6 \left (\left (2+x \right )^{2}-7-4 x \right )}+\frac {4 \,{\mathrm e}^{2+x} \left (2 x +3\right )}{3 \left (\left (2+x \right )^{2}-7-4 x \right )}-\frac {3 \,{\mathrm e}^{2+x} \left (12+7 x \right )}{\left (2+x \right )^{2}-7-4 x}+\frac {4 \,{\mathrm e}^{2+x} \left (45+26 x \right )}{3 \left (\left (2+x \right )^{2}-7-4 x \right )}+{\mathrm e}^{2+x}-\frac {{\mathrm e}^{2+x} \left (168+97 x \right )}{6 \left (\left (2+x \right )^{2}-7-4 x \right )}\)	\(212\)
default	\(-\frac {13 x}{12 \left (\left (2+x \right )^{2}-7-4 x \right )}+\frac {\frac {22 x}{3}+11}{\left (2+x \right )^{2}-7-4 x}+\frac {-90-\frac {105 x}{2}}{\left (2+x \right )^{2}-7-4 x}-\frac {20 \left (-\frac {15}{2}-\frac {13 x}{3}\right )}{\left (2+x \right )^{2}-7-4 x}+5+\frac {5 x}{2}+\frac {-70-\frac {485 x}{12}}{\left (2+x \right )^{2}-7-4 x}-\frac {{\mathrm e}^{2+x} x}{6 \left (\left (2+x \right )^{2}-7-4 x \right )}+\frac {4 \,{\mathrm e}^{2+x} \left (2 x +3\right )}{3 \left (\left (2+x \right )^{2}-7-4 x \right )}-\frac {3 \,{\mathrm e}^{2+x} \left (12+7 x \right )}{\left (2+x \right )^{2}-7-4 x}+\frac {4 \,{\mathrm e}^{2+x} \left (45+26 x \right )}{3 \left (\left (2+x \right )^{2}-7-4 x \right )}+{\mathrm e}^{2+x}-\frac {{\mathrm e}^{2+x} \left (168+97 x \right )}{6 \left (\left (2+x \right )^{2}-7-4 x \right )}\)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^4-12*x^2+18)*exp(2+x)+5*x^4-30*x^2-4*x+45)/(2*x^4-12*x^2+18),x,method=_RETURNVERBOSE)

[Out]

5/2*x+1/(x^2-3)+exp(2+x)

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maxima [A]  time = 0.50, size = 15, normalized size = 0.79 \begin {gather*} \frac {5}{2} \, x + \frac {1}{x^{2} - 3} + e^{\left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4-12*x^2+18)*exp(2+x)+5*x^4-30*x^2-4*x+45)/(2*x^4-12*x^2+18),x, algorithm="maxima")

[Out]

5/2*x + 1/(x^2 - 3) + e^(x + 2)

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mupad [B]  time = 4.30, size = 19, normalized size = 1.00 \begin {gather*} \frac {5\,x}{2}+{\mathrm {e}}^{x+2}+\frac {2}{2\,x^2-6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + 2)*(2*x^4 - 12*x^2 + 18) - 4*x - 30*x^2 + 5*x^4 + 45)/(2*x^4 - 12*x^2 + 18),x)

[Out]

(5*x)/2 + exp(x + 2) + 2/(2*x^2 - 6)

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sympy [A]  time = 0.14, size = 15, normalized size = 0.79 \begin {gather*} \frac {5 x}{2} + e^{x + 2} + \frac {1}{x^{2} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**4-12*x**2+18)*exp(2+x)+5*x**4-30*x**2-4*x+45)/(2*x**4-12*x**2+18),x)

[Out]

5*x/2 + exp(x + 2) + 1/(x**2 - 3)

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