3.11.83 \(\int \frac {x-x^2+3 x^3+e^x x (1-x+3 x^2) (4-6 x+34 x^2+x^3-3 x^4+e^5 (1-x+8 x^2+3 x^3))}{x-x^2+3 x^3} \, dx\)

Optimal. Leaf size=26 \[ -3+x+e^x \left (4+e^5-x\right ) x \left (1-x+3 x^2\right ) \]

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Rubi [B]  time = 0.60, antiderivative size = 59, normalized size of antiderivative = 2.27, number of steps used = 32, number of rules used = 5, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {1594, 6688, 2196, 2194, 2176} \begin {gather*} -3 e^x x^4+13 e^x x^3+3 e^{x+5} x^3-5 e^x x^2-e^{x+5} x^2+4 e^x x+e^{x+5} x+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x - x^2 + 3*x^3 + E^x*x*(1 - x + 3*x^2)*(4 - 6*x + 34*x^2 + x^3 - 3*x^4 + E^5*(1 - x + 8*x^2 + 3*x^3)))/(
x - x^2 + 3*x^3),x]

[Out]

x + 4*E^x*x + E^(5 + x)*x - 5*E^x*x^2 - E^(5 + x)*x^2 + 13*E^x*x^3 + 3*E^(5 + x)*x^3 - 3*E^x*x^4

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x-x^2+3 x^3+e^x x \left (1-x+3 x^2\right ) \left (4-6 x+34 x^2+x^3-3 x^4+e^5 \left (1-x+8 x^2+3 x^3\right )\right )}{x \left (1-x+3 x^2\right )} \, dx\\ &=\int \left (1+e^{5+x} \left (1-x+8 x^2+3 x^3\right )+e^x \left (4-6 x+34 x^2+x^3-3 x^4\right )\right ) \, dx\\ &=x+\int e^{5+x} \left (1-x+8 x^2+3 x^3\right ) \, dx+\int e^x \left (4-6 x+34 x^2+x^3-3 x^4\right ) \, dx\\ &=x+\int \left (e^{5+x}-e^{5+x} x+8 e^{5+x} x^2+3 e^{5+x} x^3\right ) \, dx+\int \left (4 e^x-6 e^x x+34 e^x x^2+e^x x^3-3 e^x x^4\right ) \, dx\\ &=x+3 \int e^{5+x} x^3 \, dx-3 \int e^x x^4 \, dx+4 \int e^x \, dx-6 \int e^x x \, dx+8 \int e^{5+x} x^2 \, dx+34 \int e^x x^2 \, dx+\int e^{5+x} \, dx-\int e^{5+x} x \, dx+\int e^x x^3 \, dx\\ &=4 e^x+e^{5+x}+x-6 e^x x-e^{5+x} x+34 e^x x^2+8 e^{5+x} x^2+e^x x^3+3 e^{5+x} x^3-3 e^x x^4-3 \int e^x x^2 \, dx+6 \int e^x \, dx-9 \int e^{5+x} x^2 \, dx+12 \int e^x x^3 \, dx-16 \int e^{5+x} x \, dx-68 \int e^x x \, dx+\int e^{5+x} \, dx\\ &=10 e^x+2 e^{5+x}+x-74 e^x x-17 e^{5+x} x+31 e^x x^2-e^{5+x} x^2+13 e^x x^3+3 e^{5+x} x^3-3 e^x x^4+6 \int e^x x \, dx+16 \int e^{5+x} \, dx+18 \int e^{5+x} x \, dx-36 \int e^x x^2 \, dx+68 \int e^x \, dx\\ &=78 e^x+18 e^{5+x}+x-68 e^x x+e^{5+x} x-5 e^x x^2-e^{5+x} x^2+13 e^x x^3+3 e^{5+x} x^3-3 e^x x^4-6 \int e^x \, dx-18 \int e^{5+x} \, dx+72 \int e^x x \, dx\\ &=72 e^x+x+4 e^x x+e^{5+x} x-5 e^x x^2-e^{5+x} x^2+13 e^x x^3+3 e^{5+x} x^3-3 e^x x^4-72 \int e^x \, dx\\ &=x+4 e^x x+e^{5+x} x-5 e^x x^2-e^{5+x} x^2+13 e^x x^3+3 e^{5+x} x^3-3 e^x x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 51, normalized size = 1.96 \begin {gather*} x+e^x \left (e^5 x-e^5 x^2+3 e^5 x^3\right )+e^x \left (4 x-5 x^2+13 x^3-3 x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x - x^2 + 3*x^3 + E^x*x*(1 - x + 3*x^2)*(4 - 6*x + 34*x^2 + x^3 - 3*x^4 + E^5*(1 - x + 8*x^2 + 3*x^
3)))/(x - x^2 + 3*x^3),x]

[Out]

x + E^x*(E^5*x - E^5*x^2 + 3*E^5*x^3) + E^x*(4*x - 5*x^2 + 13*x^3 - 3*x^4)

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fricas [A]  time = 0.83, size = 27, normalized size = 1.04 \begin {gather*} -{\left (x - e^{5} - 4\right )} e^{\left (x + \log \left (3 \, x^{2} - x + 1\right ) + \log \relax (x)\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^3+8*x^2-x+1)*exp(5)-3*x^4+x^3+34*x^2-6*x+4)*exp(log(3*x^2-x+1)+x+log(x))+3*x^3-x^2+x)/(3*x^3-
x^2+x),x, algorithm="fricas")

[Out]

-(x - e^5 - 4)*e^(x + log(3*x^2 - x + 1) + log(x)) + x

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giac [A]  time = 0.38, size = 52, normalized size = 2.00 \begin {gather*} -3 \, x^{4} e^{x} + 3 \, x^{3} e^{\left (x + 5\right )} + 13 \, x^{3} e^{x} - x^{2} e^{\left (x + 5\right )} - 5 \, x^{2} e^{x} + x e^{\left (x + 5\right )} + 4 \, x e^{x} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^3+8*x^2-x+1)*exp(5)-3*x^4+x^3+34*x^2-6*x+4)*exp(log(3*x^2-x+1)+x+log(x))+3*x^3-x^2+x)/(3*x^3-
x^2+x),x, algorithm="giac")

[Out]

-3*x^4*e^x + 3*x^3*e^(x + 5) + 13*x^3*e^x - x^2*e^(x + 5) - 5*x^2*e^x + x*e^(x + 5) + 4*x*e^x + x

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maple [A]  time = 0.04, size = 43, normalized size = 1.65

method	result	size
risch	\(x +\left (3 x^{3} {\mathrm e}^{5}-3 x^{4}-x^{2} {\mathrm e}^{5}+13 x^{3}+x \,{\mathrm e}^{5}-5 x^{2}+4 x \right ) {\mathrm e}^{x}\)	\(43\)
default	\(x +13 \,{\mathrm e}^{x} x^{3}-5 \,{\mathrm e}^{x} x^{2}+{\mathrm e}^{5} {\mathrm e}^{x}+4 \,{\mathrm e}^{x} x -3 \,{\mathrm e}^{x} x^{4}+3 \,{\mathrm e}^{5} \left ({\mathrm e}^{x} x^{3}-3 \,{\mathrm e}^{x} x^{2}+6 \,{\mathrm e}^{x} x -6 \,{\mathrm e}^{x}\right )-{\mathrm e}^{5} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+8 \,{\mathrm e}^{5} \left ({\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x^3+8*x^2-x+1)*exp(5)-3*x^4+x^3+34*x^2-6*x+4)*exp(ln(3*x^2-x+1)+x+ln(x))+3*x^3-x^2+x)/(3*x^3-x^2+x),x
,method=_RETURNVERBOSE)

[Out]

x+(3*x^3*exp(5)-3*x^4-x^2*exp(5)+13*x^3+x*exp(5)-5*x^2+4*x)*exp(x)

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maxima [A]  time = 0.55, size = 38, normalized size = 1.46 \begin {gather*} -{\left (3 \, x^{4} - x^{3} {\left (3 \, e^{5} + 13\right )} + x^{2} {\left (e^{5} + 5\right )} - x {\left (e^{5} + 4\right )}\right )} e^{x} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^3+8*x^2-x+1)*exp(5)-3*x^4+x^3+34*x^2-6*x+4)*exp(log(3*x^2-x+1)+x+log(x))+3*x^3-x^2+x)/(3*x^3-
x^2+x),x, algorithm="maxima")

[Out]

-(3*x^4 - x^3*(3*e^5 + 13) + x^2*(e^5 + 5) - x*(e^5 + 4))*e^x + x

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mupad [B]  time = 1.01, size = 52, normalized size = 2.00 \begin {gather*} x+x\,{\mathrm {e}}^{x+5}-5\,x^2\,{\mathrm {e}}^x+13\,x^3\,{\mathrm {e}}^x-3\,x^4\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^{x+5}+3\,x^3\,{\mathrm {e}}^{x+5}+4\,x\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + exp(x + log(3*x^2 - x + 1) + log(x))*(exp(5)*(8*x^2 - x + 3*x^3 + 1) - 6*x + 34*x^2 + x^3 - 3*x^4 + 4
) - x^2 + 3*x^3)/(x - x^2 + 3*x^3),x)

[Out]

x + x*exp(x + 5) - 5*x^2*exp(x) + 13*x^3*exp(x) - 3*x^4*exp(x) - x^2*exp(x + 5) + 3*x^3*exp(x + 5) + 4*x*exp(x
)

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sympy [A]  time = 0.16, size = 42, normalized size = 1.62 \begin {gather*} x + \left (- 3 x^{4} + 13 x^{3} + 3 x^{3} e^{5} - x^{2} e^{5} - 5 x^{2} + 4 x + x e^{5}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x**3+8*x**2-x+1)*exp(5)-3*x**4+x**3+34*x**2-6*x+4)*exp(ln(3*x**2-x+1)+x+ln(x))+3*x**3-x**2+x)/(
3*x**3-x**2+x),x)

[Out]

x + (-3*x**4 + 13*x**3 + 3*x**3*exp(5) - x**2*exp(5) - 5*x**2 + 4*x + x*exp(5))*exp(x)

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