3.88.34 \(\int \frac {-5 x^2+e^{\frac {9+3 x-20 x^2-3 x^3+e^2 (-3-x+7 x^2+x^3)}{x}} (-9-20 x^2-6 x^3+e^2 (3+7 x^2+2 x^3))}{5 x^2} \, dx\)

Optimal. Leaf size=35 \[ \frac {1}{5} e^{x-\left (3-e^2\right ) \left (4+(-1+x) \left (8+\frac {3}{x}+x\right )\right )}-x \]

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Rubi [A]  time = 0.78, antiderivative size = 45, normalized size of antiderivative = 1.29, number of steps used = 4, number of rules used = 3, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {12, 14, 6706} \begin {gather*} \frac {1}{5} \exp \left (-3 x^2-e^2 \left (-x^2-7 x+\frac {3}{x}+1\right )-20 x+\frac {9}{x}+3\right )-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*x^2 + E^((9 + 3*x - 20*x^2 - 3*x^3 + E^2*(-3 - x + 7*x^2 + x^3))/x)*(-9 - 20*x^2 - 6*x^3 + E^2*(3 + 7*
x^2 + 2*x^3)))/(5*x^2),x]

[Out]

E^(3 + 9/x - 20*x - 3*x^2 - E^2*(1 + 3/x - 7*x - x^2))/5 - 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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-5 x^2+\exp \left (\frac {9+3 x-20 x^2-3 x^3+e^2 \left (-3-x+7 x^2+x^3\right )}{x}\right ) \left (-9-20 x^2-6 x^3+e^2 \left (3+7 x^2+2 x^3\right )\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \left (-5+\frac {\exp \left (3+\frac {9}{x}-20 x-3 x^2+e^2 \left (-1-\frac {3}{x}+7 x+x^2\right )\right ) \left (-3 \left (3-e^2\right )-\left (20-7 e^2\right ) x^2-2 \left (3-e^2\right ) x^3\right )}{x^2}\right ) \, dx\\ &=-x+\frac {1}{5} \int \frac {\exp \left (3+\frac {9}{x}-20 x-3 x^2+e^2 \left (-1-\frac {3}{x}+7 x+x^2\right )\right ) \left (-3 \left (3-e^2\right )-\left (20-7 e^2\right ) x^2-2 \left (3-e^2\right ) x^3\right )}{x^2} \, dx\\ &=\frac {1}{5} e^{3+\frac {9}{x}-20 x-3 x^2-e^2 \left (1+\frac {3}{x}-7 x-x^2\right )}-x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.42, size = 45, normalized size = 1.29 \begin {gather*} \frac {1}{5} e^{3-e^2-\frac {3 \left (-3+e^2\right )}{x}+\left (-20+7 e^2\right ) x+\left (-3+e^2\right ) x^2}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*x^2 + E^((9 + 3*x - 20*x^2 - 3*x^3 + E^2*(-3 - x + 7*x^2 + x^3))/x)*(-9 - 20*x^2 - 6*x^3 + E^2*(
3 + 7*x^2 + 2*x^3)))/(5*x^2),x]

[Out]

E^(3 - E^2 - (3*(-3 + E^2))/x + (-20 + 7*E^2)*x + (-3 + E^2)*x^2)/5 - x

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fricas [A]  time = 0.47, size = 44, normalized size = 1.26 \begin {gather*} -x + \frac {1}{5} \, e^{\left (-\frac {3 \, x^{3} + 20 \, x^{2} - {\left (x^{3} + 7 \, x^{2} - x - 3\right )} e^{2} - 3 \, x - 9}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*x^3+7*x^2+3)*exp(2)-6*x^3-20*x^2-9)*exp(((x^3+7*x^2-x-3)*exp(2)-3*x^3-20*x^2+3*x+9)/x)-5*x^
2)/x^2,x, algorithm="fricas")

[Out]

-x + 1/5*e^(-(3*x^3 + 20*x^2 - (x^3 + 7*x^2 - x - 3)*e^2 - 3*x - 9)/x)

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giac [A]  time = 0.21, size = 48, normalized size = 1.37 \begin {gather*} -x + \frac {1}{5} \, e^{\left (\frac {x^{3} e^{2} - 3 \, x^{3} + 7 \, x^{2} e^{2} - 20 \, x^{2} - x e^{2} + 3 \, x - 3 \, e^{2} + 9}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*x^3+7*x^2+3)*exp(2)-6*x^3-20*x^2-9)*exp(((x^3+7*x^2-x-3)*exp(2)-3*x^3-20*x^2+3*x+9)/x)-5*x^
2)/x^2,x, algorithm="giac")

[Out]

-x + 1/5*e^((x^3*e^2 - 3*x^3 + 7*x^2*e^2 - 20*x^2 - x*e^2 + 3*x - 3*e^2 + 9)/x)

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maple [A]  time = 0.42, size = 49, normalized size = 1.40

method	result	size
risch	\(-x +\frac {{\mathrm e}^{\frac {x^{3} {\mathrm e}^{2}+7 x^{2} {\mathrm e}^{2}-3 x^{3}-{\mathrm e}^{2} x -20 x^{2}-3 \,{\mathrm e}^{2}+3 x +9}{x}}}{5}\)	\(49\)
norman	\(\frac {-x^{2}+\frac {x \,{\mathrm e}^{\frac {\left (x^{3}+7 x^{2}-x -3\right ) {\mathrm e}^{2}-3 x^{3}-20 x^{2}+3 x +9}{x}}}{5}}{x}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(((2*x^3+7*x^2+3)*exp(2)-6*x^3-20*x^2-9)*exp(((x^3+7*x^2-x-3)*exp(2)-3*x^3-20*x^2+3*x+9)/x)-5*x^2)/x^2
,x,method=_RETURNVERBOSE)

[Out]

-x+1/5*exp((x^3*exp(2)+7*x^2*exp(2)-3*x^3-exp(2)*x-20*x^2-3*exp(2)+3*x+9)/x)

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maxima [A]  time = 0.49, size = 44, normalized size = 1.26 \begin {gather*} -x + \frac {1}{5} \, e^{\left (x^{2} e^{2} - 3 \, x^{2} + 7 \, x e^{2} - 20 \, x - \frac {3 \, e^{2}}{x} + \frac {9}{x} - e^{2} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*x^3+7*x^2+3)*exp(2)-6*x^3-20*x^2-9)*exp(((x^3+7*x^2-x-3)*exp(2)-3*x^3-20*x^2+3*x+9)/x)-5*x^
2)/x^2,x, algorithm="maxima")

[Out]

-x + 1/5*e^(x^2*e^2 - 3*x^2 + 7*x*e^2 - 20*x - 3*e^2/x + 9/x - e^2 + 3)

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mupad [B]  time = 5.79, size = 50, normalized size = 1.43 \begin {gather*} \frac {{\mathrm {e}}^{x^2\,{\mathrm {e}}^2}\,{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^2}{x}}\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-20\,x}\,{\mathrm {e}}^3\,{\mathrm {e}}^{-3\,x^2}\,{\mathrm {e}}^{9/x}\,{\mathrm {e}}^{7\,x\,{\mathrm {e}}^2}}{5}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(-(exp(2)*(x - 7*x^2 - x^3 + 3) - 3*x + 20*x^2 + 3*x^3 - 9)/x)*(20*x^2 - exp(2)*(7*x^2 + 2*x^3 + 3)
+ 6*x^3 + 9))/5 + x^2)/x^2,x)

[Out]

(exp(x^2*exp(2))*exp(-(3*exp(2))/x)*exp(-exp(2))*exp(-20*x)*exp(3)*exp(-3*x^2)*exp(9/x)*exp(7*x*exp(2)))/5 - x

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sympy [A]  time = 0.26, size = 36, normalized size = 1.03 \begin {gather*} - x + \frac {e^{\frac {- 3 x^{3} - 20 x^{2} + 3 x + \left (x^{3} + 7 x^{2} - x - 3\right ) e^{2} + 9}{x}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*x**3+7*x**2+3)*exp(2)-6*x**3-20*x**2-9)*exp(((x**3+7*x**2-x-3)*exp(2)-3*x**3-20*x**2+3*x+9)
/x)-5*x**2)/x**2,x)

[Out]

-x + exp((-3*x**3 - 20*x**2 + 3*x + (x**3 + 7*x**2 - x - 3)*exp(2) + 9)/x)/5

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