3.33.52 \(\int \frac {9 x^3+e^{\frac {2+x+2 x^2}{x}} (-2-x+2 x^2)+e^{\frac {2+x+2 x^2}{2 x}} (-6 x+6 x^3)}{4 x^3} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{4} \left (3+\frac {e^{x+\frac {2+x}{2 x}}}{x}\right )^2 x \]

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Rubi [B]  time = 0.23, antiderivative size = 55, normalized size of antiderivative = 2.04, number of steps used = 5, number of rules used = 4, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 14, 6706, 2288} \begin {gather*} -\frac {e^{2 x+\frac {2}{x}+1} \left (1-x^2\right )}{4 \left (1-\frac {1}{x^2}\right ) x^3}+\frac {9 x}{4}+\frac {3}{2} e^{x+\frac {1}{x}+\frac {1}{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9*x^3 + E^((2 + x + 2*x^2)/x)*(-2 - x + 2*x^2) + E^((2 + x + 2*x^2)/(2*x))*(-6*x + 6*x^3))/(4*x^3),x]

[Out]

(3*E^(1/2 + x^(-1) + x))/2 + (9*x)/4 - (E^(1 + 2/x + 2*x)*(1 - x^2))/(4*(1 - x^(-2))*x^3)

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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}{4} \int \frac {9 x^3+e^{\frac {2+x+2 x^2}{x}} \left (-2-x+2 x^2\right )+e^{\frac {2+x+2 x^2}{2 x}} \left (-6 x+6 x^3\right )}{x^3} \, dx\\ &=\frac {1}{4} \int \left (9+\frac {6 e^{\frac {1}{2}+\frac {1}{x}+x} \left (-1+x^2\right )}{x^2}+\frac {e^{1+\frac {2}{x}+2 x} \left (-2-x+2 x^2\right )}{x^3}\right ) \, dx\\ &=\frac {9 x}{4}+\frac {1}{4} \int \frac {e^{1+\frac {2}{x}+2 x} \left (-2-x+2 x^2\right )}{x^3} \, dx+\frac {3}{2} \int \frac {e^{\frac {1}{2}+\frac {1}{x}+x} \left (-1+x^2\right )}{x^2} \, dx\\ &=\frac {3}{2} e^{\frac {1}{2}+\frac {1}{x}+x}+\frac {9 x}{4}-\frac {e^{1+\frac {2}{x}+2 x} \left (1-x^2\right )}{4 \left (1-\frac {1}{x^2}\right ) x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 36, normalized size = 1.33 \begin {gather*} \frac {1}{4} \left (6 e^{\frac {1}{2}+\frac {1}{x}+x}+\frac {e^{1+\frac {2}{x}+2 x}}{x}+9 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9*x^3 + E^((2 + x + 2*x^2)/x)*(-2 - x + 2*x^2) + E^((2 + x + 2*x^2)/(2*x))*(-6*x + 6*x^3))/(4*x^3),
x]

[Out]

(6*E^(1/2 + x^(-1) + x) + E^(1 + 2/x + 2*x)/x + 9*x)/4

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fricas [A]  time = 0.48, size = 41, normalized size = 1.52 \begin {gather*} \frac {9 \, x^{2} + 6 \, x e^{\left (\frac {2 \, x^{2} + x + 2}{2 \, x}\right )} + e^{\left (\frac {2 \, x^{2} + x + 2}{x}\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2*x^2-x-2)*exp(1/2*(2*x^2+x+2)/x)^2+(6*x^3-6*x)*exp(1/2*(2*x^2+x+2)/x)+9*x^3)/x^3,x, algorithm
="fricas")

[Out]

1/4*(9*x^2 + 6*x*e^(1/2*(2*x^2 + x + 2)/x) + e^((2*x^2 + x + 2)/x))/x

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giac [A]  time = 0.25, size = 41, normalized size = 1.52 \begin {gather*} \frac {9 \, x^{2} + 6 \, x e^{\left (\frac {2 \, x^{2} + x + 2}{2 \, x}\right )} + e^{\left (\frac {2 \, x^{2} + x + 2}{x}\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2*x^2-x-2)*exp(1/2*(2*x^2+x+2)/x)^2+(6*x^3-6*x)*exp(1/2*(2*x^2+x+2)/x)+9*x^3)/x^3,x, algorithm
="giac")

[Out]

1/4*(9*x^2 + 6*x*e^(1/2*(2*x^2 + x + 2)/x) + e^((2*x^2 + x + 2)/x))/x

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maple [A]  time = 0.30, size = 39, normalized size = 1.44

method	result	size
risch	\(\frac {9 x}{4}+\frac {{\mathrm e}^{\frac {2 x^{2}+x +2}{x}}}{4 x}+\frac {3 \,{\mathrm e}^{\frac {2 x^{2}+x +2}{2 x}}}{2}\)	\(39\)
norman	\(\frac {\frac {9 x^{3}}{4}+\frac {3 \,{\mathrm e}^{\frac {2 x^{2}+x +2}{2 x}} x^{2}}{2}+\frac {{\mathrm e}^{\frac {2 x^{2}+x +2}{x}} x}{4}}{x^{2}}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((2*x^2-x-2)*exp(1/2*(2*x^2+x+2)/x)^2+(6*x^3-6*x)*exp(1/2*(2*x^2+x+2)/x)+9*x^3)/x^3,x,method=_RETURNVE
RBOSE)

[Out]

9/4*x+1/4/x*exp((2*x^2+x+2)/x)+3/2*exp(1/2*(2*x^2+x+2)/x)

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maxima [A]  time = 0.71, size = 31, normalized size = 1.15 \begin {gather*} \frac {9}{4} \, x + \frac {6 \, x e^{\left (x + \frac {1}{x} + \frac {1}{2}\right )} + e^{\left (2 \, x + \frac {2}{x} + 1\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2*x^2-x-2)*exp(1/2*(2*x^2+x+2)/x)^2+(6*x^3-6*x)*exp(1/2*(2*x^2+x+2)/x)+9*x^3)/x^3,x, algorithm
="maxima")

[Out]

9/4*x + 1/4*(6*x*e^(x + 1/x + 1/2) + e^(2*x + 2/x + 1))/x

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mupad [B]  time = 2.01, size = 18, normalized size = 0.67 \begin {gather*} \frac {{\left (3\,x+{\mathrm {e}}^{x+\frac {1}{x}+\frac {1}{2}}\right )}^2}{4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp((2*(x/2 + x^2 + 1))/x)*(x - 2*x^2 + 2))/4 + (exp((x/2 + x^2 + 1)/x)*(6*x - 6*x^3))/4 - (9*x^3)/4)/x
^3,x)

[Out]

(3*x + exp(x + 1/x + 1/2))^2/(4*x)

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sympy [B]  time = 0.20, size = 36, normalized size = 1.33 \begin {gather*} \frac {9 x}{4} + \frac {12 x e^{\frac {x^{2} + \frac {x}{2} + 1}{x}} + 2 e^{\frac {2 \left (x^{2} + \frac {x}{2} + 1\right )}{x}}}{8 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2*x**2-x-2)*exp(1/2*(2*x**2+x+2)/x)**2+(6*x**3-6*x)*exp(1/2*(2*x**2+x+2)/x)+9*x**3)/x**3,x)

[Out]

9*x/4 + (12*x*exp((x**2 + x/2 + 1)/x) + 2*exp(2*(x**2 + x/2 + 1)/x))/(8*x)

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