3.18.28 \(\int \frac {4 x^2+32 x^3+e^{\frac {2+5 x^2-4 x^3}{4 x}} (-2+5 x^2-8 x^3)}{4 x^2} \, dx\)

Optimal. Leaf size=30 \[ 2+e^{-x^2+\frac {2+5 x^2}{4 x}}+x+4 x^2 \]

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Rubi [A]  time = 0.20, antiderivative size = 33, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 3, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {12, 14, 6706} \begin {gather*} e^{\frac {-4 x^3+5 x^2+2}{4 x}}+\frac {1}{16} (8 x+1)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x^2 + 32*x^3 + E^((2 + 5*x^2 - 4*x^3)/(4*x))*(-2 + 5*x^2 - 8*x^3))/(4*x^2),x]

[Out]

E^((2 + 5*x^2 - 4*x^3)/(4*x)) + (1 + 8*x)^2/16

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}{4} \int \frac {4 x^2+32 x^3+e^{\frac {2+5 x^2-4 x^3}{4 x}} \left (-2+5 x^2-8 x^3\right )}{x^2} \, dx\\ &=\frac {1}{4} \int \left (4 (1+8 x)+\frac {e^{\frac {2+5 x^2-4 x^3}{4 x}} \left (-2+5 x^2-8 x^3\right )}{x^2}\right ) \, dx\\ &=\frac {1}{16} (1+8 x)^2+\frac {1}{4} \int \frac {e^{\frac {2+5 x^2-4 x^3}{4 x}} \left (-2+5 x^2-8 x^3\right )}{x^2} \, dx\\ &=e^{\frac {2+5 x^2-4 x^3}{4 x}}+\frac {1}{16} (1+8 x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 27, normalized size = 0.90 \begin {gather*} e^{\frac {1}{2 x}+\frac {5 x}{4}-x^2}+x+4 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 + 32*x^3 + E^((2 + 5*x^2 - 4*x^3)/(4*x))*(-2 + 5*x^2 - 8*x^3))/(4*x^2),x]

[Out]

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

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fricas [A]  time = 0.83, size = 25, normalized size = 0.83 \begin {gather*} 4 \, x^{2} + x + e^{\left (-\frac {4 \, x^{3} - 5 \, x^{2} - 2}{4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8*x^3+5*x^2-2)*exp(1/4*(-4*x^3+5*x^2+2)/x)+32*x^3+4*x^2)/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 25, normalized size = 0.83 \begin {gather*} 4 \, x^{2} + x + e^{\left (-\frac {4 \, x^{3} - 5 \, x^{2} - 2}{4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8*x^3+5*x^2-2)*exp(1/4*(-4*x^3+5*x^2+2)/x)+32*x^3+4*x^2)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 26, normalized size = 0.87

method	result	size
risch	\(4 x^{2}+x +{\mathrm e}^{-\frac {4 x^{3}-5 x^{2}-2}{4 x}}\)	\(26\)
norman	\(\frac {x^{2}+x \,{\mathrm e}^{\frac {-4 x^{3}+5 x^{2}+2}{4 x}}+4 x^{3}}{x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-8*x^3+5*x^2-2)*exp(1/4*(-4*x^3+5*x^2+2)/x)+32*x^3+4*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

4*x^2+x+exp(-1/4*(4*x^3-5*x^2-2)/x)

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maxima [A]  time = 0.51, size = 22, normalized size = 0.73 \begin {gather*} 4 \, x^{2} + x + e^{\left (-x^{2} + \frac {5}{4} \, x + \frac {1}{2 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8*x^3+5*x^2-2)*exp(1/4*(-4*x^3+5*x^2+2)/x)+32*x^3+4*x^2)/x^2,x, algorithm="maxima")

[Out]

4*x^2 + x + e^(-x^2 + 5/4*x + 1/2/x)

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mupad [B]  time = 1.09, size = 22, normalized size = 0.73 \begin {gather*} x+{\mathrm {e}}^{\frac {5\,x}{4}+\frac {1}{2\,x}-x^2}+4\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp((5*x)/4 + 1/(2*x) - x^2) + 4*x^2

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sympy [A]  time = 0.14, size = 22, normalized size = 0.73 \begin {gather*} 4 x^{2} + x + e^{\frac {- x^{3} + \frac {5 x^{2}}{4} + \frac {1}{2}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8*x**3+5*x**2-2)*exp(1/4*(-4*x**3+5*x**2+2)/x)+32*x**3+4*x**2)/x**2,x)

[Out]

4*x**2 + x + exp((-x**3 + 5*x**2/4 + 1/2)/x)

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