3.54.50 \(\int \frac {-5 x+5 x^2+10 x^3+e^{\frac {4+6 x+x^2}{x}} (-4+x^2)}{5 x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {1}{5} e^{\frac {4+x+x (5+x)}{x}}+x+x^2-\log (3 x) \]

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Rubi [A]  time = 0.19, antiderivative size = 23, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 3, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {12, 14, 6706} \begin {gather*} x^2+x+\frac {1}{5} e^{x+\frac {4}{x}+6}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*x + 5*x^2 + 10*x^3 + E^((4 + 6*x + x^2)/x)*(-4 + x^2))/(5*x^2),x]

[Out]

E^(6 + 4/x + x)/5 + x + x^2 - Log[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+5 x^2+10 x^3+e^{\frac {4+6 x+x^2}{x}} \left (-4+x^2\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^{6+\frac {4}{x}+x} (-2+x) (2+x)}{x^2}+\frac {5 \left (-1+x+2 x^2\right )}{x}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{6+\frac {4}{x}+x} (-2+x) (2+x)}{x^2} \, dx+\int \frac {-1+x+2 x^2}{x} \, dx\\ &=\frac {1}{5} e^{6+\frac {4}{x}+x}+\int \left (1-\frac {1}{x}+2 x\right ) \, dx\\ &=\frac {1}{5} e^{6+\frac {4}{x}+x}+x+x^2-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 27, normalized size = 0.93 \begin {gather*} \frac {1}{5} \left (e^{6+\frac {4}{x}+x}+5 x+5 x^2-5 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*x + 5*x^2 + 10*x^3 + E^((4 + 6*x + x^2)/x)*(-4 + x^2))/(5*x^2),x]

[Out]

(E^(6 + 4/x + x) + 5*x + 5*x^2 - 5*Log[x])/5

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fricas [A]  time = 0.61, size = 24, normalized size = 0.83 \begin {gather*} x^{2} + x + \frac {1}{5} \, e^{\left (\frac {x^{2} + 6 \, x + 4}{x}\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x^2-4)*exp((x^2+6*x+4)/x)+10*x^3+5*x^2-5*x)/x^2,x, algorithm="fricas")

[Out]

x^2 + x + 1/5*e^((x^2 + 6*x + 4)/x) - log(x)

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giac [A]  time = 0.15, size = 24, normalized size = 0.83 \begin {gather*} x^{2} + x + \frac {1}{5} \, e^{\left (\frac {x^{2} + 6 \, x + 4}{x}\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x^2-4)*exp((x^2+6*x+4)/x)+10*x^3+5*x^2-5*x)/x^2,x, algorithm="giac")

[Out]

x^2 + x + 1/5*e^((x^2 + 6*x + 4)/x) - log(x)

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maple [A]  time = 0.19, size = 25, normalized size = 0.86

method	result	size
risch	\(x^{2}-\ln \relax (x )+x +\frac {{\mathrm e}^{\frac {x^{2}+6 x +4}{x}}}{5}\)	\(25\)
norman	\(\frac {x^{2}+x^{3}+\frac {x \,{\mathrm e}^{\frac {x^{2}+6 x +4}{x}}}{5}}{x}-\ln \relax (x )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((x^2-4)*exp((x^2+6*x+4)/x)+10*x^3+5*x^2-5*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

x^2-ln(x)+x+1/5*exp((x^2+6*x+4)/x)

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maxima [A]  time = 0.44, size = 20, normalized size = 0.69 \begin {gather*} x^{2} + x + \frac {1}{5} \, e^{\left (x + \frac {4}{x} + 6\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x^2-4)*exp((x^2+6*x+4)/x)+10*x^3+5*x^2-5*x)/x^2,x, algorithm="maxima")

[Out]

x^2 + x + 1/5*e^(x + 4/x + 6) - log(x)

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mupad [B]  time = 3.50, size = 20, normalized size = 0.69 \begin {gather*} x+\frac {{\mathrm {e}}^{x+\frac {4}{x}+6}}{5}-\ln \relax (x)+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp(x + 4/x + 6)/5 - log(x) + x^2

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sympy [A]  time = 0.14, size = 20, normalized size = 0.69 \begin {gather*} x^{2} + x + \frac {e^{\frac {x^{2} + 6 x + 4}{x}}}{5} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x**2-4)*exp((x**2+6*x+4)/x)+10*x**3+5*x**2-5*x)/x**2,x)

[Out]

x**2 + x + exp((x**2 + 6*x + 4)/x)/5 - log(x)

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