3.73.71 \(\int \frac {1+x+16 x^4+e^x (2 x^2+x^3)}{x} \, dx\)

Optimal. Leaf size=21 \[ -5+x^2 \left (e^x+4 x^2\right )+\log \left (e^x x\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 16, normalized size of antiderivative = 0.76, number of steps used = 11, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {14, 2196, 2176, 2194} \begin {gather*} 4 x^4+e^x x^2+x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x + 16*x^4 + E^x*(2*x^2 + x^3))/x,x]

[Out]

x + E^x*x^2 + 4*x^4 + Log[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 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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^x x (2+x)+\frac {1+x+16 x^4}{x}\right ) \, dx\\ &=\int e^x x (2+x) \, dx+\int \frac {1+x+16 x^4}{x} \, dx\\ &=\int \left (2 e^x x+e^x x^2\right ) \, dx+\int \left (1+\frac {1}{x}+16 x^3\right ) \, dx\\ &=x+4 x^4+\log (x)+2 \int e^x x \, dx+\int e^x x^2 \, dx\\ &=x+2 e^x x+e^x x^2+4 x^4+\log (x)-2 \int e^x \, dx-2 \int e^x x \, dx\\ &=-2 e^x+x+e^x x^2+4 x^4+\log (x)+2 \int e^x \, dx\\ &=x+e^x x^2+4 x^4+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.76 \begin {gather*} x+e^x x^2+4 x^4+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x + 16*x^4 + E^x*(2*x^2 + x^3))/x,x]

[Out]

x + E^x*x^2 + 4*x^4 + Log[x]

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fricas [A]  time = 0.63, size = 15, normalized size = 0.71 \begin {gather*} 4 \, x^{4} + x^{2} e^{x} + x + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+2*x^2)*exp(x)+16*x^4+x+1)/x,x, algorithm="fricas")

[Out]

4*x^4 + x^2*e^x + x + log(x)

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giac [A]  time = 0.12, size = 15, normalized size = 0.71 \begin {gather*} 4 \, x^{4} + x^{2} e^{x} + x + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+2*x^2)*exp(x)+16*x^4+x+1)/x,x, algorithm="giac")

[Out]

4*x^4 + x^2*e^x + x + log(x)

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maple [A]  time = 0.02, size = 16, normalized size = 0.76

method	result	size
default	\(x +\ln \relax (x )+{\mathrm e}^{x} x^{2}+4 x^{4}\)	\(16\)
norman	\(x +\ln \relax (x )+{\mathrm e}^{x} x^{2}+4 x^{4}\)	\(16\)
risch	\(x +\ln \relax (x )+{\mathrm e}^{x} x^{2}+4 x^{4}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3+2*x^2)*exp(x)+16*x^4+x+1)/x,x,method=_RETURNVERBOSE)

[Out]

x+ln(x)+exp(x)*x^2+4*x^4

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maxima [A]  time = 0.36, size = 27, normalized size = 1.29 \begin {gather*} 4 \, x^{4} + {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 2 \, {\left (x - 1\right )} e^{x} + x + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+2*x^2)*exp(x)+16*x^4+x+1)/x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 15, normalized size = 0.71 \begin {gather*} x+\ln \relax (x)+x^2\,{\mathrm {e}}^x+4\,x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + exp(x)*(2*x^2 + x^3) + 16*x^4 + 1)/x,x)

[Out]

x + log(x) + x^2*exp(x) + 4*x^4

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sympy [A]  time = 0.12, size = 15, normalized size = 0.71 \begin {gather*} 4 x^{4} + x^{2} e^{x} + x + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3+2*x**2)*exp(x)+16*x**4+x+1)/x,x)

[Out]

4*x**4 + x**2*exp(x) + x + log(x)

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