3.15.24 \(\int (-1+e^x (8+e^2+x+4 x^3+x^4-\log (2))) \, dx\)

Optimal. Leaf size=22 \[ 2-x+e^x \left (7+e^2+x+x^4-\log (2)\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 34, normalized size of antiderivative = 1.55, number of steps used = 15, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2196, 2176, 2194} \begin {gather*} e^x x^4+e^x x-x-e^x+e^x \left (8+e^2-\log (2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-1 + E^x*(8 + E^2 + x + 4*x^3 + x^4 - Log[2]),x]

[Out]

-E^x - x + E^x*x + E^x*x^4 + E^x*(8 + E^2 - Log[2])

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} &=-x+\int e^x \left (8+e^2+x+4 x^3+x^4-\log (2)\right ) \, dx\\ &=-x+\int \left (e^x x+4 e^x x^3+e^x x^4+8 e^x \left (1+\frac {1}{8} \left (e^2-\log (2)\right )\right )\right ) \, dx\\ &=-x+4 \int e^x x^3 \, dx+\left (8+e^2-\log (2)\right ) \int e^x \, dx+\int e^x x \, dx+\int e^x x^4 \, dx\\ &=-x+e^x x+4 e^x x^3+e^x x^4+e^x \left (8+e^2-\log (2)\right )-4 \int e^x x^3 \, dx-12 \int e^x x^2 \, dx-\int e^x \, dx\\ &=-e^x-x+e^x x-12 e^x x^2+e^x x^4+e^x \left (8+e^2-\log (2)\right )+12 \int e^x x^2 \, dx+24 \int e^x x \, dx\\ &=-e^x-x+25 e^x x+e^x x^4+e^x \left (8+e^2-\log (2)\right )-24 \int e^x \, dx-24 \int e^x x \, dx\\ &=-25 e^x-x+e^x x+e^x x^4+e^x \left (8+e^2-\log (2)\right )+24 \int e^x \, dx\\ &=-e^x-x+e^x x+e^x x^4+e^x \left (8+e^2-\log (2)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 21, normalized size = 0.95 \begin {gather*} -x+e^x \left (7+e^2+x+x^4-\log (2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-1 + E^x*(8 + E^2 + x + 4*x^3 + x^4 - Log[2]),x]

[Out]

-x + E^x*(7 + E^2 + x + x^4 - Log[2])

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fricas [A]  time = 0.71, size = 19, normalized size = 0.86 \begin {gather*} {\left (x^{4} + x + e^{2} - \log \relax (2) + 7\right )} e^{x} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^4 + x + e^2 - log(2) + 7)*e^x - x

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giac [A]  time = 0.37, size = 21, normalized size = 0.95 \begin {gather*} {\left (x^{4} + x - \log \relax (2) + 7\right )} e^{x} - x + e^{\left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^4 + x - log(2) + 7)*e^x - x + e^(x + 2)

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maple [A]  time = 0.03, size = 20, normalized size = 0.91

method	result	size
risch	\({\mathrm e}^{x} \left (7+{\mathrm e}^{2}+x^{4}-\ln \relax (2)+x \right )-x\)	\(20\)
norman	\(\left (7+{\mathrm e}^{2}-\ln \relax (2)\right ) {\mathrm e}^{x}+{\mathrm e}^{x} x +{\mathrm e}^{x} x^{4}-x\)	\(26\)
default	\(-x +{\mathrm e}^{x} x +7 \,{\mathrm e}^{x}+{\mathrm e}^{x} x^{4}+{\mathrm e}^{2} {\mathrm e}^{x}-{\mathrm e}^{x} \ln \relax (2)\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)*(7+exp(2)+x^4-ln(2)+x)-x

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maxima [B]  time = 0.42, size = 62, normalized size = 2.82 \begin {gather*} {\left (x^{4} - 4 \, x^{3} + 12 \, x^{2} - 24 \, x + 24\right )} e^{x} + 4 \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} + {\left (x - 1\right )} e^{x} - e^{x} \log \relax (2) - x + e^{\left (x + 2\right )} + 8 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^4 - 4*x^3 + 12*x^2 - 24*x + 24)*e^x + 4*(x^3 - 3*x^2 + 6*x - 6)*e^x + (x - 1)*e^x - e^x*log(2) - x + e^(x +
2) + 8*e^x

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mupad [B]  time = 0.07, size = 25, normalized size = 1.14 \begin {gather*} x^4\,{\mathrm {e}}^x-x+x\,{\mathrm {e}}^x+{\mathrm {e}}^x\,\left ({\mathrm {e}}^2-\ln \relax (2)+7\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^4*exp(x) - x + x*exp(x) + exp(x)*(exp(2) - log(2) + 7)

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sympy [A]  time = 0.10, size = 17, normalized size = 0.77 \begin {gather*} - x + \left (x^{4} + x - \log {\relax (2 )} + 7 + e^{2}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(2)+exp(2)+x**4+4*x**3+x+8)*exp(x)-1,x)

[Out]

-x + (x**4 + x - log(2) + 7 + exp(2))*exp(x)

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