3.13.87 \(\int \frac {1-x-4 e^5 x^2+e^{5+x} (-10 x^5-2 x^6)}{x} \, dx\)

Optimal. Leaf size=25 \[ -x-2 e^5 x^2 \left (1+e^x x^3\right )+\log (3)+\log (x) \]

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Rubi [A]  time = 0.18, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 17, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {14, 2196, 2176, 2194} \begin {gather*} -2 e^{x+5} x^5-2 e^5 x^2-x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x - 4*E^5*x^2 + E^(5 + x)*(-10*x^5 - 2*x^6))/x,x]

[Out]

-x - 2*E^5*x^2 - 2*E^(5 + x)*x^5 + 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 (-2 e^{5+x} x^4 (5+x)+\frac {1-x-4 e^5 x^2}{x}\right ) \, dx\\ &=-\left (2 \int e^{5+x} x^4 (5+x) \, dx\right )+\int \frac {1-x-4 e^5 x^2}{x} \, dx\\ &=-\left (2 \int \left (5 e^{5+x} x^4+e^{5+x} x^5\right ) \, dx\right )+\int \left (-1+\frac {1}{x}-4 e^5 x\right ) \, dx\\ &=-x-2 e^5 x^2+\log (x)-2 \int e^{5+x} x^5 \, dx-10 \int e^{5+x} x^4 \, dx\\ &=-x-2 e^5 x^2-10 e^{5+x} x^4-2 e^{5+x} x^5+\log (x)+10 \int e^{5+x} x^4 \, dx+40 \int e^{5+x} x^3 \, dx\\ &=-x-2 e^5 x^2+40 e^{5+x} x^3-2 e^{5+x} x^5+\log (x)-40 \int e^{5+x} x^3 \, dx-120 \int e^{5+x} x^2 \, dx\\ &=-x-2 e^5 x^2-120 e^{5+x} x^2-2 e^{5+x} x^5+\log (x)+120 \int e^{5+x} x^2 \, dx+240 \int e^{5+x} x \, dx\\ &=-x+240 e^{5+x} x-2 e^5 x^2-2 e^{5+x} x^5+\log (x)-240 \int e^{5+x} \, dx-240 \int e^{5+x} x \, dx\\ &=-240 e^{5+x}-x-2 e^5 x^2-2 e^{5+x} x^5+\log (x)+240 \int e^{5+x} \, dx\\ &=-x-2 e^5 x^2-2 e^{5+x} x^5+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 0.96 \begin {gather*} -x \left (1+2 e^5 x+2 e^{5+x} x^4\right )+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x - 4*E^5*x^2 + E^(5 + x)*(-10*x^5 - 2*x^6))/x,x]

[Out]

-(x*(1 + 2*E^5*x + 2*E^(5 + x)*x^4)) + Log[x]

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fricas [A]  time = 0.67, size = 22, normalized size = 0.88 \begin {gather*} -2 \, x^{5} e^{\left (x + 5\right )} - 2 \, x^{2} e^{5} - x + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x^5*e^(x + 5) - 2*x^2*e^5 - x + log(x)

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giac [A]  time = 0.28, size = 22, normalized size = 0.88 \begin {gather*} -2 \, x^{5} e^{\left (x + 5\right )} - 2 \, x^{2} e^{5} - x + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x^5*e^(x + 5) - 2*x^2*e^5 - x + log(x)

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maple [A]  time = 0.03, size = 23, normalized size = 0.92

method	result	size
norman	\(-x -2 x^{2} {\mathrm e}^{5}-2 x^{5} {\mathrm e}^{5} {\mathrm e}^{x}+\ln \relax (x )\)	\(23\)
risch	\(-x -2 x^{2} {\mathrm e}^{5}-2 x^{5} {\mathrm e}^{5+x}+\ln \relax (x )\)	\(23\)
default	\(\ln \relax (x )-x -2 x^{2} {\mathrm e}^{5}-10 \,{\mathrm e}^{5} \left ({\mathrm e}^{x} x^{4}-4 \,{\mathrm e}^{x} x^{3}+12 \,{\mathrm e}^{x} x^{2}-24 \,{\mathrm e}^{x} x +24 \,{\mathrm e}^{x}\right )-2 \,{\mathrm e}^{5} \left (x^{5} {\mathrm e}^{x}-5 \,{\mathrm e}^{x} x^{4}+20 \,{\mathrm e}^{x} x^{3}-60 \,{\mathrm e}^{x} x^{2}+120 \,{\mathrm e}^{x} x -120 \,{\mathrm e}^{x}\right )\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x-2*x^2*exp(5)-2*x^5*exp(5)*exp(x)+ln(x)

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maxima [B]  time = 0.44, size = 88, normalized size = 3.52 \begin {gather*} -2 \, x^{2} e^{5} - 2 \, {\left (x^{5} e^{5} - 5 \, x^{4} e^{5} + 20 \, x^{3} e^{5} - 60 \, x^{2} e^{5} + 120 \, x e^{5} - 120 \, e^{5}\right )} e^{x} - 10 \, {\left (x^{4} e^{5} - 4 \, x^{3} e^{5} + 12 \, x^{2} e^{5} - 24 \, x e^{5} + 24 \, e^{5}\right )} e^{x} - x + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x^2*e^5 - 2*(x^5*e^5 - 5*x^4*e^5 + 20*x^3*e^5 - 60*x^2*e^5 + 120*x*e^5 - 120*e^5)*e^x - 10*(x^4*e^5 - 4*x^3
*e^5 + 12*x^2*e^5 - 24*x*e^5 + 24*e^5)*e^x - x + log(x)

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mupad [B]  time = 0.89, size = 22, normalized size = 0.88 \begin {gather*} \ln \relax (x)-x-2\,x^5\,{\mathrm {e}}^{x+5}-2\,x^2\,{\mathrm {e}}^5 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - x - 2*x^5*exp(x + 5) - 2*x^2*exp(5)

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sympy [A]  time = 0.14, size = 24, normalized size = 0.96 \begin {gather*} - 2 x^{5} e^{5} e^{x} - 2 x^{2} e^{5} - x + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x**5*exp(5)*exp(x) - 2*x**2*exp(5) - x + log(x)

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