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

Optimal. Leaf size=31 \[ 3-e^x+2 x-\frac {e^5 \left (\frac {e^{2 x}}{x^2}+x^2\right )}{x} \]

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Rubi [A]  time = 0.06, antiderivative size = 27, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6, 14, 2194, 2197} \begin {gather*} -\frac {e^{2 x+5}}{x^3}+\left (2-e^5\right ) x-e^x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^x - E^(5 + 2*x)/x^3 + (2 - E^5)*x

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 27, normalized size = 0.87 \begin {gather*} -e^x-\frac {e^{5+2 x}}{x^3}+2 x-e^5 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^x - E^(5 + 2*x)/x^3 + 2*x - E^5*x

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fricas [A]  time = 0.53, size = 29, normalized size = 0.94 \begin {gather*} -\frac {x^{4} e^{5} - 2 \, x^{4} + x^{3} e^{x} + e^{\left (2 \, x + 5\right )}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.28, size = 29, normalized size = 0.94 \begin {gather*} -\frac {x^{4} e^{5} - 2 \, x^{4} + x^{3} e^{x} + e^{\left (2 \, x + 5\right )}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 25, normalized size = 0.81

method	result	size
risch	\(-x \,{\mathrm e}^{5}+2 x -\frac {{\mathrm e}^{5+2 x}}{x^{3}}-{\mathrm e}^{x}\)	\(25\)
norman	\(\frac {\left (2-{\mathrm e}^{5}\right ) x^{4}-{\mathrm e}^{5} {\mathrm e}^{2 x}-{\mathrm e}^{x} x^{3}}{x^{3}}\)	\(31\)
default	\(2 x +3 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{2 x}}{3 x^{3}}-\frac {{\mathrm e}^{2 x}}{3 x^{2}}-\frac {2 \,{\mathrm e}^{2 x}}{3 x}-\frac {4 \expIntegralEi \left (1, -2 x \right )}{3}\right )-2 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{2 x}}{2 x^{2}}-\frac {{\mathrm e}^{2 x}}{x}-2 \expIntegralEi \left (1, -2 x \right )\right )-x \,{\mathrm e}^{5}-{\mathrm e}^{x}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.37, size = 31, normalized size = 1.00 \begin {gather*} -x e^{5} + 8 \, e^{5} \Gamma \left (-2, -2 \, x\right ) + 24 \, e^{5} \Gamma \left (-3, -2 \, x\right ) + 2 \, x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*e^5 + 8*e^5*gamma(-2, -2*x) + 24*e^5*gamma(-3, -2*x) + 2*x - e^x

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mupad [B]  time = 6.83, size = 23, normalized size = 0.74 \begin {gather*} -{\mathrm {e}}^x-x\,\left ({\mathrm {e}}^5-2\right )-\frac {{\mathrm {e}}^{2\,x+5}}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.12, size = 26, normalized size = 0.84 \begin {gather*} x \left (2 - e^{5}\right ) + \frac {- x^{3} e^{x} - e^{5} e^{2 x}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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