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

Optimal. Leaf size=28 \[ \frac {e^{5+e^{10}} \left (-4+\frac {e^x}{x}\right )-x}{5 x} \]

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Rubi [A]  time = 0.04, antiderivative size = 30, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 14, 2197} \begin {gather*} \frac {e^{x+e^{10}+5}}{5 x^2}-\frac {4 e^{5+e^{10}}}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.01, size = 24, normalized size = 0.86 \begin {gather*} \frac {1}{5} e^{5+e^{10}} \left (\frac {e^x}{x^2}-\frac {4}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(5 + E^10)*(E^x/x^2 - 4/x))/5

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fricas [A]  time = 0.44, size = 20, normalized size = 0.71 \begin {gather*} -\frac {{\left (4 \, x e^{5} - e^{\left (x + 5\right )}\right )} e^{\left (e^{10}\right )}}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(4*x*e^5 - e^(x + 5))*e^(e^10)/x^2

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giac [A]  time = 0.14, size = 20, normalized size = 0.71 \begin {gather*} -\frac {{\left (4 \, x e^{5} - e^{\left (x + 5\right )}\right )} e^{\left (e^{10}\right )}}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(4*x*e^5 - e^(x + 5))*e^(e^10)/x^2

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maple [A]  time = 0.08, size = 23, normalized size = 0.82

method	result	size
norman	\(\frac {-\frac {4 \,{\mathrm e}^{5} {\mathrm e}^{{\mathrm e}^{10}} x}{5}+\frac {{\mathrm e}^{5} {\mathrm e}^{{\mathrm e}^{10}} {\mathrm e}^{x}}{5}}{x^{2}}\)	\(23\)
risch	\(-\frac {4 \,{\mathrm e}^{{\mathrm e}^{10}+5}}{5 x}+\frac {{\mathrm e}^{{\mathrm e}^{10}+5+x}}{5 x^{2}}\)	\(23\)
default	\(\frac {{\mathrm e}^{{\mathrm e}^{10}} \left ({\mathrm e}^{5} \left (-\frac {{\mathrm e}^{x}}{x}-\expIntegralEi \left (1, -x \right )\right )-\frac {4 \,{\mathrm e}^{5}}{x}-2 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{x}}{2 x^{2}}-\frac {{\mathrm e}^{x}}{2 x}-\frac {\expIntegralEi \left (1, -x \right )}{2}\right )\right )}{5}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.40, size = 30, normalized size = 1.07 \begin {gather*} \frac {1}{5} \, {\left (e^{5} \Gamma \left (-1, -x\right ) + 2 \, e^{5} \Gamma \left (-2, -x\right ) - \frac {4 \, e^{5}}{x}\right )} e^{\left (e^{10}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(e^5*gamma(-1, -x) + 2*e^5*gamma(-2, -x) - 4*e^5/x)*e^(e^10)

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mupad [B]  time = 6.52, size = 18, normalized size = 0.64 \begin {gather*} -\frac {{\mathrm {e}}^5\,{\mathrm {e}}^{{\mathrm {e}}^{10}}\,\left (4\,x-{\mathrm {e}}^x\right )}{5\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(10))*(4*x*exp(5) + exp(5)*exp(x)*(x - 2)))/(5*x^3),x)

[Out]

-(exp(5)*exp(exp(10))*(4*x - exp(x)))/(5*x^2)

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sympy [A]  time = 0.12, size = 29, normalized size = 1.04 \begin {gather*} - \frac {4 e^{5} e^{e^{10}}}{5 x} + \frac {e^{5} e^{x} e^{e^{10}}}{5 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*exp(5)*exp(exp(10))/(5*x) + exp(5)*exp(x)*exp(exp(10))/(5*x**2)

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