3.97.47 \(\int \frac {e^{\frac {8 x+5 x^2+e^5 (8+4 x)}{x^2}} (e^5 (-16-4 x)-8 x)+x^3}{x^3} \, dx\)

Optimal. Leaf size=19 \[ e^{1+\frac {4 (2+x) \left (e^5+x\right )}{x^2}}+x \]

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Rubi [A]  time = 0.33, antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 2, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {14, 6706} \begin {gather*} e^{\frac {4 e^5 (x+2)}{x^2}+\frac {8}{x}+5}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((8*x + 5*x^2 + E^5*(8 + 4*x))/x^2)*(E^5*(-16 - 4*x) - 8*x) + x^3)/x^3,x]

[Out]

E^(5 + 8/x + (4*E^5*(2 + x))/x^2) + 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 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+\frac {4 e^{5+\frac {8}{x}+\frac {4 e^5 (2+x)}{x^2}} \left (-4 e^5-\left (2+e^5\right ) x\right )}{x^3}\right ) \, dx\\ &=x+4 \int \frac {e^{5+\frac {8}{x}+\frac {4 e^5 (2+x)}{x^2}} \left (-4 e^5-\left (2+e^5\right ) x\right )}{x^3} \, dx\\ &=e^{5+\frac {8}{x}+\frac {4 e^5 (2+x)}{x^2}}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 24, normalized size = 1.26 \begin {gather*} e^{5+\frac {8 e^5}{x^2}+\frac {4 \left (2+e^5\right )}{x}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((8*x + 5*x^2 + E^5*(8 + 4*x))/x^2)*(E^5*(-16 - 4*x) - 8*x) + x^3)/x^3,x]

[Out]

E^(5 + (8*E^5)/x^2 + (4*(2 + E^5))/x) + x

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fricas [A]  time = 0.56, size = 23, normalized size = 1.21 \begin {gather*} x + e^{\left (\frac {5 \, x^{2} + 4 \, {\left (x + 2\right )} e^{5} + 8 \, x}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16-4*x)*exp(5)-8*x)*exp(((4*x+8)*exp(5)+5*x^2+8*x)/x^2)+x^3)/x^3,x, algorithm="fricas")

[Out]

x + e^((5*x^2 + 4*(x + 2)*e^5 + 8*x)/x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} - 4 \, {\left ({\left (x + 4\right )} e^{5} + 2 \, x\right )} e^{\left (\frac {5 \, x^{2} + 4 \, {\left (x + 2\right )} e^{5} + 8 \, x}{x^{2}}\right )}}{x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16-4*x)*exp(5)-8*x)*exp(((4*x+8)*exp(5)+5*x^2+8*x)/x^2)+x^3)/x^3,x, algorithm="giac")

[Out]

integrate((x^3 - 4*((x + 4)*e^5 + 2*x)*e^((5*x^2 + 4*(x + 2)*e^5 + 8*x)/x^2))/x^3, x)

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maple [A]  time = 0.08, size = 26, normalized size = 1.37

method	result	size
risch	\(x +{\mathrm e}^{\frac {4 x \,{\mathrm e}^{5}+5 x^{2}+8 \,{\mathrm e}^{5}+8 x}{x^{2}}}\)	\(26\)
norman	\(\frac {x^{3}+x^{2} {\mathrm e}^{\frac {\left (4 x +8\right ) {\mathrm e}^{5}+5 x^{2}+8 x}{x^{2}}}}{x^{2}}\)	\(35\)
derivativedivides	\(x -i \sqrt {\pi }\, {\mathrm e}^{5-\frac {\left (4 \,{\mathrm e}^{5}+8\right )^{2} {\mathrm e}^{-5}}{32}} \sqrt {2}\, {\mathrm e}^{-\frac {5}{2}} \erf \left (\frac {2 i \sqrt {2}\, {\mathrm e}^{\frac {5}{2}}}{x}+\frac {i \left (4 \,{\mathrm e}^{5}+8\right ) \sqrt {2}\, {\mathrm e}^{-\frac {5}{2}}}{8}\right )-\frac {i \sqrt {\pi }\, {\mathrm e}^{10-\frac {\left (4 \,{\mathrm e}^{5}+8\right )^{2} {\mathrm e}^{-5}}{32}} \sqrt {2}\, {\mathrm e}^{-\frac {5}{2}} \erf \left (\frac {2 i \sqrt {2}\, {\mathrm e}^{\frac {5}{2}}}{x}+\frac {i \left (4 \,{\mathrm e}^{5}+8\right ) \sqrt {2}\, {\mathrm e}^{-\frac {5}{2}}}{8}\right )}{2}+{\mathrm e}^{-5} {\mathrm e}^{10+\frac {8 \,{\mathrm e}^{5}}{x^{2}}+\frac {4 \,{\mathrm e}^{5}+8}{x}}+\frac {i \left (4 \,{\mathrm e}^{5}+8\right ) {\mathrm e}^{-\frac {15}{2}} \sqrt {\pi }\, {\mathrm e}^{10-\frac {\left (4 \,{\mathrm e}^{5}+8\right )^{2} {\mathrm e}^{-5}}{32}} \sqrt {2}\, \erf \left (\frac {2 i \sqrt {2}\, {\mathrm e}^{\frac {5}{2}}}{x}+\frac {i \left (4 \,{\mathrm e}^{5}+8\right ) \sqrt {2}\, {\mathrm e}^{-\frac {5}{2}}}{8}\right )}{8}\)	\(217\)
default	\(x -i \sqrt {\pi }\, {\mathrm e}^{5-\frac {\left (4 \,{\mathrm e}^{5}+8\right )^{2} {\mathrm e}^{-5}}{32}} \sqrt {2}\, {\mathrm e}^{-\frac {5}{2}} \erf \left (\frac {2 i \sqrt {2}\, {\mathrm e}^{\frac {5}{2}}}{x}+\frac {i \left (4 \,{\mathrm e}^{5}+8\right ) \sqrt {2}\, {\mathrm e}^{-\frac {5}{2}}}{8}\right )-\frac {i \sqrt {\pi }\, {\mathrm e}^{10-\frac {\left (4 \,{\mathrm e}^{5}+8\right )^{2} {\mathrm e}^{-5}}{32}} \sqrt {2}\, {\mathrm e}^{-\frac {5}{2}} \erf \left (\frac {2 i \sqrt {2}\, {\mathrm e}^{\frac {5}{2}}}{x}+\frac {i \left (4 \,{\mathrm e}^{5}+8\right ) \sqrt {2}\, {\mathrm e}^{-\frac {5}{2}}}{8}\right )}{2}+{\mathrm e}^{-5} {\mathrm e}^{10+\frac {8 \,{\mathrm e}^{5}}{x^{2}}+\frac {4 \,{\mathrm e}^{5}+8}{x}}+\frac {i \left (4 \,{\mathrm e}^{5}+8\right ) {\mathrm e}^{-\frac {15}{2}} \sqrt {\pi }\, {\mathrm e}^{10-\frac {\left (4 \,{\mathrm e}^{5}+8\right )^{2} {\mathrm e}^{-5}}{32}} \sqrt {2}\, \erf \left (\frac {2 i \sqrt {2}\, {\mathrm e}^{\frac {5}{2}}}{x}+\frac {i \left (4 \,{\mathrm e}^{5}+8\right ) \sqrt {2}\, {\mathrm e}^{-\frac {5}{2}}}{8}\right )}{8}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-16-4*x)*exp(5)-8*x)*exp(((4*x+8)*exp(5)+5*x^2+8*x)/x^2)+x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

x+exp((4*x*exp(5)+5*x^2+8*exp(5)+8*x)/x^2)

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maxima [A]  time = 0.53, size = 24, normalized size = 1.26 \begin {gather*} x + e^{\left (\frac {4 \, e^{5}}{x} + \frac {8}{x} + \frac {8 \, e^{5}}{x^{2}} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16-4*x)*exp(5)-8*x)*exp(((4*x+8)*exp(5)+5*x^2+8*x)/x^2)+x^3)/x^3,x, algorithm="maxima")

[Out]

x + e^(4*e^5/x + 8/x + 8*e^5/x^2 + 5)

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mupad [B]  time = 5.61, size = 27, normalized size = 1.42 \begin {gather*} x+{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^5}{x}}\,{\mathrm {e}}^{\frac {8\,{\mathrm {e}}^5}{x^2}}\,{\mathrm {e}}^5\,{\mathrm {e}}^{8/x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((8*x + 5*x^2 + exp(5)*(4*x + 8))/x^2)*(8*x + exp(5)*(4*x + 16)) - x^3)/x^3,x)

[Out]

x + exp((4*exp(5))/x)*exp((8*exp(5))/x^2)*exp(5)*exp(8/x)

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sympy [A]  time = 0.21, size = 22, normalized size = 1.16 \begin {gather*} x + e^{\frac {5 x^{2} + 8 x + \left (4 x + 8\right ) e^{5}}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16-4*x)*exp(5)-8*x)*exp(((4*x+8)*exp(5)+5*x**2+8*x)/x**2)+x**3)/x**3,x)

[Out]

x + exp((5*x**2 + 8*x + (4*x + 8)*exp(5))/x**2)

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