3.96.28 \(\int \frac {e^{\frac {225+9 x+45 e^{5+x} x-5 x^2}{5 x}} (-45-x^2+9 e^{5+x} x^2)}{x^2} \, dx\)

Optimal. Leaf size=25 \[ e^{-x+\frac {9 \left (5+\frac {x}{5}+e^{5+x} x\right )}{x}} \]

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Rubi [A]  time = 0.48, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 1, number of rules used = 1, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6706} \begin {gather*} e^{\frac {-5 x^2+45 e^{x+5} x+9 x+225}{5 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((225 + 9*x + 45*E^(5 + x)*x - 5*x^2)/(5*x))*(-45 - x^2 + 9*E^(5 + x)*x^2))/x^2,x]

[Out]

E^((225 + 9*x + 45*E^(5 + x)*x - 5*x^2)/(5*x))

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} &=e^{\frac {225+9 x+45 e^{5+x} x-5 x^2}{5 x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 21, normalized size = 0.84 \begin {gather*} e^{\frac {9}{5}+9 e^{5+x}+\frac {45}{x}-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((225 + 9*x + 45*E^(5 + x)*x - 5*x^2)/(5*x))*(-45 - x^2 + 9*E^(5 + x)*x^2))/x^2,x]

[Out]

E^(9/5 + 9*E^(5 + x) + 45/x - x)

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fricas [A]  time = 0.55, size = 23, normalized size = 0.92 \begin {gather*} e^{\left (-\frac {5 \, x^{2} - 45 \, x e^{\left (x + 5\right )} - 9 \, x - 225}{5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*x^2*exp(5+x)-x^2-45)*exp(1/5*(45*x*exp(5+x)-5*x^2+9*x+225)/x)/x^2,x, algorithm="fricas")

[Out]

e^(-1/5*(5*x^2 - 45*x*e^(x + 5) - 9*x - 225)/x)

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giac [A]  time = 0.19, size = 17, normalized size = 0.68 \begin {gather*} e^{\left (-x + \frac {45}{x} + 9 \, e^{\left (x + 5\right )} + \frac {9}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*x^2*exp(5+x)-x^2-45)*exp(1/5*(45*x*exp(5+x)-5*x^2+9*x+225)/x)/x^2,x, algorithm="giac")

[Out]

e^(-x + 45/x + 9*e^(x + 5) + 9/5)

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maple [A]  time = 0.12, size = 24, normalized size = 0.96

method	result	size
norman	\({\mathrm e}^{\frac {45 x \,{\mathrm e}^{5+x}-5 x^{2}+9 x +225}{5 x}}\)	\(24\)
risch	\({\mathrm e}^{-\frac {-45 x \,{\mathrm e}^{5+x}+5 x^{2}-9 x -225}{5 x}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*x^2*exp(5+x)-x^2-45)*exp(1/5*(45*x*exp(5+x)-5*x^2+9*x+225)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(1/5*(45*x*exp(5+x)-5*x^2+9*x+225)/x)

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maxima [A]  time = 0.42, size = 17, normalized size = 0.68 \begin {gather*} e^{\left (-x + \frac {45}{x} + 9 \, e^{\left (x + 5\right )} + \frac {9}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*x^2*exp(5+x)-x^2-45)*exp(1/5*(45*x*exp(5+x)-5*x^2+9*x+225)/x)/x^2,x, algorithm="maxima")

[Out]

e^(-x + 45/x + 9*e^(x + 5) + 9/5)

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mupad [B]  time = 9.30, size = 20, normalized size = 0.80 \begin {gather*} {\mathrm {e}}^{9\,{\mathrm {e}}^5\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{9/5}\,{\mathrm {e}}^{45/x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(((9*x)/5 + 9*x*exp(x + 5) - x^2 + 45)/x)*(x^2 - 9*x^2*exp(x + 5) + 45))/x^2,x)

[Out]

exp(9*exp(5)*exp(x))*exp(-x)*exp(9/5)*exp(45/x)

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sympy [A]  time = 0.19, size = 20, normalized size = 0.80 \begin {gather*} e^{\frac {- x^{2} + 9 x e^{x + 5} + \frac {9 x}{5} + 45}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*x**2*exp(5+x)-x**2-45)*exp(1/5*(45*x*exp(5+x)-5*x**2+9*x+225)/x)/x**2,x)

[Out]

exp((-x**2 + 9*x*exp(x + 5) + 9*x/5 + 45)/x)

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