3.66.15 \(\int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{e^2 x^2} \, dx\)

Optimal. Leaf size=17 \[ 5+\frac {5 e^{14+2 x}}{x}-x \]

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Rubi [A]  time = 0.05, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2197} \begin {gather*} \frac {5 e^{2 x+14}}{x}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^2*x^2) + E^(16 + 2*x)*(-5 + 10*x))/(E^2*x^2),x]

[Out]

(5*E^(14 + 2*x))/x - 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 {\int \frac {-e^2 x^2+e^{16+2 x} (-5+10 x)}{x^2} \, dx}{e^2}\\ &=\frac {\int \left (-e^2+\frac {5 e^{16+2 x} (-1+2 x)}{x^2}\right ) \, dx}{e^2}\\ &=-x+\frac {5 \int \frac {e^{16+2 x} (-1+2 x)}{x^2} \, dx}{e^2}\\ &=\frac {5 e^{14+2 x}}{x}-x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 0.94 \begin {gather*} \frac {5 e^{2 (7+x)}}{x}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^2*x^2) + E^(16 + 2*x)*(-5 + 10*x))/(E^2*x^2),x]

[Out]

(5*E^(2*(7 + x)))/x - x

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fricas [A]  time = 0.58, size = 22, normalized size = 1.29 \begin {gather*} -\frac {{\left (x^{2} e^{2} - 5 \, e^{\left (2 \, x + 16\right )}\right )} e^{\left (-2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-5)*exp(16)*exp(2*x)-x^2*exp(2))/x^2/exp(2),x, algorithm="fricas")

[Out]

-(x^2*e^2 - 5*e^(2*x + 16))*e^(-2)/x

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giac [A]  time = 0.14, size = 22, normalized size = 1.29 \begin {gather*} -\frac {{\left (x^{2} e^{2} - 5 \, e^{\left (2 \, x + 16\right )}\right )} e^{\left (-2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-5)*exp(16)*exp(2*x)-x^2*exp(2))/x^2/exp(2),x, algorithm="giac")

[Out]

-(x^2*e^2 - 5*e^(2*x + 16))*e^(-2)/x

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maple [A]  time = 0.17, size = 16, normalized size = 0.94

method	result	size
risch	\(-x +\frac {5 \,{\mathrm e}^{14+2 x}}{x}\)	\(16\)
norman	\(\frac {-x^{2}+5 \,{\mathrm e}^{-2} {\mathrm e}^{16} {\mathrm e}^{2 x}}{x}\)	\(23\)
default	\({\mathrm e}^{-2} \left (-10 \,{\mathrm e}^{16} \left (-\frac {{\mathrm e}^{2 x}}{2 x}-\expIntegralEi \left (1, -2 x \right )\right )-10 \,{\mathrm e}^{16} \expIntegralEi \left (1, -2 x \right )-{\mathrm e}^{2} x \right )\)	\(42\)
derivativedivides	\(\frac {{\mathrm e}^{-2} \left (-20 \,{\mathrm e}^{16} \left (-\frac {{\mathrm e}^{2 x}}{2 x}-\expIntegralEi \left (1, -2 x \right )\right )-20 \,{\mathrm e}^{16} \expIntegralEi \left (1, -2 x \right )-2 \,{\mathrm e}^{2} x \right )}{2}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x-5)*exp(16)*exp(2*x)-x^2*exp(2))/x^2/exp(2),x,method=_RETURNVERBOSE)

[Out]

-x+5/x*exp(14+2*x)

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maxima [C]  time = 0.44, size = 26, normalized size = 1.53 \begin {gather*} {\left (10 \, {\rm Ei}\left (2 \, x\right ) e^{16} - x e^{2} - 10 \, e^{16} \Gamma \left (-1, -2 \, x\right )\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-5)*exp(16)*exp(2*x)-x^2*exp(2))/x^2/exp(2),x, algorithm="maxima")

[Out]

(10*Ei(2*x)*e^16 - x*e^2 - 10*e^16*gamma(-1, -2*x))*e^(-2)

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mupad [B]  time = 0.06, size = 15, normalized size = 0.88 \begin {gather*} \frac {5\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{14}}{x}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2)*(x^2*exp(2) - exp(2*x)*exp(16)*(10*x - 5)))/x^2,x)

[Out]

(5*exp(2*x)*exp(14))/x - x

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sympy [A]  time = 0.11, size = 12, normalized size = 0.71 \begin {gather*} - x + \frac {5 e^{14} e^{2 x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-5)*exp(16)*exp(2*x)-x**2*exp(2))/x**2/exp(2),x)

[Out]

-x + 5*exp(14)*exp(2*x)/x

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