3.30.73 \(\int \frac {509+e^{2 x} (2-4 x)+63 x^2+2 x^3-6 x^4+e^x (-64+64 x-4 x^2-4 x^3)}{x^2} \, dx\) [2973]

Optimal. Leaf size=28 \[ x^2+\frac {3+x-x^2-2 \left (-16+e^x+x^2\right )^2}{x} \]

[Out]

(x-x^2-2*(x^2+exp(x)-16)^2+3)/x+x^2

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Rubi [A]
time = 0.14, antiderivative size = 41, normalized size of antiderivative = 1.46, number of steps used = 13, number of rules used = 7, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.137, Rules used = {14, 2228, 2230, 2225, 2208, 2209, 2207} \begin {gather*} -2 x^3+x^2-4 e^x x+63 x+\frac {64 e^x}{x}-\frac {2 e^{2 x}}{x}-\frac {509}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(509 + E^(2*x)*(2 - 4*x) + 63*x^2 + 2*x^3 - 6*x^4 + E^x*(-64 + 64*x - 4*x^2 - 4*x^3))/x^2,x]

[Out]

-509/x + (64*E^x)/x - (2*E^(2*x))/x + 63*x - 4*E^x*x + x^2 - 2*x^3

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 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2225

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 2228

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
]

Rule 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 e^{2 x} (-1+2 x)}{x^2}-\frac {4 e^x \left (16-16 x+x^2+x^3\right )}{x^2}+\frac {509+63 x^2+2 x^3-6 x^4}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{2 x} (-1+2 x)}{x^2} \, dx\right )-4 \int \frac {e^x \left (16-16 x+x^2+x^3\right )}{x^2} \, dx+\int \frac {509+63 x^2+2 x^3-6 x^4}{x^2} \, dx\\ &=-\frac {2 e^{2 x}}{x}-4 \int \left (e^x+\frac {16 e^x}{x^2}-\frac {16 e^x}{x}+e^x x\right ) \, dx+\int \left (63+\frac {509}{x^2}+2 x-6 x^2\right ) \, dx\\ &=-\frac {509}{x}-\frac {2 e^{2 x}}{x}+63 x+x^2-2 x^3-4 \int e^x \, dx-4 \int e^x x \, dx-64 \int \frac {e^x}{x^2} \, dx+64 \int \frac {e^x}{x} \, dx\\ &=-4 e^x-\frac {509}{x}+\frac {64 e^x}{x}-\frac {2 e^{2 x}}{x}+63 x-4 e^x x+x^2-2 x^3+64 \text {Ei}(x)+4 \int e^x \, dx-64 \int \frac {e^x}{x} \, dx\\ &=-\frac {509}{x}+\frac {64 e^x}{x}-\frac {2 e^{2 x}}{x}+63 x-4 e^x x+x^2-2 x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.96, size = 36, normalized size = 1.29 \begin {gather*} \frac {-509-2 e^{2 x}+63 x^2+x^3-2 x^4-4 e^x \left (-16+x^2\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(509 + E^(2*x)*(2 - 4*x) + 63*x^2 + 2*x^3 - 6*x^4 + E^x*(-64 + 64*x - 4*x^2 - 4*x^3))/x^2,x]

[Out]

(-509 - 2*E^(2*x) + 63*x^2 + x^3 - 2*x^4 - 4*E^x*(-16 + x^2))/x

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Maple [A]
time = 0.12, size = 39, normalized size = 1.39

method	result	size
norman	\(\frac {-509+x^{3}+63 x^{2}-2 x^{4}-2 \,{\mathrm e}^{2 x}-4 \,{\mathrm e}^{x} x^{2}+64 \,{\mathrm e}^{x}}{x}\)	\(37\)
default	\(x^{2}+63 x -\frac {509}{x}-2 x^{3}+\frac {64 \,{\mathrm e}^{x}}{x}-4 \,{\mathrm e}^{x} x -\frac {2 \,{\mathrm e}^{2 x}}{x}\)	\(39\)
risch	\(x^{2}+63 x -\frac {509}{x}-2 x^{3}-\frac {2 \,{\mathrm e}^{2 x}}{x}-\frac {4 \left (x^{2}-16\right ) {\mathrm e}^{x}}{x}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x+2)*exp(x)^2+(-4*x^3-4*x^2+64*x-64)*exp(x)-6*x^4+2*x^3+63*x^2+509)/x^2,x,method=_RETURNVERBOSE)

[Out]

x^2+63*x-509/x-2*x^3+64*exp(x)/x-4*exp(x)*x-2*exp(x)^2/x

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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.30, size = 52, normalized size = 1.86 \begin {gather*} -2 \, x^{3} + x^{2} - 4 \, {\left (x - 1\right )} e^{x} + 63 \, x - \frac {509}{x} - 4 \, {\rm Ei}\left (2 \, x\right ) + 64 \, {\rm Ei}\left (x\right ) - 4 \, e^{x} - 64 \, \Gamma \left (-1, -x\right ) + 4 \, \Gamma \left (-1, -2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+2)*exp(x)^2+(-4*x^3-4*x^2+64*x-64)*exp(x)-6*x^4+2*x^3+63*x^2+509)/x^2,x, algorithm="maxima")

[Out]

-2*x^3 + x^2 - 4*(x - 1)*e^x + 63*x - 509/x - 4*Ei(2*x) + 64*Ei(x) - 4*e^x - 64*gamma(-1, -x) + 4*gamma(-1, -2
*x)

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Fricas [A]
time = 0.37, size = 37, normalized size = 1.32 \begin {gather*} -\frac {2 \, x^{4} - x^{3} - 63 \, x^{2} + 4 \, {\left (x^{2} - 16\right )} e^{x} + 2 \, e^{\left (2 \, x\right )} + 509}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+2)*exp(x)^2+(-4*x^3-4*x^2+64*x-64)*exp(x)-6*x^4+2*x^3+63*x^2+509)/x^2,x, algorithm="fricas")

[Out]

-(2*x^4 - x^3 - 63*x^2 + 4*(x^2 - 16)*e^x + 2*e^(2*x) + 509)/x

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Sympy [A]
time = 0.05, size = 37, normalized size = 1.32 \begin {gather*} - 2 x^{3} + x^{2} + 63 x - \frac {509}{x} + \frac {- 2 x e^{2 x} + \left (- 4 x^{3} + 64 x\right ) e^{x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+2)*exp(x)**2+(-4*x**3-4*x**2+64*x-64)*exp(x)-6*x**4+2*x**3+63*x**2+509)/x**2,x)

[Out]

-2*x**3 + x**2 + 63*x - 509/x + (-2*x*exp(2*x) + (-4*x**3 + 64*x)*exp(x))/x**2

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Giac [A]
time = 0.38, size = 39, normalized size = 1.39 \begin {gather*} -\frac {2 \, x^{4} - x^{3} + 4 \, x^{2} e^{x} - 63 \, x^{2} + 2 \, e^{\left (2 \, x\right )} - 64 \, e^{x} + 509}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+2)*exp(x)^2+(-4*x^3-4*x^2+64*x-64)*exp(x)-6*x^4+2*x^3+63*x^2+509)/x^2,x, algorithm="giac")

[Out]

-(2*x^4 - x^3 + 4*x^2*e^x - 63*x^2 + 2*e^(2*x) - 64*e^x + 509)/x

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Mupad [B]
time = 0.08, size = 32, normalized size = 1.14 \begin {gather*} x\,\left (x-4\,{\mathrm {e}}^x-2\,x^2+63\right )-\frac {2\,{\mathrm {e}}^{2\,x}-64\,{\mathrm {e}}^x+509}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x)*(4*x - 2) - 63*x^2 - 2*x^3 + 6*x^4 + exp(x)*(4*x^2 - 64*x + 4*x^3 + 64) - 509)/x^2,x)

[Out]

x*(x - 4*exp(x) - 2*x^2 + 63) - (2*exp(2*x) - 64*exp(x) + 509)/x

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