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

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

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Rubi [A]  time = 0.13, 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, 2197, 2199, 2194, 2177, 2178, 2176} $$\begin{array}{c}-2x^3+x^2-4e^{x}x+63x+\frac{64e^x}{x}-\frac{2e^{2x}}{x}-\frac{509}{x}\end{array}$$

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 2176

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] &&  !$UseGamma === True

Rule 2177

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] &&  !$UseGamma ===
True

Rule 2178

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

Rule 2194

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 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
]

Rule 2199

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] &&  !$UseGamma === True

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-\frac{2e^{2x}(-1+2x)}{x^2}-\frac{4e^x(16-16x+x^2+x^3)}{x^2}+\frac{509+63x^2+2x^3-6x^4}{x^2})dx\\ & =-(2\int \frac{e^{2x}(-1+2x)}{x^2}dx)-4\int \frac{e^x(16-16x+x^2+x^3)}{x^2}dx+\int \frac{509+63x^2+2x^3-6x^4}{x^2}dx\\ & =-\frac{2e^{2x}}{x}-4\int (e^x+\frac{16e^x}{x^2}-\frac{16e^x}{x}+e^{x}x)dx+\int (63+\frac{509}{x^2}+2x-6x^2)dx\\ & =-\frac{509}{x}-\frac{2e^{2x}}{x}+63x+x^2-2x^3-4\int e^{x}dx-4\int e^{x}xdx-64\int \frac{e^x}{x^2}dx+64\int \frac{e^x}{x}dx\\ & =-4e^x-\frac{509}{x}+\frac{64e^x}{x}-\frac{2e^{2x}}{x}+63x-4e^{x}x+x^2-2x^3+64\text{Ei}(x)+4\int e^{x}dx-64\int \frac{e^x}{x}dx\\ & =-\frac{509}{x}+\frac{64e^x}{x}-\frac{2e^{2x}}{x}+63x-4e^{x}x+x^2-2x^3\end{array}\end{array}$$

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Mathematica [A]  time = 0.05, size = 36, normalized size = 1.29 $$\begin{array}{c}\frac{-509-2e^{2x}+63x^2+x^3-2x^4-4e^x(-16+x^2)}{x}\end{array}$$

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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fricas [A]  time = 0.55, size = 37, normalized size = 1.32 $$\begin{array}{c}-\frac{2x^4-x^3-63x^2+4(x^2-16)e^x+2e^{\left(2x\right)}+509}{x}\end{array}$$

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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giac [A]  time = 0.21, size = 39, normalized size = 1.39 $$\begin{array}{c}-\frac{2x^4-x^3+4x^2{e}^x-63x^2+2e^{\left(2x\right)}-64e^x+509}{x}\end{array}$$

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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maple [A]  time = 0.03, size = 37, normalized size = 1.32

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

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]

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

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maxima [C]  time = 0.53, size = 52, normalized size = 1.86 $$\begin{array}{c}-2x^3+x^2-4(x-1)e^x+63x-\frac{509}{x}-4\mathrm{E}\mathrm{i}\left(2x\right)+64\mathrm{E}\mathrm{i}(x)-4e^x-64\mathrm{\Gamma }(-1,-x)+4\mathrm{\Gamma }(-1,-2x)\end{array}$$

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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mupad [B]  time = 0.08, size = 32, normalized size = 1.14 $$\begin{array}{c}x(x-4{\mathrm{e}}^x-2x^2+63)-\frac{2{\mathrm{e}}^{2x}-64{\mathrm{e}}^x+509}{x}\end{array}$$

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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sympy [A]  time = 0.13, size = 37, normalized size = 1.32 $$\begin{array}{c}-2x^3+x^2+63x-\frac{509}{x}+\frac{-2x{e}^{2x}+(-4x^3+64x)e^x}{x^2}\end{array}$$

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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