3.71.4 \(\int \frac {-80 x+e^x (1064-912 x+270 x^2-27 x^3)}{-1000+900 x-270 x^2+27 x^3} \, dx\) [7004]

Optimal. Leaf size=27 \[ 5-e^x-\frac {e^x-x^2}{\left (5-\frac {3 x}{2}\right )^2} \]

[Out]

5-exp(x)-(exp(x)-x^2)/(5-3/2*x)^2

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Rubi [A]
time = 0.14, antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {6874, 37, 2230, 2225, 2208, 2209} \begin {gather*} \frac {4 x^2}{(10-3 x)^2}-e^x-\frac {4 e^x}{(10-3 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-80*x + E^x*(1064 - 912*x + 270*x^2 - 27*x^3))/(-1000 + 900*x - 270*x^2 + 27*x^3),x]

[Out]

-E^x - (4*E^x)/(10 - 3*x)^2 + (4*x^2)/(10 - 3*x)^2

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {80 x}{(-10+3 x)^3}-\frac {e^x \left (-1064+912 x-270 x^2+27 x^3\right )}{(-10+3 x)^3}\right ) \, dx\\ &=-\left (80 \int \frac {x}{(-10+3 x)^3} \, dx\right )-\int \frac {e^x \left (-1064+912 x-270 x^2+27 x^3\right )}{(-10+3 x)^3} \, dx\\ &=\frac {4 x^2}{(10-3 x)^2}-\int \left (e^x-\frac {24 e^x}{(-10+3 x)^3}+\frac {4 e^x}{(-10+3 x)^2}\right ) \, dx\\ &=\frac {4 x^2}{(10-3 x)^2}-4 \int \frac {e^x}{(-10+3 x)^2} \, dx+24 \int \frac {e^x}{(-10+3 x)^3} \, dx-\int e^x \, dx\\ &=-e^x-\frac {4 e^x}{(10-3 x)^2}-\frac {4 e^x}{3 (10-3 x)}+\frac {4 x^2}{(10-3 x)^2}-\frac {4}{3} \int \frac {e^x}{-10+3 x} \, dx+4 \int \frac {e^x}{(-10+3 x)^2} \, dx\\ &=-e^x-\frac {4 e^x}{(10-3 x)^2}+\frac {4 x^2}{(10-3 x)^2}-\frac {4}{9} e^{10/3} \text {Ei}\left (\frac {1}{3} (-10+3 x)\right )+\frac {4}{3} \int \frac {e^x}{-10+3 x} \, dx\\ &=-e^x-\frac {4 e^x}{(10-3 x)^2}+\frac {4 x^2}{(10-3 x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.39, size = 34, normalized size = 1.26 \begin {gather*} \frac {80 (-5+3 x)-9 e^x \left (104-60 x+9 x^2\right )}{9 (10-3 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-80*x + E^x*(1064 - 912*x + 270*x^2 - 27*x^3))/(-1000 + 900*x - 270*x^2 + 27*x^3),x]

[Out]

(80*(-5 + 3*x) - 9*E^x*(104 - 60*x + 9*x^2))/(9*(10 - 3*x)^2)

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Maple [A]
time = 0.13, size = 33, normalized size = 1.22

method	result	size
norman	\(\frac {\frac {80 x}{3}+60 \,{\mathrm e}^{x} x -9 \,{\mathrm e}^{x} x^{2}-104 \,{\mathrm e}^{x}-\frac {400}{9}}{\left (3 x -10\right )^{2}}\)	\(30\)
default	\(\frac {400}{9 \left (3 x -10\right )^{2}}+\frac {80}{9 \left (3 x -10\right )}-\frac {4 \,{\mathrm e}^{x}}{9 \left (x -\frac {10}{3}\right )^{2}}-{\mathrm e}^{x}\)	\(33\)
risch	\(\frac {\frac {80 x}{27}-\frac {400}{81}}{x^{2}-\frac {20}{3} x +\frac {100}{9}}-\frac {\left (9 x^{2}-60 x +104\right ) {\mathrm e}^{x}}{\left (3 x -10\right )^{2}}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-27*x^3+270*x^2-912*x+1064)*exp(x)-80*x)/(27*x^3-270*x^2+900*x-1000),x,method=_RETURNVERBOSE)

[Out]

400/9/(3*x-10)^2+80/9/(3*x-10)-4/9*exp(x)/(x-10/3)^2-exp(x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-27*x^3+270*x^2-912*x+1064)*exp(x)-80*x)/(27*x^3-270*x^2+900*x-1000),x, algorithm="maxima")

[Out]

-3*(9*x^3 - 90*x^2 + 304*x)*e^x/(27*x^3 - 270*x^2 + 900*x - 1000) + 80/9*(3*x - 5)/(9*x^2 - 60*x + 100) - 1064
/3*e^(10/3)*exp_integral_e(3, -x + 10/3)/(3*x - 10)^2 - integrate(24*(3*x + 380)*e^x/(81*x^4 - 1080*x^3 + 5400
*x^2 - 12000*x + 10000), x)

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Fricas [A]
time = 0.39, size = 33, normalized size = 1.22 \begin {gather*} -\frac {9 \, {\left (9 \, x^{2} - 60 \, x + 104\right )} e^{x} - 240 \, x + 400}{9 \, {\left (9 \, x^{2} - 60 \, x + 100\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-27*x^3+270*x^2-912*x+1064)*exp(x)-80*x)/(27*x^3-270*x^2+900*x-1000),x, algorithm="fricas")

[Out]

-1/9*(9*(9*x^2 - 60*x + 104)*e^x - 240*x + 400)/(9*x^2 - 60*x + 100)

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Sympy [A]
time = 0.07, size = 37, normalized size = 1.37 \begin {gather*} - \frac {80 \cdot \left (5 - 3 x\right )}{81 x^{2} - 540 x + 900} + \frac {\left (- 9 x^{2} + 60 x - 104\right ) e^{x}}{9 x^{2} - 60 x + 100} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-27*x**3+270*x**2-912*x+1064)*exp(x)-80*x)/(27*x**3-270*x**2+900*x-1000),x)

[Out]

-80*(5 - 3*x)/(81*x**2 - 540*x + 900) + (-9*x**2 + 60*x - 104)*exp(x)/(9*x**2 - 60*x + 100)

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Giac [A]
time = 0.41, size = 35, normalized size = 1.30 \begin {gather*} -\frac {81 \, x^{2} e^{x} - 540 \, x e^{x} - 240 \, x + 936 \, e^{x} + 400}{9 \, {\left (9 \, x^{2} - 60 \, x + 100\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-27*x^3+270*x^2-912*x+1064)*exp(x)-80*x)/(27*x^3-270*x^2+900*x-1000),x, algorithm="giac")

[Out]

-1/9*(81*x^2*e^x - 540*x*e^x - 240*x + 936*e^x + 400)/(9*x^2 - 60*x + 100)

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Mupad [B]
time = 4.09, size = 23, normalized size = 0.85 \begin {gather*} -{\mathrm {e}}^x-\frac {4\,{\mathrm {e}}^x-\frac {80\,x}{3}+\frac {400}{9}}{{\left (3\,x-10\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(80*x + exp(x)*(912*x - 270*x^2 + 27*x^3 - 1064))/(900*x - 270*x^2 + 27*x^3 - 1000),x)

[Out]

- exp(x) - (4*exp(x) - (80*x)/3 + 400/9)/(3*x - 10)^2

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