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

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

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Rubi [A]  time = 0.24, 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 = {6742, 37, 2199, 2194, 2177, 2178} \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 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 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

Rule 6742

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 = 0.11, size = 28, normalized size = 1.04 \begin {gather*} \frac {4 x^2+e^x \left (-104+60 x-9 x^2\right )}{(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]

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

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fricas [A]  time = 0.53, 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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giac [A]  time = 0.16, 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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maple [A]  time = 0.04, size = 30, normalized size = 1.11

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]

(80/3*x+60*exp(x)*x-9*exp(x)*x^2-104*exp(x)-400/9)/(3*x-10)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {3 \, {\left (9 \, x^{3} - 90 \, x^{2} + 304 \, x\right )} e^{x}}{27 \, x^{3} - 270 \, x^{2} + 900 \, x - 1000} + \frac {80 \, {\left (3 \, x - 5\right )}}{9 \, {\left (9 \, x^{2} - 60 \, x + 100\right )}} - \frac {1064 \, e^{\frac {10}{3}} E_{3}\left (-x + \frac {10}{3}\right )}{3 \, {\left (3 \, x - 10\right )}^{2}} - \int \frac {24 \, {\left (3 \, x + 380\right )} e^{x}}{81 \, x^{4} - 1080 \, x^{3} + 5400 \, x^{2} - 12000 \, x + 10000}\,{d x} \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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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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sympy [A]  time = 0.14, size = 37, normalized size = 1.37 \begin {gather*} - \frac {80 \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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