3.37.85 \(\int \frac {e^{-x} (2 x-6 x^2+5 x^3+e^x (1-11 x+20 x^2+25 x^3))}{1-10 x+25 x^2} \, dx\)

Optimal. Leaf size=31 \[ 5-e^5+x+\left (\frac {-1+e^{-x}}{-5+\frac {1}{x}}+\frac {x}{2}\right ) x \]

________________________________________________________________________________________

Rubi [A]  time = 0.44, antiderivative size = 59, normalized size of antiderivative = 1.90, number of steps used = 13, number of rules used = 8, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.157, Rules used = {27, 6742, 2199, 2194, 2176, 2177, 2178, 1850} \begin {gather*} \frac {x^2}{2}-\frac {e^{-x} x}{5}+\frac {6 x}{5}-\frac {e^{-x}}{25}+\frac {e^{-x}}{25 (1-5 x)}-\frac {1}{25 (1-5 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x - 6*x^2 + 5*x^3 + E^x*(1 - 11*x + 20*x^2 + 25*x^3))/(E^x*(1 - 10*x + 25*x^2)),x]

[Out]

-1/25*1/E^x - 1/(25*(1 - 5*x)) + 1/(25*E^x*(1 - 5*x)) + (6*x)/5 - x/(5*E^x) + x^2/2

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

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 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 \frac {e^{-x} \left (2 x-6 x^2+5 x^3+e^x \left (1-11 x+20 x^2+25 x^3\right )\right )}{(-1+5 x)^2} \, dx\\ &=\int \left (\frac {e^{-x} x \left (2-6 x+5 x^2\right )}{(-1+5 x)^2}+\frac {1-11 x+20 x^2+25 x^3}{(-1+5 x)^2}\right ) \, dx\\ &=\int \frac {e^{-x} x \left (2-6 x+5 x^2\right )}{(-1+5 x)^2} \, dx+\int \frac {1-11 x+20 x^2+25 x^3}{(-1+5 x)^2} \, dx\\ &=\int \left (\frac {6}{5}+x-\frac {1}{5 (-1+5 x)^2}\right ) \, dx+\int \left (-\frac {4 e^{-x}}{25}+\frac {e^{-x} x}{5}+\frac {e^{-x}}{5 (-1+5 x)^2}+\frac {e^{-x}}{25 (-1+5 x)}\right ) \, dx\\ &=-\frac {1}{25 (1-5 x)}+\frac {6 x}{5}+\frac {x^2}{2}+\frac {1}{25} \int \frac {e^{-x}}{-1+5 x} \, dx-\frac {4}{25} \int e^{-x} \, dx+\frac {1}{5} \int e^{-x} x \, dx+\frac {1}{5} \int \frac {e^{-x}}{(-1+5 x)^2} \, dx\\ &=\frac {4 e^{-x}}{25}-\frac {1}{25 (1-5 x)}+\frac {e^{-x}}{25 (1-5 x)}+\frac {6 x}{5}-\frac {e^{-x} x}{5}+\frac {x^2}{2}+\frac {\text {Ei}\left (\frac {1}{5} (1-5 x)\right )}{125 \sqrt [5]{e}}-\frac {1}{25} \int \frac {e^{-x}}{-1+5 x} \, dx+\frac {1}{5} \int e^{-x} \, dx\\ &=-\frac {e^{-x}}{25}-\frac {1}{25 (1-5 x)}+\frac {e^{-x}}{25 (1-5 x)}+\frac {6 x}{5}-\frac {e^{-x} x}{5}+\frac {x^2}{2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 32, normalized size = 1.03 \begin {gather*} -\frac {2-60 x+\left (275-50 e^{-x}\right ) x^2+125 x^3}{50-250 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x - 6*x^2 + 5*x^3 + E^x*(1 - 11*x + 20*x^2 + 25*x^3))/(E^x*(1 - 10*x + 25*x^2)),x]

[Out]

-((2 - 60*x + (275 - 50/E^x)*x^2 + 125*x^3)/(50 - 250*x))

________________________________________________________________________________________

fricas [A]  time = 0.74, size = 38, normalized size = 1.23 \begin {gather*} -\frac {{\left (50 \, x^{2} - {\left (125 \, x^{3} + 275 \, x^{2} - 60 \, x + 2\right )} e^{x}\right )} e^{\left (-x\right )}}{50 \, {\left (5 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x^3+20*x^2-11*x+1)*exp(x)+5*x^3-6*x^2+2*x)/(25*x^2-10*x+1)/exp(x),x, algorithm="fricas")

[Out]

-1/50*(50*x^2 - (125*x^3 + 275*x^2 - 60*x + 2)*e^x)*e^(-x)/(5*x - 1)

________________________________________________________________________________________

giac [A]  time = 0.13, size = 33, normalized size = 1.06 \begin {gather*} \frac {125 \, x^{3} - 50 \, x^{2} e^{\left (-x\right )} + 275 \, x^{2} - 60 \, x + 2}{50 \, {\left (5 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x^3+20*x^2-11*x+1)*exp(x)+5*x^3-6*x^2+2*x)/(25*x^2-10*x+1)/exp(x),x, algorithm="giac")

[Out]

1/50*(125*x^3 - 50*x^2*e^(-x) + 275*x^2 - 60*x + 2)/(5*x - 1)

________________________________________________________________________________________

maple [A]  time = 0.13, size = 33, normalized size = 1.06

method	result	size
risch	\(\frac {x^{2}}{2}+\frac {6 x}{5}+\frac {1}{125 x -25}-\frac {x^{2} {\mathrm e}^{-x}}{5 x -1}\)	\(33\)
norman	\(\frac {\left (-{\mathrm e}^{x} x -x^{2}+\frac {11 \,{\mathrm e}^{x} x^{2}}{2}+\frac {5 \,{\mathrm e}^{x} x^{3}}{2}\right ) {\mathrm e}^{-x}}{5 x -1}\)	\(38\)
default	\(\frac {1}{125 x -25}+\frac {6 x}{5}+\frac {x^{2}}{2}-\frac {{\mathrm e}^{-x}}{25 \left (5 x -1\right )}+\frac {6 \,{\mathrm e}^{-x}}{25}-\frac {\left (5 x +7\right ) {\mathrm e}^{-x}}{25}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((25*x^3+20*x^2-11*x+1)*exp(x)+5*x^3-6*x^2+2*x)/(25*x^2-10*x+1)/exp(x),x,method=_RETURNVERBOSE)

[Out]

1/2*x^2+6/5*x+1/125/(x-1/5)-x^2/(5*x-1)*exp(-x)

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 33, normalized size = 1.06 \begin {gather*} \frac {125 \, x^{3} - 50 \, x^{2} e^{\left (-x\right )} + 275 \, x^{2} - 60 \, x + 2}{50 \, {\left (5 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x^3+20*x^2-11*x+1)*exp(x)+5*x^3-6*x^2+2*x)/(25*x^2-10*x+1)/exp(x),x, algorithm="maxima")

[Out]

1/50*(125*x^3 - 50*x^2*e^(-x) + 275*x^2 - 60*x + 2)/(5*x - 1)

________________________________________________________________________________________

mupad [B]  time = 0.12, size = 37, normalized size = 1.19 \begin {gather*} \frac {x^2}{{\mathrm {e}}^x-5\,x\,{\mathrm {e}}^x}+\frac {\frac {5\,x^3}{2}+\frac {11\,x^2}{2}-x}{5\,x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(2*x - 6*x^2 + 5*x^3 + exp(x)*(20*x^2 - 11*x + 25*x^3 + 1)))/(25*x^2 - 10*x + 1),x)

[Out]

x^2/(exp(x) - 5*x*exp(x)) + ((11*x^2)/2 - x + (5*x^3)/2)/(5*x - 1)

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 27, normalized size = 0.87 \begin {gather*} \frac {x^{2}}{2} - \frac {x^{2} e^{- x}}{5 x - 1} + \frac {6 x}{5} + \frac {1}{125 x - 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x**3+20*x**2-11*x+1)*exp(x)+5*x**3-6*x**2+2*x)/(25*x**2-10*x+1)/exp(x),x)

[Out]

x**2/2 - x**2*exp(-x)/(5*x - 1) + 6*x/5 + 1/(125*x - 25)

________________________________________________________________________________________