3.25.86 \(\int \frac {1}{25} (-175+25 x+4 e^{-8 x} (-70+580 x-80 x^2)+16 e^{-16 x} (4 x-32 x^2)) \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{2} \left (7-x-\frac {8}{5} e^{-8 x} x\right )^2 \]

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Rubi [B]  time = 0.13, antiderivative size = 45, normalized size of antiderivative = 2.14, number of steps used = 18, number of rules used = 5, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {12, 2196, 2194, 2176, 1593} \begin {gather*} \frac {32}{25} e^{-16 x} x^2+\frac {8}{5} e^{-8 x} x^2+\frac {x^2}{2}-\frac {56}{5} e^{-8 x} x-7 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-175 + 25*x + (4*(-70 + 580*x - 80*x^2))/E^(8*x) + (16*(4*x - 32*x^2))/E^(16*x))/25,x]

[Out]

-7*x - (56*x)/(5*E^(8*x)) + x^2/2 + (32*x^2)/(25*E^(16*x)) + (8*x^2)/(5*E^(8*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx\\ &=-7 x+\frac {x^2}{2}+\frac {4}{25} \int e^{-8 x} \left (-70+580 x-80 x^2\right ) \, dx+\frac {16}{25} \int e^{-16 x} \left (4 x-32 x^2\right ) \, dx\\ &=-7 x+\frac {x^2}{2}+\frac {4}{25} \int \left (-70 e^{-8 x}+580 e^{-8 x} x-80 e^{-8 x} x^2\right ) \, dx+\frac {16}{25} \int e^{-16 x} (4-32 x) x \, dx\\ &=-7 x+\frac {x^2}{2}+\frac {16}{25} \int \left (4 e^{-16 x} x-32 e^{-16 x} x^2\right ) \, dx-\frac {56}{5} \int e^{-8 x} \, dx-\frac {64}{5} \int e^{-8 x} x^2 \, dx+\frac {464}{5} \int e^{-8 x} x \, dx\\ &=\frac {7 e^{-8 x}}{5}-7 x-\frac {58}{5} e^{-8 x} x+\frac {x^2}{2}+\frac {8}{5} e^{-8 x} x^2+\frac {64}{25} \int e^{-16 x} x \, dx-\frac {16}{5} \int e^{-8 x} x \, dx+\frac {58}{5} \int e^{-8 x} \, dx-\frac {512}{25} \int e^{-16 x} x^2 \, dx\\ &=-\frac {1}{20} e^{-8 x}-7 x-\frac {4}{25} e^{-16 x} x-\frac {56}{5} e^{-8 x} x+\frac {x^2}{2}+\frac {32}{25} e^{-16 x} x^2+\frac {8}{5} e^{-8 x} x^2+\frac {4}{25} \int e^{-16 x} \, dx-\frac {2}{5} \int e^{-8 x} \, dx-\frac {64}{25} \int e^{-16 x} x \, dx\\ &=-\frac {1}{100} e^{-16 x}-7 x-\frac {56}{5} e^{-8 x} x+\frac {x^2}{2}+\frac {32}{25} e^{-16 x} x^2+\frac {8}{5} e^{-8 x} x^2-\frac {4}{25} \int e^{-16 x} \, dx\\ &=-7 x-\frac {56}{5} e^{-8 x} x+\frac {x^2}{2}+\frac {32}{25} e^{-16 x} x^2+\frac {8}{5} e^{-8 x} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 41, normalized size = 1.95 \begin {gather*} -7 x+\frac {x^2}{2}+\frac {32}{25} e^{-16 x} x^2-\frac {8}{5} e^{-8 x} \left (7 x-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-175 + 25*x + (4*(-70 + 580*x - 80*x^2))/E^(8*x) + (16*(4*x - 32*x^2))/E^(16*x))/25,x]

[Out]

-7*x + x^2/2 + (32*x^2)/(25*E^(16*x)) - (8*(7*x - x^2))/(5*E^(8*x))

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fricas [A]  time = 0.57, size = 41, normalized size = 1.95 \begin {gather*} \frac {2}{25} \, x^{2} e^{\left (-16 \, x + 4 \, \log \relax (2)\right )} + \frac {1}{2} \, x^{2} + \frac {2}{5} \, {\left (x^{2} - 7 \, x\right )} e^{\left (-8 \, x + 2 \, \log \relax (2)\right )} - 7 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-32*x^2+4*x)*exp(log(2)-4*x)^4+1/25*(-80*x^2+580*x-70)*exp(log(2)-4*x)^2+x-7,x, algorithm="fri
cas")

[Out]

2/25*x^2*e^(-16*x + 4*log(2)) + 1/2*x^2 + 2/5*(x^2 - 7*x)*e^(-8*x + 2*log(2)) - 7*x

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giac [A]  time = 0.21, size = 41, normalized size = 1.95 \begin {gather*} \frac {2}{25} \, x^{2} e^{\left (-16 \, x + 4 \, \log \relax (2)\right )} + \frac {1}{2} \, x^{2} + \frac {2}{5} \, {\left (x^{2} - 7 \, x\right )} e^{\left (-8 \, x + 2 \, \log \relax (2)\right )} - 7 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-32*x^2+4*x)*exp(log(2)-4*x)^4+1/25*(-80*x^2+580*x-70)*exp(log(2)-4*x)^2+x-7,x, algorithm="gia
c")

[Out]

2/25*x^2*e^(-16*x + 4*log(2)) + 1/2*x^2 + 2/5*(x^2 - 7*x)*e^(-8*x + 2*log(2)) - 7*x

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maple [A]  time = 0.05, size = 34, normalized size = 1.62

method	result	size
risch	\(\frac {32 \,{\mathrm e}^{-16 x} x^{2}}{25}+\frac {4 \left (10 x^{2}-70 x \right ) {\mathrm e}^{-8 x}}{25}+\frac {x^{2}}{2}-7 x\)	\(34\)
norman	\(-7 x +\frac {x^{2}}{2}-\frac {56 \,{\mathrm e}^{-8 x} x}{5}+\frac {8 \,{\mathrm e}^{-8 x} x^{2}}{5}+\frac {32 \,{\mathrm e}^{-16 x} x^{2}}{25}\)	\(50\)
default	\(-7 x +\frac {x^{2}}{2}+\frac {2 \,{\mathrm e}^{-16 x} \left (\ln \relax (2)-4 x \right )^{2}}{25}+\frac {2 \,{\mathrm e}^{-16 x} \ln \relax (2)^{2}}{25}-\frac {4 \,{\mathrm e}^{-16 x} \left (\ln \relax (2)-4 x \right ) \ln \relax (2)}{25}+\frac {14 \,{\mathrm e}^{-8 x} \left (\ln \relax (2)-4 x \right )}{5}+\frac {{\mathrm e}^{-8 x} \left (\ln \relax (2)-4 x \right )^{2}}{10}-\frac {14 \,{\mathrm e}^{-8 x} \ln \relax (2)}{5}+\frac {{\mathrm e}^{-8 x} \ln \relax (2)^{2}}{10}-\frac {{\mathrm e}^{-8 x} \left (\ln \relax (2)-4 x \right ) \ln \relax (2)}{5}\)	\(146\)
derivativedivides	\(\frac {7 \ln \relax (2)}{4}-7 x +\frac {\left (\ln \relax (2)-4 x \right )^{2}}{32}+\frac {2 \,{\mathrm e}^{-16 x} \left (\ln \relax (2)-4 x \right )^{2}}{25}+\frac {2 \,{\mathrm e}^{-16 x} \ln \relax (2)^{2}}{25}-\frac {4 \,{\mathrm e}^{-16 x} \left (\ln \relax (2)-4 x \right ) \ln \relax (2)}{25}+\frac {14 \,{\mathrm e}^{-8 x} \left (\ln \relax (2)-4 x \right )}{5}+\frac {{\mathrm e}^{-8 x} \left (\ln \relax (2)-4 x \right )^{2}}{10}-\frac {14 \,{\mathrm e}^{-8 x} \ln \relax (2)}{5}+\frac {{\mathrm e}^{-8 x} \ln \relax (2)^{2}}{10}-\frac {{\mathrm e}^{-8 x} \left (\ln \relax (2)-4 x \right ) \ln \relax (2)}{5}-\frac {\left (\ln \relax (2)-4 x \right ) \ln \relax (2)}{16}\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(-32*x^2+4*x)*exp(ln(2)-4*x)^4+1/25*(-80*x^2+580*x-70)*exp(ln(2)-4*x)^2+x-7,x,method=_RETURNVERBOSE)

[Out]

32/25*exp(-16*x)*x^2+4/25*(10*x^2-70*x)*exp(-8*x)+1/2*x^2-7*x

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maxima [A]  time = 0.39, size = 31, normalized size = 1.48 \begin {gather*} \frac {32}{25} \, x^{2} e^{\left (-16 \, x\right )} + \frac {1}{2} \, x^{2} + \frac {8}{5} \, {\left (x^{2} - 7 \, x\right )} e^{\left (-8 \, x\right )} - 7 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-32*x^2+4*x)*exp(log(2)-4*x)^4+1/25*(-80*x^2+580*x-70)*exp(log(2)-4*x)^2+x-7,x, algorithm="max
ima")

[Out]

32/25*x^2*e^(-16*x) + 1/2*x^2 + 8/5*(x^2 - 7*x)*e^(-8*x) - 7*x

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mupad [B]  time = 0.07, size = 32, normalized size = 1.52 \begin {gather*} \frac {x\,{\mathrm {e}}^{-16\,x}\,\left (5\,{\mathrm {e}}^{8\,x}+8\right )\,\left (8\,x-70\,{\mathrm {e}}^{8\,x}+5\,x\,{\mathrm {e}}^{8\,x}\right )}{50} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x + (exp(4*log(2) - 16*x)*(4*x - 32*x^2))/25 - (exp(2*log(2) - 8*x)*(80*x^2 - 580*x + 70))/25 - 7,x)

[Out]

(x*exp(-16*x)*(5*exp(8*x) + 8)*(8*x - 70*exp(8*x) + 5*x*exp(8*x)))/50

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sympy [B]  time = 0.14, size = 34, normalized size = 1.62 \begin {gather*} \frac {x^{2}}{2} + \frac {32 x^{2} e^{- 16 x}}{25} - 7 x + \frac {\left (200 x^{2} - 1400 x\right ) e^{- 8 x}}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-32*x**2+4*x)*exp(ln(2)-4*x)**4+1/25*(-80*x**2+580*x-70)*exp(ln(2)-4*x)**2+x-7,x)

[Out]

x**2/2 + 32*x**2*exp(-16*x)/25 - 7*x + (200*x**2 - 1400*x)*exp(-8*x)/125

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