3.81.24 \(\int \frac {1}{8} (200 x+75 x^2+4 x^3+e^{2-2 x} (40 x-28 x^2-8 x^3)+(-40 x-12 x^2) \log (5)) \, dx\) [8024]

Optimal. Leaf size=28 \[ \frac {1}{2} x^2 (5+x) \left (5+e^{2-2 x}+\frac {x}{4}-\log (5)\right ) \]

[Out]

1/2*(5+x)*x^2*(1/4*x+exp(1-x)^2-ln(5)+5)

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(68\) vs. \(2(28)=56\).
time = 0.09, antiderivative size = 68, normalized size of antiderivative = 2.43, number of steps used = 15, number of rules used = 5, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {12, 1608, 2227, 2207, 2225} \begin {gather*} \frac {x^4}{8}+\frac {1}{2} e^{2-2 x} x^3+\frac {25 x^3}{8}-\frac {1}{2} x^3 \log (5)+\frac {5}{2} e^{2-2 x} x^2+\frac {25 x^2}{2}-\frac {5}{2} x^2 \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(200*x + 75*x^2 + 4*x^3 + E^(2 - 2*x)*(40*x - 28*x^2 - 8*x^3) + (-40*x - 12*x^2)*Log[5])/8,x]

[Out]

(25*x^2)/2 + (5*E^(2 - 2*x)*x^2)/2 + (25*x^3)/8 + (E^(2 - 2*x)*x^3)/2 + x^4/8 - (5*x^2*Log[5])/2 - (x^3*Log[5]
)/2

Rule 12

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

Rule 1608

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

Rule 2207

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] &&  !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 2227

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] &&  !TrueQ[$UseGamma]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \left (200 x+75 x^2+4 x^3+e^{2-2 x} \left (40 x-28 x^2-8 x^3\right )+\left (-40 x-12 x^2\right ) \log (5)\right ) \, dx\\ &=\frac {25 x^2}{2}+\frac {25 x^3}{8}+\frac {x^4}{8}+\frac {1}{8} \int e^{2-2 x} \left (40 x-28 x^2-8 x^3\right ) \, dx+\frac {1}{8} \log (5) \int \left (-40 x-12 x^2\right ) \, dx\\ &=\frac {25 x^2}{2}+\frac {25 x^3}{8}+\frac {x^4}{8}-\frac {5}{2} x^2 \log (5)-\frac {1}{2} x^3 \log (5)+\frac {1}{8} \int e^{2-2 x} x \left (40-28 x-8 x^2\right ) \, dx\\ &=\frac {25 x^2}{2}+\frac {25 x^3}{8}+\frac {x^4}{8}-\frac {5}{2} x^2 \log (5)-\frac {1}{2} x^3 \log (5)+\frac {1}{8} \int \left (40 e^{2-2 x} x-28 e^{2-2 x} x^2-8 e^{2-2 x} x^3\right ) \, dx\\ &=\frac {25 x^2}{2}+\frac {25 x^3}{8}+\frac {x^4}{8}-\frac {5}{2} x^2 \log (5)-\frac {1}{2} x^3 \log (5)-\frac {7}{2} \int e^{2-2 x} x^2 \, dx+5 \int e^{2-2 x} x \, dx-\int e^{2-2 x} x^3 \, dx\\ &=-\frac {5}{2} e^{2-2 x} x+\frac {25 x^2}{2}+\frac {7}{4} e^{2-2 x} x^2+\frac {25 x^3}{8}+\frac {1}{2} e^{2-2 x} x^3+\frac {x^4}{8}-\frac {5}{2} x^2 \log (5)-\frac {1}{2} x^3 \log (5)-\frac {3}{2} \int e^{2-2 x} x^2 \, dx+\frac {5}{2} \int e^{2-2 x} \, dx-\frac {7}{2} \int e^{2-2 x} x \, dx\\ &=-\frac {5}{4} e^{2-2 x}-\frac {3}{4} e^{2-2 x} x+\frac {25 x^2}{2}+\frac {5}{2} e^{2-2 x} x^2+\frac {25 x^3}{8}+\frac {1}{2} e^{2-2 x} x^3+\frac {x^4}{8}-\frac {5}{2} x^2 \log (5)-\frac {1}{2} x^3 \log (5)-\frac {3}{2} \int e^{2-2 x} x \, dx-\frac {7}{4} \int e^{2-2 x} \, dx\\ &=-\frac {3}{8} e^{2-2 x}+\frac {25 x^2}{2}+\frac {5}{2} e^{2-2 x} x^2+\frac {25 x^3}{8}+\frac {1}{2} e^{2-2 x} x^3+\frac {x^4}{8}-\frac {5}{2} x^2 \log (5)-\frac {1}{2} x^3 \log (5)-\frac {3}{4} \int e^{2-2 x} \, dx\\ &=\frac {25 x^2}{2}+\frac {5}{2} e^{2-2 x} x^2+\frac {25 x^3}{8}+\frac {1}{2} e^{2-2 x} x^3+\frac {x^4}{8}-\frac {5}{2} x^2 \log (5)-\frac {1}{2} x^3 \log (5)\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.10, size = 34, normalized size = 1.21 \begin {gather*} \frac {1}{8} e^{-2 x} x^2 (5+x) \left (4 e^2+e^{2 x} (20+x-4 \log (5))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(200*x + 75*x^2 + 4*x^3 + E^(2 - 2*x)*(40*x - 28*x^2 - 8*x^3) + (-40*x - 12*x^2)*Log[5])/8,x]

[Out]

(x^2*(5 + x)*(4*E^2 + E^(2*x)*(20 + x - 4*Log[5])))/(8*E^(2*x))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(89\) vs. \(2(25)=50\).
time = 0.79, size = 90, normalized size = 3.21

method	result	size
risch	\(\frac {\left (4 x^{3}+20 x^{2}\right ) {\mathrm e}^{-2 x +2}}{8}-\frac {x^{3} \ln \left (5\right )}{2}-\frac {5 x^{2} \ln \left (5\right )}{2}+\frac {x^{4}}{8}+\frac {25 x^{3}}{8}+\frac {25 x^{2}}{2}\)	\(50\)
norman	\(\left (-\frac {5 \ln \left (5\right )}{2}+\frac {25}{2}\right ) x^{2}+\left (-\frac {\ln \left (5\right )}{2}+\frac {25}{8}\right ) x^{3}+\frac {x^{4}}{8}+\frac {5 x^{2} {\mathrm e}^{-2 x +2}}{2}+\frac {x^{3} {\mathrm e}^{-2 x +2}}{2}\)	\(53\)
default	\(\frac {25 x^{2}}{2}+\frac {25 x^{3}}{8}+\frac {x^{4}}{8}-\frac {x^{3} \ln \left (5\right )}{2}-\frac {5 x^{2} \ln \left (5\right )}{2}+3 \,{\mathrm e}^{-2 x +2}+4 \,{\mathrm e}^{-2 x +2} \left (1-x \right )^{2}-\frac {13 \,{\mathrm e}^{-2 x +2} \left (1-x \right )}{2}-\frac {{\mathrm e}^{-2 x +2} \left (1-x \right )^{3}}{2}\)	\(90\)
derivativedivides	\(-25+25 x +\frac {25 \left (1-x \right )^{2}}{2}+\frac {25 x^{3}}{8}+\frac {x^{4}}{8}+\frac {\ln \left (5\right ) \left (1-x \right )^{3}}{2}-4 \ln \left (5\right ) \left (1-x \right )^{2}+\frac {13 \left (1-x \right ) \ln \left (5\right )}{2}+3 \,{\mathrm e}^{-2 x +2}-\frac {13 \,{\mathrm e}^{-2 x +2} \left (1-x \right )}{2}+4 \,{\mathrm e}^{-2 x +2} \left (1-x \right )^{2}-\frac {{\mathrm e}^{-2 x +2} \left (1-x \right )^{3}}{2}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*(-8*x^3-28*x^2+40*x)*exp(1-x)^2+1/8*(-12*x^2-40*x)*ln(5)+1/2*x^3+75/8*x^2+25*x,x,method=_RETURNVERBOSE
)

[Out]

25/2*x^2+25/8*x^3+1/8*x^4-1/2*x^3*ln(5)-5/2*x^2*ln(5)+3*exp(1-x)^2+4*exp(1-x)^2*(1-x)^2-13/2*exp(1-x)^2*(1-x)-
1/2*exp(1-x)^2*(1-x)^3

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (23) = 46\).
time = 0.29, size = 49, normalized size = 1.75 \begin {gather*} \frac {1}{8} \, x^{4} + \frac {25}{8} \, x^{3} + \frac {25}{2} \, x^{2} + \frac {1}{2} \, {\left (x^{3} e^{2} + 5 \, x^{2} e^{2}\right )} e^{\left (-2 \, x\right )} - \frac {1}{2} \, {\left (x^{3} + 5 \, x^{2}\right )} \log \left (5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-8*x^3-28*x^2+40*x)*exp(1-x)^2+1/8*(-12*x^2-40*x)*log(5)+1/2*x^3+75/8*x^2+25*x,x, algorithm="ma
xima")

[Out]

1/8*x^4 + 25/8*x^3 + 25/2*x^2 + 1/2*(x^3*e^2 + 5*x^2*e^2)*e^(-2*x) - 1/2*(x^3 + 5*x^2)*log(5)

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Fricas [A]
time = 0.39, size = 46, normalized size = 1.64 \begin {gather*} \frac {1}{8} \, x^{4} + \frac {25}{8} \, x^{3} + \frac {25}{2} \, x^{2} + \frac {1}{2} \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (-2 \, x + 2\right )} - \frac {1}{2} \, {\left (x^{3} + 5 \, x^{2}\right )} \log \left (5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-8*x^3-28*x^2+40*x)*exp(1-x)^2+1/8*(-12*x^2-40*x)*log(5)+1/2*x^3+75/8*x^2+25*x,x, algorithm="fr
icas")

[Out]

1/8*x^4 + 25/8*x^3 + 25/2*x^2 + 1/2*(x^3 + 5*x^2)*e^(-2*x + 2) - 1/2*(x^3 + 5*x^2)*log(5)

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Sympy [A]
time = 0.06, size = 46, normalized size = 1.64 \begin {gather*} \frac {x^{4}}{8} + x^{3} \cdot \left (\frac {25}{8} - \frac {\log {\left (5 \right )}}{2}\right ) + x^{2} \cdot \left (\frac {25}{2} - \frac {5 \log {\left (5 \right )}}{2}\right ) + \frac {\left (x^{3} + 5 x^{2}\right ) e^{2 - 2 x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-8*x**3-28*x**2+40*x)*exp(1-x)**2+1/8*(-12*x**2-40*x)*ln(5)+1/2*x**3+75/8*x**2+25*x,x)

[Out]

x**4/8 + x**3*(25/8 - log(5)/2) + x**2*(25/2 - 5*log(5)/2) + (x**3 + 5*x**2)*exp(2 - 2*x)/2

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Giac [A]
time = 0.41, size = 46, normalized size = 1.64 \begin {gather*} \frac {1}{8} \, x^{4} + \frac {25}{8} \, x^{3} + \frac {25}{2} \, x^{2} + \frac {1}{2} \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (-2 \, x + 2\right )} - \frac {1}{2} \, {\left (x^{3} + 5 \, x^{2}\right )} \log \left (5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-8*x^3-28*x^2+40*x)*exp(1-x)^2+1/8*(-12*x^2-40*x)*log(5)+1/2*x^3+75/8*x^2+25*x,x, algorithm="gi
ac")

[Out]

1/8*x^4 + 25/8*x^3 + 25/2*x^2 + 1/2*(x^3 + 5*x^2)*e^(-2*x + 2) - 1/2*(x^3 + 5*x^2)*log(5)

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Mupad [B]
time = 5.67, size = 50, normalized size = 1.79 \begin {gather*} \frac {5\,x^2\,{\mathrm {e}}^{2-2\,x}}{2}-x^3\,\left (\frac {\ln \left (5\right )}{2}-\frac {25}{8}\right )-x^2\,\left (\frac {5\,\ln \left (5\right )}{2}-\frac {25}{2}\right )+\frac {x^3\,{\mathrm {e}}^{2-2\,x}}{2}+\frac {x^4}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(25*x - (log(5)*(40*x + 12*x^2))/8 - (exp(2 - 2*x)*(28*x^2 - 40*x + 8*x^3))/8 + (75*x^2)/8 + x^3/2,x)

[Out]

(5*x^2*exp(2 - 2*x))/2 - x^3*(log(5)/2 - 25/8) - x^2*((5*log(5))/2 - 25/2) + (x^3*exp(2 - 2*x))/2 + x^4/8

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