3.11.12 \(\int \frac {1}{4} (8-10 x-18 x^2-24 x^3-5 x^4+e^{2 x^2} (10 x+3 x^2+20 x^3+4 x^4)) \, dx\)

Optimal. Leaf size=29 \[ x \left (2+\frac {1}{4} x (5+x) \left (-1+e^{2 x^2}-x-x^2\right )\right ) \]

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Rubi [B]  time = 0.17, antiderivative size = 60, normalized size of antiderivative = 2.07, number of steps used = 12, number of rules used = 5, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {12, 2226, 2209, 2212, 2204} \begin {gather*} -\frac {x^5}{4}-\frac {3 x^4}{2}-\frac {3 x^3}{2}+\frac {5}{4} e^{2 x^2} x^2-\frac {5 x^2}{4}+\frac {1}{4} e^{2 x^2} x^3+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 - 10*x - 18*x^2 - 24*x^3 - 5*x^4 + E^(2*x^2)*(10*x + 3*x^2 + 20*x^3 + 4*x^4))/4,x]

[Out]

2*x - (5*x^2)/4 + (5*E^(2*x^2)*x^2)/4 - (3*x^3)/2 + (E^(2*x^2)*x^3)/4 - (3*x^4)/2 - x^5/4

Rule 12

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

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2209

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

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \left (8-10 x-18 x^2-24 x^3-5 x^4+e^{2 x^2} \left (10 x+3 x^2+20 x^3+4 x^4\right )\right ) \, dx\\ &=2 x-\frac {5 x^2}{4}-\frac {3 x^3}{2}-\frac {3 x^4}{2}-\frac {x^5}{4}+\frac {1}{4} \int e^{2 x^2} \left (10 x+3 x^2+20 x^3+4 x^4\right ) \, dx\\ &=2 x-\frac {5 x^2}{4}-\frac {3 x^3}{2}-\frac {3 x^4}{2}-\frac {x^5}{4}+\frac {1}{4} \int \left (10 e^{2 x^2} x+3 e^{2 x^2} x^2+20 e^{2 x^2} x^3+4 e^{2 x^2} x^4\right ) \, dx\\ &=2 x-\frac {5 x^2}{4}-\frac {3 x^3}{2}-\frac {3 x^4}{2}-\frac {x^5}{4}+\frac {3}{4} \int e^{2 x^2} x^2 \, dx+\frac {5}{2} \int e^{2 x^2} x \, dx+5 \int e^{2 x^2} x^3 \, dx+\int e^{2 x^2} x^4 \, dx\\ &=\frac {5 e^{2 x^2}}{8}+2 x+\frac {3}{16} e^{2 x^2} x-\frac {5 x^2}{4}+\frac {5}{4} e^{2 x^2} x^2-\frac {3 x^3}{2}+\frac {1}{4} e^{2 x^2} x^3-\frac {3 x^4}{2}-\frac {x^5}{4}-\frac {3}{16} \int e^{2 x^2} \, dx-\frac {3}{4} \int e^{2 x^2} x^2 \, dx-\frac {5}{2} \int e^{2 x^2} x \, dx\\ &=2 x-\frac {5 x^2}{4}+\frac {5}{4} e^{2 x^2} x^2-\frac {3 x^3}{2}+\frac {1}{4} e^{2 x^2} x^3-\frac {3 x^4}{2}-\frac {x^5}{4}-\frac {3}{32} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} x\right )+\frac {3}{16} \int e^{2 x^2} \, dx\\ &=2 x-\frac {5 x^2}{4}+\frac {5}{4} e^{2 x^2} x^2-\frac {3 x^3}{2}+\frac {1}{4} e^{2 x^2} x^3-\frac {3 x^4}{2}-\frac {x^5}{4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 51, normalized size = 1.76 \begin {gather*} \frac {1}{4} \left (8 x-5 x^2+5 e^{2 x^2} x^2-6 x^3+e^{2 x^2} x^3-6 x^4-x^5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 - 10*x - 18*x^2 - 24*x^3 - 5*x^4 + E^(2*x^2)*(10*x + 3*x^2 + 20*x^3 + 4*x^4))/4,x]

[Out]

(8*x - 5*x^2 + 5*E^(2*x^2)*x^2 - 6*x^3 + E^(2*x^2)*x^3 - 6*x^4 - x^5)/4

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fricas [A]  time = 0.91, size = 41, normalized size = 1.41 \begin {gather*} -\frac {1}{4} \, x^{5} - \frac {3}{2} \, x^{4} - \frac {3}{2} \, x^{3} - \frac {5}{4} \, x^{2} + \frac {1}{4} \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (2 \, x^{2}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*x^4+20*x^3+3*x^2+10*x)*exp(x^2)^2-5/4*x^4-6*x^3-9/2*x^2-5/2*x+2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.41, size = 41, normalized size = 1.41 \begin {gather*} -\frac {1}{4} \, x^{5} - \frac {3}{2} \, x^{4} - \frac {3}{2} \, x^{3} - \frac {5}{4} \, x^{2} + \frac {1}{4} \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (2 \, x^{2}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*x^4+20*x^3+3*x^2+10*x)*exp(x^2)^2-5/4*x^4-6*x^3-9/2*x^2-5/2*x+2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 42, normalized size = 1.45

method	result	size
risch	\(\frac {\left (x^{3}+5 x^{2}\right ) {\mathrm e}^{2 x^{2}}}{4}-\frac {x^{5}}{4}-\frac {3 x^{4}}{2}-\frac {3 x^{3}}{2}-\frac {5 x^{2}}{4}+2 x\)	\(42\)
default	\(2 x -\frac {5 x^{2}}{4}-\frac {3 x^{3}}{2}-\frac {3 x^{4}}{2}-\frac {x^{5}}{4}+\frac {5 x^{2} {\mathrm e}^{2 x^{2}}}{4}+\frac {x^{3} {\mathrm e}^{2 x^{2}}}{4}\)	\(47\)
norman	\(2 x -\frac {5 x^{2}}{4}-\frac {3 x^{3}}{2}-\frac {3 x^{4}}{2}-\frac {x^{5}}{4}+\frac {5 x^{2} {\mathrm e}^{2 x^{2}}}{4}+\frac {x^{3} {\mathrm e}^{2 x^{2}}}{4}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(4*x^4+20*x^3+3*x^2+10*x)*exp(x^2)^2-5/4*x^4-6*x^3-9/2*x^2-5/2*x+2,x,method=_RETURNVERBOSE)

[Out]

1/4*(x^3+5*x^2)*exp(2*x^2)-1/4*x^5-3/2*x^4-3/2*x^3-5/4*x^2+2*x

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maxima [A]  time = 0.51, size = 41, normalized size = 1.41 \begin {gather*} -\frac {1}{4} \, x^{5} - \frac {3}{2} \, x^{4} - \frac {3}{2} \, x^{3} - \frac {5}{4} \, x^{2} + \frac {1}{4} \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (2 \, x^{2}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*x^4+20*x^3+3*x^2+10*x)*exp(x^2)^2-5/4*x^4-6*x^3-9/2*x^2-5/2*x+2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.74, size = 41, normalized size = 1.41 \begin {gather*} -\frac {x\,\left (5\,x-5\,x\,{\mathrm {e}}^{2\,x^2}-x^2\,{\mathrm {e}}^{2\,x^2}+6\,x^2+6\,x^3+x^4-8\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x^2)*(10*x + 3*x^2 + 20*x^3 + 4*x^4))/4 - (5*x)/2 - (9*x^2)/2 - 6*x^3 - (5*x^4)/4 + 2,x)

[Out]

-(x*(5*x - 5*x*exp(2*x^2) - x^2*exp(2*x^2) + 6*x^2 + 6*x^3 + x^4 - 8))/4

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sympy [A]  time = 0.12, size = 44, normalized size = 1.52 \begin {gather*} - \frac {x^{5}}{4} - \frac {3 x^{4}}{2} - \frac {3 x^{3}}{2} - \frac {5 x^{2}}{4} + 2 x + \frac {\left (x^{3} + 5 x^{2}\right ) e^{2 x^{2}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*x**4+20*x**3+3*x**2+10*x)*exp(x**2)**2-5/4*x**4-6*x**3-9/2*x**2-5/2*x+2,x)

[Out]

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

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