3.29.82 \(\int \frac {1}{25} (240+288 x+48 x^2+e^{x^2} (32+64 x+64 x^2+32 x^3)) \, dx\)

Optimal. Leaf size=19 \[ \frac {1}{25} \left (7+e^{x^2}+x\right ) (4+4 x)^2 \]

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Rubi [B]  time = 0.08, antiderivative size = 51, normalized size of antiderivative = 2.68, number of steps used = 10, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2226, 2204, 2209, 2212} \begin {gather*} \frac {16 x^3}{25}+\frac {16}{25} e^{x^2} x^2+\frac {144 x^2}{25}+\frac {32 e^{x^2} x}{25}+\frac {16 e^{x^2}}{25}+\frac {48 x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(240 + 288*x + 48*x^2 + E^x^2*(32 + 64*x + 64*x^2 + 32*x^3))/25,x]

[Out]

(16*E^x^2)/25 + (48*x)/5 + (32*E^x^2*x)/25 + (144*x^2)/25 + (16*E^x^2*x^2)/25 + (16*x^3)/25

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}{25} \int \left (240+288 x+48 x^2+e^{x^2} \left (32+64 x+64 x^2+32 x^3\right )\right ) \, dx\\ &=\frac {48 x}{5}+\frac {144 x^2}{25}+\frac {16 x^3}{25}+\frac {1}{25} \int e^{x^2} \left (32+64 x+64 x^2+32 x^3\right ) \, dx\\ &=\frac {48 x}{5}+\frac {144 x^2}{25}+\frac {16 x^3}{25}+\frac {1}{25} \int \left (32 e^{x^2}+64 e^{x^2} x+64 e^{x^2} x^2+32 e^{x^2} x^3\right ) \, dx\\ &=\frac {48 x}{5}+\frac {144 x^2}{25}+\frac {16 x^3}{25}+\frac {32}{25} \int e^{x^2} \, dx+\frac {32}{25} \int e^{x^2} x^3 \, dx+\frac {64}{25} \int e^{x^2} x \, dx+\frac {64}{25} \int e^{x^2} x^2 \, dx\\ &=\frac {32 e^{x^2}}{25}+\frac {48 x}{5}+\frac {32 e^{x^2} x}{25}+\frac {144 x^2}{25}+\frac {16}{25} e^{x^2} x^2+\frac {16 x^3}{25}+\frac {16}{25} \sqrt {\pi } \text {erfi}(x)-\frac {32}{25} \int e^{x^2} \, dx-\frac {32}{25} \int e^{x^2} x \, dx\\ &=\frac {16 e^{x^2}}{25}+\frac {48 x}{5}+\frac {32 e^{x^2} x}{25}+\frac {144 x^2}{25}+\frac {16}{25} e^{x^2} x^2+\frac {16 x^3}{25}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 26, normalized size = 1.37 \begin {gather*} \frac {16}{25} \left (e^{x^2} (1+x)^2+x \left (15+9 x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(240 + 288*x + 48*x^2 + E^x^2*(32 + 64*x + 64*x^2 + 32*x^3))/25,x]

[Out]

(16*(E^x^2*(1 + x)^2 + x*(15 + 9*x + x^2)))/25

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fricas [A]  time = 1.05, size = 28, normalized size = 1.47 \begin {gather*} \frac {16}{25} \, x^{3} + \frac {144}{25} \, x^{2} + \frac {16}{25} \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (x^{2}\right )} + \frac {48}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(32*x^3+64*x^2+64*x+32)*exp(x^2)+48/25*x^2+288/25*x+48/5,x, algorithm="fricas")

[Out]

16/25*x^3 + 144/25*x^2 + 16/25*(x^2 + 2*x + 1)*e^(x^2) + 48/5*x

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giac [A]  time = 0.18, size = 28, normalized size = 1.47 \begin {gather*} \frac {16}{25} \, x^{3} + \frac {144}{25} \, x^{2} + \frac {16}{25} \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (x^{2}\right )} + \frac {48}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(32*x^3+64*x^2+64*x+32)*exp(x^2)+48/25*x^2+288/25*x+48/5,x, algorithm="giac")

[Out]

16/25*x^3 + 144/25*x^2 + 16/25*(x^2 + 2*x + 1)*e^(x^2) + 48/5*x

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maple [A]  time = 0.02, size = 31, normalized size = 1.63

method	result	size
risch	\(\frac {\left (16 x^{2}+32 x +16\right ) {\mathrm e}^{x^{2}}}{25}+\frac {16 x^{3}}{25}+\frac {144 x^{2}}{25}+\frac {48 x}{5}\)	\(31\)
default	\(\frac {48 x}{5}+\frac {144 x^{2}}{25}+\frac {16 x^{3}}{25}+\frac {16 x^{2} {\mathrm e}^{x^{2}}}{25}+\frac {16 \,{\mathrm e}^{x^{2}}}{25}+\frac {32 \,{\mathrm e}^{x^{2}} x}{25}\)	\(37\)
norman	\(\frac {48 x}{5}+\frac {144 x^{2}}{25}+\frac {16 x^{3}}{25}+\frac {16 x^{2} {\mathrm e}^{x^{2}}}{25}+\frac {16 \,{\mathrm e}^{x^{2}}}{25}+\frac {32 \,{\mathrm e}^{x^{2}} x}{25}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(32*x^3+64*x^2+64*x+32)*exp(x^2)+48/25*x^2+288/25*x+48/5,x,method=_RETURNVERBOSE)

[Out]

1/25*(16*x^2+32*x+16)*exp(x^2)+16/25*x^3+144/25*x^2+48/5*x

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maxima [A]  time = 0.36, size = 28, normalized size = 1.47 \begin {gather*} \frac {16}{25} \, x^{3} + \frac {144}{25} \, x^{2} + \frac {16}{25} \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (x^{2}\right )} + \frac {48}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(32*x^3+64*x^2+64*x+32)*exp(x^2)+48/25*x^2+288/25*x+48/5,x, algorithm="maxima")

[Out]

16/25*x^3 + 144/25*x^2 + 16/25*(x^2 + 2*x + 1)*e^(x^2) + 48/5*x

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mupad [B]  time = 0.07, size = 36, normalized size = 1.89 \begin {gather*} \frac {48\,x}{5}+\frac {16\,{\mathrm {e}}^{x^2}}{25}+\frac {32\,x\,{\mathrm {e}}^{x^2}}{25}+\frac {16\,x^2\,{\mathrm {e}}^{x^2}}{25}+\frac {144\,x^2}{25}+\frac {16\,x^3}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((288*x)/25 + (exp(x^2)*(64*x + 64*x^2 + 32*x^3 + 32))/25 + (48*x^2)/25 + 48/5,x)

[Out]

(48*x)/5 + (16*exp(x^2))/25 + (32*x*exp(x^2))/25 + (16*x^2*exp(x^2))/25 + (144*x^2)/25 + (16*x^3)/25

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sympy [A]  time = 0.11, size = 34, normalized size = 1.79 \begin {gather*} \frac {16 x^{3}}{25} + \frac {144 x^{2}}{25} + \frac {48 x}{5} + \frac {\left (16 x^{2} + 32 x + 16\right ) e^{x^{2}}}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(32*x**3+64*x**2+64*x+32)*exp(x**2)+48/25*x**2+288/25*x+48/5,x)

[Out]

16*x**3/25 + 144*x**2/25 + 48*x/5 + (16*x**2 + 32*x + 16)*exp(x**2)/25

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