3.1.48 \(\int \frac {15-15 e^4-20 e x+e^{x^2} (-3-6 x^2+e^4 (3+6 x^2)+e (4 x+4 x^3))}{e} \, dx\)

Optimal. Leaf size=25 \[ \left (5-e^{x^2}\right ) x \left (x-3 \left (-\frac {1}{e}+e^3+x\right )\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 46, normalized size of antiderivative = 1.84, number of steps used = 10, number of rules used = 5, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 2226, 2204, 2209, 2212} \begin {gather*} 2 e^{x^2} x^2-10 x^2-3 \left (1-e^4\right ) e^{x^2-1} x+\frac {15 \left (1-e^4\right ) x}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 - 15*E^4 - 20*E*x + E^x^2*(-3 - 6*x^2 + E^4*(3 + 6*x^2) + E*(4*x + 4*x^3)))/E,x]

[Out]

(15*(1 - E^4)*x)/E - 3*E^(-1 + x^2)*(1 - E^4)*x - 10*x^2 + 2*E^x^2*x^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 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 {\int \left (15-15 e^4-20 e x+e^{x^2} \left (-3-6 x^2+e^4 \left (3+6 x^2\right )+e \left (4 x+4 x^3\right )\right )\right ) \, dx}{e}\\ &=\frac {15 \left (1-e^4\right ) x}{e}-10 x^2+\frac {\int e^{x^2} \left (-3-6 x^2+e^4 \left (3+6 x^2\right )+e \left (4 x+4 x^3\right )\right ) \, dx}{e}\\ &=\frac {15 \left (1-e^4\right ) x}{e}-10 x^2+\frac {\int \left (3 e^{x^2} \left (-1+e^4\right )+4 e^{1+x^2} x+6 e^{x^2} \left (-1+e^4\right ) x^2+4 e^{1+x^2} x^3\right ) \, dx}{e}\\ &=\frac {15 \left (1-e^4\right ) x}{e}-10 x^2+\frac {4 \int e^{1+x^2} x \, dx}{e}+\frac {4 \int e^{1+x^2} x^3 \, dx}{e}-\frac {\left (3 \left (1-e^4\right )\right ) \int e^{x^2} \, dx}{e}-\frac {\left (6 \left (1-e^4\right )\right ) \int e^{x^2} x^2 \, dx}{e}\\ &=2 e^{x^2}+\frac {15 \left (1-e^4\right ) x}{e}-3 e^{-1+x^2} \left (1-e^4\right ) x-10 x^2+2 e^{x^2} x^2-\frac {3 \left (1-e^4\right ) \sqrt {\pi } \text {erfi}(x)}{2 e}-\frac {4 \int e^{1+x^2} x \, dx}{e}+\frac {\left (3 \left (1-e^4\right )\right ) \int e^{x^2} \, dx}{e}\\ &=\frac {15 \left (1-e^4\right ) x}{e}-3 e^{-1+x^2} \left (1-e^4\right ) x-10 x^2+2 e^{x^2} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 23, normalized size = 0.92 \begin {gather*} \frac {\left (-5+e^{x^2}\right ) x \left (-3+3 e^4+2 e x\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 - 15*E^4 - 20*E*x + E^x^2*(-3 - 6*x^2 + E^4*(3 + 6*x^2) + E*(4*x + 4*x^3)))/E,x]

[Out]

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

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fricas [B]  time = 0.59, size = 42, normalized size = 1.68 \begin {gather*} -{\left (10 \, x^{2} e + 15 \, x e^{4} - {\left (2 \, x^{2} e + 3 \, x e^{4} - 3 \, x\right )} e^{\left (x^{2}\right )} - 15 \, x\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^2+3)*exp(1)*exp(3)+(4*x^3+4*x)*exp(1)-6*x^2-3)*exp(x^2)-15*exp(1)*exp(3)-20*x*exp(1)+15)/exp(
1),x, algorithm="fricas")

[Out]

-(10*x^2*e + 15*x*e^4 - (2*x^2*e + 3*x*e^4 - 3*x)*e^(x^2) - 15*x)*e^(-1)

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giac [B]  time = 0.25, size = 47, normalized size = 1.88 \begin {gather*} -{\left (10 \, x^{2} e - 2 \, x^{2} e^{\left (x^{2} + 1\right )} + 15 \, x e^{4} - 3 \, x e^{\left (x^{2} + 4\right )} + 3 \, x e^{\left (x^{2}\right )} - 15 \, x\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^2+3)*exp(1)*exp(3)+(4*x^3+4*x)*exp(1)-6*x^2-3)*exp(x^2)-15*exp(1)*exp(3)-20*x*exp(1)+15)/exp(
1),x, algorithm="giac")

[Out]

-(10*x^2*e - 2*x^2*e^(x^2 + 1) + 15*x*e^4 - 3*x*e^(x^2 + 4) + 3*x*e^(x^2) - 15*x)*e^(-1)

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maple [A]  time = 0.03, size = 42, normalized size = 1.68

method	result	size
risch	\(-15 x \,{\mathrm e}^{3}-10 x^{2}+15 \,{\mathrm e}^{-1} x +\left (3 x \,{\mathrm e}^{4}+2 x^{2} {\mathrm e}-3 x \right ) {\mathrm e}^{\left (x -1\right ) \left (x +1\right )}\)	\(42\)
norman	\(-10 x^{2}+2 x^{2} {\mathrm e}^{x^{2}}-15 \left ({\mathrm e} \,{\mathrm e}^{3}-1\right ) {\mathrm e}^{-1} x +3 \left ({\mathrm e} \,{\mathrm e}^{3}-1\right ) {\mathrm e}^{-1} x \,{\mathrm e}^{x^{2}}\)	\(48\)
default	\({\mathrm e}^{-1} \left (15 x -3 \,{\mathrm e}^{x^{2}} x +2 \,{\mathrm e} \,{\mathrm e}^{x^{2}}+4 \,{\mathrm e} \left (\frac {x^{2} {\mathrm e}^{x^{2}}}{2}-\frac {{\mathrm e}^{x^{2}}}{2}\right )+\frac {3 \,{\mathrm e} \,{\mathrm e}^{3} \sqrt {\pi }\, \erfi \relax (x )}{2}+6 \,{\mathrm e} \,{\mathrm e}^{3} \left (\frac {{\mathrm e}^{x^{2}} x}{2}-\frac {\sqrt {\pi }\, \erfi \relax (x )}{4}\right )-10 x^{2} {\mathrm e}-15 x \,{\mathrm e} \,{\mathrm e}^{3}\right )\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((6*x^2+3)*exp(1)*exp(3)+(4*x^3+4*x)*exp(1)-6*x^2-3)*exp(x^2)-15*exp(1)*exp(3)-20*x*exp(1)+15)/exp(1),x,m
ethod=_RETURNVERBOSE)

[Out]

-15*x*exp(3)-10*x^2+15*exp(-1)*x+(3*x*exp(4)+2*x^2*exp(1)-3*x)*exp((x-1)*(x+1))

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maxima [B]  time = 0.59, size = 41, normalized size = 1.64 \begin {gather*} -{\left (10 \, x^{2} e + 15 \, x e^{4} - {\left (2 \, x^{2} e + 3 \, x {\left (e^{4} - 1\right )}\right )} e^{\left (x^{2}\right )} - 15 \, x\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^2+3)*exp(1)*exp(3)+(4*x^3+4*x)*exp(1)-6*x^2-3)*exp(x^2)-15*exp(1)*exp(3)-20*x*exp(1)+15)/exp(
1),x, algorithm="maxima")

[Out]

-(10*x^2*e + 15*x*e^4 - (2*x^2*e + 3*x*(e^4 - 1))*e^(x^2) - 15*x)*e^(-1)

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mupad [B]  time = 0.10, size = 21, normalized size = 0.84 \begin {gather*} x\,{\mathrm {e}}^{-1}\,\left ({\mathrm {e}}^{x^2}-5\right )\,\left (3\,{\mathrm {e}}^4+2\,x\,\mathrm {e}-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-1)*(15*exp(4) + 20*x*exp(1) - exp(x^2)*(exp(1)*(4*x + 4*x^3) + exp(4)*(6*x^2 + 3) - 6*x^2 - 3) - 15)
,x)

[Out]

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

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sympy [B]  time = 0.14, size = 42, normalized size = 1.68 \begin {gather*} - 10 x^{2} + \frac {x \left (15 - 15 e^{4}\right )}{e} + \frac {\left (2 e x^{2} - 3 x + 3 x e^{4}\right ) e^{x^{2}}}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x**2+3)*exp(1)*exp(3)+(4*x**3+4*x)*exp(1)-6*x**2-3)*exp(x**2)-15*exp(1)*exp(3)-20*x*exp(1)+15)/
exp(1),x)

[Out]

-10*x**2 + x*(15 - 15*exp(4))*exp(-1) + (2*E*x**2 - 3*x + 3*x*exp(4))*exp(-1)*exp(x**2)

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