∫ −48+3x4+6e2x2 x4+ 3√_e (−12x+3x3 )+ex2 (12x−27x3−6 3√_ex4 −6x5)________________________________________________

3.52.99 \(\int \frac {-48+3 x^4+6 e^{2 x^2} x^4+\sqrt [3]{e} (-12 x+3 x^3)+e^{x^2} (12 x-27 x^3-6 \sqrt [3]{e} x^4-6 x^5)}{8 x^3} \, dx\)

Optimal. Leaf size=26 \[ 3 \left (\frac {1}{x}+\frac {1}{4} \left (\sqrt [3]{e}-e^{x^2}+x\right )\right )^2 \]

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Rubi [B] time = 0.24, antiderivative size = 68, normalized size of antiderivative = 2.62, number of steps used = 13, number of rules used = 8, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.113, Rules used = {12, 14, 2209, 1590, 6742, 2204, 2214, 2212} \begin {gather*} \frac {3 \left (x^2+\sqrt [3]{e} x+4\right )^2}{16 x^2}+\frac {3 e^{2 x^2}}{16}-\frac {3}{8} e^{x^2+\frac {1}{3}}-\frac {3 e^{x^2} x}{8}-\frac {3 e^{x^2}}{2 x} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(-48 + 3*x^4 + 6*E^(2*x^2)*x^4 + E^(1/3)*(-12*x + 3*x^3) + E^x^2*(12*x - 27*x^3 - 6*E^(1/3)*x^4 - 6*x^5))/

(8*x^3),x]

[Out]

(3*E^(2*x^2))/16 - (3*E^(1/3 + x^2))/8 - (3*E^x^2)/(2*x) - (3*E^x^2*x)/8 + (3*(4 + E^(1/3)*x + x^2)^2)/(16*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]

&& !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1590

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S

imp[(Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*Rr^(n + 1))/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x

, r]), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp

, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n

}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

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 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m

+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), 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[-4, (m + 1)/n, 5] && IntegerQ[n

] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \frac {-48+3 x^4+6 e^{2 x^2} x^4+\sqrt [3]{e} \left (-12 x+3 x^3\right )+e^{x^2} \left (12 x-27 x^3-6 \sqrt [3]{e} x^4-6 x^5\right )}{x^3} \, dx\\ &=\frac {1}{8} \int \left (6 e^{2 x^2} x+\frac {3 \left (-4+x^2\right ) \left (4+\sqrt [3]{e} x+x^2\right )}{x^3}-\frac {3 e^{x^2} \left (-4+9 x^2+2 \sqrt [3]{e} x^3+2 x^4\right )}{x^2}\right ) \, dx\\ &=\frac {3}{8} \int \frac {\left (-4+x^2\right ) \left (4+\sqrt [3]{e} x+x^2\right )}{x^3} \, dx-\frac {3}{8} \int \frac {e^{x^2} \left (-4+9 x^2+2 \sqrt [3]{e} x^3+2 x^4\right )}{x^2} \, dx+\frac {3}{4} \int e^{2 x^2} x \, dx\\ &=\frac {3 e^{2 x^2}}{16}+\frac {3 \left (4+\sqrt [3]{e} x+x^2\right )^2}{16 x^2}-\frac {3}{8} \int \left (9 e^{x^2}-\frac {4 e^{x^2}}{x^2}+2 e^{\frac {1}{3}+x^2} x+2 e^{x^2} x^2\right ) \, dx\\ &=\frac {3 e^{2 x^2}}{16}+\frac {3 \left (4+\sqrt [3]{e} x+x^2\right )^2}{16 x^2}-\frac {3}{4} \int e^{\frac {1}{3}+x^2} x \, dx-\frac {3}{4} \int e^{x^2} x^2 \, dx+\frac {3}{2} \int \frac {e^{x^2}}{x^2} \, dx-\frac {27}{8} \int e^{x^2} \, dx\\ &=\frac {3 e^{2 x^2}}{16}-\frac {3}{8} e^{\frac {1}{3}+x^2}-\frac {3 e^{x^2}}{2 x}-\frac {3 e^{x^2} x}{8}+\frac {3 \left (4+\sqrt [3]{e} x+x^2\right )^2}{16 x^2}-\frac {27}{16} \sqrt {\pi } \text {erfi}(x)+\frac {3}{8} \int e^{x^2} \, dx+3 \int e^{x^2} \, dx\\ &=\frac {3 e^{2 x^2}}{16}-\frac {3}{8} e^{\frac {1}{3}+x^2}-\frac {3 e^{x^2}}{2 x}-\frac {3 e^{x^2} x}{8}+\frac {3 \left (4+\sqrt [3]{e} x+x^2\right )^2}{16 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [B] time = 0.14, size = 67, normalized size = 2.58 \begin {gather*} -\frac {3}{8} \left (-\frac {1}{2} e^{2 x^2}+e^{\frac {1}{3}+x^2}+e^{x^2} \left (\frac {4}{x}+x\right )-\frac {\left (-4+x^2\right )^2}{2 x^2}-\frac {\sqrt [3]{e} \left (4+x^2\right )}{x}\right ) \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(-48 + 3*x^4 + 6*E^(2*x^2)*x^4 + E^(1/3)*(-12*x + 3*x^3) + E^x^2*(12*x - 27*x^3 - 6*E^(1/3)*x^4 - 6*

x^5))/(8*x^3),x]

[Out]

(-3*(-1/2*E^(2*x^2) + E^(1/3 + x^2) + E^x^2*(4/x + x) - (-4 + x^2)^2/(2*x^2) - (E^(1/3)*(4 + x^2))/x))/8

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fricas [B] time = 0.57, size = 50, normalized size = 1.92 \begin {gather*} \frac {3 \, {\left (x^{4} + x^{2} e^{\left (2 \, x^{2}\right )} + 2 \, {\left (x^{3} + 4 \, x\right )} e^{\frac {1}{3}} - 2 \, {\left (x^{3} + x^{2} e^{\frac {1}{3}} + 4 \, x\right )} e^{\left (x^{2}\right )} + 16\right )}}{16 \, x^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(6*x^4*exp(x^2)^2+(-6*x^4*exp(1/3)-6*x^5-27*x^3+12*x)*exp(x^2)+(3*x^3-12*x)*exp(1/3)+3*x^4-48)/x

^3,x, algorithm="fricas")

[Out]

3/16*(x^4 + x^2*e^(2*x^2) + 2*(x^3 + 4*x)*e^(1/3) - 2*(x^3 + x^2*e^(1/3) + 4*x)*e^(x^2) + 16)/x^2

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giac [B] time = 0.19, size = 59, normalized size = 2.27 \begin {gather*} \frac {3 \, {\left (x^{4} + 2 \, x^{3} e^{\frac {1}{3}} - 2 \, x^{3} e^{\left (x^{2}\right )} + x^{2} e^{\left (2 \, x^{2}\right )} - 2 \, x^{2} e^{\left (x^{2} + \frac {1}{3}\right )} + 8 \, x e^{\frac {1}{3}} - 8 \, x e^{\left (x^{2}\right )} + 16\right )}}{16 \, x^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(6*x^4*exp(x^2)^2+(-6*x^4*exp(1/3)-6*x^5-27*x^3+12*x)*exp(x^2)+(3*x^3-12*x)*exp(1/3)+3*x^4-48)/x

^3,x, algorithm="giac")

[Out]

3/16*(x^4 + 2*x^3*e^(1/3) - 2*x^3*e^(x^2) + x^2*e^(2*x^2) - 2*x^2*e^(x^2 + 1/3) + 8*x*e^(1/3) - 8*x*e^(x^2) +

16)/x^2

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maple [B] time = 0.06, size = 50, normalized size = 1.92


method	result	size

risch	\(\frac {3 \,{\mathrm e}^{\frac {1}{3}} x}{8}+\frac {3 x^{2}}{16}+\frac {12 \,{\mathrm e}^{\frac {1}{3}} x +24}{8 x^{2}}+\frac {3 \,{\mathrm e}^{2 x^{2}}}{16}-\frac {3 \left ({\mathrm e}^{\frac {1}{3}} x +x^{2}+4\right ) {\mathrm e}^{x^{2}}}{8 x}\)	\(50\)
default	\(\frac {3 x^{2}}{16}+\frac {3}{x^{2}}+\frac {3 \,{\mathrm e}^{\frac {1}{3}}}{2 x}+\frac {3 \,{\mathrm e}^{2 x^{2}}}{16}-\frac {3 \,{\mathrm e}^{x^{2}} x}{8}-\frac {3 \,{\mathrm e}^{x^{2}}}{2 x}-\frac {3 \,{\mathrm e}^{\frac {1}{3}} {\mathrm e}^{x^{2}}}{8}+\frac {3 \,{\mathrm e}^{\frac {1}{3}} x}{8}\)	\(56\)
norman	\(\frac {3+\frac {3 x^{4}}{16}+\frac {3 x^{2} {\mathrm e}^{2 x^{2}}}{16}+\frac {3 x^{3} {\mathrm e}^{\frac {1}{3}}}{8}-\frac {3 x^{3} {\mathrm e}^{x^{2}}}{8}+\frac {3 \,{\mathrm e}^{\frac {1}{3}} x}{2}-\frac {3 \,{\mathrm e}^{x^{2}} x}{2}-\frac {3 x^{2} {\mathrm e}^{\frac {1}{3}} {\mathrm e}^{x^{2}}}{8}}{x^{2}}\)	\(62\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*(6*x^4*exp(x^2)^2+(-6*x^4*exp(1/3)-6*x^5-27*x^3+12*x)*exp(x^2)+(3*x^3-12*x)*exp(1/3)+3*x^4-48)/x^3,x,m

ethod=_RETURNVERBOSE)

[Out]

3/8*exp(1/3)*x+3/16*x^2+1/8*(12*exp(1/3)*x+24)/x^2+3/16*exp(2*x^2)-3/8*(exp(1/3)*x+x^2+4)/x*exp(x^2)

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maxima [C] time = 0.38, size = 74, normalized size = 2.85 \begin {gather*} \frac {3}{16} \, x^{2} + \frac {3}{8} \, x e^{\frac {1}{3}} - \frac {3}{8} \, x e^{\left (x^{2}\right )} + \frac {3}{2} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right ) - \frac {3 \, \sqrt {-x^{2}} \Gamma \left (-\frac {1}{2}, -x^{2}\right )}{4 \, x} + \frac {3 \, e^{\frac {1}{3}}}{2 \, x} + \frac {3}{x^{2}} + \frac {3}{16} \, e^{\left (2 \, x^{2}\right )} - \frac {3}{8} \, e^{\left (x^{2} + \frac {1}{3}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(6*x^4*exp(x^2)^2+(-6*x^4*exp(1/3)-6*x^5-27*x^3+12*x)*exp(x^2)+(3*x^3-12*x)*exp(1/3)+3*x^4-48)/x

^3,x, algorithm="maxima")

[Out]

3/16*x^2 + 3/8*x*e^(1/3) - 3/8*x*e^(x^2) + 3/2*I*sqrt(pi)*erf(I*x) - 3/4*sqrt(-x^2)*gamma(-1/2, -x^2)/x + 3/2*

e^(1/3)/x + 3/x^2 + 3/16*e^(2*x^2) - 3/8*e^(x^2 + 1/3)

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mupad [B] time = 3.27, size = 56, normalized size = 2.15 \begin {gather*} \frac {3\,{\mathrm {e}}^{2\,x^2}}{16}-\frac {3\,{\mathrm {e}}^{x^2+\frac {1}{3}}}{8}-\frac {x\,\left (\frac {3\,{\mathrm {e}}^{x^2}}{2}-\frac {3\,{\mathrm {e}}^{1/3}}{2}\right )-3}{x^2}-x\,\left (\frac {3\,{\mathrm {e}}^{x^2}}{8}-\frac {3\,{\mathrm {e}}^{1/3}}{8}\right )+\frac {3\,x^2}{16} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(1/3)*(12*x - 3*x^3))/8 - (3*x^4*exp(2*x^2))/4 - (3*x^4)/8 + (exp(x^2)*(6*x^4*exp(1/3) - 12*x + 27*x

^3 + 6*x^5))/8 + 6)/x^3,x)

[Out]

(3*exp(2*x^2))/16 - (3*exp(x^2 + 1/3))/8 - (x*((3*exp(x^2))/2 - (3*exp(1/3))/2) - 3)/x^2 - x*((3*exp(x^2))/8 -

(3*exp(1/3))/8) + (3*x^2)/16

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sympy [B] time = 0.20, size = 66, normalized size = 2.54 \begin {gather*} \frac {3 x^{2}}{16} + \frac {3 x e^{\frac {1}{3}}}{8} + \frac {24 x e^{2 x^{2}} + \left (- 48 x^{2} - 48 x e^{\frac {1}{3}} - 192\right ) e^{x^{2}}}{128 x} + \frac {12 x e^{\frac {1}{3}} + 24}{8 x^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(6*x**4*exp(x**2)**2+(-6*x**4*exp(1/3)-6*x**5-27*x**3+12*x)*exp(x**2)+(3*x**3-12*x)*exp(1/3)+3*x

**4-48)/x**3,x)

[Out]

3*x**2/16 + 3*x*exp(1/3)/8 + (24*x*exp(2*x**2) + (-48*x**2 - 48*x*exp(1/3) - 192)*exp(x**2))/(128*x) + (12*x*e

xp(1/3) + 24)/(8*x**2)

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