3.71.71 \(\int \frac {-18+e^{x^2} (10 x^4-3 x^5+10 x^6-2 x^7+e^x (-5 x^3-3 x^4-9 x^5+2 x^6))}{3 x^3} \, dx\)

Optimal. Leaf size=32 \[ \frac {3}{x^2}+\frac {1}{3} \left (1-e^{x^2} (5-x) \left (e^x-x\right ) x\right ) \]

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Rubi [A]  time = 1.09, antiderivative size = 56, normalized size of antiderivative = 1.75, number of steps used = 34, number of rules used = 9, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {12, 14, 6742, 2234, 2204, 2209, 2240, 2212, 2241} \begin {gather*} \frac {5}{3} e^{x^2} x^2+\frac {1}{3} e^{x^2+x} x^2-\frac {5}{3} e^{x^2+x} x+\frac {3}{x^2}-\frac {1}{3} e^{x^2} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-18 + E^x^2*(10*x^4 - 3*x^5 + 10*x^6 - 2*x^7 + E^x*(-5*x^3 - 3*x^4 - 9*x^5 + 2*x^6)))/(3*x^3),x]

[Out]

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

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 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 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2240

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] - Dist[(b*e - 2*c*d)/(2*c), Int[F^(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
NeQ[b*e - 2*c*d, 0]

Rule 2241

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*F^(a + b*x + c*x^2))/(2*c*Log[F]), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*F^(a + b*x + c*x^2)
, x], x] - Dist[((m - 1)*e^2)/(2*c*Log[F]), Int[(d + e*x)^(m - 2)*F^(a + b*x + c*x^2), x], x]) /; FreeQ[{F, a,
b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && GtQ[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}{3} \int \frac {-18+e^{x^2} \left (10 x^4-3 x^5+10 x^6-2 x^7+e^x \left (-5 x^3-3 x^4-9 x^5+2 x^6\right )\right )}{x^3} \, dx\\ &=\frac {1}{3} \int \left (-\frac {18}{x^3}+e^{x^2} \left (-5 e^x+10 x-3 e^x x-3 x^2-9 e^x x^2+10 x^3+2 e^x x^3-2 x^4\right )\right ) \, dx\\ &=\frac {3}{x^2}+\frac {1}{3} \int e^{x^2} \left (-5 e^x+10 x-3 e^x x-3 x^2-9 e^x x^2+10 x^3+2 e^x x^3-2 x^4\right ) \, dx\\ &=\frac {3}{x^2}+\frac {1}{3} \int \left (-5 e^{x+x^2}+10 e^{x^2} x-3 e^{x+x^2} x-3 e^{x^2} x^2-9 e^{x+x^2} x^2+10 e^{x^2} x^3+2 e^{x+x^2} x^3-2 e^{x^2} x^4\right ) \, dx\\ &=\frac {3}{x^2}+\frac {2}{3} \int e^{x+x^2} x^3 \, dx-\frac {2}{3} \int e^{x^2} x^4 \, dx-\frac {5}{3} \int e^{x+x^2} \, dx-3 \int e^{x+x^2} x^2 \, dx+\frac {10}{3} \int e^{x^2} x \, dx+\frac {10}{3} \int e^{x^2} x^3 \, dx-\int e^{x+x^2} x \, dx-\int e^{x^2} x^2 \, dx\\ &=\frac {5 e^{x^2}}{3}-\frac {e^{x+x^2}}{2}+\frac {3}{x^2}-\frac {e^{x^2} x}{2}-\frac {3}{2} e^{x+x^2} x+\frac {5}{3} e^{x^2} x^2+\frac {1}{3} e^{x+x^2} x^2-\frac {1}{3} e^{x^2} x^3-\frac {1}{3} \int e^{x+x^2} x^2 \, dx+\frac {1}{2} \int e^{x^2} \, dx+\frac {1}{2} \int e^{x+x^2} \, dx-\frac {2}{3} \int e^{x+x^2} x \, dx+\frac {3}{2} \int e^{x+x^2} \, dx+\frac {3}{2} \int e^{x+x^2} x \, dx-\frac {10}{3} \int e^{x^2} x \, dx-\frac {5 \int e^{\frac {1}{4} (1+2 x)^2} \, dx}{3 \sqrt [4]{e}}+\int e^{x^2} x^2 \, dx\\ &=-\frac {1}{12} e^{x+x^2}+\frac {3}{x^2}-\frac {5}{3} e^{x+x^2} x+\frac {5}{3} e^{x^2} x^2+\frac {1}{3} e^{x+x^2} x^2-\frac {1}{3} e^{x^2} x^3+\frac {1}{4} \sqrt {\pi } \text {erfi}(x)-\frac {5 \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (1+2 x)\right )}{6 \sqrt [4]{e}}+\frac {1}{6} \int e^{x+x^2} \, dx+\frac {1}{6} \int e^{x+x^2} x \, dx+\frac {1}{3} \int e^{x+x^2} \, dx-\frac {1}{2} \int e^{x^2} \, dx-\frac {3}{4} \int e^{x+x^2} \, dx+\frac {\int e^{\frac {1}{4} (1+2 x)^2} \, dx}{2 \sqrt [4]{e}}+\frac {3 \int e^{\frac {1}{4} (1+2 x)^2} \, dx}{2 \sqrt [4]{e}}\\ &=\frac {3}{x^2}-\frac {5}{3} e^{x+x^2} x+\frac {5}{3} e^{x^2} x^2+\frac {1}{3} e^{x+x^2} x^2-\frac {1}{3} e^{x^2} x^3+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} (1+2 x)\right )}{6 \sqrt [4]{e}}-\frac {1}{12} \int e^{x+x^2} \, dx+\frac {\int e^{\frac {1}{4} (1+2 x)^2} \, dx}{6 \sqrt [4]{e}}+\frac {\int e^{\frac {1}{4} (1+2 x)^2} \, dx}{3 \sqrt [4]{e}}-\frac {3 \int e^{\frac {1}{4} (1+2 x)^2} \, dx}{4 \sqrt [4]{e}}\\ &=\frac {3}{x^2}-\frac {5}{3} e^{x+x^2} x+\frac {5}{3} e^{x^2} x^2+\frac {1}{3} e^{x+x^2} x^2-\frac {1}{3} e^{x^2} x^3+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} (1+2 x)\right )}{24 \sqrt [4]{e}}-\frac {\int e^{\frac {1}{4} (1+2 x)^2} \, dx}{12 \sqrt [4]{e}}\\ &=\frac {3}{x^2}-\frac {5}{3} e^{x+x^2} x+\frac {5}{3} e^{x^2} x^2+\frac {1}{3} e^{x+x^2} x^2-\frac {1}{3} e^{x^2} x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 37, normalized size = 1.16 \begin {gather*} \frac {3}{x^2}+\frac {1}{3} e^{x^2} \left (5 x^2-x^3+e^x \left (-5 x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-18 + E^x^2*(10*x^4 - 3*x^5 + 10*x^6 - 2*x^7 + E^x*(-5*x^3 - 3*x^4 - 9*x^5 + 2*x^6)))/(3*x^3),x]

[Out]

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

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fricas [A]  time = 0.85, size = 34, normalized size = 1.06 \begin {gather*} -\frac {{\left (x^{5} - 5 \, x^{4} - {\left (x^{4} - 5 \, x^{3}\right )} e^{x}\right )} e^{\left (x^{2}\right )} - 9}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((2*x^6-9*x^5-3*x^4-5*x^3)*exp(x)-2*x^7+10*x^6-3*x^5+10*x^4)*exp(x^2)-18)/x^3,x, algorithm="fri
cas")

[Out]

-1/3*((x^5 - 5*x^4 - (x^4 - 5*x^3)*e^x)*e^(x^2) - 9)/x^2

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giac [A]  time = 0.23, size = 46, normalized size = 1.44 \begin {gather*} -\frac {x^{5} e^{\left (x^{2}\right )} - x^{4} e^{\left (x^{2} + x\right )} - 5 \, x^{4} e^{\left (x^{2}\right )} + 5 \, x^{3} e^{\left (x^{2} + x\right )} - 9}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((2*x^6-9*x^5-3*x^4-5*x^3)*exp(x)-2*x^7+10*x^6-3*x^5+10*x^4)*exp(x^2)-18)/x^3,x, algorithm="gia
c")

[Out]

-1/3*(x^5*e^(x^2) - x^4*e^(x^2 + x) - 5*x^4*e^(x^2) + 5*x^3*e^(x^2 + x) - 9)/x^2

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maple [A]  time = 0.04, size = 35, normalized size = 1.09

method	result	size
risch	\(\frac {3}{x^{2}}+\frac {\left (-x^{3}+{\mathrm e}^{x} x^{2}+5 x^{2}-5 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{x^{2}}}{3}\)	\(35\)
default	\(\frac {3}{x^{2}}+\frac {5 x^{2} {\mathrm e}^{x^{2}}}{3}-\frac {x^{3} {\mathrm e}^{x^{2}}}{3}-\frac {5 x \,{\mathrm e}^{x^{2}+x}}{3}+\frac {x^{2} {\mathrm e}^{x^{2}+x}}{3}\)	\(45\)
norman	\(\frac {3+\frac {5 x^{4} {\mathrm e}^{x^{2}}}{3}-\frac {{\mathrm e}^{x^{2}} x^{5}}{3}-\frac {5 \,{\mathrm e}^{x} {\mathrm e}^{x^{2}} x^{3}}{3}+\frac {{\mathrm e}^{x} {\mathrm e}^{x^{2}} x^{4}}{3}}{x^{2}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(((2*x^6-9*x^5-3*x^4-5*x^3)*exp(x)-2*x^7+10*x^6-3*x^5+10*x^4)*exp(x^2)-18)/x^3,x,method=_RETURNVERBOSE
)

[Out]

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

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maxima [C]  time = 0.45, size = 294, normalized size = 9.19 \begin {gather*} \frac {5}{6} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x + \frac {1}{2} i\right ) e^{\left (-\frac {1}{4}\right )} + \frac {1}{24} \, {\left (\frac {12 \, {\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{\left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} - \frac {\sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} + 6 \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} - 8 \, \Gamma \left (2, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )\right )} e^{\left (-\frac {1}{4}\right )} + \frac {3}{8} \, {\left (\frac {4 \, {\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{\left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} - \frac {\sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} + 4 \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}\right )} e^{\left (-\frac {1}{4}\right )} + \frac {1}{4} \, {\left (\frac {\sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} - 2 \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}\right )} e^{\left (-\frac {1}{4}\right )} - \frac {1}{6} \, {\left (2 \, x^{3} - 3 \, x\right )} e^{\left (x^{2}\right )} + \frac {5}{3} \, {\left (x^{2} - 1\right )} e^{\left (x^{2}\right )} - \frac {1}{2} \, x e^{\left (x^{2}\right )} + \frac {3}{x^{2}} + \frac {5}{3} \, e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((2*x^6-9*x^5-3*x^4-5*x^3)*exp(x)-2*x^7+10*x^6-3*x^5+10*x^4)*exp(x^2)-18)/x^3,x, algorithm="max
ima")

[Out]

5/6*I*sqrt(pi)*erf(I*x + 1/2*I)*e^(-1/4) + 1/24*(12*(2*x + 1)^3*gamma(3/2, -1/4*(2*x + 1)^2)/(-(2*x + 1)^2)^(3
/2) - sqrt(pi)*(2*x + 1)*(erf(1/2*sqrt(-(2*x + 1)^2)) - 1)/sqrt(-(2*x + 1)^2) + 6*e^(1/4*(2*x + 1)^2) - 8*gamm
a(2, -1/4*(2*x + 1)^2))*e^(-1/4) + 3/8*(4*(2*x + 1)^3*gamma(3/2, -1/4*(2*x + 1)^2)/(-(2*x + 1)^2)^(3/2) - sqrt
(pi)*(2*x + 1)*(erf(1/2*sqrt(-(2*x + 1)^2)) - 1)/sqrt(-(2*x + 1)^2) + 4*e^(1/4*(2*x + 1)^2))*e^(-1/4) + 1/4*(s
qrt(pi)*(2*x + 1)*(erf(1/2*sqrt(-(2*x + 1)^2)) - 1)/sqrt(-(2*x + 1)^2) - 2*e^(1/4*(2*x + 1)^2))*e^(-1/4) - 1/6
*(2*x^3 - 3*x)*e^(x^2) + 5/3*(x^2 - 1)*e^(x^2) - 1/2*x*e^(x^2) + 3/x^2 + 5/3*e^(x^2)

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mupad [B]  time = 4.22, size = 34, normalized size = 1.06 \begin {gather*} {\mathrm {e}}^{x^2}\,\left (\frac {x^2\,{\mathrm {e}}^x}{3}-\frac {5\,x\,{\mathrm {e}}^x}{3}+\frac {5\,x^2}{3}-\frac {x^3}{3}\right )+\frac {3}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(x^2)*(exp(x)*(5*x^3 + 3*x^4 + 9*x^5 - 2*x^6) - 10*x^4 + 3*x^5 - 10*x^6 + 2*x^7))/3 + 6)/x^3,x)

[Out]

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

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sympy [A]  time = 0.24, size = 32, normalized size = 1.00 \begin {gather*} \frac {\left (- x^{3} + x^{2} e^{x} + 5 x^{2} - 5 x e^{x}\right ) e^{x^{2}}}{3} + \frac {3}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((2*x**6-9*x**5-3*x**4-5*x**3)*exp(x)-2*x**7+10*x**6-3*x**5+10*x**4)*exp(x**2)-18)/x**3,x)

[Out]

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

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