3.5.29 \(\int (36+18 x+\frac {25 e^{-8-x+2 x^2} (-9 x^2+36 x^3)}{x^2}+e^{\frac {1}{2} (-8-x+2 x^2+2 \log (\frac {5}{x}))} (63 x^2+36 x^3)) \, dx\)

Optimal. Leaf size=26 \[ 9 \left (2+5 e^{-\frac {x}{2}+x \left (-\frac {4}{x}+x\right )}+x\right )^2 \]

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Rubi [A]  time = 0.63, antiderivative size = 52, normalized size of antiderivative = 2.00, number of steps used = 19, number of rules used = 11, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.159, Rules used = {1584, 2236, 1593, 2274, 12, 6742, 2244, 2240, 2234, 2204, 2241} \begin {gather*} 9 x^2+90 e^{x^2-\frac {x}{2}-4} x+180 e^{x^2-\frac {x}{2}-4}+225 e^{2 x^2-x-8}+36 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[36 + 18*x + (25*E^(-8 - x + 2*x^2)*(-9*x^2 + 36*x^3))/x^2 + E^((-8 - x + 2*x^2 + 2*Log[5/x])/2)*(63*x^2 +
36*x^3),x]

[Out]

180*E^(-4 - x/2 + x^2) + 225*E^(-8 - x + 2*x^2) + 36*x + 90*E^(-4 - x/2 + x^2)*x + 9*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 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
 p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 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 2236

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] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

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 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=36 x+9 x^2+25 \int \frac {e^{-8-x+2 x^2} \left (-9 x^2+36 x^3\right )}{x^2} \, dx+\int e^{\frac {1}{2} \left (-8-x+2 x^2+2 \log \left (\frac {5}{x}\right )\right )} \left (63 x^2+36 x^3\right ) \, dx\\ &=36 x+9 x^2+25 \int e^{-8-x+2 x^2} (-9+36 x) \, dx+\int e^{\frac {1}{2} \left (-8-x+2 x^2+2 \log \left (\frac {5}{x}\right )\right )} x^2 (63+36 x) \, dx\\ &=225 e^{-8-x+2 x^2}+36 x+9 x^2+\int 5 e^{\frac {1}{2} \left (-8-x+2 x^2\right )} x (63+36 x) \, dx\\ &=225 e^{-8-x+2 x^2}+36 x+9 x^2+5 \int e^{\frac {1}{2} \left (-8-x+2 x^2\right )} x (63+36 x) \, dx\\ &=225 e^{-8-x+2 x^2}+36 x+9 x^2+5 \int \left (63 e^{\frac {1}{2} \left (-8-x+2 x^2\right )} x+36 e^{\frac {1}{2} \left (-8-x+2 x^2\right )} x^2\right ) \, dx\\ &=225 e^{-8-x+2 x^2}+36 x+9 x^2+180 \int e^{\frac {1}{2} \left (-8-x+2 x^2\right )} x^2 \, dx+315 \int e^{\frac {1}{2} \left (-8-x+2 x^2\right )} x \, dx\\ &=225 e^{-8-x+2 x^2}+36 x+9 x^2+180 \int e^{-4-\frac {x}{2}+x^2} x^2 \, dx+315 \int e^{-4-\frac {x}{2}+x^2} x \, dx\\ &=\frac {315}{2} e^{-4-\frac {x}{2}+x^2}+225 e^{-8-x+2 x^2}+36 x+90 e^{-4-\frac {x}{2}+x^2} x+9 x^2+45 \int e^{-4-\frac {x}{2}+x^2} x \, dx+\frac {315}{4} \int e^{-4-\frac {x}{2}+x^2} \, dx-90 \int e^{-4-\frac {x}{2}+x^2} \, dx\\ &=180 e^{-4-\frac {x}{2}+x^2}+225 e^{-8-x+2 x^2}+36 x+90 e^{-4-\frac {x}{2}+x^2} x+9 x^2+\frac {45}{4} \int e^{-4-\frac {x}{2}+x^2} \, dx+\frac {315 \int e^{\frac {1}{4} \left (-\frac {1}{2}+2 x\right )^2} \, dx}{4 e^{65/16}}-\frac {90 \int e^{\frac {1}{4} \left (-\frac {1}{2}+2 x\right )^2} \, dx}{e^{65/16}}\\ &=180 e^{-4-\frac {x}{2}+x^2}+225 e^{-8-x+2 x^2}+36 x+90 e^{-4-\frac {x}{2}+x^2} x+9 x^2-\frac {45 \sqrt {\pi } \text {erfi}\left (\frac {1}{4} (-1+4 x)\right )}{8 e^{65/16}}+\frac {45 \int e^{\frac {1}{4} \left (-\frac {1}{2}+2 x\right )^2} \, dx}{4 e^{65/16}}\\ &=180 e^{-4-\frac {x}{2}+x^2}+225 e^{-8-x+2 x^2}+36 x+90 e^{-4-\frac {x}{2}+x^2} x+9 x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 48, normalized size = 1.85 \begin {gather*} 9 \left (25 e^{-8-x+2 x^2}+4 x+x^2+5 e^{-\frac {x}{2}+x^2} \left (\frac {4}{e^4}+\frac {2 x}{e^4}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[36 + 18*x + (25*E^(-8 - x + 2*x^2)*(-9*x^2 + 36*x^3))/x^2 + E^((-8 - x + 2*x^2 + 2*Log[5/x])/2)*(63*
x^2 + 36*x^3),x]

[Out]

9*(25*E^(-8 - x + 2*x^2) + 4*x + x^2 + 5*E^(-1/2*x + x^2)*(4/E^4 + (2*x)/E^4))

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fricas [A]  time = 0.65, size = 57, normalized size = 2.19 \begin {gather*} 9 \, x^{2} e^{\left (2 \, x^{2} - x + 2 \, \log \left (\frac {5}{x}\right ) - 8\right )} + 9 \, x^{2} + 18 \, {\left (x^{2} + 2 \, x\right )} e^{\left (x^{2} - \frac {1}{2} \, x + \log \left (\frac {5}{x}\right ) - 4\right )} + 36 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*x^3-9*x^2)*exp(log(5/x)+x^2-1/2*x-4)^2+(36*x^3+63*x^2)*exp(log(5/x)+x^2-1/2*x-4)+18*x+36,x, algo
rithm="fricas")

[Out]

9*x^2*e^(2*x^2 - x + 2*log(5/x) - 8) + 9*x^2 + 18*(x^2 + 2*x)*e^(x^2 - 1/2*x + log(5/x) - 4) + 36*x

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giac [A]  time = 0.44, size = 47, normalized size = 1.81 \begin {gather*} 9 \, x^{2} + 90 \, {\left (x e^{\left (x^{2} - \frac {1}{2} \, x\right )} + 2 \, e^{\left (x^{2} - \frac {1}{2} \, x\right )}\right )} e^{\left (-4\right )} + 36 \, x + 225 \, e^{\left (2 \, x^{2} - x - 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*x^3-9*x^2)*exp(log(5/x)+x^2-1/2*x-4)^2+(36*x^3+63*x^2)*exp(log(5/x)+x^2-1/2*x-4)+18*x+36,x, algo
rithm="giac")

[Out]

9*x^2 + 90*(x*e^(x^2 - 1/2*x) + 2*e^(x^2 - 1/2*x))*e^(-4) + 36*x + 225*e^(2*x^2 - x - 8)

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maple [A]  time = 0.09, size = 39, normalized size = 1.50




method result size



risch \(225 \,{\mathrm e}^{2 x^{2}-x -8}+5 \left (36+18 x \right ) {\mathrm e}^{-4+x^{2}-\frac {1}{2} x}+9 x^{2}+36 x\) \(39\)
norman \(36 x +9 x^{2}+36 x \,{\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+18 x^{2} {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+225 \,{\mathrm e}^{2 x^{2}-x -8}\) \(70\)
default \(36 x +9 \,{\mathrm e}^{2 x^{2}-x +2 \ln \left (\frac {5}{x}\right )+2 \ln \relax (x )-8}+36 \,{\mathrm e}^{-4+\ln \left (\frac {5}{x}\right )+\ln \relax (x )+x^{2}-\frac {x}{2}}+18 x \,{\mathrm e}^{-4+\ln \left (\frac {5}{x}\right )+\ln \relax (x )+x^{2}-\frac {x}{2}}+9 x^{2}\) \(74\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((36*x^3-9*x^2)*exp(ln(5/x)+x^2-1/2*x-4)^2+(36*x^3+63*x^2)*exp(ln(5/x)+x^2-1/2*x-4)+18*x+36,x,method=_RETUR
NVERBOSE)

[Out]

225*exp(2*x^2-x-8)+5*(36+18*x)*exp(-4+x^2-1/2*x)+9*x^2+36*x

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maxima [A]  time = 0.72, size = 36, normalized size = 1.38 \begin {gather*} 9 \, x^{2} + 90 \, {\left (x + 2\right )} e^{\left (x^{2} - \frac {1}{2} \, x - 4\right )} + 36 \, x + 225 \, e^{\left (2 \, x^{2} - x - 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*x^3-9*x^2)*exp(log(5/x)+x^2-1/2*x-4)^2+(36*x^3+63*x^2)*exp(log(5/x)+x^2-1/2*x-4)+18*x+36,x, algo
rithm="maxima")

[Out]

9*x^2 + 90*(x + 2)*e^(x^2 - 1/2*x - 4) + 36*x + 225*e^(2*x^2 - x - 8)

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mupad [B]  time = 0.51, size = 45, normalized size = 1.73 \begin {gather*} 36\,x+180\,{\mathrm {e}}^{x^2-\frac {x}{2}-4}+225\,{\mathrm {e}}^{2\,x^2-x-8}+90\,x\,{\mathrm {e}}^{x^2-\frac {x}{2}-4}+9\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(18*x - exp(2*log(5/x) - x + 2*x^2 - 8)*(9*x^2 - 36*x^3) + exp(log(5/x) - x/2 + x^2 - 4)*(63*x^2 + 36*x^3)
+ 36,x)

[Out]

36*x + 180*exp(x^2 - x/2 - 4) + 225*exp(2*x^2 - x - 8) + 90*x*exp(x^2 - x/2 - 4) + 9*x^2

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sympy [A]  time = 0.14, size = 34, normalized size = 1.31 \begin {gather*} 9 x^{2} + 36 x + \left (90 x + 180\right ) e^{x^{2} - \frac {x}{2} - 4} + 225 e^{2 x^{2} - x - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*x**3-9*x**2)*exp(ln(5/x)+x**2-1/2*x-4)**2+(36*x**3+63*x**2)*exp(ln(5/x)+x**2-1/2*x-4)+18*x+36,x)

[Out]

9*x**2 + 36*x + (90*x + 180)*exp(x**2 - x/2 - 4) + 225*exp(2*x**2 - x - 8)

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