3.61.37 \(\int \frac {e^{51 x^2} (-1+2 x+101 x^2-204 x^3+102 x^4)+e^{x^2} (-10 x^3+5 x^4-10 x^5+10 x^6)}{x^2-2 x^3+x^4} \, dx\)

Optimal. Leaf size=27 \[ e^{x^2} x \left (\frac {e^{50 x^2}}{x^2}+\frac {5 x}{-1+x}\right ) \]

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Rubi [A]  time = 0.37, antiderivative size = 35, normalized size of antiderivative = 1.30, number of steps used = 6, number of rules used = 4, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1594, 27, 6688, 2288} \begin {gather*} \frac {e^{51 x^2}}{x}-\frac {5 e^{x^2} \left (x^2-x^3\right )}{(1-x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(51*x^2)*(-1 + 2*x + 101*x^2 - 204*x^3 + 102*x^4) + E^x^2*(-10*x^3 + 5*x^4 - 10*x^5 + 10*x^6))/(x^2 - 2
*x^3 + x^4),x]

[Out]

E^(51*x^2)/x - (5*E^x^2*(x^2 - x^3))/(1 - x)^2

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{51 x^2} \left (-1+2 x+101 x^2-204 x^3+102 x^4\right )+e^{x^2} \left (-10 x^3+5 x^4-10 x^5+10 x^6\right )}{x^2 \left (1-2 x+x^2\right )} \, dx\\ &=\int \frac {e^{51 x^2} \left (-1+2 x+101 x^2-204 x^3+102 x^4\right )+e^{x^2} \left (-10 x^3+5 x^4-10 x^5+10 x^6\right )}{(-1+x)^2 x^2} \, dx\\ &=\int \left (e^{51 x^2} \left (102-\frac {1}{x^2}\right )+\frac {5 e^{x^2} x \left (-2+x-2 x^2+2 x^3\right )}{(-1+x)^2}\right ) \, dx\\ &=5 \int \frac {e^{x^2} x \left (-2+x-2 x^2+2 x^3\right )}{(-1+x)^2} \, dx+\int e^{51 x^2} \left (102-\frac {1}{x^2}\right ) \, dx\\ &=\frac {e^{51 x^2}}{x}-\frac {5 e^{x^2} \left (x^2-x^3\right )}{(1-x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 37, normalized size = 1.37 \begin {gather*} \frac {e^{51 x^2}}{x}+\frac {5 e^{x^2} \left (-2 x^2+2 x^3\right )}{2 (-1+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(51*x^2)*(-1 + 2*x + 101*x^2 - 204*x^3 + 102*x^4) + E^x^2*(-10*x^3 + 5*x^4 - 10*x^5 + 10*x^6))/(x
^2 - 2*x^3 + x^4),x]

[Out]

E^(51*x^2)/x + (5*E^x^2*(-2*x^2 + 2*x^3))/(2*(-1 + x)^2)

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fricas [A]  time = 0.56, size = 30, normalized size = 1.11 \begin {gather*} \frac {5 \, x^{3} e^{\left (x^{2}\right )} + {\left (x - 1\right )} e^{\left (51 \, x^{2}\right )}}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((102*x^4-204*x^3+101*x^2+2*x-1)*exp(1/2*x^2)^2*exp(25*x^2)^2+(10*x^6-10*x^5+5*x^4-10*x^3)*exp(1/2*x
^2)^2)/(x^4-2*x^3+x^2),x, algorithm="fricas")

[Out]

(5*x^3*e^(x^2) + (x - 1)*e^(51*x^2))/(x^2 - x)

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giac [A]  time = 0.12, size = 36, normalized size = 1.33 \begin {gather*} \frac {5 \, x^{3} e^{\left (x^{2}\right )} + x e^{\left (51 \, x^{2}\right )} - e^{\left (51 \, x^{2}\right )}}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((102*x^4-204*x^3+101*x^2+2*x-1)*exp(1/2*x^2)^2*exp(25*x^2)^2+(10*x^6-10*x^5+5*x^4-10*x^3)*exp(1/2*x
^2)^2)/(x^4-2*x^3+x^2),x, algorithm="giac")

[Out]

(5*x^3*e^(x^2) + x*e^(51*x^2) - e^(51*x^2))/(x^2 - x)

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maple [A]  time = 0.05, size = 26, normalized size = 0.96

method	result	size
risch	\(\frac {{\mathrm e}^{51 x^{2}}}{x}+\frac {5 x^{2} {\mathrm e}^{x^{2}}}{x -1}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((102*x^4-204*x^3+101*x^2+2*x-1)*exp(1/2*x^2)^2*exp(25*x^2)^2+(10*x^6-10*x^5+5*x^4-10*x^3)*exp(1/2*x^2)^2)
/(x^4-2*x^3+x^2),x,method=_RETURNVERBOSE)

[Out]

1/x*exp(51*x^2)+5*x^2/(x-1)*exp(x^2)

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maxima [A]  time = 0.38, size = 30, normalized size = 1.11 \begin {gather*} \frac {5 \, x^{3} e^{\left (x^{2}\right )} + {\left (x - 1\right )} e^{\left (51 \, x^{2}\right )}}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((102*x^4-204*x^3+101*x^2+2*x-1)*exp(1/2*x^2)^2*exp(25*x^2)^2+(10*x^6-10*x^5+5*x^4-10*x^3)*exp(1/2*x
^2)^2)/(x^4-2*x^3+x^2),x, algorithm="maxima")

[Out]

(5*x^3*e^(x^2) + (x - 1)*e^(51*x^2))/(x^2 - x)

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mupad [B]  time = 4.53, size = 35, normalized size = 1.30 \begin {gather*} \frac {{\mathrm {e}}^{x^2}\,\left (x\,{\mathrm {e}}^{50\,x^2}-{\mathrm {e}}^{50\,x^2}+5\,x^3\right )}{x\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(51*x^2)*(2*x + 101*x^2 - 204*x^3 + 102*x^4 - 1) - exp(x^2)*(10*x^3 - 5*x^4 + 10*x^5 - 10*x^6))/(x^2 -
2*x^3 + x^4),x)

[Out]

(exp(x^2)*(x*exp(50*x^2) - exp(50*x^2) + 5*x^3))/(x*(x - 1))

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sympy [A]  time = 0.44, size = 24, normalized size = 0.89 \begin {gather*} \frac {5 x^{3} e^{x^{2}} + \left (x - 1\right ) e^{51 x^{2}}}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((102*x**4-204*x**3+101*x**2+2*x-1)*exp(1/2*x**2)**2*exp(25*x**2)**2+(10*x**6-10*x**5+5*x**4-10*x**3
)*exp(1/2*x**2)**2)/(x**4-2*x**3+x**2),x)

[Out]

(5*x**3*exp(x**2) + (x - 1)*exp(51*x**2))/(x**2 - x)

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