3.282 \(\int \frac{3 x^2+2 x^3}{\sqrt{-3+2 x+x^2} (-3+x+2 x^2)} \, dx\)

Optimal. Leaf size=36 \[ \frac{\sqrt{x^2+2 x-3}}{2 (1-x)}+\sqrt{x^2+2 x-3} \]

[Out]

Sqrt[-3 + 2*x + x^2] + Sqrt[-3 + 2*x + x^2]/(2*(1 - x))

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Rubi [A]  time = 0.127551, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {1593, 1586, 1638, 650} \[ \frac{\sqrt{x^2+2 x-3}}{2 (1-x)}+\sqrt{x^2+2 x-3} \]

Antiderivative was successfully verified.

[In]

Int[(3*x^2 + 2*x^3)/(Sqrt[-3 + 2*x + x^2]*(-3 + x + 2*x^2)),x]

[Out]

Sqrt[-3 + 2*x + x^2] + Sqrt[-3 + 2*x + x^2]/(2*(1 - x))

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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1638

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q + e*f*(m + p + q)*(d + e*x)^(q - 2)*(b*d - 2*a*e +
(2*c*d - b*e)*x), x], x], x] /; NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] &&
NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps

\begin{align*} \int \frac{3 x^2+2 x^3}{\sqrt{-3+2 x+x^2} \left (-3+x+2 x^2\right )} \, dx &=\int \frac{x^2 (3+2 x)}{\sqrt{-3+2 x+x^2} \left (-3+x+2 x^2\right )} \, dx\\ &=\int \frac{x^2}{(-1+x) \sqrt{-3+2 x+x^2}} \, dx\\ &=\sqrt{-3+2 x+x^2}+\int \frac{1}{(-1+x) \sqrt{-3+2 x+x^2}} \, dx\\ &=\sqrt{-3+2 x+x^2}+\frac{\sqrt{-3+2 x+x^2}}{2 (1-x)}\\ \end{align*}

Mathematica [A]  time = 0.0111896, size = 26, normalized size = 0.72 \[ \frac{2 x^2+3 x-9}{2 \sqrt{x^2+2 x-3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(3*x^2 + 2*x^3)/(Sqrt[-3 + 2*x + x^2]*(-3 + x + 2*x^2)),x]

[Out]

(-9 + 3*x + 2*x^2)/(2*Sqrt[-3 + 2*x + x^2])

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Maple [A]  time = 0.005, size = 21, normalized size = 0.6 \begin{align*}{\frac{ \left ( -3+2\,x \right ) \left ( 3+x \right ) }{2}{\frac{1}{\sqrt{{x}^{2}+2\,x-3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3+3*x^2)/(2*x^2+x-3)/(x^2+2*x-3)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.46946, size = 38, normalized size = 1.06 \begin{align*} \sqrt{x^{2} + 2 \, x - 3} - \frac{\sqrt{x^{2} + 2 \, x - 3}}{2 \,{\left (x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2)/(2*x^2+x-3)/(x^2+2*x-3)^(1/2),x, algorithm="maxima")

[Out]

sqrt(x^2 + 2*x - 3) - 1/2*sqrt(x^2 + 2*x - 3)/(x - 1)

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Fricas [A]  time = 1.88616, size = 58, normalized size = 1.61 \begin{align*} \frac{\sqrt{x^{2} + 2 \, x - 3}{\left (2 \, x - 3\right )}}{2 \,{\left (x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2)/(2*x^2+x-3)/(x^2+2*x-3)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(x^2 + 2*x - 3)*(2*x - 3)/(x - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{\left (x - 1\right ) \left (x + 3\right )} \left (x - 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3+3*x**2)/(2*x**2+x-3)/(x**2+2*x-3)**(1/2),x)

[Out]

Integral(x**2/(sqrt((x - 1)*(x + 3))*(x - 1)), x)

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Giac [A]  time = 1.07088, size = 41, normalized size = 1.14 \begin{align*} \sqrt{x^{2} + 2 \, x - 3} + \frac{2}{x - \sqrt{x^{2} + 2 \, x - 3} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2)/(2*x^2+x-3)/(x^2+2*x-3)^(1/2),x, algorithm="giac")

[Out]

sqrt(x^2 + 2*x - 3) + 2/(x - sqrt(x^2 + 2*x - 3) - 1)