3.863 \(\int \frac {\sqrt {1-x} \sqrt {2+3 x}}{\sqrt {1-x^2}} \, dx\)

Optimal. Leaf size=35 \[ \sqrt {x+1} \sqrt {3 x+2}-\frac {\sinh ^{-1}\left (\sqrt {3 x+2}\right )}{\sqrt {3}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {26, 50, 54, 215} \[ \sqrt {x+1} \sqrt {3 x+2}-\frac {\sinh ^{-1}\left (\sqrt {3 x+2}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 + x]*Sqrt[2 + 3*x] - ArcSinh[Sqrt[2 + 3*x]]/Sqrt[3]

Rule 26

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(j_))^(p_.), x_Symbol] :> Dist[(-(b^2/d))^m, Int[
u/(a - b*x^n)^m, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[j, 2*n] && EqQ[p, -m] && EqQ[b^2*c + a^2*d,
0] && GtQ[a, 0] && LtQ[d, 0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 49, normalized size = 1.40 \[ \frac {3 \sqrt {x+1} (3 x+2)-\sqrt {9 x+6} \sinh ^{-1}\left (\sqrt {3 x+2}\right )}{3 \sqrt {3 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[1 + x]*(2 + 3*x) - Sqrt[6 + 9*x]*ArcSinh[Sqrt[2 + 3*x]])/(3*Sqrt[2 + 3*x])

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fricas [B]  time = 0.45, size = 96, normalized size = 2.74 \[ \frac {\sqrt {3} {\left (x - 1\right )} \log \left (-\frac {72 \, x^{3} + 4 \, \sqrt {3} \sqrt {-x^{2} + 1} {\left (6 \, x + 5\right )} \sqrt {3 \, x + 2} \sqrt {-x + 1} + 48 \, x^{2} - 71 \, x - 49}{x - 1}\right ) - 12 \, \sqrt {-x^{2} + 1} \sqrt {3 \, x + 2} \sqrt {-x + 1}}{12 \, {\left (x - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(sqrt(3)*(x - 1)*log(-(72*x^3 + 4*sqrt(3)*sqrt(-x^2 + 1)*(6*x + 5)*sqrt(3*x + 2)*sqrt(-x + 1) + 48*x^2 -
71*x - 49)/(x - 1)) - 12*sqrt(-x^2 + 1)*sqrt(3*x + 2)*sqrt(-x + 1))/(x - 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {3 \, x + 2} \sqrt {-x + 1}}{\sqrt {-x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 86, normalized size = 2.46 \[ \frac {\sqrt {-x +1}\, \sqrt {3 x +2}\, \sqrt {-x^{2}+1}\, \left (\sqrt {3}\, \ln \left (\sqrt {3}\, x +\frac {5 \sqrt {3}}{6}+\sqrt {3 x^{2}+5 x +2}\right )-6 \sqrt {3 x^{2}+5 x +2}\right )}{6 \left (x -1\right ) \sqrt {3 x^{2}+5 x +2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(-x+1)^(1/2)*(3*x+2)^(1/2)*(-x^2+1)^(1/2)*(ln(5/6*3^(1/2)+3^(1/2)*x+(3*x^2+5*x+2)^(1/2))*3^(1/2)-6*(3*x^2+
5*x+2)^(1/2))/(x-1)/(3*x^2+5*x+2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {3 \, x + 2} \sqrt {-x + 1}}{\sqrt {-x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\sqrt {3\,x+2}\,\sqrt {1-x}}{\sqrt {1-x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {1 - x} \sqrt {3 x + 2}}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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