3.2267 \(\int \frac{\sqrt{1-2 x} \sqrt{3+5 x}}{2+3 x} \, dx\)

Optimal. Leaf size=84 \[ \frac{1}{3} \sqrt{1-2 x} \sqrt{5 x+3}+\frac{37 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{9 \sqrt{10}}+\frac{2}{9} \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right ) \]

[Out]

(Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/3 + (37*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(9*Sqrt[10]) + (2*Sqrt[7]*ArcTan[Sqrt[
1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/9

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Rubi [A]  time = 0.031629, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {101, 157, 54, 216, 93, 204} \[ \frac{1}{3} \sqrt{1-2 x} \sqrt{5 x+3}+\frac{37 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{9 \sqrt{10}}+\frac{2}{9} \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(2 + 3*x),x]

[Out]

(Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/3 + (37*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(9*Sqrt[10]) + (2*Sqrt[7]*ArcTan[Sqrt[
1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/9

Rule 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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 216

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{\sqrt{1-2 x} \sqrt{3+5 x}}{2+3 x} \, dx &=\frac{1}{3} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{1}{3} \int \frac{-10-\frac{37 x}{2}}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=\frac{1}{3} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{7}{9} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx+\frac{37}{18} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=\frac{1}{3} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{14}{9} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )+\frac{37 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{9 \sqrt{5}}\\ &=\frac{1}{3} \sqrt{1-2 x} \sqrt{3+5 x}+\frac{37 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )}{9 \sqrt{10}}+\frac{2}{9} \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0288792, size = 82, normalized size = 0.98 \[ \frac{1}{90} \left (30 \sqrt{1-2 x} \sqrt{5 x+3}-37 \sqrt{10} \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )+20 \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(2 + 3*x),x]

[Out]

(30*Sqrt[1 - 2*x]*Sqrt[3 + 5*x] - 37*Sqrt[10]*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]] + 20*Sqrt[7]*ArcTan[Sqrt[1 - 2*
x]/(Sqrt[7]*Sqrt[3 + 5*x])])/90

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Maple [A]  time = 0.01, size = 83, normalized size = 1. \begin{align*} -{\frac{1}{180}\sqrt{1-2\,x}\sqrt{3+5\,x} \left ( 20\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) -37\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) -60\,\sqrt{-10\,{x}^{2}-x+3} \right ){\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/180*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(20*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-37*10^(1/2)*
arcsin(20/11*x+1/11)-60*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)

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Maxima [A]  time = 2.12778, size = 73, normalized size = 0.87 \begin{align*} \frac{37}{180} \, \sqrt{10} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) - \frac{1}{9} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) + \frac{1}{3} \, \sqrt{-10 \, x^{2} - x + 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x),x, algorithm="maxima")

[Out]

37/180*sqrt(10)*arcsin(20/11*x + 1/11) - 1/9*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 1/3*s
qrt(-10*x^2 - x + 3)

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Fricas [A]  time = 1.91092, size = 302, normalized size = 3.6 \begin{align*} \frac{1}{9} \, \sqrt{7} \arctan \left (\frac{\sqrt{7}{\left (37 \, x + 20\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{14 \,{\left (10 \, x^{2} + x - 3\right )}}\right ) - \frac{37}{180} \, \sqrt{10} \arctan \left (\frac{\sqrt{10}{\left (20 \, x + 1\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{20 \,{\left (10 \, x^{2} + x - 3\right )}}\right ) + \frac{1}{3} \, \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x),x, algorithm="fricas")

[Out]

1/9*sqrt(7)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 37/180*sqrt(10)*a
rctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 1/3*sqrt(5*x + 3)*sqrt(-2*x +
1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{1 - 2 x} \sqrt{5 x + 3}}{3 x + 2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)**(1/2)*(3+5*x)**(1/2)/(2+3*x),x)

[Out]

Integral(sqrt(1 - 2*x)*sqrt(5*x + 3)/(3*x + 2), x)

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Giac [B]  time = 2.40807, size = 219, normalized size = 2.61 \begin{align*} -\frac{1}{180} \, \sqrt{5}{\left (2 \, \sqrt{70} \sqrt{2}{\left (\pi + 2 \, \arctan \left (-\frac{\sqrt{70} \sqrt{5 \, x + 3}{\left (\frac{{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}\right )\right )} - 37 \, \sqrt{2}{\left (\pi + 2 \, \arctan \left (-\frac{\sqrt{5 \, x + 3}{\left (\frac{{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}\right )\right )} - 12 \, \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x),x, algorithm="giac")

[Out]

-1/180*sqrt(5)*(2*sqrt(70)*sqrt(2)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sq
rt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 37*sqrt(2)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)
*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 12*sqrt(5*x +
3)*sqrt(-10*x + 5))