3.2269 \(\int \frac{\sqrt{1-2 x} \sqrt{3+5 x}}{(2+3 x)^3} \, dx\)

Optimal. Leaf size=93 \[ \frac{\sqrt{1-2 x} (5 x+3)^{3/2}}{2 (3 x+2)^2}-\frac{11 \sqrt{1-2 x} \sqrt{5 x+3}}{28 (3 x+2)}-\frac{121 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{28 \sqrt{7}} \]

[Out]

(-11*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(28*(2 + 3*x)) + (Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(2*(2 + 3*x)^2) - (121*ArcT
an[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(28*Sqrt[7])

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Rubi [A]  time = 0.0215352, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {94, 93, 204} \[ \frac{\sqrt{1-2 x} (5 x+3)^{3/2}}{2 (3 x+2)^2}-\frac{11 \sqrt{1-2 x} \sqrt{5 x+3}}{28 (3 x+2)}-\frac{121 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{28 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(28*(2 + 3*x)) + (Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(2*(2 + 3*x)^2) - (121*ArcT
an[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(28*Sqrt[7])

Rule 94

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

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)^3} \, dx &=\frac{\sqrt{1-2 x} (3+5 x)^{3/2}}{2 (2+3 x)^2}+\frac{11}{4} \int \frac{\sqrt{3+5 x}}{\sqrt{1-2 x} (2+3 x)^2} \, dx\\ &=-\frac{11 \sqrt{1-2 x} \sqrt{3+5 x}}{28 (2+3 x)}+\frac{\sqrt{1-2 x} (3+5 x)^{3/2}}{2 (2+3 x)^2}+\frac{121}{56} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{11 \sqrt{1-2 x} \sqrt{3+5 x}}{28 (2+3 x)}+\frac{\sqrt{1-2 x} (3+5 x)^{3/2}}{2 (2+3 x)^2}+\frac{121}{28} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )\\ &=-\frac{11 \sqrt{1-2 x} \sqrt{3+5 x}}{28 (2+3 x)}+\frac{\sqrt{1-2 x} (3+5 x)^{3/2}}{2 (2+3 x)^2}-\frac{121 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{28 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0360224, size = 69, normalized size = 0.74 \[ \frac{1}{196} \left (\frac{7 \sqrt{1-2 x} \sqrt{5 x+3} (37 x+20)}{(3 x+2)^2}-121 \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(20 + 37*x))/(2 + 3*x)^2 - 121*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 +
5*x])])/196

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Maple [B]  time = 0.01, size = 154, normalized size = 1.7 \begin{align*}{\frac{1}{392\, \left ( 2+3\,x \right ) ^{2}}\sqrt{1-2\,x}\sqrt{3+5\,x} \left ( 1089\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}+1452\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+484\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +518\,x\sqrt{-10\,{x}^{2}-x+3}+280\,\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)^3,x)

[Out]

1/392*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(1089*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2+1452*7^
(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+484*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-
x+3)^(1/2))+518*x*(-10*x^2-x+3)^(1/2)+280*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(2+3*x)^2

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Maxima [A]  time = 3.4633, size = 122, normalized size = 1.31 \begin{align*} \frac{121}{392} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) + \frac{5}{21} \, \sqrt{-10 \, x^{2} - x + 3} + \frac{3 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{14 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac{37 \, \sqrt{-10 \, x^{2} - x + 3}}{84 \,{\left (3 \, x + 2\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)^3,x, algorithm="maxima")

[Out]

121/392*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 5/21*sqrt(-10*x^2 - x + 3) + 3/14*(-10*x^2
- x + 3)^(3/2)/(9*x^2 + 12*x + 4) - 37/84*sqrt(-10*x^2 - x + 3)/(3*x + 2)

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Fricas [A]  time = 1.80934, size = 250, normalized size = 2.69 \begin{align*} -\frac{121 \, \sqrt{7}{\left (9 \, x^{2} + 12 \, x + 4\right )} \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 ) - 14 \,{\left (37 \, x + 20\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{392 \,{\left (9 \, x^{2} + 12 \, x + 4\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)^3,x, algorithm="fricas")

[Out]

-1/392*(121*sqrt(7)*(9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 +
x - 3)) - 14*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)

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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}}{\left (3 x + 2\right )^{3}}\, 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)**3,x)

[Out]

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

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Giac [B]  time = 1.77022, size = 343, normalized size = 3.69 \begin{align*} \frac{121}{3920} \, \sqrt{5}{\left (\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 )} - \frac{280 \, \sqrt{2}{\left ({\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )}^{3} - \frac{280 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}{\sqrt{5 \, x + 3}} + \frac{1120 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )}}{{\left ({\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )}^{2} + 280\right )}^{2}}\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)^3,x, algorithm="giac")

[Out]

121/3920*sqrt(5)*(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)))) - 280*sqrt(2)*(((sqrt(2)*sqrt(-10*x + 5) - sqr
t(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 - 280*(sqrt(2)*sqrt(-10*x + 5)
- sqrt(22))/sqrt(5*x + 3) + 1120*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x + 5
) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)^2)