3.2321 \(\int \frac{\sqrt{1-2 x} (2+3 x)}{(3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=74 \[ -\frac{2 (1-2 x)^{3/2}}{165 (5 x+3)^{3/2}}-\frac{6 \sqrt{1-2 x}}{25 \sqrt{5 x+3}}-\frac{6}{25} \sqrt{\frac{2}{5}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right ) \]

[Out]

(-2*(1 - 2*x)^(3/2))/(165*(3 + 5*x)^(3/2)) - (6*Sqrt[1 - 2*x])/(25*Sqrt[3 + 5*x]) - (6*Sqrt[2/5]*ArcSin[Sqrt[2
/11]*Sqrt[3 + 5*x]])/25

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Rubi [A]  time = 0.0158633, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {78, 47, 54, 216} \[ -\frac{2 (1-2 x)^{3/2}}{165 (5 x+3)^{3/2}}-\frac{6 \sqrt{1-2 x}}{25 \sqrt{5 x+3}}-\frac{6}{25} \sqrt{\frac{2}{5}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 - 2*x)^(3/2))/(165*(3 + 5*x)^(3/2)) - (6*Sqrt[1 - 2*x])/(25*Sqrt[3 + 5*x]) - (6*Sqrt[2/5]*ArcSin[Sqrt[2
/11]*Sqrt[3 + 5*x]])/25

Rule 78

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

Rule 47

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

Rubi steps

\begin{align*} \int \frac{\sqrt{1-2 x} (2+3 x)}{(3+5 x)^{5/2}} \, dx &=-\frac{2 (1-2 x)^{3/2}}{165 (3+5 x)^{3/2}}+\frac{3}{5} \int \frac{\sqrt{1-2 x}}{(3+5 x)^{3/2}} \, dx\\ &=-\frac{2 (1-2 x)^{3/2}}{165 (3+5 x)^{3/2}}-\frac{6 \sqrt{1-2 x}}{25 \sqrt{3+5 x}}-\frac{6}{25} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{2 (1-2 x)^{3/2}}{165 (3+5 x)^{3/2}}-\frac{6 \sqrt{1-2 x}}{25 \sqrt{3+5 x}}-\frac{12 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{25 \sqrt{5}}\\ &=-\frac{2 (1-2 x)^{3/2}}{165 (3+5 x)^{3/2}}-\frac{6 \sqrt{1-2 x}}{25 \sqrt{3+5 x}}-\frac{6}{25} \sqrt{\frac{2}{5}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0780631, size = 73, normalized size = 0.99 \[ \frac{10 \left (970 x^2+119 x-302\right )+198 \sqrt{10-20 x} (5 x+3)^{3/2} \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{4125 \sqrt{1-2 x} (5 x+3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*(-302 + 119*x + 970*x^2) + 198*Sqrt[10 - 20*x]*(3 + 5*x)^(3/2)*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]])/(4125*Sqr
t[1 - 2*x]*(3 + 5*x)^(3/2))

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Maple [A]  time = 0.012, size = 96, normalized size = 1.3 \begin{align*} -{\frac{1}{4125} \left ( 2475\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}+2970\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x+891\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) +4850\,x\sqrt{-10\,{x}^{2}-x+3}+3020\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{1-2\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}} \left ( 3+5\,x \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4125*(2475*10^(1/2)*arcsin(20/11*x+1/11)*x^2+2970*10^(1/2)*arcsin(20/11*x+1/11)*x+891*10^(1/2)*arcsin(20/11
*x+1/11)+4850*x*(-10*x^2-x+3)^(1/2)+3020*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)

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Maxima [A]  time = 2.34406, size = 65, normalized size = 0.88 \begin{align*} -\frac{4 \, \sqrt{-10 \, x^{2} - x + 3}}{15 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac{8 \, \sqrt{-10 \, x^{2} - x + 3}}{165 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*sqrt(-10*x^2 - x + 3)/(25*x^2 + 30*x + 9) + 8/165*sqrt(-10*x^2 - x + 3)/(5*x + 3)

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Fricas [A]  time = 1.74078, size = 274, normalized size = 3.7 \begin{align*} \frac{99 \, \sqrt{5} \sqrt{2}{\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac{\sqrt{5} \sqrt{2}{\left (20 \, x + 1\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{20 \,{\left (10 \, x^{2} + x - 3\right )}}\right ) - 10 \,{\left (485 \, x + 302\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{4125 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4125*(99*sqrt(5)*sqrt(2)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x
+ 1)/(10*x^2 + x - 3)) - 10*(485*x + 302)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.14845, size = 194, normalized size = 2.62 \begin{align*} -\frac{\sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}}{66000 \,{\left (5 \, x + 3\right )}^{\frac{3}{2}}} - \frac{6}{125} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) - \frac{13 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}{1100 \, \sqrt{5 \, x + 3}} + \frac{{\left (\frac{195 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} + 4 \, \sqrt{10}\right )}{\left (5 \, x + 3\right )}^{\frac{3}{2}}}{4125 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/66000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) - 6/125*sqrt(10)*arcsin(1/11*sqrt(22)
*sqrt(5*x + 3)) - 13/1100*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 1/4125*(195*sqrt(10)*(
sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4*sqrt(10))*(5*x + 3)^(3/2)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(
22))^3