3.2382 \(\int \frac{(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=96 \[ -\frac{2 (1-2 x)^{5/2}}{165 (5 x+3)^{3/2}}-\frac{38 (1-2 x)^{3/2}}{165 \sqrt{5 x+3}}-\frac{38}{275} \sqrt{5 x+3} \sqrt{1-2 x}-\frac{19}{25} \sqrt{\frac{2}{5}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right ) \]

[Out]

(-2*(1 - 2*x)^(5/2))/(165*(3 + 5*x)^(3/2)) - (38*(1 - 2*x)^(3/2))/(165*Sqrt[3 + 5*x]) - (38*Sqrt[1 - 2*x]*Sqrt
[3 + 5*x])/275 - (19*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/25

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Rubi [A]  time = 0.0206144, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {78, 47, 50, 54, 216} \[ -\frac{2 (1-2 x)^{5/2}}{165 (5 x+3)^{3/2}}-\frac{38 (1-2 x)^{3/2}}{165 \sqrt{5 x+3}}-\frac{38}{275} \sqrt{5 x+3} \sqrt{1-2 x}-\frac{19}{25} \sqrt{\frac{2}{5}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 - 2*x)^(5/2))/(165*(3 + 5*x)^(3/2)) - (38*(1 - 2*x)^(3/2))/(165*Sqrt[3 + 5*x]) - (38*Sqrt[1 - 2*x]*Sqrt
[3 + 5*x])/275 - (19*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 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 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{(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx &=-\frac{2 (1-2 x)^{5/2}}{165 (3+5 x)^{3/2}}+\frac{19}{33} \int \frac{(1-2 x)^{3/2}}{(3+5 x)^{3/2}} \, dx\\ &=-\frac{2 (1-2 x)^{5/2}}{165 (3+5 x)^{3/2}}-\frac{38 (1-2 x)^{3/2}}{165 \sqrt{3+5 x}}-\frac{38}{55} \int \frac{\sqrt{1-2 x}}{\sqrt{3+5 x}} \, dx\\ &=-\frac{2 (1-2 x)^{5/2}}{165 (3+5 x)^{3/2}}-\frac{38 (1-2 x)^{3/2}}{165 \sqrt{3+5 x}}-\frac{38}{275} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{19}{25} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{2 (1-2 x)^{5/2}}{165 (3+5 x)^{3/2}}-\frac{38 (1-2 x)^{3/2}}{165 \sqrt{3+5 x}}-\frac{38}{275} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{38 \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)^{5/2}}{165 (3+5 x)^{3/2}}-\frac{38 (1-2 x)^{3/2}}{165 \sqrt{3+5 x}}-\frac{38}{275} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{19}{25} \sqrt{\frac{2}{5}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )\\ \end{align*}

Mathematica [C]  time = 0.0323211, size = 59, normalized size = 0.61 \[ -\frac{2 (1-2 x)^{5/2} \left (19 \sqrt{22} (5 x+3)^{3/2} \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};\frac{5}{11} (1-2 x)\right )+121\right )}{19965 (5 x+3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 - 2*x)^(5/2)*(121 + 19*Sqrt[22]*(3 + 5*x)^(3/2)*Hypergeometric2F1[3/2, 5/2, 7/2, (5*(1 - 2*x))/11]))/(1
9965*(3 + 5*x)^(3/2))

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Maple [A]  time = 0.01, size = 113, normalized size = 1.2 \begin{align*} -{\frac{1}{750} \left ( 1425\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}+1710\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x+900\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}+513\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) +2900\,x\sqrt{-10\,{x}^{2}-x+3}+1460\,\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((1-2*x)^(3/2)*(2+3*x)/(3+5*x)^(5/2),x)

[Out]

-1/750*(1425*10^(1/2)*arcsin(20/11*x+1/11)*x^2+1710*10^(1/2)*arcsin(20/11*x+1/11)*x+900*x^2*(-10*x^2-x+3)^(1/2
)+513*10^(1/2)*arcsin(20/11*x+1/11)+2900*x*(-10*x^2-x+3)^(1/2)+1460*(-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 = 1.92077, size = 161, normalized size = 1.68 \begin{align*} -\frac{19}{250} \, \sqrt{5} \sqrt{2} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) - \frac{{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{75 \,{\left (125 \, x^{3} + 225 \, x^{2} + 135 \, x + 27\right )}} + \frac{3 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{25 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac{11 \, \sqrt{-10 \, x^{2} - x + 3}}{375 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac{283 \, \sqrt{-10 \, x^{2} - x + 3}}{375 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-19/250*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 1/75*(-10*x^2 - x + 3)^(3/2)/(125*x^3 + 225*x^2 + 135*x + 27)
+ 3/25*(-10*x^2 - x + 3)^(3/2)/(25*x^2 + 30*x + 9) - 11/375*sqrt(-10*x^2 - x + 3)/(25*x^2 + 30*x + 9) - 283/3
75*sqrt(-10*x^2 - x + 3)/(5*x + 3)

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Fricas [A]  time = 1.52198, size = 284, normalized size = 2.96 \begin{align*} \frac{57 \, \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 ) - 20 \,{\left (45 \, x^{2} + 145 \, x + 73\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{750 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/750*(57*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)) - 20*(45*x^2 + 145*x + 73)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 2.5855, size = 220, normalized size = 2.29 \begin{align*} -\frac{\sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}}{30000 \,{\left (5 \, x + 3\right )}^{\frac{3}{2}}} - \frac{6}{625} \, \sqrt{5} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5} - \frac{19}{125} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) - \frac{61 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}{2500 \, \sqrt{5 \, x + 3}} + \frac{{\left (\frac{183 \, \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}}}{1875 \,{\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((1-2*x)^(3/2)*(2+3*x)/(3+5*x)^(5/2),x, algorithm="giac")

[Out]

-1/30000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) - 6/625*sqrt(5)*sqrt(5*x + 3)*sqrt(-1
0*x + 5) - 19/125*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 61/2500*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - s
qrt(22))/sqrt(5*x + 3) + 1/1875*(183*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