3.2292 \(\int \frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{(2+3 x)^2} \, dx\)

Optimal. Leaf size=137 \[ -\frac{\sqrt{1-2 x} (5 x+3)^{5/2}}{3 (3 x+2)}+\frac{5}{6} \sqrt{1-2 x} (5 x+3)^{3/2}-\frac{95}{72} \sqrt{1-2 x} \sqrt{5 x+3}+\frac{155}{216} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )-\frac{59 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{27 \sqrt{7}} \]

[Out]

(-95*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/72 + (5*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/6 - (Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(
3*(2 + 3*x)) + (155*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/216 - (59*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3
+ 5*x])])/(27*Sqrt[7])

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Rubi [A]  time = 0.0512806, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {97, 154, 157, 54, 216, 93, 204} \[ -\frac{\sqrt{1-2 x} (5 x+3)^{5/2}}{3 (3 x+2)}+\frac{5}{6} \sqrt{1-2 x} (5 x+3)^{3/2}-\frac{95}{72} \sqrt{1-2 x} \sqrt{5 x+3}+\frac{155}{216} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )-\frac{59 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{27 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-95*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/72 + (5*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/6 - (Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(
3*(2 + 3*x)) + (155*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/216 - (59*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3
+ 5*x])])/(27*Sqrt[7])

Rule 97

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)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 154

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

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} (3+5 x)^{5/2}}{(2+3 x)^2} \, dx &=-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{3 (2+3 x)}+\frac{1}{3} \int \frac{\left (\frac{19}{2}-30 x\right ) (3+5 x)^{3/2}}{\sqrt{1-2 x} (2+3 x)} \, dx\\ &=\frac{5}{6} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{3 (2+3 x)}-\frac{1}{36} \int \frac{(-72-285 x) \sqrt{3+5 x}}{\sqrt{1-2 x} (2+3 x)} \, dx\\ &=-\frac{95}{72} \sqrt{1-2 x} \sqrt{3+5 x}+\frac{5}{6} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{3 (2+3 x)}+\frac{1}{216} \int \frac{1011+\frac{2325 x}{2}}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{95}{72} \sqrt{1-2 x} \sqrt{3+5 x}+\frac{5}{6} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{3 (2+3 x)}+\frac{59}{54} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx+\frac{775}{432} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{95}{72} \sqrt{1-2 x} \sqrt{3+5 x}+\frac{5}{6} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{3 (2+3 x)}+\frac{59}{27} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )+\frac{1}{216} \left (155 \sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )\\ &=-\frac{95}{72} \sqrt{1-2 x} \sqrt{3+5 x}+\frac{5}{6} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{3 (2+3 x)}+\frac{155}{216} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )-\frac{59 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{27 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.122055, size = 122, normalized size = 0.89 \[ \frac{-42 \sqrt{5 x+3} \left (600 x^3-30 x^2-227 x+46\right )-1085 \sqrt{10-20 x} (3 x+2) \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )-944 \sqrt{7-14 x} (3 x+2) \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{3024 \sqrt{1-2 x} (3 x+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-42*Sqrt[3 + 5*x]*(46 - 227*x - 30*x^2 + 600*x^3) - 1085*Sqrt[10 - 20*x]*(2 + 3*x)*ArcSin[Sqrt[5/11]*Sqrt[1 -
2*x]] - 944*Sqrt[7 - 14*x]*(2 + 3*x)*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(3024*Sqrt[1 - 2*x]*(2 +
3*x))

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Maple [A]  time = 0.011, size = 163, normalized size = 1.2 \begin{align*}{\frac{1}{12096+18144\,x}\sqrt{1-2\,x}\sqrt{3+5\,x} \left ( 3255\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x+2832\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+25200\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}+2170\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) +1888\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +11340\,x\sqrt{-10\,{x}^{2}-x+3}-3864\,\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((3+5*x)^(5/2)*(1-2*x)^(1/2)/(2+3*x)^2,x)

[Out]

1/6048*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(3255*10^(1/2)*arcsin(20/11*x+1/11)*x+2832*7^(1/2)*arctan(1/14*(37*x+20)*7^
(1/2)/(-10*x^2-x+3)^(1/2))*x+25200*x^2*(-10*x^2-x+3)^(1/2)+2170*10^(1/2)*arcsin(20/11*x+1/11)+1888*7^(1/2)*arc
tan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+11340*x*(-10*x^2-x+3)^(1/2)-3864*(-10*x^2-x+3)^(1/2))/(-10*x^2
-x+3)^(1/2)/(2+3*x)

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Maxima [A]  time = 3.43311, size = 122, normalized size = 0.89 \begin{align*} \frac{25}{18} \, \sqrt{-10 \, x^{2} - x + 3} x + \frac{155}{864} \, \sqrt{10} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) + \frac{59}{378} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) - \frac{65}{216} \, \sqrt{-10 \, x^{2} - x + 3} - \frac{\sqrt{-10 \, x^{2} - x + 3}}{27 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/18*sqrt(-10*x^2 - x + 3)*x + 155/864*sqrt(10)*arcsin(20/11*x + 1/11) + 59/378*sqrt(7)*arcsin(37/11*x/abs(3*
x + 2) + 20/11/abs(3*x + 2)) - 65/216*sqrt(-10*x^2 - x + 3) - 1/27*sqrt(-10*x^2 - x + 3)/(3*x + 2)

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Fricas [A]  time = 1.70085, size = 402, normalized size = 2.93 \begin{align*} -\frac{1085 \, \sqrt{5} \sqrt{2}{\left (3 \, x + 2\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 ) + 944 \, \sqrt{7}{\left (3 \, x + 2\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 ) - 84 \,{\left (300 \, x^{2} + 135 \, x - 46\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{6048 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6048*(1085*sqrt(5)*sqrt(2)*(3*x + 2)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(1
0*x^2 + x - 3)) + 944*sqrt(7)*(3*x + 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 +
x - 3)) - 84*(300*x^2 + 135*x - 46)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(3*x + 2)

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

[Out]

Timed out

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Giac [B]  time = 2.82238, size = 394, normalized size = 2.88 \begin{align*} \frac{59}{3780} \, \sqrt{70} \sqrt{10}{\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{1}{216} \,{\left (12 \, \sqrt{5}{\left (5 \, x + 3\right )} - 49 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5} + \frac{155}{864} \, \sqrt{10}{\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 )} - \frac{22 \, \sqrt{10}{\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 )}}{27 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

59/3780*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*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)))) + 1/216*(12*sqrt(5)*(5*x + 3) - 49*sqrt(5))*sqrt(5*x +
3)*sqrt(-10*x + 5) + 155/864*sqrt(10)*(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)))) - 22/27*sqrt(10)*((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)))/(((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)