3.2448 \(\int \frac{(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=164 \[ -\frac{376 (1-2 x)^{3/2} (3 x+2)^3}{75 \sqrt{5 x+3}}-\frac{2 (1-2 x)^{5/2} (3 x+2)^3}{15 (5 x+3)^{3/2}}+\frac{741}{250} (1-2 x)^{3/2} \sqrt{5 x+3} (3 x+2)^2+\frac{21 (1-2 x)^{3/2} \sqrt{5 x+3} (4392 x+3185)}{40000}+\frac{69713 \sqrt{1-2 x} \sqrt{5 x+3}}{400000}+\frac{766843 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{400000 \sqrt{10}} \]

[Out]

(-2*(1 - 2*x)^(5/2)*(2 + 3*x)^3)/(15*(3 + 5*x)^(3/2)) - (376*(1 - 2*x)^(3/2)*(2 + 3*x)^3)/(75*Sqrt[3 + 5*x]) +
(69713*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/400000 + (741*(1 - 2*x)^(3/2)*(2 + 3*x)^2*Sqrt[3 + 5*x])/250 + (21*(1 - 2
*x)^(3/2)*Sqrt[3 + 5*x]*(3185 + 4392*x))/40000 + (766843*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(400000*Sqrt[10])

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Rubi [A]  time = 0.0507194, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {97, 150, 153, 147, 50, 54, 216} \[ -\frac{376 (1-2 x)^{3/2} (3 x+2)^3}{75 \sqrt{5 x+3}}-\frac{2 (1-2 x)^{5/2} (3 x+2)^3}{15 (5 x+3)^{3/2}}+\frac{741}{250} (1-2 x)^{3/2} \sqrt{5 x+3} (3 x+2)^2+\frac{21 (1-2 x)^{3/2} \sqrt{5 x+3} (4392 x+3185)}{40000}+\frac{69713 \sqrt{1-2 x} \sqrt{5 x+3}}{400000}+\frac{766843 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{400000 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 - 2*x)^(5/2)*(2 + 3*x)^3)/(15*(3 + 5*x)^(3/2)) - (376*(1 - 2*x)^(3/2)*(2 + 3*x)^3)/(75*Sqrt[3 + 5*x]) +
(69713*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/400000 + (741*(1 - 2*x)^(3/2)*(2 + 3*x)^2*Sqrt[3 + 5*x])/250 + (21*(1 - 2
*x)^(3/2)*Sqrt[3 + 5*x]*(3185 + 4392*x))/40000 + (766843*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(400000*Sqrt[10])

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 150

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

Rule 153

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] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

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)^{5/2} (2+3 x)^3}{(3+5 x)^{5/2}} \, dx &=-\frac{2 (1-2 x)^{5/2} (2+3 x)^3}{15 (3+5 x)^{3/2}}+\frac{2}{15} \int \frac{(-1-33 x) (1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx\\ &=-\frac{2 (1-2 x)^{5/2} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac{376 (1-2 x)^{3/2} (2+3 x)^3}{75 \sqrt{3+5 x}}+\frac{4}{75} \int \frac{\left (\frac{249}{2}-2223 x\right ) \sqrt{1-2 x} (2+3 x)^2}{\sqrt{3+5 x}} \, dx\\ &=-\frac{2 (1-2 x)^{5/2} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac{376 (1-2 x)^{3/2} (2+3 x)^3}{75 \sqrt{3+5 x}}+\frac{741}{250} (1-2 x)^{3/2} (2+3 x)^2 \sqrt{3+5 x}-\frac{1}{750} \int \frac{\sqrt{1-2 x} (2+3 x) \left (1155+\frac{34587 x}{2}\right )}{\sqrt{3+5 x}} \, dx\\ &=-\frac{2 (1-2 x)^{5/2} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac{376 (1-2 x)^{3/2} (2+3 x)^3}{75 \sqrt{3+5 x}}+\frac{741}{250} (1-2 x)^{3/2} (2+3 x)^2 \sqrt{3+5 x}+\frac{21 (1-2 x)^{3/2} \sqrt{3+5 x} (3185+4392 x)}{40000}+\frac{69713 \int \frac{\sqrt{1-2 x}}{\sqrt{3+5 x}} \, dx}{80000}\\ &=-\frac{2 (1-2 x)^{5/2} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac{376 (1-2 x)^{3/2} (2+3 x)^3}{75 \sqrt{3+5 x}}+\frac{69713 \sqrt{1-2 x} \sqrt{3+5 x}}{400000}+\frac{741}{250} (1-2 x)^{3/2} (2+3 x)^2 \sqrt{3+5 x}+\frac{21 (1-2 x)^{3/2} \sqrt{3+5 x} (3185+4392 x)}{40000}+\frac{766843 \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx}{800000}\\ &=-\frac{2 (1-2 x)^{5/2} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac{376 (1-2 x)^{3/2} (2+3 x)^3}{75 \sqrt{3+5 x}}+\frac{69713 \sqrt{1-2 x} \sqrt{3+5 x}}{400000}+\frac{741}{250} (1-2 x)^{3/2} (2+3 x)^2 \sqrt{3+5 x}+\frac{21 (1-2 x)^{3/2} \sqrt{3+5 x} (3185+4392 x)}{40000}+\frac{766843 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{400000 \sqrt{5}}\\ &=-\frac{2 (1-2 x)^{5/2} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac{376 (1-2 x)^{3/2} (2+3 x)^3}{75 \sqrt{3+5 x}}+\frac{69713 \sqrt{1-2 x} \sqrt{3+5 x}}{400000}+\frac{741}{250} (1-2 x)^{3/2} (2+3 x)^2 \sqrt{3+5 x}+\frac{21 (1-2 x)^{3/2} \sqrt{3+5 x} (3185+4392 x)}{40000}+\frac{766843 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )}{400000 \sqrt{10}}\\ \end{align*}

Mathematica [A]  time = 0.0536917, size = 93, normalized size = 0.57 \[ \frac{-10 \left (12960000 x^6-4536000 x^5-16421400 x^4+13874190 x^3+12677675 x^2-3232208 x-2322001\right )-2300529 \sqrt{10-20 x} (5 x+3)^{3/2} \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{12000000 \sqrt{1-2 x} (5 x+3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*(-2322001 - 3232208*x + 12677675*x^2 + 13874190*x^3 - 16421400*x^4 - 4536000*x^5 + 12960000*x^6) - 230052
9*Sqrt[10 - 20*x]*(3 + 5*x)^(3/2)*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]])/(12000000*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))

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Maple [A]  time = 0.013, size = 164, normalized size = 1. \begin{align*}{\frac{1}{24000000} \left ( 129600000\,{x}^{5}\sqrt{-10\,{x}^{2}-x+3}[phone]\,{x}^{4}\sqrt{-10\,{x}^{2}-x+3}[phone]\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}-154494000\,{x}^{3}\sqrt{-10\,{x}^{2}-x+3}[phone]\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x+61494900\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}[phone]\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) [phone]\,x\sqrt{-10\,{x}^{2}-x+3}[phone]\,\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)^(5/2)*(2+3*x)^3/(3+5*x)^(5/2),x)

[Out]

1/24000000*(129600000*x^5*(-10*x^2-x+3)^(1/2)[phone]*x^4*(-10*x^2-x+3)^(1/2)[phone]*10^(1/2)*arcsin(20/11*
x+1/11)*x^2-154494000*x^3*(-10*x^2-x+3)^(1/2)[phone]*10^(1/2)*arcsin(20/11*x+1/11)*x+61494900*x^2*(-10*x^2-x
+3)^(1/2)[phone]*10^(1/2)*arcsin(20/11*x+1/11)[phone]*x*(-10*x^2-x+3)^(1/2)[phone]*(-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 [C]  time = 2.2545, size = 439, normalized size = 2.68 \begin{align*} -\frac{395307}{8000000} i \, \sqrt{5} \sqrt{2} \arcsin \left (\frac{20}{11} \, x + \frac{23}{11}\right ) + \frac{23221}{500000} \, \sqrt{5} \sqrt{2} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) + \frac{99}{5000} \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}} + \frac{{\left (-10 \, x^{2} - x + 3\right )}^{\frac{5}{2}}}{625 \,{\left (625 \, x^{4} + 1500 \, x^{3} + 1350 \, x^{2} + 540 \, x + 81\right )}} + \frac{9 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{5}{2}}}{1250 \,{\left (125 \, x^{3} + 225 \, x^{2} + 135 \, x + 27\right )}} + \frac{9 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{5}{2}}}{625 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac{27 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{5}{2}}}{2500 \,{\left (5 \, x + 3\right )}} + \frac{3267}{20000} \, \sqrt{10 \, x^{2} + 23 \, x + \frac{51}{5}} x + \frac{75141}{400000} \, \sqrt{10 \, x^{2} + 23 \, x + \frac{51}{5}} + \frac{3267}{25000} \, \sqrt{-10 \, x^{2} - x + 3} - \frac{11 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{3750 \,{\left (125 \, x^{3} + 225 \, x^{2} + 135 \, x + 27\right )}} + \frac{99 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{2500 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac{99 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{2500 \,{\left (5 \, x + 3\right )}} - \frac{121 \, \sqrt{-10 \, x^{2} - x + 3}}{18750 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac{9493 \, \sqrt{-10 \, x^{2} - x + 3}}{37500 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-395307/8000000*I*sqrt(5)*sqrt(2)*arcsin(20/11*x + 23/11) + 23221/500000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11
) + 99/5000*(-10*x^2 - x + 3)^(3/2) + 1/625*(-10*x^2 - x + 3)^(5/2)/(625*x^4 + 1500*x^3 + 1350*x^2 + 540*x + 8
1) + 9/1250*(-10*x^2 - x + 3)^(5/2)/(125*x^3 + 225*x^2 + 135*x + 27) + 9/625*(-10*x^2 - x + 3)^(5/2)/(25*x^2 +
30*x + 9) + 27/2500*(-10*x^2 - x + 3)^(5/2)/(5*x + 3) + 3267/20000*sqrt(10*x^2 + 23*x + 51/5)*x + 75141/40000
0*sqrt(10*x^2 + 23*x + 51/5) + 3267/25000*sqrt(-10*x^2 - x + 3) - 11/3750*(-10*x^2 - x + 3)^(3/2)/(125*x^3 + 2
25*x^2 + 135*x + 27) + 99/2500*(-10*x^2 - x + 3)^(3/2)/(25*x^2 + 30*x + 9) + 99/2500*(-10*x^2 - x + 3)^(3/2)/(
5*x + 3) - 121/18750*sqrt(-10*x^2 - x + 3)/(25*x^2 + 30*x + 9) - 9493/37500*sqrt(-10*x^2 - x + 3)/(5*x + 3)

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Fricas [A]  time = 1.80311, size = 354, normalized size = 2.16 \begin{align*} -\frac{2300529 \, \sqrt{10}{\left (25 \, x^{2} + 30 \, x + 9\right )} \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 ) - 20 \,{\left (6480000 \, x^{5} + 972000 \, x^{4} - 7724700 \, x^{3} + 3074745 \, x^{2} + 7876210 \, x + 2322001\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{24000000 \,{\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)^(5/2)*(2+3*x)^3/(3+5*x)^(5/2),x, algorithm="fricas")

[Out]

-1/24000000*(2300529*sqrt(10)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)
/(10*x^2 + x - 3)) - 20*(6480000*x^5 + 972000*x^4 - 7724700*x^3 + 3074745*x^2 + 7876210*x + 2322001)*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)**(5/2)*(2+3*x)**3/(3+5*x)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 2.61839, size = 273, normalized size = 1.66 \begin{align*} \frac{1}{10000000} \,{\left (36 \,{\left (24 \,{\left (4 \, \sqrt{5}{\left (5 \, x + 3\right )} - 57 \, \sqrt{5}\right )}{\left (5 \, x + 3\right )} + 4915 \, \sqrt{5}\right )}{\left (5 \, x + 3\right )} + 338795 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5} - \frac{11 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}}{3750000 \,{\left (5 \, x + 3\right )}^{\frac{3}{2}}} + \frac{766843}{4000000} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) - \frac{2079 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}{312500 \, \sqrt{5 \, x + 3}} + \frac{11 \,{\left (\frac{567 \, \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}}}{234375 \,{\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)^(5/2)*(2+3*x)^3/(3+5*x)^(5/2),x, algorithm="giac")

[Out]

1/10000000*(36*(24*(4*sqrt(5)*(5*x + 3) - 57*sqrt(5))*(5*x + 3) + 4915*sqrt(5))*(5*x + 3) + 338795*sqrt(5))*sq
rt(5*x + 3)*sqrt(-10*x + 5) - 11/3750000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 766
843/4000000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 2079/312500*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqr
t(22))/sqrt(5*x + 3) + 11/234375*(567*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