3.2442 \(\int \frac{(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^{3/2}} \, dx\)

Optimal. Leaf size=115 \[ \frac{7 (1-2 x)^{3/2}}{3 (3 x+2) \sqrt{5 x+3}}-\frac{1111 \sqrt{1-2 x}}{15 \sqrt{5 x+3}}-\frac{8}{45} \sqrt{\frac{2}{5}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )+\frac{665}{9} \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right ) \]

[Out]

(-1111*Sqrt[1 - 2*x])/(15*Sqrt[3 + 5*x]) + (7*(1 - 2*x)^(3/2))/(3*(2 + 3*x)*Sqrt[3 + 5*x]) - (8*Sqrt[2/5]*ArcS
in[Sqrt[2/11]*Sqrt[3 + 5*x]])/45 + (665*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/9

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Rubi [A]  time = 0.0405006, antiderivative size = 115, 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 = {98, 150, 157, 54, 216, 93, 204} \[ \frac{7 (1-2 x)^{3/2}}{3 (3 x+2) \sqrt{5 x+3}}-\frac{1111 \sqrt{1-2 x}}{15 \sqrt{5 x+3}}-\frac{8}{45} \sqrt{\frac{2}{5}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )+\frac{665}{9} \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1111*Sqrt[1 - 2*x])/(15*Sqrt[3 + 5*x]) + (7*(1 - 2*x)^(3/2))/(3*(2 + 3*x)*Sqrt[3 + 5*x]) - (8*Sqrt[2/5]*ArcS
in[Sqrt[2/11]*Sqrt[3 + 5*x]])/45 + (665*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/9

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(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 - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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 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{(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^{3/2}} \, dx &=\frac{7 (1-2 x)^{3/2}}{3 (2+3 x) \sqrt{3+5 x}}+\frac{1}{3} \int \frac{\sqrt{1-2 x} \left (\frac{227}{2}+4 x\right )}{(2+3 x) (3+5 x)^{3/2}} \, dx\\ &=-\frac{1111 \sqrt{1-2 x}}{15 \sqrt{3+5 x}}+\frac{7 (1-2 x)^{3/2}}{3 (2+3 x) \sqrt{3+5 x}}+\frac{2}{15} \int \frac{-\frac{7769}{4}-4 x}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{1111 \sqrt{1-2 x}}{15 \sqrt{3+5 x}}+\frac{7 (1-2 x)^{3/2}}{3 (2+3 x) \sqrt{3+5 x}}-\frac{8}{45} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx-\frac{4655}{18} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{1111 \sqrt{1-2 x}}{15 \sqrt{3+5 x}}+\frac{7 (1-2 x)^{3/2}}{3 (2+3 x) \sqrt{3+5 x}}-\frac{4655}{9} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )-\frac{16 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{45 \sqrt{5}}\\ &=-\frac{1111 \sqrt{1-2 x}}{15 \sqrt{3+5 x}}+\frac{7 (1-2 x)^{3/2}}{3 (2+3 x) \sqrt{3+5 x}}-\frac{8}{45} \sqrt{\frac{2}{5}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )+\frac{665}{9} \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.235379, size = 185, normalized size = 1.61 \[ \frac{121 \left (15 \sqrt{5 x+3} \left (6180 x^3+256544 x^2+34809 x-82313\right )+10477 \sqrt{10-20 x} \left (15 x^2+19 x+6\right ) \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )+581875 \sqrt{7-14 x} \left (15 x^2+19 x+6\right ) \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )\right )-15450 \sqrt{22} (2 x-1)^3 \left (15 x^2+19 x+6\right ) \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};\frac{5}{11} (1-2 x)\right )}{952875 \sqrt{1-2 x} (3 x+2) (5 x+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(121*(15*Sqrt[3 + 5*x]*(-82313 + 34809*x + 256544*x^2 + 6180*x^3) + 10477*Sqrt[10 - 20*x]*(6 + 19*x + 15*x^2)*
ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]] + 581875*Sqrt[7 - 14*x]*(6 + 19*x + 15*x^2)*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqr
t[3 + 5*x])]) - 15450*Sqrt[22]*(-1 + 2*x)^3*(6 + 19*x + 15*x^2)*Hypergeometric2F1[3/2, 5/2, 7/2, (5*(1 - 2*x))
/11])/(952875*Sqrt[1 - 2*x]*(2 + 3*x)*(3 + 5*x))

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Maple [B]  time = 0.012, size = 191, normalized size = 1.7 \begin{align*} -{\frac{1}{900+1350\,x} \left ( 120\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}+249375\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}+152\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x+315875\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+48\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) +99750\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +102090\,x\sqrt{-10\,{x}^{2}-x+3}+65610\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{1-2\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}}{\frac{1}{\sqrt{3+5\,x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/450*(120*10^(1/2)*arcsin(20/11*x+1/11)*x^2+249375*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2)
)*x^2+152*10^(1/2)*arcsin(20/11*x+1/11)*x+315875*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+
48*10^(1/2)*arcsin(20/11*x+1/11)+99750*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+102090*x*(-1
0*x^2-x+3)^(1/2)+65610*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(2+3*x)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)

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Maxima [A]  time = 1.78597, size = 139, normalized size = 1.21 \begin{align*} -\frac{4}{225} \, \sqrt{10} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) - \frac{665}{18} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) + \frac{6806 \, x}{45 \, \sqrt{-10 \, x^{2} - x + 3}} - \frac{10699}{135 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{343}{27 \,{\left (3 \, \sqrt{-10 \, x^{2} - x + 3} x + 2 \, \sqrt{-10 \, x^{2} - 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)^2/(3+5*x)^(3/2),x, algorithm="maxima")

[Out]

-4/225*sqrt(10)*arcsin(20/11*x + 1/11) - 665/18*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 68
06/45*x/sqrt(-10*x^2 - x + 3) - 10699/135/sqrt(-10*x^2 - x + 3) + 343/27/(3*sqrt(-10*x^2 - x + 3)*x + 2*sqrt(-
10*x^2 - x + 3))

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Fricas [A]  time = 1.77791, size = 429, normalized size = 3.73 \begin{align*} \frac{8 \, \sqrt{5} \sqrt{2}{\left (15 \, x^{2} + 19 \, x + 6\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 ) + 16625 \, \sqrt{7}{\left (15 \, x^{2} + 19 \, x + 6\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 ) - 30 \,{\left (3403 \, x + 2187\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{450 \,{\left (15 \, x^{2} + 19 \, x + 6\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/450*(8*sqrt(5)*sqrt(2)*(15*x^2 + 19*x + 6)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x +
1)/(10*x^2 + x - 3)) + 16625*sqrt(7)*(15*x^2 + 19*x + 6)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2
*x + 1)/(10*x^2 + x - 3)) - 30*(3403*x + 2187)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(15*x^2 + 19*x + 6)

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

[Out]

Timed out

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Giac [B]  time = 2.90337, size = 431, normalized size = 3.75 \begin{align*} -\frac{133}{36} \, \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{4}{225} \, \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{121}{50} \, \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 )} - \frac{1078 \, \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 )}}{3 \,{\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((1-2*x)^(5/2)/(2+3*x)^2/(3+5*x)^(3/2),x, algorithm="giac")

[Out]

-133/36*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)))) - 4/225*sqrt(10)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)*((sq
rt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 121/50*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))) - 10
78/3*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)