3.2281 \(\int \frac{\sqrt{1-2 x} (3+5 x)^{3/2}}{(2+3 x)^3} \, dx\)

Optimal. Leaf size=120 \[ -\frac{\sqrt{1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}-\frac{107 \sqrt{1-2 x} \sqrt{5 x+3}}{252 (3 x+2)}-\frac{10}{27} \sqrt{10} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )-\frac{4091 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{756 \sqrt{7}} \]

[Out]

(-107*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(252*(2 + 3*x)) - (Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(6*(2 + 3*x)^2) - (10*Sqr
t[10]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/27 - (4091*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(756*Sqrt[7]
)

________________________________________________________________________________________

Rubi [A]  time = 0.0402038, antiderivative size = 120, 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, 149, 157, 54, 216, 93, 204} \[ -\frac{\sqrt{1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}-\frac{107 \sqrt{1-2 x} \sqrt{5 x+3}}{252 (3 x+2)}-\frac{10}{27} \sqrt{10} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )-\frac{4091 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{756 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-107*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(252*(2 + 3*x)) - (Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(6*(2 + 3*x)^2) - (10*Sqr
t[10]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/27 - (4091*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(756*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 149

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] && IntegerQ[m]

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)^{3/2}}{(2+3 x)^3} \, dx &=-\frac{\sqrt{1-2 x} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac{1}{6} \int \frac{\left (\frac{9}{2}-20 x\right ) \sqrt{3+5 x}}{\sqrt{1-2 x} (2+3 x)^2} \, dx\\ &=-\frac{107 \sqrt{1-2 x} \sqrt{3+5 x}}{252 (2+3 x)}-\frac{\sqrt{1-2 x} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac{1}{126} \int \frac{-\frac{503}{4}-700 x}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{107 \sqrt{1-2 x} \sqrt{3+5 x}}{252 (2+3 x)}-\frac{\sqrt{1-2 x} (3+5 x)^{3/2}}{6 (2+3 x)^2}-\frac{50}{27} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx+\frac{4091 \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx}{1512}\\ &=-\frac{107 \sqrt{1-2 x} \sqrt{3+5 x}}{252 (2+3 x)}-\frac{\sqrt{1-2 x} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac{4091}{756} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )-\frac{1}{27} \left (20 \sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )\\ &=-\frac{107 \sqrt{1-2 x} \sqrt{3+5 x}}{252 (2+3 x)}-\frac{\sqrt{1-2 x} (3+5 x)^{3/2}}{6 (2+3 x)^2}-\frac{10}{27} \sqrt{10} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )-\frac{4091 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{756 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.117544, size = 121, normalized size = 1.01 \[ \frac{21 \sqrt{5 x+3} \left (1062 x^2+149 x-340\right )+1960 \sqrt{10-20 x} (3 x+2)^2 \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )-4091 \sqrt{7-14 x} (3 x+2)^2 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{5292 \sqrt{1-2 x} (3 x+2)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(21*Sqrt[3 + 5*x]*(-340 + 149*x + 1062*x^2) + 1960*Sqrt[10 - 20*x]*(2 + 3*x)^2*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]
] - 4091*Sqrt[7 - 14*x]*(2 + 3*x)^2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(5292*Sqrt[1 - 2*x]*(2 + 3*
x)^2)

________________________________________________________________________________________

Maple [B]  time = 0.011, size = 191, normalized size = 1.6 \begin{align*}{\frac{1}{10584\, \left ( 2+3\,x \right ) ^{2}}\sqrt{1-2\,x}\sqrt{3+5\,x} \left ( 36819\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}-17640\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}+49092\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x-23520\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x+16364\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) -7840\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) -22302\,x\sqrt{-10\,{x}^{2}-x+3}-14280\,\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)^(3/2)*(1-2*x)^(1/2)/(2+3*x)^3,x)

[Out]

1/10584*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(36819*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2-1764
0*10^(1/2)*arcsin(20/11*x+1/11)*x^2+49092*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x-23520*1
0^(1/2)*arcsin(20/11*x+1/11)*x+16364*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-7840*10^(1/2)*
arcsin(20/11*x+1/11)-22302*x*(-10*x^2-x+3)^(1/2)-14280*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(2+3*x)^2

________________________________________________________________________________________

Maxima [A]  time = 3.59721, size = 136, normalized size = 1.13 \begin{align*} -\frac{5}{27} \, \sqrt{10} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) + \frac{4091}{10584} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) - \frac{5}{63} \, \sqrt{-10 \, x^{2} - x + 3} - \frac{{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{14 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac{103 \, \sqrt{-10 \, x^{2} - x + 3}}{252 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/27*sqrt(10)*arcsin(20/11*x + 1/11) + 4091/10584*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) -
5/63*sqrt(-10*x^2 - x + 3) - 1/14*(-10*x^2 - x + 3)^(3/2)/(9*x^2 + 12*x + 4) - 103/252*sqrt(-10*x^2 - x + 3)/
(3*x + 2)

________________________________________________________________________________________

Fricas [A]  time = 1.86134, size = 410, normalized size = 3.42 \begin{align*} -\frac{4091 \, \sqrt{7}{\left (9 \, x^{2} + 12 \, x + 4\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 ) - 1960 \, \sqrt{10}{\left (9 \, x^{2} + 12 \, x + 4\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 ) + 42 \,{\left (531 \, x + 340\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{10584 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10584*(4091*sqrt(7)*(9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2
+ x - 3)) - 1960*sqrt(10)*(9*x^2 + 12*x + 4)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10
*x^2 + x - 3)) + 42*(531*x + 340)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 3.1403, size = 437, normalized size = 3.64 \begin{align*} \frac{4091}{105840} \, \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{5}{27} \, \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{11 \,{\left (107 \, \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} + 48440 \, \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 )}\right )}}{126 \,{\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 )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4091/105840*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)))) - 5/27*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)))) - 11/126*(107*sqr
t(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
)))^3 + 48440*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)^2