\(\int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^2} \, dx\) [871]

Optimal result

Integrand size = 26, antiderivative size = 91 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^2} \, dx=-\frac {\sqrt {1-2 x} \sqrt {3+5 x}}{3 (2+3 x)}-\frac {2}{9} \sqrt {10} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {37 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{9 \sqrt {7}} \] Output:

-1/3*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)-2/9*arcsin(1/11*22^(1/2)*(3+5*x)^ 
(1/2))*10^(1/2)-37/63*7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/ 
2))
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^2} \, dx=-\frac {\sqrt {1-2 x} \sqrt {3+5 x}}{6+9 x}+\frac {2}{9} \sqrt {10} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )-\frac {37 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{9 \sqrt {7}} \] Input:

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

Output:

-((Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(6 + 9*x)) + (2*Sqrt[10]*ArcTan[Sqrt[5/2 - 
 5*x]/Sqrt[3 + 5*x]])/9 - (37*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x]) 
])/(9*Sqrt[7])
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {108, 27, 175, 64, 104, 217, 223}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {1}{3} \int -\frac {20 x+1}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{6} \int \frac {20 x+1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{6} \left (\frac {37}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {20}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{6} \left (\frac {37}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {8}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{6} \left (\frac {74}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {8}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{6} \left (-\frac {8}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {74 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3 \sqrt {7}}\right )-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{6} \left (-\frac {4}{3} \sqrt {10} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {74 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3 \sqrt {7}}\right )-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\)

Input:

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

Output:

-1/3*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(2 + 3*x) + ((-4*Sqrt[10]*ArcSin[Sqrt[2 
/11]*Sqrt[3 + 5*x]])/3 - (74*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])] 
)/(3*Sqrt[7]))/6
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 64

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp 
[2/b   Subst[Int[1/Sqrt[c - a*(d/b) + d*(x^2/b)], x], x, Sqrt[a + b*x]], x] 
 /; FreeQ[{a, b, c, d}, x] && GtQ[c - a*(d/b), 0] && ( !GtQ[a - c*(b/d), 0] 
 || PosQ[b])
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[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] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 108

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) 
, x] - Simp[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 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 223

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]
 

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.41

method	result	size
risch	\(\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{3 \left (2+3 x \right ) \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}-\frac {\left (\frac {\sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )}{9}-\frac {37 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right )}{126}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(128\)
default	\(-\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (42 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -111 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +28 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-74 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+42 \sqrt {-10 x^{2}-x +3}\right )}{126 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )}\)	\(131\)

Input:

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

Output:

1/3*(-1+2*x)/(2+3*x)*(3+5*x)^(1/2)/(-(-1+2*x)*(3+5*x))^(1/2)*((1-2*x)*(3+5 
*x))^(1/2)/(1-2*x)^(1/2)-(1/9*10^(1/2)*arcsin(20/11*x+1/11)-37/126*7^(1/2) 
*arctan(9/14*(20/3+37/3*x)*7^(1/2)/(-90*(2/3+x)^2+67+111*x)^(1/2)))*((1-2* 
x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)/(3+5*x)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^2} \, dx=-\frac {37 \, \sqrt {7} {\left (3 \, x + 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, \sqrt {10} {\left (3 \, x + 2\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 42 \, \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{126 \, {\left (3 \, x + 2\right )}} \] Input:

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

Output:

-1/126*(37*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)) - 14*sqrt(10)*(3*x + 2)*arctan(1/20*sqrt 
(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 42*sqrt(5 
*x + 3)*sqrt(-2*x + 1))/(3*x + 2)
 

Sympy [F]

\[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^2} \, dx=\int \frac {\sqrt {1 - 2 x} \sqrt {5 x + 3}}{\left (3 x + 2\right )^{2}}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^2} \, dx=-\frac {1}{9} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {37}{126} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {\sqrt {-10 \, x^{2} - x + 3}}{3 \, {\left (3 \, x + 2\right )}} \] Input:

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

Output:

-1/9*sqrt(10)*arcsin(20/11*x + 1/11) + 37/126*sqrt(7)*arcsin(37/11*x/abs(3 
*x + 2) + 20/11/abs(3*x + 2)) - 1/3*sqrt(-10*x^2 - x + 3)/(3*x + 2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (67) = 134\).

Time = 0.20 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.86 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^2} \, dx=\frac {37}{1260} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {1}{9} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {22 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{3 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} \] Input:

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

Output:

37/1260*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((s 
qrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 
5) - sqrt(22)))) - 1/9*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)))) - 22/3*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)*s 
qrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(- 
10*x + 5) - sqrt(22)))^2 + 280)
 

Mupad [B] (verification not implemented)

Time = 3.76 (sec) , antiderivative size = 752, normalized size of antiderivative = 8.26 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(37*7^(1/2)*atan((31139252096*7^(1/2)*((1 - 2*x)^(1/2) - 1))/(38759765625* 
(3^(1/2) - (5*x + 3)^(1/2))*((1076274944*((1 - 2*x)^(1/2) - 1)^2)/(1550390 
625*(3^(1/2) - (5*x + 3)^(1/2))^2) + (14849685376*3^(1/2)*((1 - 2*x)^(1/2) 
 - 1))/(12919921875*(3^(1/2) - (5*x + 3)^(1/2))) - 2152549888/7751953125)) 
 - (901743872*3^(1/2)*7^(1/2))/(4306640625*((1076274944*((1 - 2*x)^(1/2) - 
 1)^2)/(1550390625*(3^(1/2) - (5*x + 3)^(1/2))^2) + (14849685376*3^(1/2)*( 
(1 - 2*x)^(1/2) - 1))/(12919921875*(3^(1/2) - (5*x + 3)^(1/2))) - 21525498 
88/7751953125)) + (450871936*3^(1/2)*7^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(861 
328125*(3^(1/2) - (5*x + 3)^(1/2))^2*((1076274944*((1 - 2*x)^(1/2) - 1)^2) 
/(1550390625*(3^(1/2) - (5*x + 3)^(1/2))^2) + (14849685376*3^(1/2)*((1 - 2 
*x)^(1/2) - 1))/(12919921875*(3^(1/2) - (5*x + 3)^(1/2))) - 2152549888/775 
1953125))))/63 - (4*10^(1/2)*atan((10^(1/2)*((1 - 2*x)^(1/2) - 1))/(2*(3^( 
1/2) - (5*x + 3)^(1/2)))))/9 - (2*((1 - 2*x)^(1/2) - 1)^3)/(3*(3^(1/2) - ( 
5*x + 3)^(1/2))^3*((14*((1 - 2*x)^(1/2) - 1)^2)/(25*(3^(1/2) - (5*x + 3)^( 
1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/2) - (5*x + 3)^(1/2))^4 + (6*3^(1 
/2)*((1 - 2*x)^(1/2) - 1)^3)/(5*(3^(1/2) - (5*x + 3)^(1/2))^3) - (12*3^(1/ 
2)*((1 - 2*x)^(1/2) - 1))/(25*(3^(1/2) - (5*x + 3)^(1/2))) + 4/25)) + (4*( 
(1 - 2*x)^(1/2) - 1))/(15*(3^(1/2) - (5*x + 3)^(1/2))*((14*((1 - 2*x)^(1/2 
) - 1)^2)/(25*(3^(1/2) - (5*x + 3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^ 
(1/2) - (5*x + 3)^(1/2))^4 + (6*3^(1/2)*((1 - 2*x)^(1/2) - 1)^3)/(5*(3^...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.15 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^2} \, dx=\frac {42 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x +28 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )+111 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +74 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-111 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -74 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-21 \sqrt {5 x +3}\, \sqrt {-2 x +1}}{189 x +126} \] Input:

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

Output:

(42*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))*x + 28*sqrt(10)*asi 
n((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) + 111*sqrt(7)*atan((sqrt(33) - sqrt 
(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x + 74*sqr 
t(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11 
))/2))/sqrt(2)) - 111*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 
2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x - 74*sqrt(7)*atan((sqrt(33) + s 
qrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2)) - 21*sq 
rt(5*x + 3)*sqrt( - 2*x + 1))/(63*(3*x + 2))