3.24.94 \(\int \frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{(2+3 x)^3} \, dx\) [2394]

3.24.94.1 Optimal result
3.24.94.2 Mathematica [A] (verified)
3.24.94.3 Rubi [A] (verified)
3.24.94.4 Maple [A] (verified)
3.24.94.5 Fricas [A] (verification not implemented)
3.24.94.6 Sympy [F]
3.24.94.7 Maxima [A] (verification not implemented)
3.24.94.8 Giac [B] (verification not implemented)
3.24.94.9 Mupad [F(-1)]

3.24.94.1 Optimal result

Integrand size = 26, antiderivative size = 144 \[ \int \frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=\frac {19}{18} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{6 (2+3 x)^2}+\frac {5 (1-2 x)^{3/2} \sqrt {3+5 x}}{4 (2+3 x)}+\frac {118}{27} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {155}{108} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \]

output
118/135*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-155/108*arctan(1/7*(1 
-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-1/6*(1-2*x)^(5/2)*(3+5*x)^(1/2) 
/(2+3*x)^2+5/4*(1-2*x)^(3/2)*(3+5*x)^(1/2)/(2+3*x)+19/18*(1-2*x)^(1/2)*(3+ 
5*x)^(1/2)
 
3.24.94.2 Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.75 \[ \int \frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=\frac {1}{540} \left (\frac {15 \sqrt {1-2 x} \left (708+2485 x+2319 x^2+240 x^3\right )}{(2+3 x)^2 \sqrt {3+5 x}}-472 \sqrt {10} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )-775 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\right ) \]

input
Integrate[((1 - 2*x)^(5/2)*Sqrt[3 + 5*x])/(2 + 3*x)^3,x]
 
output
((15*Sqrt[1 - 2*x]*(708 + 2485*x + 2319*x^2 + 240*x^3))/((2 + 3*x)^2*Sqrt[ 
3 + 5*x]) - 472*Sqrt[10]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]] - 775*Sqrt[ 
7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/540
 
3.24.94.3 Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {108, 27, 166, 27, 171, 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.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {1}{6} \int -\frac {5 (1-2 x)^{3/2} (12 x+5)}{2 (3 x+2)^2 \sqrt {5 x+3}}dx-\frac {(1-2 x)^{5/2} \sqrt {5 x+3}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {5}{12} \int \frac {(1-2 x)^{3/2} (12 x+5)}{(3 x+2)^2 \sqrt {5 x+3}}dx-\frac {\sqrt {5 x+3} (1-2 x)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 166

\(\displaystyle -\frac {5}{12} \left (-\frac {1}{3} \int \frac {3 \sqrt {1-2 x} (76 x+61)}{2 (3 x+2) \sqrt {5 x+3}}dx-\frac {3 \sqrt {5 x+3} (1-2 x)^{3/2}}{3 x+2}\right )-\frac {\sqrt {5 x+3} (1-2 x)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {5}{12} \left (-\frac {1}{2} \int \frac {\sqrt {1-2 x} (76 x+61)}{(3 x+2) \sqrt {5 x+3}}dx-\frac {3 \sqrt {5 x+3} (1-2 x)^{3/2}}{3 x+2}\right )-\frac {\sqrt {5 x+3} (1-2 x)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 171

\(\displaystyle -\frac {5}{12} \left (\frac {1}{2} \left (-\frac {1}{15} \int \frac {944 x+991}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {76}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3 (1-2 x)^{3/2} \sqrt {5 x+3}}{3 x+2}\right )-\frac {\sqrt {5 x+3} (1-2 x)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 175

\(\displaystyle -\frac {5}{12} \left (\frac {1}{2} \left (\frac {1}{15} \left (-\frac {944}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1085}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )-\frac {76}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3 (1-2 x)^{3/2} \sqrt {5 x+3}}{3 x+2}\right )-\frac {\sqrt {5 x+3} (1-2 x)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 64

\(\displaystyle -\frac {5}{12} \left (\frac {1}{2} \left (\frac {1}{15} \left (-\frac {1085}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {1888}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {76}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3 (1-2 x)^{3/2} \sqrt {5 x+3}}{3 x+2}\right )-\frac {\sqrt {5 x+3} (1-2 x)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {5}{12} \left (\frac {1}{2} \left (\frac {1}{15} \left (-\frac {2170}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {1888}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {76}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3 (1-2 x)^{3/2} \sqrt {5 x+3}}{3 x+2}\right )-\frac {\sqrt {5 x+3} (1-2 x)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {5}{12} \left (\frac {1}{2} \left (\frac {1}{15} \left (\frac {310}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-\frac {1888}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {76}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3 (1-2 x)^{3/2} \sqrt {5 x+3}}{3 x+2}\right )-\frac {\sqrt {5 x+3} (1-2 x)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 223

\(\displaystyle -\frac {5}{12} \left (\frac {1}{2} \left (\frac {1}{15} \left (\frac {310}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-\frac {944}{3} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )\right )-\frac {76}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3 (1-2 x)^{3/2} \sqrt {5 x+3}}{3 x+2}\right )-\frac {\sqrt {5 x+3} (1-2 x)^{5/2}}{6 (3 x+2)^2}\)

input
Int[((1 - 2*x)^(5/2)*Sqrt[3 + 5*x])/(2 + 3*x)^3,x]
 
output
-1/6*((1 - 2*x)^(5/2)*Sqrt[3 + 5*x])/(2 + 3*x)^2 - (5*((-3*(1 - 2*x)^(3/2) 
*Sqrt[3 + 5*x])/(2 + 3*x) + ((-76*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/15 + ((-944 
*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/3 + (310*Sqrt[7]*ArcTan[Sqrt[ 
1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/3)/15)/2))/12
 

3.24.94.3.1 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 166
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> 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] - Simp[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] /; FreeQ[{a, b, c, d, 
e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
 

rule 171
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[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] /; Fre 
eQ[{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 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]
 
3.24.94.4 Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.96

method result size
risch \(-\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \left (48 x^{2}+435 x +236\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{36 \left (2+3 x \right )^{2} \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}-\frac {\left (-\frac {59 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )}{135}-\frac {155 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right )}{216}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\sqrt {1-2 x}\, \sqrt {3+5 x}}\) \(138\)
default \(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (4248 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+6975 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+5664 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +9300 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +1440 x^{2} \sqrt {-10 x^{2}-x +3}+1888 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+3100 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+13050 x \sqrt {-10 x^{2}-x +3}+7080 \sqrt {-10 x^{2}-x +3}\right )}{1080 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )^{2}}\) \(208\)

input
int((1-2*x)^(5/2)*(3+5*x)^(1/2)/(2+3*x)^3,x,method=_RETURNVERBOSE)
 
output
-1/36*(-1+2*x)*(3+5*x)^(1/2)*(48*x^2+435*x+236)/(2+3*x)^2/(-(-1+2*x)*(3+5* 
x))^(1/2)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)-(-59/135*10^(1/2)*arcsin(2 
0/11*x+1/11)-155/216*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)
 
3.24.94.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.02 \[ \int \frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=-\frac {472 \, \sqrt {5} \sqrt {2} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 775 \, \sqrt {7} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 30 \, {\left (48 \, x^{2} + 435 \, x + 236\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1080 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

input
integrate((1-2*x)^(5/2)*(3+5*x)^(1/2)/(2+3*x)^3,x, algorithm="fricas")
 
output
-1/1080*(472*sqrt(5)*sqrt(2)*(9*x^2 + 12*x + 4)*arctan(1/20*sqrt(5)*sqrt(2 
)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 775*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)) - 30*(48*x^2 + 435*x + 236)*sqrt(5*x + 3)*sqrt(-2* 
x + 1))/(9*x^2 + 12*x + 4)
 
3.24.94.6 Sympy [F]

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

input
integrate((1-2*x)**(5/2)*(3+5*x)**(1/2)/(2+3*x)**3,x)
 
output
Integral((1 - 2*x)**(5/2)*sqrt(5*x + 3)/(3*x + 2)**3, x)
 
3.24.94.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=\frac {59}{135} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {155}{216} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {13}{9} \, \sqrt {-10 \, x^{2} - x + 3} + \frac {7 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{6 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {49 \, \sqrt {-10 \, x^{2} - x + 3}}{36 \, {\left (3 \, x + 2\right )}} \]

input
integrate((1-2*x)^(5/2)*(3+5*x)^(1/2)/(2+3*x)^3,x, algorithm="maxima")
 
output
59/135*sqrt(10)*arcsin(20/11*x + 1/11) + 155/216*sqrt(7)*arcsin(37/11*x/ab 
s(3*x + 2) + 20/11/abs(3*x + 2)) + 13/9*sqrt(-10*x^2 - x + 3) + 7/6*(-10*x 
^2 - x + 3)^(3/2)/(9*x^2 + 12*x + 4) - 49/36*sqrt(-10*x^2 - x + 3)/(3*x + 
2)
 
3.24.94.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (106) = 212\).

Time = 0.50 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.38 \[ \int \frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=\frac {31}{432} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {59}{135} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {4}{135} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {77 \, {\left (17 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} - 13720 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}\right )}}{54 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]

input
integrate((1-2*x)^(5/2)*(3+5*x)^(1/2)/(2+3*x)^3,x, algorithm="giac")
 
output
31/432*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*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)))) + 59/135*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)))) + 4/135*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 77/54*(17*sq 
rt(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 - 13720*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
 
3.24.94.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}\,\sqrt {5\,x+3}}{{\left (3\,x+2\right )}^3} \,d x \]

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