\(\int \frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{(2+3 x)^2} \, dx\) [945]

Optimal result

Integrand size = 26, antiderivative size = 135 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{(2+3 x)^2} \, dx=\frac {41}{36} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {5}{6} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{3 (2+3 x)}+\frac {1649 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{108 \sqrt {10}}+\frac {37}{27} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \] Output:

41/36*(1-2*x)^(1/2)*(3+5*x)^(1/2)+5/6*(1-2*x)^(3/2)*(3+5*x)^(1/2)-(1-2*x)^ 
(3/2)*(3+5*x)^(3/2)/(6+9*x)+1649/1080*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))* 
10^(1/2)+37/27*7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.80 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{(2+3 x)^2} \, dx=\frac {\frac {30 \sqrt {1-2 x} \left (318+845 x+345 x^2-300 x^3\right )}{(2+3 x) \sqrt {3+5 x}}-1649 \sqrt {10} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )+1480 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1080} \] Input:

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

Output:

((30*Sqrt[1 - 2*x]*(318 + 845*x + 345*x^2 - 300*x^3))/((2 + 3*x)*Sqrt[3 + 
5*x]) - 1649*Sqrt[10]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]] + 1480*Sqrt[7] 
*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/1080
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {108, 27, 171, 27, 171, 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.

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

\(\Big \downarrow \) 108

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

\(\displaystyle \frac {1}{2} \left (-\frac {1}{30} \int -\frac {10 (15-107 x) \sqrt {5 x+3}}{\sqrt {1-2 x} (3 x+2)}dx-\frac {2}{3} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^{3/2}}{3 (3 x+2)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {(15-107 x) \sqrt {5 x+3}}{\sqrt {1-2 x} (3 x+2)}dx-\frac {2}{3} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^{3/2}}{3 (3 x+2)}\)

\(\Big \downarrow \) 171

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {107}{6} \sqrt {1-2 x} \sqrt {5 x+3}-\frac {1}{6} \int -\frac {1649 x+754}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )-\frac {2}{3} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^{3/2}}{3 (3 x+2)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{12} \int \frac {1649 x+754}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {107}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2}{3} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^{3/2}}{3 (3 x+2)}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{12} \left (\frac {1649}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1036}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {107}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2}{3} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^{3/2}}{3 (3 x+2)}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{12} \left (\frac {3298}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1036}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {107}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2}{3} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^{3/2}}{3 (3 x+2)}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{12} \left (\frac {3298}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {2072}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}\right )+\frac {107}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2}{3} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^{3/2}}{3 (3 x+2)}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{12} \left (\frac {3298}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {296}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )+\frac {107}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2}{3} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^{3/2}}{3 (3 x+2)}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{12} \left (\frac {1649}{3} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {296}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )+\frac {107}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2}{3} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^{3/2}}{3 (3 x+2)}\)

Input:

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

Output:

-1/3*((1 - 2*x)^(3/2)*(3 + 5*x)^(3/2))/(2 + 3*x) + ((-2*Sqrt[1 - 2*x]*(3 + 
 5*x)^(3/2))/3 + ((107*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/6 + ((1649*Sqrt[2/5]*A 
rcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/3 + (296*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sq 
rt[7]*Sqrt[3 + 5*x])])/3)/12)/3)/2
 

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 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]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01

method	result	size
risch	\(\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \left (60 x^{2}-105 x -106\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{36 \left (2+3 x \right ) \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {\left (\frac {1649 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )}{2160}-\frac {37 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right )}{54}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(137\)
default	\(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (4947 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -4440 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x -3600 x^{2} \sqrt {-10 x^{2}-x +3}+3298 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-2960 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+6300 x \sqrt {-10 x^{2}-x +3}+6360 \sqrt {-10 x^{2}-x +3}\right )}{2160 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )}\)	\(163\)

Input:

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

Output:

1/36*(-1+2*x)*(3+5*x)^(1/2)*(60*x^2-105*x-106)/(2+3*x)/(-(-1+2*x)*(3+5*x)) 
^(1/2)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)+(1649/2160*10^(1/2)*arcsin(20 
/11*x+1/11)-37/54*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 = 0.09 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.93 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{(2+3 x)^2} \, dx=\frac {1480 \, \sqrt {7} {\left (3 \, x + 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 1649 \, \sqrt {10} {\left (3 \, x + 2\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 60 \, {\left (60 \, x^{2} - 105 \, x - 106\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{2160 \, {\left (3 \, x + 2\right )}} \] Input:

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

Output:

1/2160*(1480*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)) - 1649*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)) - 60*(6 
0*x^2 - 105*x - 106)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(3*x + 2)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.67 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{(2+3 x)^2} \, dx=-\frac {5}{3} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {1649}{2160} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {37}{54} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {71}{36} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{3 \, {\left (3 \, x + 2\right )}} \] Input:

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

Output:

-5/3*sqrt(-10*x^2 - x + 3)*x + 1649/2160*sqrt(10)*arcsin(20/11*x + 1/11) - 
 37/54*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 71/36*s 
qrt(-10*x^2 - x + 3) - 1/3*(-10*x^2 - x + 3)^(3/2)/(3*x + 2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (99) = 198\).

Time = 0.27 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.16 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{(2+3 x)^2} \, dx=-\frac {37}{540} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {1}{540} \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} - 181 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {1649}{2160} \, \sqrt {10} {\left (\pi - 2 \, \arctan \left (\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {154 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{27 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} \] Input:

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

Output:

-37/540*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)))) - 1/540*(12*sqrt(5)*(5*x + 3) - 181*sqrt(5))*sqrt(5*x + 3 
)*sqrt(-10*x + 5) + 1649/2160*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)))) + 154/27*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)))/(((s 
qrt(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)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.69 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{(2+3 x)^2} \, dx=\frac {-4947 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x -3298 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )-4440 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -2960 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+4440 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +2960 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-1800 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x^{2}+3150 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x +3180 \sqrt {5 x +3}\, \sqrt {-2 x +1}}{3240 x +2160} \] Input:

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

Output:

( - 4947*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))*x - 3298*sqrt( 
10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) - 4440*sqrt(7)*atan((sqrt(33 
) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x 
- 2960*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5 
))/sqrt(11))/2))/sqrt(2)) + 4440*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asi 
n((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x + 2960*sqrt(7)*atan( 
(sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqr 
t(2)) - 1800*sqrt(5*x + 3)*sqrt( - 2*x + 1)*x**2 + 3150*sqrt(5*x + 3)*sqrt 
( - 2*x + 1)*x + 3180*sqrt(5*x + 3)*sqrt( - 2*x + 1))/(1080*(3*x + 2))