\(\int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^3} \, dx\) [884]

Optimal result

Integrand size = 26, antiderivative size = 120 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^3} \, dx=-\frac {107 \sqrt {1-2 x} \sqrt {3+5 x}}{252 (2+3 x)}-\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{6 (2+3 x)^2}-\frac {10}{27} \sqrt {10} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {4091 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{756 \sqrt {7}} \] Output:

-107*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(504+756*x)-1/6*(1-2*x)^(1/2)*(3+5*x)^(3/ 
2)/(2+3*x)^2-10/27*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-4091/5292* 
7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^3} \, dx=\frac {-\frac {21 \sqrt {1-2 x} \sqrt {3+5 x} (340+531 x)}{(2+3 x)^2}+1960 \sqrt {10} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )-4091 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{5292} \] Input:

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

Output:

((-21*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(340 + 531*x))/(2 + 3*x)^2 + 1960*Sqrt[1 
0]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]] - 4091*Sqrt[7]*ArcTan[Sqrt[1 - 2* 
x]/(Sqrt[7]*Sqrt[3 + 5*x])])/5292
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {108, 27, 166, 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.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {1}{6} \int \frac {(9-40 x) \sqrt {5 x+3}}{2 \sqrt {1-2 x} (3 x+2)^2}dx-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{12} \int \frac {(9-40 x) \sqrt {5 x+3}}{\sqrt {1-2 x} (3 x+2)^2}dx-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 166

\(\displaystyle \frac {1}{12} \left (\frac {1}{21} \int -\frac {2800 x+503}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {107 \sqrt {1-2 x} \sqrt {5 x+3}}{21 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{12} \left (-\frac {1}{42} \int \frac {2800 x+503}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {107 \sqrt {1-2 x} \sqrt {5 x+3}}{21 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{12} \left (\frac {1}{42} \left (\frac {4091}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {2800}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )-\frac {107 \sqrt {1-2 x} \sqrt {5 x+3}}{21 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{12} \left (\frac {1}{42} \left (\frac {4091}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {1120}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {107 \sqrt {1-2 x} \sqrt {5 x+3}}{21 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{12} \left (\frac {1}{42} \left (\frac {8182}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {1120}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {107 \sqrt {1-2 x} \sqrt {5 x+3}}{21 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{12} \left (\frac {1}{42} \left (-\frac {1120}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {8182 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3 \sqrt {7}}\right )-\frac {107 \sqrt {1-2 x} \sqrt {5 x+3}}{21 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{12} \left (\frac {1}{42} \left (-\frac {560}{3} \sqrt {10} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {8182 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3 \sqrt {7}}\right )-\frac {107 \sqrt {1-2 x} \sqrt {5 x+3}}{21 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}\)

Input:

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

Output:

-1/6*(Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(2 + 3*x)^2 + ((-107*Sqrt[1 - 2*x]*Sq 
rt[3 + 5*x])/(21*(2 + 3*x)) + ((-560*Sqrt[10]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5 
*x]])/3 - (8182*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(3*Sqrt[7]) 
)/42)/12
 

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 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 = 0.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.11

method	result	size
risch	\(\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \left (340+531 x \right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{252 \left (2+3 x \right )^{2} \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}-\frac {\left (\frac {5 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )}{27}-\frac {4091 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right )}{10584}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(133\)
default	\(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (36819 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}-17640 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+49092 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x -23520 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +16364 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-7840 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-22302 x \sqrt {-10 x^{2}-x +3}-14280 \sqrt {-10 x^{2}-x +3}\right )}{10584 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )^{2}}\)	\(191\)

Input:

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

Output:

1/252*(-1+2*x)*(3+5*x)^(1/2)*(340+531*x)/(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)-(5/27*10^(1/2)*arcsin(20/11*x+1/1 
1)-4091/10584*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 = 0.08 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^3} \, dx=-\frac {4091 \, \sqrt {7} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 1960 \, \sqrt {10} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 42 \, {\left (531 \, x + 340\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{10584 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \] Input:

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

Output:

-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]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^3} \, dx=-\frac {5}{27} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {4091}{10584} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {5}{63} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{14 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {103 \, \sqrt {-10 \, x^{2} - x + 3}}{252 \, {\left (3 \, x + 2\right )}} \] Input:

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

Output:

-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*(-1 
0*x^2 - x + 3)^(3/2)/(9*x^2 + 12*x + 4) - 103/252*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. 319 vs. \(2 (90) = 180\).

Time = 0.26 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.66 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^3} \, dx=\frac {4091}{105840} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {5}{27} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {11 \, \sqrt {10} {\left (107 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {48440 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {193760 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{126 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \] Input:

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

Output:

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)*((s 
qrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 
5) - sqrt(22)))) - 11/126*sqrt(10)*(107*((sqrt(2)*sqrt(-10*x + 5) - sqrt(2 
2))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^ 
3 + 48440*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 193760*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) - sq 
rt(22)))^2 + 280)^2
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.58 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^3} \, dx=\frac {17640 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x^{2}+23520 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x +7840 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )+36819 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}+49092 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +16364 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-36819 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}-49092 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -16364 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-11151 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x -7140 \sqrt {5 x +3}\, \sqrt {-2 x +1}}{47628 x^{2}+63504 x +21168} \] Input:

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

Output:

(17640*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))*x**2 + 23520*sqr 
t(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))*x + 7840*sqrt(10)*asin((sq 
rt( - 2*x + 1)*sqrt(5))/sqrt(11)) + 36819*sqrt(7)*atan((sqrt(33) - sqrt(35 
)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x**2 + 49092* 
sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt 
(11))/2))/sqrt(2))*x + 16364*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((s 
qrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2)) - 36819*sqrt(7)*atan((sqrt 
(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2)) 
*x**2 - 49092*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1) 
*sqrt(5))/sqrt(11))/2))/sqrt(2))*x - 16364*sqrt(7)*atan((sqrt(33) + sqrt(3 
5)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2)) - 11151*sqrt 
(5*x + 3)*sqrt( - 2*x + 1)*x - 7140*sqrt(5*x + 3)*sqrt( - 2*x + 1))/(5292* 
(9*x**2 + 12*x + 4))