\(\int \frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{(2+3 x)^3} \, dx\) [896]

Optimal result

Integrand size = 26, antiderivative size = 144 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{(2+3 x)^3} \, dx=\frac {215}{84} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {59 \sqrt {1-2 x} (3+5 x)^{3/2}}{84 (2+3 x)}-\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{6 (2+3 x)^2}+\frac {25}{9} \sqrt {\frac {5}{2}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )+\frac {2119 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{252 \sqrt {7}} \] Output:

215/84*(1-2*x)^(1/2)*(3+5*x)^(1/2)-59*(1-2*x)^(1/2)*(3+5*x)^(3/2)/(168+252 
*x)-1/6*(1-2*x)^(1/2)*(3+5*x)^(5/2)/(2+3*x)^2+25/18*arcsin(1/11*22^(1/2)*( 
3+5*x)^(1/2))*10^(1/2)+2119/1764*7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/ 
(3+5*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{(2+3 x)^3} \, dx=\frac {\frac {21 \sqrt {1-2 x} \sqrt {3+5 x} \left (380+1039 x+700 x^2\right )}{(2+3 x)^2}-2450 \sqrt {10} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )+2119 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1764} \] Input:

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

Output:

((21*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(380 + 1039*x + 700*x^2))/(2 + 3*x)^2 - 2 
450*Sqrt[10]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]] + 2119*Sqrt[7]*ArcTan[S 
qrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/1764
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {108, 27, 166, 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.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {1}{6} \int \frac {(19-60 x) (5 x+3)^{3/2}}{2 \sqrt {1-2 x} (3 x+2)^2}dx-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{12} \int \frac {(19-60 x) (5 x+3)^{3/2}}{\sqrt {1-2 x} (3 x+2)^2}dx-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 166

\(\displaystyle \frac {1}{12} \left (\frac {1}{21} \int \frac {9 (133-860 x) \sqrt {5 x+3}}{2 \sqrt {1-2 x} (3 x+2)}dx-\frac {59 \sqrt {1-2 x} (5 x+3)^{3/2}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{12} \left (\frac {3}{14} \int \frac {(133-860 x) \sqrt {5 x+3}}{\sqrt {1-2 x} (3 x+2)}dx-\frac {59 \sqrt {1-2 x} (5 x+3)^{3/2}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 171

\(\displaystyle \frac {1}{12} \left (\frac {3}{14} \left (\frac {430}{3} \sqrt {1-2 x} \sqrt {5 x+3}-\frac {1}{6} \int -\frac {2 (3500 x+1627)}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )-\frac {59 \sqrt {1-2 x} (5 x+3)^{3/2}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{12} \left (\frac {3}{14} \left (\frac {1}{3} \int \frac {3500 x+1627}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {430}{3} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {59 \sqrt {1-2 x} (5 x+3)^{3/2}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{12} \left (\frac {3}{14} \left (\frac {1}{3} \left (\frac {3500}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {2119}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {430}{3} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {59 \sqrt {1-2 x} (5 x+3)^{3/2}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{12} \left (\frac {3}{14} \left (\frac {1}{3} \left (\frac {1400}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {2119}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {430}{3} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {59 \sqrt {1-2 x} (5 x+3)^{3/2}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{12} \left (\frac {3}{14} \left (\frac {1}{3} \left (\frac {1400}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {4238}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}\right )+\frac {430}{3} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {59 \sqrt {1-2 x} (5 x+3)^{3/2}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{12} \left (\frac {3}{14} \left (\frac {1}{3} \left (\frac {1400}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {4238 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3 \sqrt {7}}\right )+\frac {430}{3} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {59 \sqrt {1-2 x} (5 x+3)^{3/2}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{12} \left (\frac {3}{14} \left (\frac {1}{3} \left (\frac {700}{3} \sqrt {10} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {4238 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3 \sqrt {7}}\right )+\frac {430}{3} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {59 \sqrt {1-2 x} (5 x+3)^{3/2}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}\)

Input:

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

Output:

-1/6*(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(2 + 3*x)^2 + ((-59*Sqrt[1 - 2*x]*(3 
+ 5*x)^(3/2))/(7*(2 + 3*x)) + (3*((430*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/3 + (( 
700*Sqrt[10]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/3 + (4238*ArcTan[Sqrt[1 - 2 
*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(3*Sqrt[7]))/3))/14)/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 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 = 0.48 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.96

method	result	size
risch	\(-\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \left (700 x^{2}+1039 x +380\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{84 \left (2+3 x \right )^{2} \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}-\frac {\left (-\frac {25 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )}{36}+\frac {2119 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right )}{3528}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(138\)
default	\(-\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (19071 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}-22050 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+25428 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x -29400 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -29400 x^{2} \sqrt {-10 x^{2}-x +3}+8476 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-9800 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-43638 x \sqrt {-10 x^{2}-x +3}-15960 \sqrt {-10 x^{2}-x +3}\right )}{3528 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )^{2}}\)	\(208\)

Input:

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

Output:

-1/84*(-1+2*x)*(3+5*x)^(1/2)*(700*x^2+1039*x+380)/(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)-(-25/36*10^(1/2)*arcsin( 
20/11*x+1/11)+2119/3528*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 = 141, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{(2+3 x)^3} \, dx=\frac {2119 \, \sqrt {7} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 4900 \, \sqrt {\frac {5}{2}} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {\frac {5}{2}} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{10 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 42 \, {\left (700 \, x^{2} + 1039 \, x + 380\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{3528 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \] Input:

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

Output:

1/3528*(2119*sqrt(7)*(9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sq 
rt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 4900*sqrt(5/2)*(9*x^2 + 12* 
x + 4)*arctan(1/10*sqrt(5/2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x 
^2 + x - 3)) + 42*(700*x^2 + 1039*x + 380)*sqrt(5*x + 3)*sqrt(-2*x + 1))/( 
9*x^2 + 12*x + 4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{(2+3 x)^3} \, dx=\frac {25}{36} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {2119}{3528} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {20}{21} \, \sqrt {-10 \, x^{2} - x + 3} + \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{42 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {9 \, \sqrt {-10 \, x^{2} - x + 3}}{28 \, {\left (3 \, x + 2\right )}} \] Input:

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

Output:

25/36*sqrt(10)*arcsin(20/11*x + 1/11) - 2119/3528*sqrt(7)*arcsin(37/11*x/a 
bs(3*x + 2) + 20/11/abs(3*x + 2)) + 20/21*sqrt(-10*x^2 - x + 3) + 1/42*(-1 
0*x^2 - x + 3)^(3/2)/(9*x^2 + 12*x + 4) + 9/28*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. 343 vs. \(2 (106) = 212\).

Time = 0.29 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.38 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{(2+3 x)^3} \, dx=-\frac {2119}{35280} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {25}{36} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {5}{27} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {11 \, {\left (247 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + 87640 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}\right )}}{378 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \] Input:

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

Output:

-2119/35280*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)))) + 25/36*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)))) + 5/27*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5) + 11/378*(2 
47*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)))^3 + 87640*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
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.27 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{(2+3 x)^3} \, dx=\frac {-22050 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x^{2}-29400 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x -9800 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )-19071 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}-25428 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -8476 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+19071 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}+25428 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +8476 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+14700 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x^{2}+21819 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x +7980 \sqrt {5 x +3}\, \sqrt {-2 x +1}}{15876 x^{2}+21168 x +7056} \] Input:

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

Output:

( - 22050*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))*x**2 - 29400* 
sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))*x - 9800*sqrt(10)*asin( 
(sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) - 19071*sqrt(7)*atan((sqrt(33) - sqrt 
(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x**2 - 254 
28*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/s 
qrt(11))/2))/sqrt(2))*x - 8476*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin( 
(sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2)) + 19071*sqrt(7)*atan((sq 
rt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2 
))*x**2 + 25428*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 
1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x + 8476*sqrt(7)*atan((sqrt(33) + sqrt( 
35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2)) + 14700*sqr 
t(5*x + 3)*sqrt( - 2*x + 1)*x**2 + 21819*sqrt(5*x + 3)*sqrt( - 2*x + 1)*x 
+ 7980*sqrt(5*x + 3)*sqrt( - 2*x + 1))/(1764*(9*x**2 + 12*x + 4))