\(\int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx\) [1035]

Optimal result

Integrand size = 26, antiderivative size = 122 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {7 (1-2 x)^{3/2} \sqrt {3+5 x}}{6 (2+3 x)^2}+\frac {637 \sqrt {1-2 x} \sqrt {3+5 x}}{36 (2+3 x)}-\frac {8}{27} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {3035}{108} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \] Output:

7/6*(1-2*x)^(3/2)*(3+5*x)^(1/2)/(2+3*x)^2+637*(1-2*x)^(1/2)*(3+5*x)^(1/2)/ 
(72+108*x)-8/135*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-3035/108*7^( 
1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {1}{540} \left (\frac {105 \sqrt {1-2 x} \sqrt {3+5 x} (188+261 x)}{(2+3 x)^2}+32 \sqrt {10} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )-15175 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\right ) \] Input:

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

Output:

((105*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(188 + 261*x))/(2 + 3*x)^2 + 32*Sqrt[10] 
*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]] - 15175*Sqrt[7]*ArcTan[Sqrt[1 - 2*x 
]/(Sqrt[7]*Sqrt[3 + 5*x])])/540
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 132, 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 = {109, 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.

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {1}{6} \int \frac {\sqrt {1-2 x} (16 x+223)}{2 (3 x+2)^2 \sqrt {5 x+3}}dx+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{12} \int \frac {\sqrt {1-2 x} (16 x+223)}{(3 x+2)^2 \sqrt {5 x+3}}dx+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 166

\(\displaystyle \frac {1}{12} \left (\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}-\frac {1}{3} \int -\frac {7039-64 x}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} \int \frac {7039-64 x}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} \left (\frac {21245}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {64}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} \left (\frac {21245}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {128}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} \left (\frac {42490}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {128}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} \left (-\frac {128}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {6070}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{12} \left (\frac {1}{6} \left (-\frac {64}{3} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {6070}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )+\frac {637 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{6 (3 x+2)^2}\)

Input:

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

Output:

(7*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/(6*(2 + 3*x)^2) + ((637*Sqrt[1 - 2*x]*Sq 
rt[3 + 5*x])/(3*(2 + 3*x)) + ((-64*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5* 
x]])/3 - (6070*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/3)/6 
)/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 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((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 - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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 = 0.26 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.09

method	result	size
risch	\(-\frac {7 \left (-1+2 x \right ) \sqrt {3+5 x}\, \left (188+261 x \right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{36 \left (2+3 x \right )^{2} \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}-\frac {\left (\frac {4 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )}{135}-\frac {3035 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right )}{216}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(133\)
default	\(-\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (288 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-136575 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+384 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -182100 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +128 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-60700 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-54810 x \sqrt {-10 x^{2}-x +3}-39480 \sqrt {-10 x^{2}-x +3}\right )}{1080 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )^{2}}\)	\(191\)

Input:

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

Output:

-7/36*(-1+2*x)*(3+5*x)^(1/2)*(188+261*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)-(4/135*10^(1/2)*arcsin(20/11*x+1/ 
11)-3035/216*7^(1/2)*arctan(9/14*(20/3+37/3*x)*7^(1/2)/(-90*(2/3+x)^2+67+1 
11*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 = 0.10 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.11 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=-\frac {3035 \, \sqrt {7} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 32 \, \sqrt {\frac {2}{5}} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {\frac {2}{5}} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{4 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 42 \, {\left (261 \, x + 188\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{216 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \] Input:

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

Output:

-1/216*(3035*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)) - 32*sqrt(2/5)*(9*x^2 + 12*x 
+ 4)*arctan(1/4*sqrt(2/5)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 
+ x - 3)) - 42*(261*x + 188)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 
 4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=-\frac {4}{135} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {3035}{216} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {49 \, \sqrt {-10 \, x^{2} - x + 3}}{18 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {203 \, \sqrt {-10 \, x^{2} - x + 3}}{12 \, {\left (3 \, x + 2\right )}} \] Input:

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

Output:

-4/135*sqrt(10)*arcsin(20/11*x + 1/11) + 3035/216*sqrt(7)*arcsin(37/11*x/a 
bs(3*x + 2) + 20/11/abs(3*x + 2)) + 49/18*sqrt(-10*x^2 - x + 3)/(9*x^2 + 1 
2*x + 4) + 203/12*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. 324 vs. \(2 (90) = 180\).

Time = 0.25 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.66 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {607}{432} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {4}{135} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {77 \, {\left (157 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + 25480 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}\right )}}{18 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \] Input:

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

Output:

607/432*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)))) - 4/135*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)))) + 77/18*(157*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 + 
25480*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqr 
t(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. \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}}{{\left (3\,x+2\right )}^3\,\sqrt {5\,x+3}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.54 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {288 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x^{2}+384 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x +128 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )+136575 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}+182100 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +60700 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-136575 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}-182100 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -60700 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+27405 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x +19740 \sqrt {5 x +3}\, \sqrt {-2 x +1}}{4860 x^{2}+6480 x +2160} \] Input:

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

Output:

(288*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))*x**2 + 384*sqrt(10 
)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))*x + 128*sqrt(10)*asin((sqrt( - 
 2*x + 1)*sqrt(5))/sqrt(11)) + 136575*sqrt(7)*atan((sqrt(33) - sqrt(35)*ta 
n(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x**2 + 182100*sqr 
t(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11 
))/2))/sqrt(2))*x + 60700*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt 
( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2)) - 136575*sqrt(7)*atan((sqrt(3 
3) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x 
**2 - 182100*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)* 
sqrt(5))/sqrt(11))/2))/sqrt(2))*x - 60700*sqrt(7)*atan((sqrt(33) + sqrt(35 
)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2)) + 27405*sqrt( 
5*x + 3)*sqrt( - 2*x + 1)*x + 19740*sqrt(5*x + 3)*sqrt( - 2*x + 1))/(540*( 
9*x**2 + 12*x + 4))