\(\int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx\) [1056]

Optimal result

Integrand size = 26, antiderivative size = 108 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=-\frac {22 (1-2 x)^{3/2}}{15 (3+5 x)^{3/2}}+\frac {814 \sqrt {1-2 x}}{25 \sqrt {3+5 x}}-\frac {8}{75} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {98}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \] Output:

-22/15*(1-2*x)^(3/2)/(3+5*x)^(3/2)+814/25*(1-2*x)^(1/2)/(3+5*x)^(1/2)-8/37 
5*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-98/3*7^(1/2)*arctan(1/7*(1- 
2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))
 

Mathematica [A] (verified)

Time = 1.15 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.53 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=\frac {2}{375} \left (\frac {55 \sqrt {1-2 x} (328+565 x)}{(3+5 x)^{3/2}}+6125 \sqrt {7} \arctan \left (\frac {\sqrt {2 \left (34+\sqrt {1155}\right )} \sqrt {3+5 x}}{-\sqrt {11}+\sqrt {5-10 x}}\right )+8 \sqrt {10} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )+6125 \sqrt {7} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {34+\sqrt {1155}} \left (-\sqrt {11}+\sqrt {5-10 x}\right )}\right )\right ) \] Input:

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

Output:

(2*((55*Sqrt[1 - 2*x]*(328 + 565*x))/(3 + 5*x)^(3/2) + 6125*Sqrt[7]*ArcTan 
[(Sqrt[2*(34 + Sqrt[1155])]*Sqrt[3 + 5*x])/(-Sqrt[11] + Sqrt[5 - 10*x])] + 
 8*Sqrt[10]*ArcTan[Sqrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 - 10*x])] + 6125*Sqrt 
[7]*ArcTan[Sqrt[6 + 10*x]/(Sqrt[34 + Sqrt[1155]]*(-Sqrt[11] + Sqrt[5 - 10* 
x]))]))/375
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {109, 27, 167, 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.

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

\(\Big \downarrow \) 109

\(\displaystyle -\frac {2}{15} \int \frac {3 (79-4 x) \sqrt {1-2 x}}{2 (3 x+2) (5 x+3)^{3/2}}dx-\frac {22 (1-2 x)^{3/2}}{15 (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{5} \int \frac {(79-4 x) \sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}dx-\frac {22 (1-2 x)^{3/2}}{15 (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 167

\(\displaystyle \frac {1}{5} \left (\frac {814 \sqrt {1-2 x}}{5 \sqrt {5 x+3}}-\frac {2}{5} \int -\frac {2853-8 x}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )-\frac {22 (1-2 x)^{3/2}}{15 (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{5} \int \frac {2853-8 x}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {814 \sqrt {1-2 x}}{5 \sqrt {5 x+3}}\right )-\frac {22 (1-2 x)^{3/2}}{15 (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{5} \left (\frac {1}{5} \left (\frac {8575}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {8}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )+\frac {814 \sqrt {1-2 x}}{5 \sqrt {5 x+3}}\right )-\frac {22 (1-2 x)^{3/2}}{15 (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{5} \left (\frac {1}{5} \left (\frac {8575}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {16}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {814 \sqrt {1-2 x}}{5 \sqrt {5 x+3}}\right )-\frac {22 (1-2 x)^{3/2}}{15 (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{5} \left (\frac {1}{5} \left (\frac {17150}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {16}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {814 \sqrt {1-2 x}}{5 \sqrt {5 x+3}}\right )-\frac {22 (1-2 x)^{3/2}}{15 (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{5} \left (\frac {1}{5} \left (-\frac {16}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {2450}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )+\frac {814 \sqrt {1-2 x}}{5 \sqrt {5 x+3}}\right )-\frac {22 (1-2 x)^{3/2}}{15 (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{5} \left (\frac {1}{5} \left (-\frac {8}{3} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {2450}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )+\frac {814 \sqrt {1-2 x}}{5 \sqrt {5 x+3}}\right )-\frac {22 (1-2 x)^{3/2}}{15 (5 x+3)^{3/2}}\)

Input:

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

Output:

(-22*(1 - 2*x)^(3/2))/(15*(3 + 5*x)^(3/2)) + ((814*Sqrt[1 - 2*x])/(5*Sqrt[ 
3 + 5*x]) + ((-8*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/3 - (2450*Sqr 
t[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/3)/5)/5
 

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 167

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] && LtQ[m, -1] && GtQ[n, 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(183\) vs. \(2(76)=152\).

Time = 0.24 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.70

method	result	size
default	\(-\frac {\left (100 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-153125 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+120 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -183750 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +36 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-55125 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-62150 x \sqrt {-10 x^{2}-x +3}-36080 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{375 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\)	\(184\)

Input:

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

Output:

-1/375*(100*10^(1/2)*arcsin(20/11*x+1/11)*x^2-153125*7^(1/2)*arctan(1/14*( 
37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2+120*10^(1/2)*arcsin(20/11*x+1/11 
)*x-183750*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+36 
*10^(1/2)*arcsin(20/11*x+1/11)-55125*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2) 
/(-10*x^2-x+3)^(1/2))-62150*x*(-10*x^2-x+3)^(1/2)-36080*(-10*x^2-x+3)^(1/2 
))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.26 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=-\frac {1225 \, \sqrt {7} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 4 \, \sqrt {\frac {2}{5}} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {\frac {2}{5}} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{4 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 22 \, {\left (565 \, x + 328\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{75 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \] Input:

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

Output:

-1/75*(1225*sqrt(7)*(25*x^2 + 30*x + 9)*arctan(1/14*sqrt(7)*(37*x + 20)*sq 
rt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 4*sqrt(2/5)*(25*x^2 + 30*x 
+ 9)*arctan(1/4*sqrt(2/5)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 
+ x - 3)) - 22*(565*x + 328)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x 
+ 9)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (76) = 152\).

Time = 0.12 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.51 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=\frac {626336 \, x^{2}}{17788815 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {16 \, x^{3}}{15 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {4}{375} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {49}{3} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {313168}{88944075} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {5905573412 \, x}{88944075 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {3286544 \, x^{2}}{735075 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {3102773174}{88944075 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {11007824 \, x}{735075 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {2075846}{245025 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \] Input:

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

Output:

626336/17788815*x^2/sqrt(-10*x^2 - x + 3) - 16/15*x^3/(-10*x^2 - x + 3)^(3 
/2) - 4/375*sqrt(10)*arcsin(20/11*x + 1/11) + 49/3*sqrt(7)*arcsin(37/11*x/ 
abs(3*x + 2) + 20/11/abs(3*x + 2)) + 313168/88944075*sqrt(-10*x^2 - x + 3) 
 - 5905573412/88944075*x/sqrt(-10*x^2 - x + 3) + 3286544/735075*x^2/(-10*x 
^2 - x + 3)^(3/2) + 3102773174/88944075/sqrt(-10*x^2 - x + 3) + 11007824/7 
35075*x/(-10*x^2 - x + 3)^(3/2) - 2075846/245025/(-10*x^2 - x + 3)^(3/2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (76) = 152\).

Time = 0.22 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.42 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=-\frac {11}{6000} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {49}{30} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {4}{375} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {407}{250} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} \] Input:

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

Output:

-11/6000*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 + 49/30*sqrt(70)*sqr 
t(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)))) - 4 
/375*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)))) + 407/ 
250*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)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.26 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=\frac {\frac {8 \sqrt {5 x +3}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x}{75}+\frac {8 \sqrt {5 x +3}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{125}+\frac {490 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x}{3}+98 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-\frac {490 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x}{3}-98 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+\frac {2486 \sqrt {-2 x +1}\, x}{15}+\frac {7216 \sqrt {-2 x +1}}{75}}{\sqrt {5 x +3}\, \left (5 x +3\right )} \] Input:

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

Output:

(2*(20*sqrt(5*x + 3)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))*x 
+ 12*sqrt(5*x + 3)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) + 30 
625*sqrt(5*x + 3)*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x 
+ 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x + 18375*sqrt(5*x + 3)*sqrt(7)*atan( 
(sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqr 
t(2)) - 30625*sqrt(5*x + 3)*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sq 
rt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x - 18375*sqrt(5*x + 3)*sqr 
t(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11 
))/2))/sqrt(2)) + 31075*sqrt( - 2*x + 1)*x + 18040*sqrt( - 2*x + 1)))/(375 
*sqrt(5*x + 3)*(5*x + 3))