3.25.41 \(\int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{3/2}} \, dx\) [2441]

3.25.41.1 Optimal result
3.25.41.2 Mathematica [A] (verified)
3.25.41.3 Rubi [A] (verified)
3.25.41.4 Maple [A] (verified)
3.25.41.5 Fricas [A] (verification not implemented)
3.25.41.6 Sympy [F]
3.25.41.7 Maxima [A] (verification not implemented)
3.25.41.8 Giac [B] (verification not implemented)
3.25.41.9 Mupad [F(-1)]

3.25.41.1 Optimal result

Integrand size = 26, antiderivative size = 108 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{3/2}} \, dx=-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {3+5 x}}-\frac {128}{75} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {338}{225} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )+\frac {98}{9} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \]

output
338/1125*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+98/9*arctan(1/7*(1-2 
*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-22/5*(1-2*x)^(3/2)/(3+5*x)^(1/2)- 
128/75*(1-2*x)^(1/2)*(3+5*x)^(1/2)
 
3.25.41.2 Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{3/2}} \, dx=\frac {30 \sqrt {1-2 x} (-357+10 x)-338 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )+12250 \sqrt {7} \sqrt {3+5 x} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1125 \sqrt {3+5 x}} \]

input
Integrate[(1 - 2*x)^(5/2)/((2 + 3*x)*(3 + 5*x)^(3/2)),x]
 
output
(30*Sqrt[1 - 2*x]*(-357 + 10*x) - 338*Sqrt[30 + 50*x]*ArcTan[Sqrt[5/2 - 5* 
x]/Sqrt[3 + 5*x]] + 12250*Sqrt[7]*Sqrt[3 + 5*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt 
[7]*Sqrt[3 + 5*x])])/(1125*Sqrt[3 + 5*x])
 
3.25.41.3 Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {109, 27, 171, 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.

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

\(\Big \downarrow \) 109

\(\displaystyle -\frac {2}{5} \int \frac {\sqrt {1-2 x} (128 x+167)}{2 (3 x+2) \sqrt {5 x+3}}dx-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{5} \int \frac {\sqrt {1-2 x} (128 x+167)}{(3 x+2) \sqrt {5 x+3}}dx-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 171

\(\displaystyle \frac {1}{5} \left (-\frac {1}{15} \int \frac {2633-338 x}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {128}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{5} \left (\frac {1}{15} \left (\frac {338}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {8575}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )-\frac {128}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{5} \left (\frac {1}{15} \left (\frac {676}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {8575}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )-\frac {128}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{5} \left (\frac {1}{15} \left (\frac {676}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {17150}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}\right )-\frac {128}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{5} \left (\frac {1}{15} \left (\frac {676}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {2450}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )-\frac {128}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{5} \left (\frac {1}{15} \left (\frac {338}{3} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {2450}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )-\frac {128}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {22 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\)

input
Int[(1 - 2*x)^(5/2)/((2 + 3*x)*(3 + 5*x)^(3/2)),x]
 
output
(-22*(1 - 2*x)^(3/2))/(5*Sqrt[3 + 5*x]) + ((-128*Sqrt[1 - 2*x]*Sqrt[3 + 5* 
x])/15 + ((338*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/3 + (2450*Sqrt[ 
7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/3)/15)/5
 

3.25.41.3.1 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 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]
 
3.25.41.4 Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.29

method result size
default \(\frac {\left (845 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -30625 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +507 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-18375 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+300 x \sqrt {-10 x^{2}-x +3}-10710 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{1125 \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) \(139\)

input
int((1-2*x)^(5/2)/(2+3*x)/(3+5*x)^(3/2),x,method=_RETURNVERBOSE)
 
output
1/1125*(845*10^(1/2)*arcsin(20/11*x+1/11)*x-30625*7^(1/2)*arctan(1/14*(37* 
x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+507*10^(1/2)*arcsin(20/11*x+1/11)-183 
75*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+300*x*(-10*x 
^2-x+3)^(1/2)-10710*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2) 
/(3+5*x)^(1/2)
 
3.25.41.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.18 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{3/2}} \, dx=-\frac {169 \, \sqrt {5} \sqrt {2} {\left (5 \, x + 3\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 6125 \, \sqrt {7} {\left (5 \, x + 3\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 30 \, {\left (10 \, x - 357\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1125 \, {\left (5 \, x + 3\right )}} \]

input
integrate((1-2*x)^(5/2)/(2+3*x)/(3+5*x)^(3/2),x, algorithm="fricas")
 
output
-1/1125*(169*sqrt(5)*sqrt(2)*(5*x + 3)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 
 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 6125*sqrt(7)*(5*x + 3 
)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x 
 - 3)) - 30*(10*x - 357)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(5*x + 3)
 
3.25.41.6 Sympy [F]

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

input
integrate((1-2*x)**(5/2)/(2+3*x)/(3+5*x)**(3/2),x)
 
output
Integral((1 - 2*x)**(5/2)/((3*x + 2)*(5*x + 3)**(3/2)), x)
 
3.25.41.7 Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{3/2}} \, dx=-\frac {8 \, x^{2}}{15 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {169}{1125} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {49}{9} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {1448 \, x}{75 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {238}{25 \, \sqrt {-10 \, x^{2} - x + 3}} \]

input
integrate((1-2*x)^(5/2)/(2+3*x)/(3+5*x)^(3/2),x, algorithm="maxima")
 
output
-8/15*x^2/sqrt(-10*x^2 - x + 3) + 169/1125*sqrt(10)*arcsin(20/11*x + 1/11) 
 - 49/9*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 1448/7 
5*x/sqrt(-10*x^2 - x + 3) - 238/25/sqrt(-10*x^2 - x + 3)
 
3.25.41.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (76) = 152\).

Time = 0.36 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.03 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^{3/2}} \, dx=-\frac {49}{90} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {169}{1125} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {4}{375} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {121}{250} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} \]

input
integrate((1-2*x)^(5/2)/(2+3*x)/(3+5*x)^(3/2),x, algorithm="giac")
 
output
-49/90*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sq 
rt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5 
) - sqrt(22)))) + 169/1125*sqrt(10)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)*((sq 
rt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5 
) - sqrt(22)))) + 4/375*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 121/250*sq 
rt(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)))
 
3.25.41.9 Mupad [F(-1)]

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

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