\(\int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx\) [988]

Optimal result

Integrand size = 26, antiderivative size = 104 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=-\frac {19 \sqrt {1-2 x}}{(3+5 x)^{3/2}}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)^{3/2}}+\frac {517 \sqrt {1-2 x}}{3 \sqrt {3+5 x}}-169 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \] Output:

-19*(1-2*x)^(1/2)/(3+5*x)^(3/2)+7/3*(1-2*x)^(1/2)/(2+3*x)/(3+5*x)^(3/2)+51 
7/3*(1-2*x)^(1/2)/(3+5*x)^(1/2)-169*7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/ 
2)/(3+5*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=\frac {\sqrt {1-2 x} \left (2995+9652 x+7755 x^2\right )-507 \sqrt {7} \sqrt {3+5 x} \left (6+19 x+15 x^2\right ) \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{3 (2+3 x) (3+5 x)^{3/2}} \] Input:

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

Output:

(Sqrt[1 - 2*x]*(2995 + 9652*x + 7755*x^2) - 507*Sqrt[7]*Sqrt[3 + 5*x]*(6 + 
 19*x + 15*x^2)*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(3*(2 + 3*x 
)*(3 + 5*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {107, 105, 105, 104, 217}

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.

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

\(\Big \downarrow \) 107

\(\displaystyle \frac {169}{14} \int \frac {(1-2 x)^{3/2}}{(3 x+2) (5 x+3)^{5/2}}dx+\frac {3 (1-2 x)^{5/2}}{7 (3 x+2) (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 105

\(\displaystyle \frac {169}{14} \left (-7 \int \frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}dx-\frac {2 (1-2 x)^{3/2}}{3 (5 x+3)^{3/2}}\right )+\frac {3 (1-2 x)^{5/2}}{7 (3 x+2) (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 105

\(\displaystyle \frac {169}{14} \left (-7 \left (-7 \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {2 \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{3 (5 x+3)^{3/2}}\right )+\frac {3 (1-2 x)^{5/2}}{7 (3 x+2) (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {169}{14} \left (-7 \left (-14 \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {2 \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{3 (5 x+3)^{3/2}}\right )+\frac {3 (1-2 x)^{5/2}}{7 (3 x+2) (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {169}{14} \left (-7 \left (2 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{3 (5 x+3)^{3/2}}\right )+\frac {3 (1-2 x)^{5/2}}{7 (3 x+2) (5 x+3)^{3/2}}\)

Input:

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

Output:

(3*(1 - 2*x)^(5/2))/(7*(2 + 3*x)*(3 + 5*x)^(3/2)) + (169*((-2*(1 - 2*x)^(3 
/2))/(3*(3 + 5*x)^(3/2)) - 7*((-2*Sqrt[1 - 2*x])/Sqrt[3 + 5*x] + 2*Sqrt[7] 
*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])))/14
 

Defintions of rubi rules used

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 105

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 + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(201\) vs. \(2(81)=162\).

Time = 0.38 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.94

method	result	size
default	\(\frac {\left (38025 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{3}+70980 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+44109 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +15510 x^{2} \sqrt {-10 x^{2}-x +3}+9126 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+19304 x \sqrt {-10 x^{2}-x +3}+5990 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{6 \left (2+3 x \right ) \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\)	\(202\)

Input:

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

Output:

1/6*(38025*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^3+ 
70980*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2+44109 
*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+15510*x^2*(- 
10*x^2-x+3)^(1/2)+9126*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3) 
^(1/2))+19304*x*(-10*x^2-x+3)^(1/2)+5990*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2 
)/(2+3*x)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.97 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=-\frac {507 \, \sqrt {7} {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 2 \, {\left (7755 \, x^{2} + 9652 \, x + 2995\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{6 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} \] Input:

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

Output:

-1/6*(507*sqrt(7)*(75*x^3 + 140*x^2 + 87*x + 18)*arctan(1/14*sqrt(7)*(37*x 
 + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 2*(7755*x^2 + 9652 
*x + 2995)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(75*x^3 + 140*x^2 + 87*x + 18)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.16 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=\frac {169}{2} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {1034 \, x}{3 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {2699}{15 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {3902 \, x}{45 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {343}{27 \, {\left (3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + 2 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}\right )}} - \frac {6343}{135 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \] Input:

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

Output:

169/2*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 1034/3*x 
/sqrt(-10*x^2 - x + 3) + 2699/15/sqrt(-10*x^2 - x + 3) + 3902/45*x/(-10*x^ 
2 - x + 3)^(3/2) + 343/27/(3*(-10*x^2 - x + 3)^(3/2)*x + 2*(-10*x^2 - x + 
3)^(3/2)) - 6343/135/(-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. 313 vs. \(2 (81) = 162\).

Time = 0.23 (sec) , antiderivative size = 313, normalized size of antiderivative = 3.01 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=-\frac {1}{240} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {169}{20} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {34}{5} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} + \frac {462 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{{\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280} \] Input:

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

Output:

-1/240*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sq 
rt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 + 169/20*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)))) + 34 
/5*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))) + 462*sqrt(10)*((sqrt(2)*sqr 
t(-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)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.85 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=\frac {7605 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}+9633 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +3042 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-7605 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}-9633 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -3042 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+7755 \sqrt {-2 x +1}\, x^{2}+9652 \sqrt {-2 x +1}\, x +2995 \sqrt {-2 x +1}}{3 \sqrt {5 x +3}\, \left (15 x^{2}+19 x +6\right )} \] Input:

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

Output:

(7605*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**2 + 9633*sqrt(5*x + 3)*sqrt(7)*a 
tan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2)) 
/sqrt(2))*x + 3042*sqrt(5*x + 3)*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asi 
n((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2)) - 7605*sqrt(5*x + 3)*s 
qrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt( 
11))/2))/sqrt(2))*x**2 - 9633*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 - 3042*sq 
rt(5*x + 3)*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*s 
qrt(5))/sqrt(11))/2))/sqrt(2)) + 7755*sqrt( - 2*x + 1)*x**2 + 9652*sqrt( - 
 2*x + 1)*x + 2995*sqrt( - 2*x + 1))/(3*sqrt(5*x + 3)*(15*x**2 + 19*x + 6) 
)