\(\int \frac {(1-2 x)^{3/2}}{(2+3 x) \sqrt {3+5 x}} \, dx\) [968]

Optimal result

Integrand size = 26, antiderivative size = 86 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=-\frac {2}{15} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {103}{45} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {14}{9} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \] Output:

-2/15*(1-2*x)^(1/2)*(3+5*x)^(1/2)-103/225*arcsin(1/11*22^(1/2)*(3+5*x)^(1/ 
2))*10^(1/2)-14/9*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 = 86, normalized size of antiderivative = 1.00 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=\frac {1}{225} \left (-30 \sqrt {1-2 x} \sqrt {3+5 x}+103 \sqrt {10} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )-350 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\right ) \] Input:

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

Output:

(-30*Sqrt[1 - 2*x]*Sqrt[3 + 5*x] + 103*Sqrt[10]*ArcTan[Sqrt[5/2 - 5*x]/Sqr 
t[3 + 5*x]] - 350*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/2 
25
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {113, 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.

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

\(\Big \downarrow \) 113

\(\displaystyle \frac {1}{15} \int \frac {13-103 x}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {2}{15} \sqrt {1-2 x} \sqrt {5 x+3}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{15} \left (\frac {245}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {103}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )-\frac {2}{15} \sqrt {1-2 x} \sqrt {5 x+3}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{15} \left (\frac {245}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {206}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {2}{15} \sqrt {1-2 x} \sqrt {5 x+3}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{15} \left (\frac {490}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {206}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {2}{15} \sqrt {1-2 x} \sqrt {5 x+3}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{15} \left (-\frac {206}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {70}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )-\frac {2}{15} \sqrt {1-2 x} \sqrt {5 x+3}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{15} \left (-\frac {103}{3} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {70}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )-\frac {2}{15} \sqrt {1-2 x} \sqrt {5 x+3}\)

Input:

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

Output:

(-2*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/15 + ((-103*Sqrt[2/5]*ArcSin[Sqrt[2/11]*S 
qrt[3 + 5*x]])/3 - (70*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x] 
)])/3)/15
 

Defintions of rubi rules used

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 113

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 
)/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1))   Int[(a + b*x) 
^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m 
 - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m 
 + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & 
& GtQ[m, 1] && NeQ[m + n + p + 1, 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 = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (103 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-350 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+60 \sqrt {-10 x^{2}-x +3}\right )}{450 \sqrt {-10 x^{2}-x +3}}\)	\(83\)
risch	\(\frac {2 \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{15 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {\left (-\frac {103 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )}{450}+\frac {7 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right )}{9}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(120\)

Input:

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

Output:

-1/450*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(103*10^(1/2)*arcsin(20/11*x+1/11)-350* 
7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+60*(-10*x^2-x+3 
)^(1/2))/(-10*x^2-x+3)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.13 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=-\frac {7}{9} \, \sqrt {7} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + \frac {103}{90} \, \sqrt {\frac {2}{5}} \arctan \left (\frac {\sqrt {\frac {2}{5}} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{4 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - \frac {2}{15} \, \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} \] Input:

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

Output:

-7/9*sqrt(7)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/ 
(10*x^2 + x - 3)) + 103/90*sqrt(2/5)*arctan(1/4*sqrt(2/5)*(20*x + 1)*sqrt( 
5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 2/15*sqrt(5*x + 3)*sqrt(-2*x + 
 1)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.63 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=-\frac {103}{450} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {7}{9} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {2}{15} \, \sqrt {-10 \, x^{2} - x + 3} \] Input:

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

Output:

-103/450*sqrt(10)*arcsin(20/11*x + 1/11) + 7/9*sqrt(7)*arcsin(37/11*x/abs( 
3*x + 2) + 20/11/abs(3*x + 2)) - 2/15*sqrt(-10*x^2 - x + 3)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (60) = 120\).

Time = 0.17 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.86 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=\frac {7}{90} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {103}{450} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {2}{75} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} \] Input:

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

Output:

7/90*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)))) - 103/450*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)))) - 2/75*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.16 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=\frac {103 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{225}+\frac {14 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )}{9}-\frac {14 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )}{9}-\frac {2 \sqrt {5 x +3}\, \sqrt {-2 x +1}}{15} \] Input:

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

Output:

(103*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) + 350*sqrt(7)*atan 
((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sq 
rt(2)) - 350*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)* 
sqrt(5))/sqrt(11))/2))/sqrt(2)) - 30*sqrt(5*x + 3)*sqrt( - 2*x + 1))/225