\(\int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^{5/2}} \, dx\) [924]

Optimal result

Integrand size = 24, antiderivative size = 74 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^{5/2}} \, dx=-\frac {2 (1-2 x)^{3/2}}{165 (3+5 x)^{3/2}}-\frac {6 \sqrt {1-2 x}}{25 \sqrt {3+5 x}}-\frac {6}{25} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right ) \] Output:

-2/165*(1-2*x)^(3/2)/(3+5*x)^(3/2)-6/25*(1-2*x)^(1/2)/(3+5*x)^(1/2)-6/125* 
arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^{5/2}} \, dx=-\frac {2 \sqrt {1-2 x} (302+485 x)}{825 (3+5 x)^{3/2}}+\frac {6}{25} \sqrt {\frac {2}{5}} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right ) \] Input:

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

Output:

(-2*Sqrt[1 - 2*x]*(302 + 485*x))/(825*(3 + 5*x)^(3/2)) + (6*Sqrt[2/5]*ArcT 
an[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/25
 

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 57, 64, 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.

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {3}{5} \int \frac {\sqrt {1-2 x}}{(5 x+3)^{3/2}}dx-\frac {2 (1-2 x)^{3/2}}{165 (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {3}{5} \left (-\frac {2}{5} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {2 \sqrt {1-2 x}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{165 (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {3}{5} \left (-\frac {4}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {2 \sqrt {1-2 x}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{165 (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {3}{5} \left (-\frac {2}{5} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {2 \sqrt {1-2 x}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{165 (5 x+3)^{3/2}}\)

Input:

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

Output:

(-2*(1 - 2*x)^(3/2))/(165*(3 + 5*x)^(3/2)) + (3*((-2*Sqrt[1 - 2*x])/(5*Sqr 
t[3 + 5*x]) - (2*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/5))/5
 

Defintions of rubi rules used

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, 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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 = 0.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.30

method	result	size
default	\(-\frac {\left (2475 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+2970 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +891 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+4850 x \sqrt {-10 x^{2}-x +3}+3020 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{4125 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\)	\(96\)

Input:

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

Output:

-1/4125*(2475*10^(1/2)*arcsin(20/11*x+1/11)*x^2+2970*10^(1/2)*arcsin(20/11 
*x+1/11)*x+891*10^(1/2)*arcsin(20/11*x+1/11)+4850*x*(-10*x^2-x+3)^(1/2)+30 
20*(-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 = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^{5/2}} \, dx=\frac {99 \, \sqrt {\frac {2}{5}} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {\frac {2}{5}} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{4 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 2 \, {\left (485 \, x + 302\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{825 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \] Input:

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

Output:

1/825*(99*sqrt(2/5)*(25*x^2 + 30*x + 9)*arctan(1/4*sqrt(2/5)*(20*x + 1)*sq 
rt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 2*(485*x + 302)*sqrt(5*x + 
3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^{5/2}} \, dx=-\frac {4 \, \sqrt {-10 \, x^{2} - x + 3}}{15 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {8 \, \sqrt {-10 \, x^{2} - x + 3}}{165 \, {\left (5 \, x + 3\right )}} \] Input:

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

Output:

-4/15*sqrt(-10*x^2 - x + 3)/(25*x^2 + 30*x + 9) + 8/165*sqrt(-10*x^2 - x + 
 3)/(5*x + 3)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (51) = 102\).

Time = 0.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.89 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^{5/2}} \, dx=-\frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{66000 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} - \frac {6}{125} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {13 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{1100 \, \sqrt {5 \, x + 3}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {195 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{4125 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \] Input:

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

Output:

-1/66000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) - 
 6/125*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 13/1100*sqrt(10)*(sq 
rt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 1/4125*sqrt(10)*(5*x + 3 
)^(3/2)*(195*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqrt(2 
)*sqrt(-10*x + 5) - sqrt(22))^3
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^{5/2}} \, dx=\frac {\frac {6 \sqrt {5 x +3}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x}{25}+\frac {18 \sqrt {5 x +3}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{125}-\frac {194 \sqrt {-2 x +1}\, x}{165}-\frac {604 \sqrt {-2 x +1}}{825}}{\sqrt {5 x +3}\, \left (5 x +3\right )} \] Input:

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

Output:

(2*(495*sqrt(5*x + 3)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))*x 
 + 297*sqrt(5*x + 3)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) - 
2425*sqrt( - 2*x + 1)*x - 1510*sqrt( - 2*x + 1)))/(4125*sqrt(5*x + 3)*(5*x 
 + 3))