\(\int \frac {\sqrt {1-2 x} (2+3 x)}{\sqrt {3+5 x}} \, dx\) [906]

Optimal result

Integrand size = 24, antiderivative size = 72 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {41}{200} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {3}{20} (1-2 x)^{3/2} \sqrt {3+5 x}+\frac {451 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{200 \sqrt {10}} \] Output:

41/200*(1-2*x)^(1/2)*(3+5*x)^(1/2)-3/20*(1-2*x)^(3/2)*(3+5*x)^(1/2)+451/20 
00*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {\sqrt {5-10 x} \sqrt {3+5 x} (11+60 x)-451 \sqrt {2} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )}{200 \sqrt {5}} \] Input:

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

Output:

(Sqrt[5 - 10*x]*Sqrt[3 + 5*x]*(11 + 60*x) - 451*Sqrt[2]*ArcTan[Sqrt[6 + 10 
*x]/(Sqrt[11] - Sqrt[5 - 10*x])])/(200*Sqrt[5])
 

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 77, 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 = {90, 60, 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.

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {41}{40} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx-\frac {3}{20} (1-2 x)^{3/2} \sqrt {5 x+3}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {41}{40} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{20} (1-2 x)^{3/2} \sqrt {5 x+3}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {41}{40} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{20} (1-2 x)^{3/2} \sqrt {5 x+3}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {41}{40} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{20} (1-2 x)^{3/2} \sqrt {5 x+3}\)

Input:

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

Output:

(-3*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/20 + (41*((Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) 
/5 + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(5*Sqrt[10])))/40
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 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 = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (1200 x \sqrt {-10 x^{2}-x +3}+451 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+220 \sqrt {-10 x^{2}-x +3}\right )}{4000 \sqrt {-10 x^{2}-x +3}}\)	\(70\)
risch	\(-\frac {\left (11+60 x \right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{200 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {451 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{4000 \sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(93\)

Input:

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

Output:

1/4000*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(1200*x*(-10*x^2-x+3)^(1/2)+451*10^(1/2 
)*arcsin(20/11*x+1/11)+220*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {1}{200} \, {\left (60 \, x + 11\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {451}{4000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \] Input:

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

Output:

1/200*(60*x + 11)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 451/4000*sqrt(10)*arctan( 
1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {451}{4000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {3}{10} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {11}{200} \, \sqrt {-10 \, x^{2} - x + 3} \] Input:

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

Output:

451/4000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 3/10*sqrt(-10*x^2 - x + 
3)*x + 11/200*sqrt(-10*x^2 - x + 3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {3}{2000} \, \sqrt {5} {\left (2 \, {\left (20 \, x - 23\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 143 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} + \frac {1}{25} \, \sqrt {5} {\left (11 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + 2 \, \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}\right )} \] Input:

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

Output:

3/2000*sqrt(5)*(2*(20*x - 23)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 143*sqrt(2)* 
arcsin(1/11*sqrt(22)*sqrt(5*x + 3))) + 1/25*sqrt(5)*(11*sqrt(2)*arcsin(1/1 
1*sqrt(22)*sqrt(5*x + 3)) + 2*sqrt(5*x + 3)*sqrt(-10*x + 5))
 

Mupad [B] (verification not implemented)

Time = 7.98 (sec) , antiderivative size = 550, normalized size of antiderivative = 7.64 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {\frac {4\,{\left (\sqrt {1-2\,x}-1\right )}^3}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {8\,\left (\sqrt {1-2\,x}-1\right )}{125\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {32\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}}{\frac {4\,{\left (\sqrt {1-2\,x}-1\right )}^2}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {4}{25}}-\frac {\frac {2427\,{\left (\sqrt {1-2\,x}-1\right )}^3}{3125\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {858\,\left (\sqrt {1-2\,x}-1\right )}{15625\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {2427\,{\left (\sqrt {1-2\,x}-1\right )}^5}{1250\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^5}+\frac {429\,{\left (\sqrt {1-2\,x}-1\right )}^7}{500\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^7}+\frac {96\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{625\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {672\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^4}{625\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {24\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^6}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^6}}{\frac {32\,{\left (\sqrt {1-2\,x}-1\right )}^2}{125\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {24\,{\left (\sqrt {1-2\,x}-1\right )}^4}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {8\,{\left (\sqrt {1-2\,x}-1\right )}^6}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^6}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^8}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^8}+\frac {16}{625}}-\frac {429\,\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,\left (\sqrt {1-2\,x}-1\right )}{2\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}\right )}{1000}+\frac {22\,\sqrt {2}\,\sqrt {5}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {5}\,\left (\sqrt {1-2\,x}-1\right )}{2\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}\right )}{25} \] Input:

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

Output:

((4*((1 - 2*x)^(1/2) - 1)^3)/(25*(3^(1/2) - (5*x + 3)^(1/2))^3) - (8*((1 - 
 2*x)^(1/2) - 1))/(125*(3^(1/2) - (5*x + 3)^(1/2))) + (32*3^(1/2)*((1 - 2* 
x)^(1/2) - 1)^2)/(25*(3^(1/2) - (5*x + 3)^(1/2))^2))/((4*((1 - 2*x)^(1/2) 
- 1)^2)/(5*(3^(1/2) - (5*x + 3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/ 
2) - (5*x + 3)^(1/2))^4 + 4/25) - ((2427*((1 - 2*x)^(1/2) - 1)^3)/(3125*(3 
^(1/2) - (5*x + 3)^(1/2))^3) - (858*((1 - 2*x)^(1/2) - 1))/(15625*(3^(1/2) 
 - (5*x + 3)^(1/2))) - (2427*((1 - 2*x)^(1/2) - 1)^5)/(1250*(3^(1/2) - (5* 
x + 3)^(1/2))^5) + (429*((1 - 2*x)^(1/2) - 1)^7)/(500*(3^(1/2) - (5*x + 3) 
^(1/2))^7) + (96*3^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(625*(3^(1/2) - (5*x + 3 
)^(1/2))^2) + (672*3^(1/2)*((1 - 2*x)^(1/2) - 1)^4)/(625*(3^(1/2) - (5*x + 
 3)^(1/2))^4) + (24*3^(1/2)*((1 - 2*x)^(1/2) - 1)^6)/(25*(3^(1/2) - (5*x + 
 3)^(1/2))^6))/((32*((1 - 2*x)^(1/2) - 1)^2)/(125*(3^(1/2) - (5*x + 3)^(1/ 
2))^2) + (24*((1 - 2*x)^(1/2) - 1)^4)/(25*(3^(1/2) - (5*x + 3)^(1/2))^4) + 
 (8*((1 - 2*x)^(1/2) - 1)^6)/(5*(3^(1/2) - (5*x + 3)^(1/2))^6) + ((1 - 2*x 
)^(1/2) - 1)^8/(3^(1/2) - (5*x + 3)^(1/2))^8 + 16/625) - (429*10^(1/2)*ata 
n((10^(1/2)*((1 - 2*x)^(1/2) - 1))/(2*(3^(1/2) - (5*x + 3)^(1/2)))))/1000 
+ (22*2^(1/2)*5^(1/2)*atan((2^(1/2)*5^(1/2)*((1 - 2*x)^(1/2) - 1))/(2*(3^( 
1/2) - (5*x + 3)^(1/2)))))/25
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{\sqrt {3+5 x}} \, dx=-\frac {451 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2000}+\frac {3 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x}{10}+\frac {11 \sqrt {5 x +3}\, \sqrt {-2 x +1}}{200} \] Input:

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

Output:

( - 451*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) + 600*sqrt(5*x 
+ 3)*sqrt( - 2*x + 1)*x + 110*sqrt(5*x + 3)*sqrt( - 2*x + 1))/2000