3.19.46 \(\int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx\) [1846]

3.19.46.1 Optimal result
3.19.46.2 Mathematica [A] (verified)
3.19.46.3 Rubi [A] (verified)
3.19.46.4 Maple [A] (verified)
3.19.46.5 Fricas [A] (verification not implemented)
3.19.46.6 Sympy [A] (verification not implemented)
3.19.46.7 Maxima [A] (verification not implemented)
3.19.46.8 Giac [A] (verification not implemented)
3.19.46.9 Mupad [B] (verification not implemented)

3.19.46.1 Optimal result

Integrand size = 24, antiderivative size = 131 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {1045 \sqrt {1-2 x}}{14 (3+5 x)}+\frac {\sqrt {1-2 x}}{2 (2+3 x)^2 (3+5 x)}+\frac {52 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)}-\frac {7209}{7} \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+1000 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

output
-7209/49*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+1000/11*arctanh(1/11 
*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1045/14*(1-2*x)^(1/2)/(3+5*x)+1/2*(1-2*x 
)^(1/2)/(2+3*x)^2/(3+5*x)+52/7*(1-2*x)^(1/2)/(2+3*x)/(3+5*x)
 
3.19.46.2 Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {\sqrt {1-2 x} \left (3965+12228 x+9405 x^2\right )}{14 (2+3 x)^2 (3+5 x)}-\frac {7209}{7} \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+1000 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

input
Integrate[Sqrt[1 - 2*x]/((2 + 3*x)^3*(3 + 5*x)^2),x]
 
output
-1/14*(Sqrt[1 - 2*x]*(3965 + 12228*x + 9405*x^2))/((2 + 3*x)^2*(3 + 5*x)) 
- (7209*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/7 + 1000*Sqrt[5/11]*Ar 
cTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]
 
3.19.46.3 Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {110, 25, 168, 168, 27, 174, 73, 219}

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.

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

\(\Big \downarrow \) 110

\(\displaystyle \frac {\sqrt {1-2 x}}{2 (3 x+2)^2 (5 x+3)}-\frac {1}{2} \int -\frac {18-25 x}{\sqrt {1-2 x} (3 x+2)^2 (5 x+3)^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \int \frac {18-25 x}{\sqrt {1-2 x} (3 x+2)^2 (5 x+3)^2}dx+\frac {\sqrt {1-2 x}}{2 (3 x+2)^2 (5 x+3)}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{2} \left (\frac {1}{7} \int \frac {1363-1560 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)^2}dx+\frac {104 \sqrt {1-2 x}}{7 (3 x+2) (5 x+3)}\right )+\frac {\sqrt {1-2 x}}{2 (3 x+2)^2 (5 x+3)}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{2} \left (\frac {1}{7} \left (-\frac {1}{11} \int \frac {11 (5119-3135 x)}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {1045 \sqrt {1-2 x}}{5 x+3}\right )+\frac {104 \sqrt {1-2 x}}{7 (3 x+2) (5 x+3)}\right )+\frac {\sqrt {1-2 x}}{2 (3 x+2)^2 (5 x+3)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {1}{7} \left (-\int \frac {5119-3135 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {1045 \sqrt {1-2 x}}{5 x+3}\right )+\frac {104 \sqrt {1-2 x}}{7 (3 x+2) (5 x+3)}\right )+\frac {\sqrt {1-2 x}}{2 (3 x+2)^2 (5 x+3)}\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{2} \left (\frac {1}{7} \left (21627 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx-35000 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {1045 \sqrt {1-2 x}}{5 x+3}\right )+\frac {104 \sqrt {1-2 x}}{7 (3 x+2) (5 x+3)}\right )+\frac {\sqrt {1-2 x}}{2 (3 x+2)^2 (5 x+3)}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (\frac {1}{7} \left (35000 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-21627 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {1045 \sqrt {1-2 x}}{5 x+3}\right )+\frac {104 \sqrt {1-2 x}}{7 (3 x+2) (5 x+3)}\right )+\frac {\sqrt {1-2 x}}{2 (3 x+2)^2 (5 x+3)}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {1}{7} \left (-14418 \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+14000 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )-\frac {1045 \sqrt {1-2 x}}{5 x+3}\right )+\frac {104 \sqrt {1-2 x}}{7 (3 x+2) (5 x+3)}\right )+\frac {\sqrt {1-2 x}}{2 (3 x+2)^2 (5 x+3)}\)

input
Int[Sqrt[1 - 2*x]/((2 + 3*x)^3*(3 + 5*x)^2),x]
 
output
Sqrt[1 - 2*x]/(2*(2 + 3*x)^2*(3 + 5*x)) + ((104*Sqrt[1 - 2*x])/(7*(2 + 3*x 
)*(3 + 5*x)) + ((-1045*Sqrt[1 - 2*x])/(3 + 5*x) - 14418*Sqrt[3/7]*ArcTanh[ 
Sqrt[3/7]*Sqrt[1 - 2*x]] + 14000*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2* 
x]])/7)/2
 

3.19.46.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 110
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[1/((m + 1)*(b*e - a*f))   Int[(a + b*x)^(m + 1) 
*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + 
p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt 
Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, 
 m + n])
 

rule 168
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(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] + S 
imp[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*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 174
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

rule 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 
3.19.46.4 Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.58

method result size
risch \(\frac {18810 x^{3}+15051 x^{2}-4298 x -3965}{14 \left (3+5 x \right ) \sqrt {1-2 x}\, \left (2+3 x \right )^{2}}-\frac {7209 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}+\frac {1000 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{11}\) \(76\)
derivativedivides \(\frac {10 \sqrt {1-2 x}}{-\frac {6}{5}-2 x}+\frac {1000 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{11}+\frac {\frac {1251 \left (1-2 x \right )^{\frac {3}{2}}}{7}-423 \sqrt {1-2 x}}{\left (-4-6 x \right )^{2}}-\frac {7209 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}\) \(82\)
default \(\frac {10 \sqrt {1-2 x}}{-\frac {6}{5}-2 x}+\frac {1000 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{11}+\frac {\frac {1251 \left (1-2 x \right )^{\frac {3}{2}}}{7}-423 \sqrt {1-2 x}}{\left (-4-6 x \right )^{2}}-\frac {7209 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}\) \(82\)
pseudoelliptic \(\frac {-158598 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right )^{2} \left (3+5 x \right ) \sqrt {21}+98000 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (2+3 x \right )^{2} \left (3+5 x \right ) \sqrt {55}-77 \sqrt {1-2 x}\, \left (9405 x^{2}+12228 x +3965\right )}{1078 \left (2+3 x \right )^{2} \left (3+5 x \right )}\) \(97\)
trager \(-\frac {\left (9405 x^{2}+12228 x +3965\right ) \sqrt {1-2 x}}{14 \left (2+3 x \right )^{2} \left (3+5 x \right )}-\frac {500 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{11}+\frac {7209 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{98}\) \(123\)

input
int((1-2*x)^(1/2)/(2+3*x)^3/(3+5*x)^2,x,method=_RETURNVERBOSE)
 
output
1/14*(18810*x^3+15051*x^2-4298*x-3965)/(3+5*x)/(1-2*x)^(1/2)/(2+3*x)^2-720 
9/49*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+1000/11*arctanh(1/11*55^ 
(1/2)*(1-2*x)^(1/2))*55^(1/2)
 
3.19.46.5 Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=\frac {49000 \, \sqrt {11} \sqrt {5} {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 79299 \, \sqrt {7} \sqrt {3} {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 77 \, {\left (9405 \, x^{2} + 12228 \, x + 3965\right )} \sqrt {-2 \, x + 1}}{1078 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )}} \]

input
integrate((1-2*x)^(1/2)/(2+3*x)^3/(3+5*x)^2,x, algorithm="fricas")
 
output
1/1078*(49000*sqrt(11)*sqrt(5)*(45*x^3 + 87*x^2 + 56*x + 12)*log(-(sqrt(11 
)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) + 79299*sqrt(7)*sqrt(3)*(45 
*x^3 + 87*x^2 + 56*x + 12)*log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1) + 3*x - 5)/ 
(3*x + 2)) - 77*(9405*x^2 + 12228*x + 3965)*sqrt(-2*x + 1))/(45*x^3 + 87*x 
^2 + 56*x + 12)
 
3.19.46.6 Sympy [A] (verification not implemented)

Time = 52.05 (sec) , antiderivative size = 488, normalized size of antiderivative = 3.73 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=\frac {505 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{7} - \frac {505 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{11} - 816 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right ) + 168 \left (\begin {cases} \frac {\sqrt {21} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )^{2}}\right )}{1029} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right ) - 1100 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right ) \]

input
integrate((1-2*x)**(1/2)/(2+3*x)**3/(3+5*x)**2,x)
 
output
505*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(2 
1)/3))/7 - 505*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2* 
x) + sqrt(55)/5))/11 - 816*Piecewise((sqrt(21)*(-log(sqrt(21)*sqrt(1 - 2*x 
)/7 - 1)/4 + log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/4 - 1/(4*(sqrt(21)*sqrt(1 - 
 2*x)/7 + 1)) - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)))/147, (sqrt(1 - 2*x) 
> -sqrt(21)/3) & (sqrt(1 - 2*x) < sqrt(21)/3))) + 168*Piecewise((sqrt(21)* 
(3*log(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/16 - 3*log(sqrt(21)*sqrt(1 - 2*x)/7 + 
 1)/16 + 3/(16*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)) + 1/(16*(sqrt(21)*sqrt(1 - 
2*x)/7 + 1)**2) + 3/(16*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)) - 1/(16*(sqrt(21)* 
sqrt(1 - 2*x)/7 - 1)**2))/1029, (sqrt(1 - 2*x) > -sqrt(21)/3) & (sqrt(1 - 
2*x) < sqrt(21)/3))) - 1100*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2* 
x)/11 - 1)/4 + log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(sqrt(55)*sqrt( 
1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 
2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))
 
3.19.46.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {500}{11} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {7209}{98} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {9405 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 43266 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 49721 \, \sqrt {-2 \, x + 1}}{7 \, {\left (45 \, {\left (2 \, x - 1\right )}^{3} + 309 \, {\left (2 \, x - 1\right )}^{2} + 1414 \, x - 168\right )}} \]

input
integrate((1-2*x)^(1/2)/(2+3*x)^3/(3+5*x)^2,x, algorithm="maxima")
 
output
-500/11*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2* 
x + 1))) + 7209/98*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 
 3*sqrt(-2*x + 1))) - 1/7*(9405*(-2*x + 1)^(5/2) - 43266*(-2*x + 1)^(3/2) 
+ 49721*sqrt(-2*x + 1))/(45*(2*x - 1)^3 + 309*(2*x - 1)^2 + 1414*x - 168)
 
3.19.46.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {500}{11} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {7209}{98} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {25 \, \sqrt {-2 \, x + 1}}{5 \, x + 3} + \frac {9 \, {\left (139 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 329 \, \sqrt {-2 \, x + 1}\right )}}{28 \, {\left (3 \, x + 2\right )}^{2}} \]

input
integrate((1-2*x)^(1/2)/(2+3*x)^3/(3+5*x)^2,x, algorithm="giac")
 
output
-500/11*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 
5*sqrt(-2*x + 1))) + 7209/98*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2* 
x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 25*sqrt(-2*x + 1)/(5*x + 3) + 9/2 
8*(139*(-2*x + 1)^(3/2) - 329*sqrt(-2*x + 1))/(3*x + 2)^2
 
3.19.46.9 Mupad [B] (verification not implemented)

Time = 1.41 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=\frac {1000\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{11}-\frac {7209\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{49}-\frac {\frac {7103\,\sqrt {1-2\,x}}{45}-\frac {14422\,{\left (1-2\,x\right )}^{3/2}}{105}+\frac {209\,{\left (1-2\,x\right )}^{5/2}}{7}}{\frac {1414\,x}{45}+\frac {103\,{\left (2\,x-1\right )}^2}{15}+{\left (2\,x-1\right )}^3-\frac {56}{15}} \]

input
int((1 - 2*x)^(1/2)/((3*x + 2)^3*(5*x + 3)^2),x)
 
output
(1000*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/11 - (7209*21^(1/2)*a 
tanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/49 - ((7103*(1 - 2*x)^(1/2))/45 - (144 
22*(1 - 2*x)^(3/2))/105 + (209*(1 - 2*x)^(5/2))/7)/((1414*x)/45 + (103*(2* 
x - 1)^2)/15 + (2*x - 1)^3 - 56/15)