3.20.92 \(\int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^3} \, dx\) [1992]

3.20.92.1 Optimal result
3.20.92.2 Mathematica [A] (verified)
3.20.92.3 Rubi [A] (verified)
3.20.92.4 Maple [A] (verified)
3.20.92.5 Fricas [A] (verification not implemented)
3.20.92.6 Sympy [A] (verification not implemented)
3.20.92.7 Maxima [A] (verification not implemented)
3.20.92.8 Giac [A] (verification not implemented)
3.20.92.9 Mupad [B] (verification not implemented)

3.20.92.1 Optimal result

Integrand size = 24, antiderivative size = 135 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^3} \, dx=\frac {5559 \sqrt {1-2 x}}{15625}+\frac {954}{875} (1-2 x)^{3/2} (2+3 x)^2-\frac {(1-2 x)^{5/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac {47 (1-2 x)^{3/2} (2+3 x)^3}{25 (3+5 x)}+\frac {3 (1-2 x)^{3/2} (1618+2403 x)}{6250}-\frac {5559 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625} \]

output
954/875*(1-2*x)^(3/2)*(2+3*x)^2-1/10*(1-2*x)^(5/2)*(2+3*x)^3/(3+5*x)^2-47/ 
25*(1-2*x)^(3/2)*(2+3*x)^3/(3+5*x)+3/6250*(1-2*x)^(3/2)*(1618+2403*x)-5559 
/78125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+5559/15625*(1-2*x)^(1 
/2)
 
3.20.92.2 Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.54 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^3} \, dx=\frac {\frac {5 \sqrt {1-2 x} \left (770444+2637795 x+1651030 x^2-1506900 x^3-27000 x^4+1350000 x^5\right )}{(3+5 x)^2}-77826 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1093750} \]

input
Integrate[((1 - 2*x)^(5/2)*(2 + 3*x)^3)/(3 + 5*x)^3,x]
 
output
((5*Sqrt[1 - 2*x]*(770444 + 2637795*x + 1651030*x^2 - 1506900*x^3 - 27000* 
x^4 + 1350000*x^5))/(3 + 5*x)^2 - 77826*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 
 - 2*x]])/1093750
 
3.20.92.3 Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.15, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {108, 25, 166, 27, 170, 27, 164, 60, 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.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {1}{10} \int -\frac {(1-2 x)^{3/2} (3 x+2)^2 (33 x+1)}{(5 x+3)^2}dx-\frac {(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{10} \int \frac {(1-2 x)^{3/2} (3 x+2)^2 (33 x+1)}{(5 x+3)^2}dx-\frac {(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}\)

\(\Big \downarrow \) 166

\(\displaystyle \frac {1}{10} \left (-\frac {1}{5} \int \frac {3 \sqrt {1-2 x} (3 x+2)^2 (636 x+11)}{5 x+3}dx-\frac {94 (1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{10} \left (-\frac {3}{5} \int \frac {\sqrt {1-2 x} (3 x+2)^2 (636 x+11)}{5 x+3}dx-\frac {94 (1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}\)

\(\Big \downarrow \) 170

\(\displaystyle \frac {1}{10} \left (-\frac {3}{5} \left (-\frac {1}{35} \int -\frac {7 \sqrt {1-2 x} (3 x+2) (801 x+110)}{5 x+3}dx-\frac {636}{35} (1-2 x)^{3/2} (3 x+2)^2\right )-\frac {94 (1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{10} \left (-\frac {3}{5} \left (\frac {1}{5} \int \frac {\sqrt {1-2 x} (3 x+2) (801 x+110)}{5 x+3}dx-\frac {636}{35} (1-2 x)^{3/2} (3 x+2)^2\right )-\frac {94 (1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {1}{10} \left (-\frac {3}{5} \left (\frac {1}{5} \left (-\frac {1853}{25} \int \frac {\sqrt {1-2 x}}{5 x+3}dx-\frac {1}{25} (2403 x+1618) (1-2 x)^{3/2}\right )-\frac {636}{35} (1-2 x)^{3/2} (3 x+2)^2\right )-\frac {94 (1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{10} \left (-\frac {3}{5} \left (\frac {1}{5} \left (-\frac {1853}{25} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )-\frac {1}{25} (2403 x+1618) (1-2 x)^{3/2}\right )-\frac {636}{35} (1-2 x)^{3/2} (3 x+2)^2\right )-\frac {94 (1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{10} \left (-\frac {3}{5} \left (\frac {1}{5} \left (-\frac {1853}{25} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {1}{25} (2403 x+1618) (1-2 x)^{3/2}\right )-\frac {636}{35} (1-2 x)^{3/2} (3 x+2)^2\right )-\frac {94 (1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{10} \left (-\frac {3}{5} \left (\frac {1}{5} \left (-\frac {1853}{25} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {1}{25} (2403 x+1618) (1-2 x)^{3/2}\right )-\frac {636}{35} (1-2 x)^{3/2} (3 x+2)^2\right )-\frac {94 (1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}\)

input
Int[((1 - 2*x)^(5/2)*(2 + 3*x)^3)/(3 + 5*x)^3,x]
 
output
-1/10*((1 - 2*x)^(5/2)*(2 + 3*x)^3)/(3 + 5*x)^2 + ((-94*(1 - 2*x)^(3/2)*(2 
 + 3*x)^3)/(5*(3 + 5*x)) - (3*((-636*(1 - 2*x)^(3/2)*(2 + 3*x)^2)/35 + (-1 
/25*((1 - 2*x)^(3/2)*(1618 + 2403*x)) - (1853*((2*Sqrt[1 - 2*x])/5 - (2*Sq 
rt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5))/25)/5))/5)/10
 

3.20.92.3.1 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 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 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 108
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/(b*(m + 1))) 
, x] - Simp[1/(b*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* 
x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 
*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 164
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - 
 b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( 
c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h 
*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 
3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3))   Int[( 
a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] 
&& NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
 

rule 166
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*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - 
a*f)*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* 
c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h 
)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, 
e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
 

rule 170
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && IntegerQ[m]
 

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])
 
3.20.92.4 Maple [A] (verified)

Time = 3.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.49

method result size
risch \(-\frac {2700000 x^{6}-1404000 x^{5}-2986800 x^{4}+4808960 x^{3}+3624560 x^{2}-1096907 x -770444}{218750 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {5559 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{78125}\) \(66\)
pseudoelliptic \(\frac {-77826 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}+5 \sqrt {1-2 x}\, \left (1350000 x^{5}-27000 x^{4}-1506900 x^{3}+1651030 x^{2}+2637795 x +770444\right )}{1093750 \left (3+5 x \right )^{2}}\) \(70\)
derivativedivides \(-\frac {27 \left (1-2 x \right )^{\frac {7}{2}}}{875}+\frac {54 \left (1-2 x \right )^{\frac {5}{2}}}{3125}+\frac {186 \left (1-2 x \right )^{\frac {3}{2}}}{3125}+\frac {46 \sqrt {1-2 x}}{125}+\frac {\frac {2079 \left (1-2 x \right )^{\frac {3}{2}}}{3125}-\frac {23111 \sqrt {1-2 x}}{15625}}{\left (-6-10 x \right )^{2}}-\frac {5559 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{78125}\) \(84\)
default \(-\frac {27 \left (1-2 x \right )^{\frac {7}{2}}}{875}+\frac {54 \left (1-2 x \right )^{\frac {5}{2}}}{3125}+\frac {186 \left (1-2 x \right )^{\frac {3}{2}}}{3125}+\frac {46 \sqrt {1-2 x}}{125}+\frac {\frac {2079 \left (1-2 x \right )^{\frac {3}{2}}}{3125}-\frac {23111 \sqrt {1-2 x}}{15625}}{\left (-6-10 x \right )^{2}}-\frac {5559 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{78125}\) \(84\)
trager \(\frac {\left (1350000 x^{5}-27000 x^{4}-1506900 x^{3}+1651030 x^{2}+2637795 x +770444\right ) \sqrt {1-2 x}}{218750 \left (3+5 x \right )^{2}}-\frac {5559 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{156250}\) \(87\)

input
int((1-2*x)^(5/2)*(2+3*x)^3/(3+5*x)^3,x,method=_RETURNVERBOSE)
 
output
-1/218750*(2700000*x^6-1404000*x^5-2986800*x^4+4808960*x^3+3624560*x^2-109 
6907*x-770444)/(3+5*x)^2/(1-2*x)^(1/2)-5559/78125*arctanh(1/11*55^(1/2)*(1 
-2*x)^(1/2))*55^(1/2)
 
3.20.92.5 Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^3} \, dx=\frac {38913 \, \sqrt {11} \sqrt {5} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 5 \, {\left (1350000 \, x^{5} - 27000 \, x^{4} - 1506900 \, x^{3} + 1651030 \, x^{2} + 2637795 \, x + 770444\right )} \sqrt {-2 \, x + 1}}{1093750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

input
integrate((1-2*x)^(5/2)*(2+3*x)^3/(3+5*x)^3,x, algorithm="fricas")
 
output
1/1093750*(38913*sqrt(11)*sqrt(5)*(25*x^2 + 30*x + 9)*log((sqrt(11)*sqrt(5 
)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 5*(1350000*x^5 - 27000*x^4 - 1506 
900*x^3 + 1651030*x^2 + 2637795*x + 770444)*sqrt(-2*x + 1))/(25*x^2 + 30*x 
 + 9)
 
3.20.92.6 Sympy [A] (verification not implemented)

Time = 133.23 (sec) , antiderivative size = 377, normalized size of antiderivative = 2.79 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^3} \, dx=- \frac {27 \left (1 - 2 x\right )^{\frac {7}{2}}}{875} + \frac {54 \left (1 - 2 x\right )^{\frac {5}{2}}}{3125} + \frac {186 \left (1 - 2 x\right )^{\frac {3}{2}}}{3125} + \frac {46 \sqrt {1 - 2 x}}{125} + \frac {537 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{15625} - \frac {45012 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{15625} + \frac {10648 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{15625} \]

input
integrate((1-2*x)**(5/2)*(2+3*x)**3/(3+5*x)**3,x)
 
output
-27*(1 - 2*x)**(7/2)/875 + 54*(1 - 2*x)**(5/2)/3125 + 186*(1 - 2*x)**(3/2) 
/3125 + 46*sqrt(1 - 2*x)/125 + 537*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/ 
5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/15625 - 45012*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)/1 
1 - 1)))/605, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5) 
))/15625 + 10648*Piecewise((sqrt(55)*(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1) 
/16 - 3*log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/16 + 3/(16*(sqrt(55)*sqrt(1 - 2 
*x)/11 + 1)) + 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)**2) + 3/(16*(sqrt(55) 
*sqrt(1 - 2*x)/11 - 1)) - 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)**2))/6655, 
 (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/15625
 
3.20.92.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^3} \, dx=-\frac {27}{875} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {54}{3125} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {186}{3125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {5559}{156250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {46}{125} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (945 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2101 \, \sqrt {-2 \, x + 1}\right )}}{15625 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]

input
integrate((1-2*x)^(5/2)*(2+3*x)^3/(3+5*x)^3,x, algorithm="maxima")
 
output
-27/875*(-2*x + 1)^(7/2) + 54/3125*(-2*x + 1)^(5/2) + 186/3125*(-2*x + 1)^ 
(3/2) + 5559/156250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) 
+ 5*sqrt(-2*x + 1))) + 46/125*sqrt(-2*x + 1) + 11/15625*(945*(-2*x + 1)^(3 
/2) - 2101*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)
 
3.20.92.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.87 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^3} \, dx=\frac {27}{875} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {54}{3125} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {186}{3125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {5559}{156250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {46}{125} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (945 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2101 \, \sqrt {-2 \, x + 1}\right )}}{62500 \, {\left (5 \, x + 3\right )}^{2}} \]

input
integrate((1-2*x)^(5/2)*(2+3*x)^3/(3+5*x)^3,x, algorithm="giac")
 
output
27/875*(2*x - 1)^3*sqrt(-2*x + 1) + 54/3125*(2*x - 1)^2*sqrt(-2*x + 1) + 1 
86/3125*(-2*x + 1)^(3/2) + 5559/156250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 
10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 46/125*sqrt(-2*x + 1) 
+ 11/62500*(945*(-2*x + 1)^(3/2) - 2101*sqrt(-2*x + 1))/(5*x + 3)^2
 
3.20.92.9 Mupad [B] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^3} \, dx=\frac {46\,\sqrt {1-2\,x}}{125}+\frac {186\,{\left (1-2\,x\right )}^{3/2}}{3125}+\frac {54\,{\left (1-2\,x\right )}^{5/2}}{3125}-\frac {27\,{\left (1-2\,x\right )}^{7/2}}{875}-\frac {\frac {23111\,\sqrt {1-2\,x}}{390625}-\frac {2079\,{\left (1-2\,x\right )}^{3/2}}{78125}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,5559{}\mathrm {i}}{78125} \]

input
int(((1 - 2*x)^(5/2)*(3*x + 2)^3)/(5*x + 3)^3,x)
 
output
(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*5559i)/78125 + (46*(1 - 2 
*x)^(1/2))/125 + (186*(1 - 2*x)^(3/2))/3125 + (54*(1 - 2*x)^(5/2))/3125 - 
(27*(1 - 2*x)^(7/2))/875 - ((23111*(1 - 2*x)^(1/2))/390625 - (2079*(1 - 2* 
x)^(3/2))/78125)/((44*x)/5 + (2*x - 1)^2 + 11/25)