3.20.51 \(\int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^5} \, dx\) [1951]

3.20.51.1 Optimal result
3.20.51.2 Mathematica [A] (verified)
3.20.51.3 Rubi [A] (verified)
3.20.51.4 Maple [A] (verified)
3.20.51.5 Fricas [A] (verification not implemented)
3.20.51.6 Sympy [F(-1)]
3.20.51.7 Maxima [A] (verification not implemented)
3.20.51.8 Giac [A] (verification not implemented)
3.20.51.9 Mupad [B] (verification not implemented)

3.20.51.1 Optimal result

Integrand size = 24, antiderivative size = 121 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^5} \, dx=\frac {24965 \sqrt {1-2 x}}{15876}-\frac {(1-2 x)^{7/2}}{252 (2+3 x)^4}+\frac {31 (1-2 x)^{7/2}}{588 (2+3 x)^3}-\frac {4993 (1-2 x)^{5/2}}{10584 (2+3 x)^2}+\frac {24965 (1-2 x)^{3/2}}{31752 (2+3 x)}-\frac {24965 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{2268 \sqrt {21}} \]

output
-1/252*(1-2*x)^(7/2)/(2+3*x)^4+31/588*(1-2*x)^(7/2)/(2+3*x)^3-4993/10584*( 
1-2*x)^(5/2)/(2+3*x)^2+24965/31752*(1-2*x)^(3/2)/(2+3*x)-24965/47628*arcta 
nh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+24965/15876*(1-2*x)^(1/2)
 
3.20.51.2 Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.58 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^5} \, dx=\frac {\frac {21 \sqrt {1-2 x} \left (134558+762598 x+1526937 x^2+1231065 x^3+302400 x^4\right )}{2 (2+3 x)^4}-24965 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{47628} \]

input
Integrate[((1 - 2*x)^(5/2)*(3 + 5*x)^2)/(2 + 3*x)^5,x]
 
output
((21*Sqrt[1 - 2*x]*(134558 + 762598*x + 1526937*x^2 + 1231065*x^3 + 302400 
*x^4))/(2*(2 + 3*x)^4) - 24965*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/ 
47628
 
3.20.51.3 Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {100, 87, 51, 51, 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.

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

\(\Big \downarrow \) 100

\(\displaystyle \frac {1}{252} \int \frac {(1-2 x)^{5/2} (2100 x+1121)}{(3 x+2)^4}dx-\frac {(1-2 x)^{7/2}}{252 (3 x+2)^4}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{252} \left (\frac {4993}{7} \int \frac {(1-2 x)^{5/2}}{(3 x+2)^3}dx+\frac {93 (1-2 x)^{7/2}}{7 (3 x+2)^3}\right )-\frac {(1-2 x)^{7/2}}{252 (3 x+2)^4}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {1}{252} \left (\frac {4993}{7} \left (-\frac {5}{6} \int \frac {(1-2 x)^{3/2}}{(3 x+2)^2}dx-\frac {(1-2 x)^{5/2}}{6 (3 x+2)^2}\right )+\frac {93 (1-2 x)^{7/2}}{7 (3 x+2)^3}\right )-\frac {(1-2 x)^{7/2}}{252 (3 x+2)^4}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {1}{252} \left (\frac {4993}{7} \left (-\frac {5}{6} \left (-\int \frac {\sqrt {1-2 x}}{3 x+2}dx-\frac {(1-2 x)^{3/2}}{3 (3 x+2)}\right )-\frac {(1-2 x)^{5/2}}{6 (3 x+2)^2}\right )+\frac {93 (1-2 x)^{7/2}}{7 (3 x+2)^3}\right )-\frac {(1-2 x)^{7/2}}{252 (3 x+2)^4}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{252} \left (\frac {4993}{7} \left (-\frac {5}{6} \left (-\frac {7}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx-\frac {(1-2 x)^{3/2}}{3 (3 x+2)}-\frac {2}{3} \sqrt {1-2 x}\right )-\frac {(1-2 x)^{5/2}}{6 (3 x+2)^2}\right )+\frac {93 (1-2 x)^{7/2}}{7 (3 x+2)^3}\right )-\frac {(1-2 x)^{7/2}}{252 (3 x+2)^4}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{252} \left (\frac {4993}{7} \left (-\frac {5}{6} \left (\frac {7}{3} \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {(1-2 x)^{3/2}}{3 (3 x+2)}-\frac {2}{3} \sqrt {1-2 x}\right )-\frac {(1-2 x)^{5/2}}{6 (3 x+2)^2}\right )+\frac {93 (1-2 x)^{7/2}}{7 (3 x+2)^3}\right )-\frac {(1-2 x)^{7/2}}{252 (3 x+2)^4}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{252} \left (\frac {4993}{7} \left (-\frac {5}{6} \left (\frac {2}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {(1-2 x)^{3/2}}{3 (3 x+2)}-\frac {2}{3} \sqrt {1-2 x}\right )-\frac {(1-2 x)^{5/2}}{6 (3 x+2)^2}\right )+\frac {93 (1-2 x)^{7/2}}{7 (3 x+2)^3}\right )-\frac {(1-2 x)^{7/2}}{252 (3 x+2)^4}\)

input
Int[((1 - 2*x)^(5/2)*(3 + 5*x)^2)/(2 + 3*x)^5,x]
 
output
-1/252*(1 - 2*x)^(7/2)/(2 + 3*x)^4 + ((93*(1 - 2*x)^(7/2))/(7*(2 + 3*x)^3) 
 + (4993*(-1/6*(1 - 2*x)^(5/2)/(2 + 3*x)^2 - (5*((-2*Sqrt[1 - 2*x])/3 - (1 
 - 2*x)^(3/2)/(3*(2 + 3*x)) + (2*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x] 
])/3))/6))/7)/252
 

3.20.51.3.1 Defintions of rubi rules used

rule 51
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, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

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 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 100
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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])
 
3.20.51.4 Maple [A] (verified)

Time = 3.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.50

method result size
risch \(-\frac {604800 x^{5}+2159730 x^{4}+1822809 x^{3}-1741 x^{2}-493482 x -134558}{4536 \left (2+3 x \right )^{4} \sqrt {1-2 x}}-\frac {24965 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{47628}\) \(61\)
pseudoelliptic \(\frac {-49930 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right )^{4} \sqrt {21}+21 \sqrt {1-2 x}\, \left (302400 x^{4}+1231065 x^{3}+1526937 x^{2}+762598 x +134558\right )}{95256 \left (2+3 x \right )^{4}}\) \(65\)
derivativedivides \(\frac {200 \sqrt {1-2 x}}{243}+\frac {-\frac {47185 \left (1-2 x \right )^{\frac {7}{2}}}{252}+\frac {129289 \left (1-2 x \right )^{\frac {5}{2}}}{108}-\frac {824705 \left (1-2 x \right )^{\frac {3}{2}}}{324}+\frac {1749055 \sqrt {1-2 x}}{972}}{\left (-4-6 x \right )^{4}}-\frac {24965 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{47628}\) \(75\)
default \(\frac {200 \sqrt {1-2 x}}{243}+\frac {-\frac {47185 \left (1-2 x \right )^{\frac {7}{2}}}{252}+\frac {129289 \left (1-2 x \right )^{\frac {5}{2}}}{108}-\frac {824705 \left (1-2 x \right )^{\frac {3}{2}}}{324}+\frac {1749055 \sqrt {1-2 x}}{972}}{\left (-4-6 x \right )^{4}}-\frac {24965 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{47628}\) \(75\)
trager \(\frac {\left (302400 x^{4}+1231065 x^{3}+1526937 x^{2}+762598 x +134558\right ) \sqrt {1-2 x}}{4536 \left (2+3 x \right )^{4}}+\frac {24965 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{95256}\) \(82\)

input
int((1-2*x)^(5/2)*(3+5*x)^2/(2+3*x)^5,x,method=_RETURNVERBOSE)
 
output
-1/4536*(604800*x^5+2159730*x^4+1822809*x^3-1741*x^2-493482*x-134558)/(2+3 
*x)^4/(1-2*x)^(1/2)-24965/47628*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/ 
2)
 
3.20.51.5 Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^5} \, dx=\frac {24965 \, \sqrt {21} {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 21 \, {\left (302400 \, x^{4} + 1231065 \, x^{3} + 1526937 \, x^{2} + 762598 \, x + 134558\right )} \sqrt {-2 \, x + 1}}{95256 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]

input
integrate((1-2*x)^(5/2)*(3+5*x)^2/(2+3*x)^5,x, algorithm="fricas")
 
output
1/95256*(24965*sqrt(21)*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log((3*x 
+ sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 21*(302400*x^4 + 1231065*x^3 + 
 1526937*x^2 + 762598*x + 134558)*sqrt(-2*x + 1))/(81*x^4 + 216*x^3 + 216* 
x^2 + 96*x + 16)
 
3.20.51.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^5} \, dx=\text {Timed out} \]

input
integrate((1-2*x)**(5/2)*(3+5*x)**2/(2+3*x)**5,x)
 
output
Timed out
 
3.20.51.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^5} \, dx=\frac {24965}{95256} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {200}{243} \, \sqrt {-2 \, x + 1} - \frac {1273995 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 8145207 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 17318805 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 12243385 \, \sqrt {-2 \, x + 1}}{6804 \, {\left (81 \, {\left (2 \, x - 1\right )}^{4} + 756 \, {\left (2 \, x - 1\right )}^{3} + 2646 \, {\left (2 \, x - 1\right )}^{2} + 8232 \, x - 1715\right )}} \]

input
integrate((1-2*x)^(5/2)*(3+5*x)^2/(2+3*x)^5,x, algorithm="maxima")
 
output
24965/95256*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt 
(-2*x + 1))) + 200/243*sqrt(-2*x + 1) - 1/6804*(1273995*(-2*x + 1)^(7/2) - 
 8145207*(-2*x + 1)^(5/2) + 17318805*(-2*x + 1)^(3/2) - 12243385*sqrt(-2*x 
 + 1))/(81*(2*x - 1)^4 + 756*(2*x - 1)^3 + 2646*(2*x - 1)^2 + 8232*x - 171 
5)
 
3.20.51.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^5} \, dx=\frac {24965}{95256} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {200}{243} \, \sqrt {-2 \, x + 1} + \frac {1273995 \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + 8145207 \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - 17318805 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 12243385 \, \sqrt {-2 \, x + 1}}{108864 \, {\left (3 \, x + 2\right )}^{4}} \]

input
integrate((1-2*x)^(5/2)*(3+5*x)^2/(2+3*x)^5,x, algorithm="giac")
 
output
24965/95256*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) 
 + 3*sqrt(-2*x + 1))) + 200/243*sqrt(-2*x + 1) + 1/108864*(1273995*(2*x - 
1)^3*sqrt(-2*x + 1) + 8145207*(2*x - 1)^2*sqrt(-2*x + 1) - 17318805*(-2*x 
+ 1)^(3/2) + 12243385*sqrt(-2*x + 1))/(3*x + 2)^4
 
3.20.51.9 Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^5} \, dx=\frac {200\,\sqrt {1-2\,x}}{243}-\frac {24965\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{47628}+\frac {\frac {1749055\,\sqrt {1-2\,x}}{78732}-\frac {824705\,{\left (1-2\,x\right )}^{3/2}}{26244}+\frac {129289\,{\left (1-2\,x\right )}^{5/2}}{8748}-\frac {47185\,{\left (1-2\,x\right )}^{7/2}}{20412}}{\frac {2744\,x}{27}+\frac {98\,{\left (2\,x-1\right )}^2}{3}+\frac {28\,{\left (2\,x-1\right )}^3}{3}+{\left (2\,x-1\right )}^4-\frac {1715}{81}} \]

input
int(((1 - 2*x)^(5/2)*(5*x + 3)^2)/(3*x + 2)^5,x)
 
output
(200*(1 - 2*x)^(1/2))/243 - (24965*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2 
))/7))/47628 + ((1749055*(1 - 2*x)^(1/2))/78732 - (824705*(1 - 2*x)^(3/2)) 
/26244 + (129289*(1 - 2*x)^(5/2))/8748 - (47185*(1 - 2*x)^(7/2))/20412)/(( 
2744*x)/27 + (98*(2*x - 1)^2)/3 + (28*(2*x - 1)^3)/3 + (2*x - 1)^4 - 1715/ 
81)