3.20.50 \(\int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^4} \, dx\) [1950]

3.20.50.1 Optimal result
3.20.50.2 Mathematica [A] (verified)
3.20.50.3 Rubi [A] (verified)
3.20.50.4 Maple [A] (verified)
3.20.50.5 Fricas [A] (verification not implemented)
3.20.50.6 Sympy [F(-1)]
3.20.50.7 Maxima [A] (verification not implemented)
3.20.50.8 Giac [A] (verification not implemented)
3.20.50.9 Mupad [B] (verification not implemented)

3.20.50.1 Optimal result

Integrand size = 24, antiderivative size = 114 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^4} \, dx=-\frac {4435}{567} \sqrt {1-2 x}-\frac {4435 (1-2 x)^{3/2}}{3969}-\frac {(1-2 x)^{7/2}}{189 (2+3 x)^3}+\frac {211 (1-2 x)^{7/2}}{2646 (2+3 x)^2}-\frac {887 (1-2 x)^{5/2}}{882 (2+3 x)}+\frac {4435 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{81 \sqrt {21}} \]

output
-4435/3969*(1-2*x)^(3/2)-1/189*(1-2*x)^(7/2)/(2+3*x)^3+211/2646*(1-2*x)^(7 
/2)/(2+3*x)^2-887/882*(1-2*x)^(5/2)/(2+3*x)+4435/1701*arctanh(1/7*21^(1/2) 
*(1-2*x)^(1/2))*21^(1/2)-4435/567*(1-2*x)^(1/2)
 
3.20.50.2 Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.60 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^4} \, dx=\frac {\sqrt {1-2 x} \left (-12212-48697 x-61353 x^2-21240 x^3+3600 x^4\right )}{162 (2+3 x)^3}+\frac {4435 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{81 \sqrt {21}} \]

input
Integrate[((1 - 2*x)^(5/2)*(3 + 5*x)^2)/(2 + 3*x)^4,x]
 
output
(Sqrt[1 - 2*x]*(-12212 - 48697*x - 61353*x^2 - 21240*x^3 + 3600*x^4))/(162 
*(2 + 3*x)^3) + (4435*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(81*Sqrt[21])
 
3.20.50.3 Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.19, 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, 60, 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.

\(\displaystyle \int \frac {(1-2 x)^{5/2} (5 x+3)^2}{(3 x+2)^4} \, dx\)

\(\Big \downarrow \) 100

\(\displaystyle \frac {1}{189} \int \frac {(1-2 x)^{5/2} (1575 x+839)}{(3 x+2)^3}dx-\frac {(1-2 x)^{7/2}}{189 (3 x+2)^3}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{189} \left (\frac {7983}{14} \int \frac {(1-2 x)^{5/2}}{(3 x+2)^2}dx+\frac {211 (1-2 x)^{7/2}}{14 (3 x+2)^2}\right )-\frac {(1-2 x)^{7/2}}{189 (3 x+2)^3}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {1}{189} \left (\frac {7983}{14} \left (-\frac {5}{3} \int \frac {(1-2 x)^{3/2}}{3 x+2}dx-\frac {(1-2 x)^{5/2}}{3 (3 x+2)}\right )+\frac {211 (1-2 x)^{7/2}}{14 (3 x+2)^2}\right )-\frac {(1-2 x)^{7/2}}{189 (3 x+2)^3}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{189} \left (\frac {7983}{14} \left (-\frac {5}{3} \left (\frac {7}{3} \int \frac {\sqrt {1-2 x}}{3 x+2}dx+\frac {2}{9} (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{5/2}}{3 (3 x+2)}\right )+\frac {211 (1-2 x)^{7/2}}{14 (3 x+2)^2}\right )-\frac {(1-2 x)^{7/2}}{189 (3 x+2)^3}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{189} \left (\frac {7983}{14} \left (-\frac {5}{3} \left (\frac {7}{3} \left (\frac {7}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx+\frac {2}{3} \sqrt {1-2 x}\right )+\frac {2}{9} (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{5/2}}{3 (3 x+2)}\right )+\frac {211 (1-2 x)^{7/2}}{14 (3 x+2)^2}\right )-\frac {(1-2 x)^{7/2}}{189 (3 x+2)^3}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{189} \left (\frac {7983}{14} \left (-\frac {5}{3} \left (\frac {7}{3} \left (\frac {2}{3} \sqrt {1-2 x}-\frac {7}{3} \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {2}{9} (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{5/2}}{3 (3 x+2)}\right )+\frac {211 (1-2 x)^{7/2}}{14 (3 x+2)^2}\right )-\frac {(1-2 x)^{7/2}}{189 (3 x+2)^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{189} \left (\frac {7983}{14} \left (-\frac {5}{3} \left (\frac {7}{3} \left (\frac {2}{3} \sqrt {1-2 x}-\frac {2}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\right )+\frac {2}{9} (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{5/2}}{3 (3 x+2)}\right )+\frac {211 (1-2 x)^{7/2}}{14 (3 x+2)^2}\right )-\frac {(1-2 x)^{7/2}}{189 (3 x+2)^3}\)

input
Int[((1 - 2*x)^(5/2)*(3 + 5*x)^2)/(2 + 3*x)^4,x]
 
output
-1/189*(1 - 2*x)^(7/2)/(2 + 3*x)^3 + ((211*(1 - 2*x)^(7/2))/(14*(2 + 3*x)^ 
2) + (7983*(-1/3*(1 - 2*x)^(5/2)/(2 + 3*x) - (5*((2*(1 - 2*x)^(3/2))/9 + ( 
7*((2*Sqrt[1 - 2*x])/3 - (2*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/3) 
)/3))/3))/14)/189
 

3.20.50.3.1 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])
 
3.20.50.4 Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.54

method result size
risch \(-\frac {7200 x^{5}-46080 x^{4}-101466 x^{3}-36041 x^{2}+24273 x +12212}{162 \left (2+3 x \right )^{3} \sqrt {1-2 x}}+\frac {4435 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1701}\) \(61\)
pseudoelliptic \(\frac {8870 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right )^{3} \sqrt {21}+21 \sqrt {1-2 x}\, \left (3600 x^{4}-21240 x^{3}-61353 x^{2}-48697 x -12212\right )}{3402 \left (2+3 x \right )^{3}}\) \(65\)
derivativedivides \(-\frac {100 \left (1-2 x \right )^{\frac {3}{2}}}{243}-\frac {1480 \sqrt {1-2 x}}{243}-\frac {4 \left (-\frac {3091 \left (1-2 x \right )^{\frac {5}{2}}}{12}+\frac {31675 \left (1-2 x \right )^{\frac {3}{2}}}{27}-\frac {144305 \sqrt {1-2 x}}{108}\right )}{9 \left (-4-6 x \right )^{3}}+\frac {4435 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1701}\) \(75\)
default \(-\frac {100 \left (1-2 x \right )^{\frac {3}{2}}}{243}-\frac {1480 \sqrt {1-2 x}}{243}-\frac {4 \left (-\frac {3091 \left (1-2 x \right )^{\frac {5}{2}}}{12}+\frac {31675 \left (1-2 x \right )^{\frac {3}{2}}}{27}-\frac {144305 \sqrt {1-2 x}}{108}\right )}{9 \left (-4-6 x \right )^{3}}+\frac {4435 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1701}\) \(75\)
trager \(\frac {\left (3600 x^{4}-21240 x^{3}-61353 x^{2}-48697 x -12212\right ) \sqrt {1-2 x}}{162 \left (2+3 x \right )^{3}}-\frac {4435 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{3402}\) \(82\)

input
int((1-2*x)^(5/2)*(3+5*x)^2/(2+3*x)^4,x,method=_RETURNVERBOSE)
 
output
-1/162*(7200*x^5-46080*x^4-101466*x^3-36041*x^2+24273*x+12212)/(2+3*x)^3/( 
1-2*x)^(1/2)+4435/1701*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)
 
3.20.50.5 Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^4} \, dx=\frac {4435 \, \sqrt {21} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 21 \, {\left (3600 \, x^{4} - 21240 \, x^{3} - 61353 \, x^{2} - 48697 \, x - 12212\right )} \sqrt {-2 \, x + 1}}{3402 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

input
integrate((1-2*x)^(5/2)*(3+5*x)^2/(2+3*x)^4,x, algorithm="fricas")
 
output
1/3402*(4435*sqrt(21)*(27*x^3 + 54*x^2 + 36*x + 8)*log((3*x - sqrt(21)*sqr 
t(-2*x + 1) - 5)/(3*x + 2)) + 21*(3600*x^4 - 21240*x^3 - 61353*x^2 - 48697 
*x - 12212)*sqrt(-2*x + 1))/(27*x^3 + 54*x^2 + 36*x + 8)
 
3.20.50.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^4} \, dx=\text {Timed out} \]

input
integrate((1-2*x)**(5/2)*(3+5*x)**2/(2+3*x)**4,x)
 
output
Timed out
 
3.20.50.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^4} \, dx=-\frac {100}{243} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {4435}{3402} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {1480}{243} \, \sqrt {-2 \, x + 1} - \frac {27819 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 126700 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 144305 \, \sqrt {-2 \, x + 1}}{243 \, {\left (27 \, {\left (2 \, x - 1\right )}^{3} + 189 \, {\left (2 \, x - 1\right )}^{2} + 882 \, x - 98\right )}} \]

input
integrate((1-2*x)^(5/2)*(3+5*x)^2/(2+3*x)^4,x, algorithm="maxima")
 
output
-100/243*(-2*x + 1)^(3/2) - 4435/3402*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2* 
x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1480/243*sqrt(-2*x + 1) - 1/243*( 
27819*(-2*x + 1)^(5/2) - 126700*(-2*x + 1)^(3/2) + 144305*sqrt(-2*x + 1))/ 
(27*(2*x - 1)^3 + 189*(2*x - 1)^2 + 882*x - 98)
 
3.20.50.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^4} \, dx=-\frac {100}{243} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {4435}{3402} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {1480}{243} \, \sqrt {-2 \, x + 1} - \frac {27819 \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - 126700 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 144305 \, \sqrt {-2 \, x + 1}}{1944 \, {\left (3 \, x + 2\right )}^{3}} \]

input
integrate((1-2*x)^(5/2)*(3+5*x)^2/(2+3*x)^4,x, algorithm="giac")
 
output
-100/243*(-2*x + 1)^(3/2) - 4435/3402*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6 
*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1480/243*sqrt(-2*x + 1) 
- 1/1944*(27819*(2*x - 1)^2*sqrt(-2*x + 1) - 126700*(-2*x + 1)^(3/2) + 144 
305*sqrt(-2*x + 1))/(3*x + 2)^3
 
3.20.50.9 Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^4} \, dx=-\frac {1480\,\sqrt {1-2\,x}}{243}-\frac {100\,{\left (1-2\,x\right )}^{3/2}}{243}-\frac {\frac {144305\,\sqrt {1-2\,x}}{6561}-\frac {126700\,{\left (1-2\,x\right )}^{3/2}}{6561}+\frac {3091\,{\left (1-2\,x\right )}^{5/2}}{729}}{\frac {98\,x}{3}+7\,{\left (2\,x-1\right )}^2+{\left (2\,x-1\right )}^3-\frac {98}{27}}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,4435{}\mathrm {i}}{1701} \]

input
int(((1 - 2*x)^(5/2)*(5*x + 3)^2)/(3*x + 2)^4,x)
 
output
- (21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*4435i)/1701 - (1480*(1 - 
 2*x)^(1/2))/243 - (100*(1 - 2*x)^(3/2))/243 - ((144305*(1 - 2*x)^(1/2))/6 
561 - (126700*(1 - 2*x)^(3/2))/6561 + (3091*(1 - 2*x)^(5/2))/729)/((98*x)/ 
3 + 7*(2*x - 1)^2 + (2*x - 1)^3 - 98/27)