\(\int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx\) [545]

Optimal result

Integrand size = 24, antiderivative size = 87 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {18}{625} \sqrt {1-2 x}-\frac {117}{250} (1-2 x)^{3/2}+\frac {27}{250} (1-2 x)^{5/2}-\frac {\sqrt {1-2 x}}{625 (3+5 x)}-\frac {196 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \] Output:

18/625*(1-2*x)^(1/2)-117/250*(1-2*x)^(3/2)+27/250*(1-2*x)^(5/2)-(1-2*x)^(1 
/2)/(1875+3125*x)-196/34375*55^(1/2)*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {\sqrt {1-2 x} \left (-622-90 x+2385 x^2+1350 x^3\right )}{625 (3+5 x)}-\frac {196 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \] Input:

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

Output:

(Sqrt[1 - 2*x]*(-622 - 90*x + 2385*x^2 + 1350*x^3))/(625*(3 + 5*x)) - (196 
*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(625*Sqrt[55])
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {108, 27, 170, 27, 90, 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.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {1}{5} \int \frac {7 (1-3 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)}dx-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {7}{5} \int \frac {(1-3 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)}dx-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\)

\(\Big \downarrow \) 170

\(\displaystyle \frac {7}{5} \left (\frac {3}{25} \sqrt {1-2 x} (3 x+2)^2-\frac {1}{25} \int -\frac {14 (3 x+2)}{\sqrt {1-2 x} (5 x+3)}dx\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {7}{5} \left (\frac {14}{25} \int \frac {3 x+2}{\sqrt {1-2 x} (5 x+3)}dx+\frac {3}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {7}{5} \left (\frac {14}{25} \left (\frac {1}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {3}{5} \sqrt {1-2 x}\right )+\frac {3}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {7}{5} \left (\frac {14}{25} \left (-\frac {1}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {3}{5} \sqrt {1-2 x}\right )+\frac {3}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {7}{5} \left (\frac {14}{25} \left (-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}-\frac {3}{5} \sqrt {1-2 x}\right )+\frac {3}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\)

Input:

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

Output:

-1/5*(Sqrt[1 - 2*x]*(2 + 3*x)^3)/(3 + 5*x) + (7*((3*Sqrt[1 - 2*x]*(2 + 3*x 
)^2)/25 + (14*((-3*Sqrt[1 - 2*x])/5 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]] 
)/(5*Sqrt[55])))/25))/5
 

Defintions of rubi rules used

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 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 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 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])
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64

method	result	size
risch	\(-\frac {2700 x^{4}+3420 x^{3}-2565 x^{2}-1154 x +622}{625 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {196 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{34375}\)	\(56\)
pseudoelliptic	\(\frac {-196 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}+55 \sqrt {1-2 x}\, \left (1350 x^{3}+2385 x^{2}-90 x -622\right )}{103125+171875 x}\)	\(57\)
derivativedivides	\(\frac {27 \left (1-2 x \right )^{\frac {5}{2}}}{250}-\frac {117 \left (1-2 x \right )^{\frac {3}{2}}}{250}+\frac {18 \sqrt {1-2 x}}{625}+\frac {2 \sqrt {1-2 x}}{3125 \left (-\frac {6}{5}-2 x \right )}-\frac {196 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{34375}\)	\(63\)
default	\(\frac {27 \left (1-2 x \right )^{\frac {5}{2}}}{250}-\frac {117 \left (1-2 x \right )^{\frac {3}{2}}}{250}+\frac {18 \sqrt {1-2 x}}{625}+\frac {2 \sqrt {1-2 x}}{3125 \left (-\frac {6}{5}-2 x \right )}-\frac {196 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{34375}\)	\(63\)
trager	\(\frac {\sqrt {1-2 x}\, \left (1350 x^{3}+2385 x^{2}-90 x -622\right )}{1875+3125 x}-\frac {98 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (-\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x -8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )-55 \sqrt {1-2 x}}{3+5 x}\right )}{34375}\)	\(78\)

Input:

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

Output:

-1/625*(2700*x^4+3420*x^3-2565*x^2-1154*x+622)/(3+5*x)/(1-2*x)^(1/2)-196/3 
4375*55^(1/2)*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {98 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (1350 \, x^{3} + 2385 \, x^{2} - 90 \, x - 622\right )} \sqrt {-2 \, x + 1}}{34375 \, {\left (5 \, x + 3\right )}} \] Input:

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

Output:

1/34375*(98*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5* 
x + 3)) + 55*(1350*x^3 + 2385*x^2 - 90*x - 622)*sqrt(-2*x + 1))/(5*x + 3)
 

Sympy [A] (verification not implemented)

Time = 43.16 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.26 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {27 \left (1 - 2 x\right )^{\frac {5}{2}}}{250} - \frac {117 \left (1 - 2 x\right )^{\frac {3}{2}}}{250} + \frac {18 \sqrt {1 - 2 x}}{625} + \frac {97 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{34375} - \frac {44 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} \] Input:

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

Output:

27*(1 - 2*x)**(5/2)/250 - 117*(1 - 2*x)**(3/2)/250 + 18*sqrt(1 - 2*x)/625 
+ 97*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt( 
55)/5))/34375 - 44*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) > -s 
qrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/625
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {27}{250} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {117}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {98}{34375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {18}{625} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \] Input:

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

Output:

27/250*(-2*x + 1)^(5/2) - 117/250*(-2*x + 1)^(3/2) + 98/34375*sqrt(55)*log 
(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 18/625*sq 
rt(-2*x + 1) - 1/625*sqrt(-2*x + 1)/(5*x + 3)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {27}{250} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {117}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {98}{34375} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {18}{625} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \] Input:

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

Output:

27/250*(2*x - 1)^2*sqrt(-2*x + 1) - 117/250*(-2*x + 1)^(3/2) + 98/34375*sq 
rt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2* 
x + 1))) + 18/625*sqrt(-2*x + 1) - 1/625*sqrt(-2*x + 1)/(5*x + 3)
 

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {18\,\sqrt {1-2\,x}}{625}-\frac {2\,\sqrt {1-2\,x}}{3125\,\left (2\,x+\frac {6}{5}\right )}-\frac {117\,{\left (1-2\,x\right )}^{3/2}}{250}+\frac {27\,{\left (1-2\,x\right )}^{5/2}}{250}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,196{}\mathrm {i}}{34375} \] Input:

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

Output:

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*196i)/34375 - (2*(1 - 2*x 
)^(1/2))/(3125*(2*x + 6/5)) + (18*(1 - 2*x)^(1/2))/625 - (117*(1 - 2*x)^(3 
/2))/250 + (27*(1 - 2*x)^(5/2))/250
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {74250 \sqrt {-2 x +1}\, x^{3}+131175 \sqrt {-2 x +1}\, x^{2}-4950 \sqrt {-2 x +1}\, x -34210 \sqrt {-2 x +1}+490 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x +294 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right )-490 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x -294 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right )}{171875 x +103125} \] Input:

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

Output:

(74250*sqrt( - 2*x + 1)*x**3 + 131175*sqrt( - 2*x + 1)*x**2 - 4950*sqrt( - 
 2*x + 1)*x - 34210*sqrt( - 2*x + 1) + 490*sqrt(55)*log(5*sqrt( - 2*x + 1) 
 - sqrt(55))*x + 294*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55)) - 490*sqr 
t(55)*log(5*sqrt( - 2*x + 1) + sqrt(55))*x - 294*sqrt(55)*log(5*sqrt( - 2* 
x + 1) + sqrt(55)))/(34375*(5*x + 3))