\(\int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx\) [554]

Optimal result

Integrand size = 24, antiderivative size = 81 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {18}{125} \sqrt {1-2 x}-\frac {\sqrt {1-2 x}}{250 (3+5 x)^2}-\frac {131 \sqrt {1-2 x}}{2750 (3+5 x)}-\frac {409 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \] Output:

18/125*(1-2*x)^(1/2)-1/250*(1-2*x)^(1/2)/(3+5*x)^2-131*(1-2*x)^(1/2)/(8250 
+13750*x)-409/15125*55^(1/2)*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {\sqrt {1-2 x} \left (632+2245 x+1980 x^2\right )}{550 (3+5 x)^2}-\frac {409 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \] Input:

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

Output:

(Sqrt[1 - 2*x]*(632 + 2245*x + 1980*x^2))/(550*(3 + 5*x)^2) - (409*ArcTanh 
[Sqrt[5/11]*Sqrt[1 - 2*x]])/(275*Sqrt[55])
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {100, 87, 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.

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

\(\Big \downarrow \) 100

\(\displaystyle \frac {1}{550} \int \frac {\sqrt {1-2 x} (990 x+727)}{(5 x+3)^2}dx-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{550} \left (\frac {2045}{11} \int \frac {\sqrt {1-2 x}}{5 x+3}dx-\frac {133 (1-2 x)^{3/2}}{11 (5 x+3)}\right )-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{550} \left (\frac {2045}{11} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )-\frac {133 (1-2 x)^{3/2}}{11 (5 x+3)}\right )-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{550} \left (\frac {2045}{11} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {133 (1-2 x)^{3/2}}{11 (5 x+3)}\right )-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{550} \left (\frac {2045}{11} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {133 (1-2 x)^{3/2}}{11 (5 x+3)}\right )-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}\)

Input:

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

Output:

-1/550*(1 - 2*x)^(3/2)/(3 + 5*x)^2 + ((-133*(1 - 2*x)^(3/2))/(11*(3 + 5*x) 
) + (2045*((2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 
 2*x]])/5))/11)/550
 

Defintions of rubi rules used

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])
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.63

method	result	size
risch	\(-\frac {3960 x^{3}+2510 x^{2}-981 x -632}{550 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {409 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{15125}\)	\(51\)
pseudoelliptic	\(\frac {-818 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}+55 \sqrt {1-2 x}\, \left (1980 x^{2}+2245 x +632\right )}{30250 \left (3+5 x \right )^{2}}\)	\(55\)
derivativedivides	\(\frac {18 \sqrt {1-2 x}}{125}+\frac {\frac {131 \left (1-2 x \right )^{\frac {3}{2}}}{275}-\frac {133 \sqrt {1-2 x}}{125}}{\left (-6-10 x \right )^{2}}-\frac {409 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{15125}\)	\(57\)
default	\(\frac {18 \sqrt {1-2 x}}{125}+\frac {\frac {131 \left (1-2 x \right )^{\frac {3}{2}}}{275}-\frac {133 \sqrt {1-2 x}}{125}}{\left (-6-10 x \right )^{2}}-\frac {409 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{15125}\)	\(57\)
trager	\(\frac {\left (1980 x^{2}+2245 x +632\right ) \sqrt {1-2 x}}{550 \left (3+5 x \right )^{2}}-\frac {409 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (-\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x -8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )-55 \sqrt {1-2 x}}{3+5 x}\right )}{30250}\)	\(73\)

Input:

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

Output:

-1/550*(3960*x^3+2510*x^2-981*x-632)/(3+5*x)^2/(1-2*x)^(1/2)-409/15125*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 = 74, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {409 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (1980 \, x^{2} + 2245 \, x + 632\right )} \sqrt {-2 \, x + 1}}{30250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \] Input:

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

Output:

1/30250*(409*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 
1) - 8)/(5*x + 3)) + 55*(1980*x^2 + 2245*x + 632)*sqrt(-2*x + 1))/(25*x^2 
+ 30*x + 9)
 

Sympy [A] (verification not implemented)

Time = 122.44 (sec) , antiderivative size = 342, normalized size of antiderivative = 4.22 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {18 \sqrt {1 - 2 x}}{125} + \frac {87 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{6875} - \frac {256 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{125} + \frac {88 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{125} \] Input:

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

Output:

18*sqrt(1 - 2*x)/125 + 87*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log( 
sqrt(1 - 2*x) + sqrt(55)/5))/6875 - 256*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*(sq 
rt(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)))/125 + 88* 
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)))/125
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {409}{30250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {18}{125} \, \sqrt {-2 \, x + 1} + \frac {655 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1463 \, \sqrt {-2 \, x + 1}}{1375 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \] Input:

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

Output:

409/30250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(- 
2*x + 1))) + 18/125*sqrt(-2*x + 1) + 1/1375*(655*(-2*x + 1)^(3/2) - 1463*s 
qrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {409}{30250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {18}{125} \, \sqrt {-2 \, x + 1} + \frac {655 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1463 \, \sqrt {-2 \, x + 1}}{5500 \, {\left (5 \, x + 3\right )}^{2}} \] Input:

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

Output:

409/30250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) 
+ 5*sqrt(-2*x + 1))) + 18/125*sqrt(-2*x + 1) + 1/5500*(655*(-2*x + 1)^(3/2 
) - 1463*sqrt(-2*x + 1))/(5*x + 3)^2
 

Mupad [B] (verification not implemented)

Time = 1.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {18\,\sqrt {1-2\,x}}{125}-\frac {409\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{15125}-\frac {\frac {133\,\sqrt {1-2\,x}}{3125}-\frac {131\,{\left (1-2\,x\right )}^{3/2}}{6875}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \] Input:

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

Output:

(18*(1 - 2*x)^(1/2))/125 - (409*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/ 
11))/15125 - ((133*(1 - 2*x)^(1/2))/3125 - (131*(1 - 2*x)^(3/2))/6875)/((4 
4*x)/5 + (2*x - 1)^2 + 11/25)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.88 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {108900 \sqrt {-2 x +1}\, x^{2}+123475 \sqrt {-2 x +1}\, x +34760 \sqrt {-2 x +1}+10225 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x^{2}+12270 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x +3681 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right )-10225 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x^{2}-12270 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x -3681 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right )}{756250 x^{2}+907500 x +272250} \] Input:

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

Output:

(108900*sqrt( - 2*x + 1)*x**2 + 123475*sqrt( - 2*x + 1)*x + 34760*sqrt( - 
2*x + 1) + 10225*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x**2 + 12270* 
sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x + 3681*sqrt(55)*log(5*sqrt( 
- 2*x + 1) - sqrt(55)) - 10225*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55)) 
*x**2 - 12270*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55))*x - 3681*sqrt(55 
)*log(5*sqrt( - 2*x + 1) + sqrt(55)))/(30250*(25*x**2 + 30*x + 9))