\(\int \frac {(1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx\) [970]

Optimal result

Integrand size = 26, antiderivative size = 93 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {(1-2 x)^{3/2} \sqrt {3+5 x}}{2 (2+3 x)^2}+\frac {33 \sqrt {1-2 x} \sqrt {3+5 x}}{4 (2+3 x)}-\frac {363 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{4 \sqrt {7}} \] Output:

1/2*(1-2*x)^(3/2)*(3+5*x)^(1/2)/(2+3*x)^2+33*(1-2*x)^(1/2)*(3+5*x)^(1/2)/( 
8+12*x)-363/28*7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.74 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {\sqrt {1-2 x} \sqrt {3+5 x} (68+95 x)}{4 (2+3 x)^2}-\frac {363 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{4 \sqrt {7}} \] Input:

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

Output:

(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(68 + 95*x))/(4*(2 + 3*x)^2) - (363*ArcTan[Sq 
rt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(4*Sqrt[7])
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {105, 105, 104, 217}

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.

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

\(\Big \downarrow \) 105

\(\displaystyle \frac {33}{4} \int \frac {\sqrt {1-2 x}}{(3 x+2)^2 \sqrt {5 x+3}}dx+\frac {\sqrt {5 x+3} (1-2 x)^{3/2}}{2 (3 x+2)^2}\)

\(\Big \downarrow \) 105

\(\displaystyle \frac {33}{4} \left (\frac {11}{2} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 x+2}\right )+\frac {\sqrt {5 x+3} (1-2 x)^{3/2}}{2 (3 x+2)^2}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {33}{4} \left (11 \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}+\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 x+2}\right )+\frac {\sqrt {5 x+3} (1-2 x)^{3/2}}{2 (3 x+2)^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {33}{4} \left (\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 x+2}-\frac {11 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}\right )+\frac {\sqrt {5 x+3} (1-2 x)^{3/2}}{2 (3 x+2)^2}\)

Input:

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

Output:

((1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/(2*(2 + 3*x)^2) + (33*((Sqrt[1 - 2*x]*Sqrt 
[3 + 5*x])/(2 + 3*x) - (11*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/ 
Sqrt[7]))/4
 

Defintions of rubi rules used

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 105

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

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.28

method	result	size
risch	\(-\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \left (68+95 x \right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{4 \left (2+3 x \right )^{2} \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {363 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{56 \sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(119\)
default	\(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (3267 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+4356 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +1452 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+1330 x \sqrt {-10 x^{2}-x +3}+952 \sqrt {-10 x^{2}-x +3}\right )}{56 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )^{2}}\)	\(154\)

Input:

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

Output:

-1/4*(-1+2*x)*(3+5*x)^(1/2)*(68+95*x)/(2+3*x)^2/(-(-1+2*x)*(3+5*x))^(1/2)* 
((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)+363/56*7^(1/2)*arctan(9/14*(20/3+37/ 
3*x)*7^(1/2)/(-90*(2/3+x)^2+67+111*x)^(1/2))*((1-2*x)*(3+5*x))^(1/2)/(1-2* 
x)^(1/2)/(3+5*x)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=-\frac {363 \, \sqrt {7} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (95 \, x + 68\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{56 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \] Input:

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

Output:

-1/56*(363*sqrt(7)*(9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt 
(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(95*x + 68)*sqrt(5*x + 3)* 
sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)
 

Sympy [F]

\[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {3}{2}}}{\left (3 x + 2\right )^{3} \sqrt {5 x + 3}}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {363}{56} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {7 \, \sqrt {-10 \, x^{2} - x + 3}}{6 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {95 \, \sqrt {-10 \, x^{2} - x + 3}}{12 \, {\left (3 \, x + 2\right )}} \] Input:

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

Output:

363/56*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 7/6*sqr 
t(-10*x^2 - x + 3)/(9*x^2 + 12*x + 4) + 95/12*sqrt(-10*x^2 - x + 3)/(3*x + 
 2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (72) = 144\).

Time = 0.20 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.69 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {363}{560} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {605 \, \sqrt {10} {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {168 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {672 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{2 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \] Input:

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

Output:

363/560*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((s 
qrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 
5) - sqrt(22)))) + 605/2*sqrt(10)*(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/s 
qrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 + 1 
68*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 672*sqrt(5*x + 3)/ 
(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22 
))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 
 + 280)^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\int \frac {{\left (1-2\,x\right )}^{3/2}}{{\left (3\,x+2\right )}^3\,\sqrt {5\,x+3}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.71 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {3267 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}+4356 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +1452 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-3267 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}-4356 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -1452 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+665 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x +476 \sqrt {5 x +3}\, \sqrt {-2 x +1}}{252 x^{2}+336 x +112} \] Input:

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

Output:

(3267*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5) 
)/sqrt(11))/2))/sqrt(2))*x**2 + 4356*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan 
(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x + 1452*sqrt(7)*a 
tan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2)) 
/sqrt(2)) - 3267*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 
 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x**2 - 4356*sqrt(7)*atan((sqrt(33) + s 
qrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x - 145 
2*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sq 
rt(11))/2))/sqrt(2)) + 665*sqrt(5*x + 3)*sqrt( - 2*x + 1)*x + 476*sqrt(5*x 
 + 3)*sqrt( - 2*x + 1))/(28*(9*x**2 + 12*x + 4))