\(\int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^3} \, dx\) [1078]

Optimal result

Integrand size = 26, antiderivative size = 93 \[ \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^3} \, dx=-\frac {33 \sqrt {1-2 x} \sqrt {3+5 x}}{196 (2+3 x)}-\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{14 (2+3 x)^2}-\frac {363 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{196 \sqrt {7}} \] Output:

-33*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(392+588*x)-1/14*(1-2*x)^(1/2)*(3+5*x)^(3/ 
2)/(2+3*x)^2-363/1372*7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/ 
2))
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.74 \[ \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^3} \, dx=\frac {-\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (108+169 x)}{(2+3 x)^2}-363 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1372} \] Input:

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

Output:

((-7*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(108 + 169*x))/(2 + 3*x)^2 - 363*Sqrt[7]* 
ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/1372
 

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05, 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.

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

\(\Big \downarrow \) 105

\(\displaystyle \frac {33}{28} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x} (3 x+2)^2}dx-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\)

\(\Big \downarrow \) 105

\(\displaystyle \frac {33}{28} \left (\frac {11}{14} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {33}{28} \left (\frac {11}{7} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {33}{28} \left (-\frac {11 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\)

Input:

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

Output:

-1/14*(Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(2 + 3*x)^2 + (33*(-1/7*(Sqrt[1 - 2* 
x]*Sqrt[3 + 5*x])/(2 + 3*x) - (11*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5 
*x])])/(7*Sqrt[7])))/28
 

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 = 0.32 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.28

method	result	size
risch	\(\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \left (169 x +108\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{196 \left (2+3 x \right )^{2} \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {363 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{2744 \sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(119\)
default	\(\frac {\sqrt {3+5 x}\, \sqrt {1-2 x}\, \left (3267 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+4356 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +1452 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-2366 x \sqrt {-10 x^{2}-x +3}-1512 \sqrt {-10 x^{2}-x +3}\right )}{2744 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )^{2}}\)	\(154\)

Input:

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

Output:

1/196*(-1+2*x)*(3+5*x)^(1/2)*(169*x+108)/(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)+363/2744*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 \[ \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^3} \, dx=-\frac {363 \, \sqrt {7} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 14 \, {\left (169 \, x + 108\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{2744 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \] Input:

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

Output:

-1/2744*(363*sqrt(7)*(9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sq 
rt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 14*(169*x + 108)*sqrt(5*x + 
 3)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^3} \, dx=\frac {363}{2744} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {\sqrt {-10 \, x^{2} - x + 3}}{42 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {169 \, \sqrt {-10 \, x^{2} - x + 3}}{588 \, {\left (3 \, x + 2\right )}} \] Input:

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

Output:

363/2744*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 1/42* 
sqrt(-10*x^2 - x + 3)/(9*x^2 + 12*x + 4) - 169/588*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. 252 vs. \(2 (72) = 144\).

Time = 0.22 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.71 \[ \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^3} \, dx=\frac {363}{27440} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {121 \, \sqrt {10} {\left (3 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {1400 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {5600 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{98 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \] Input:

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

Output:

363/27440*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*( 
(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x 
+ 5) - sqrt(22)))) - 121/98*sqrt(10)*(3*((sqrt(2)*sqrt(-10*x + 5) - sqrt(2 
2))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^ 
3 + 1400*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 5600*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. \[ \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^3} \, dx=\int \frac {{\left (5\,x+3\right )}^{3/2}}{\sqrt {1-2\,x}\,{\left (3\,x+2\right )}^3} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.71 \[ \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^3} \, dx=\frac {3267 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}+4356 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +1452 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-3267 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}-4356 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -1452 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-1183 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x -756 \sqrt {5 x +3}\, \sqrt {-2 x +1}}{12348 x^{2}+16464 x +5488} \] Input:

int((3+5*x)^(3/2)/(1-2*x)^(1/2)/(2+3*x)^3,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)) - 1183*sqrt(5*x + 3)*sqrt( - 2*x + 1)*x - 756*sqrt(5* 
x + 3)*sqrt( - 2*x + 1))/(1372*(9*x**2 + 12*x + 4))