\(\int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{(f+g x)^{7/2}} \, dx\) [307]

Optimal result

Integrand size = 35, antiderivative size = 97 \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{(f+g x)^{7/2}} \, dx=\frac {\sqrt {\frac {2}{3}} (3 f-2 g) (3-2 x)^{3/2}}{5 g (2 f+3 g) (f+g x)^{5/2}}-\frac {\sqrt {\frac {2}{3}} (18 f+53 g) (3-2 x)^{3/2}}{15 g (2 f+3 g)^2 (f+g x)^{3/2}} \] Output:

1/15*6^(1/2)*(3*f-2*g)*(3-2*x)^(3/2)/g/(2*f+3*g)/(g*x+f)^(5/2)-1/45*6^(1/2 
)*(18*f+53*g)*(3-2*x)^(3/2)/g/(2*f+3*g)^2/(g*x+f)^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{(f+g x)^{7/2}} \, dx=\frac {(-3+2 x) \sqrt {4+\frac {10 x}{3}-4 x^2} (2 f (19+9 x)+g (18+53 x))}{15 (2 f+3 g)^2 \sqrt {2+3 x} (f+g x)^{5/2}} \] Input:

Integrate[(Sqrt[2 + 3*x]*Sqrt[1 + (5*x)/6 - x^2])/(f + g*x)^(7/2),x]
 

Output:

((-3 + 2*x)*Sqrt[4 + (10*x)/3 - 4*x^2]*(2*f*(19 + 9*x) + g*(18 + 53*x)))/( 
15*(2*f + 3*g)^2*Sqrt[2 + 3*x]*(f + g*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1245, 87, 27, 48}

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.

\(\displaystyle \int \frac {\sqrt {3 x+2} \sqrt {-x^2+\frac {5 x}{6}+1}}{(f+g x)^{7/2}} \, dx\)

\(\Big \downarrow \) 1245

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {(18 f+53 g) \int \frac {\sqrt {3-2 x}}{\sqrt {6} (f+g x)^{5/2}}dx}{5 g (2 f+3 g)}+\frac {\sqrt {\frac {2}{3}} (3-2 x)^{3/2} (3 f-2 g)}{5 g (2 f+3 g) (f+g x)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(18 f+53 g) \int \frac {\sqrt {3-2 x}}{(f+g x)^{5/2}}dx}{5 \sqrt {6} g (2 f+3 g)}+\frac {\sqrt {\frac {2}{3}} (3-2 x)^{3/2} (3 f-2 g)}{5 g (2 f+3 g) (f+g x)^{5/2}}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {\sqrt {\frac {2}{3}} (3-2 x)^{3/2} (3 f-2 g)}{5 g (2 f+3 g) (f+g x)^{5/2}}-\frac {\sqrt {\frac {2}{3}} (3-2 x)^{3/2} (18 f+53 g)}{15 g (2 f+3 g)^2 (f+g x)^{3/2}}\)

Input:

Int[(Sqrt[2 + 3*x]*Sqrt[1 + (5*x)/6 - x^2])/(f + g*x)^(7/2),x]
 

Output:

(Sqrt[2/3]*(3*f - 2*g)*(3 - 2*x)^(3/2))/(5*g*(2*f + 3*g)*(f + g*x)^(5/2)) 
- (Sqrt[2/3]*(18*f + 53*g)*(3 - 2*x)^(3/2))/(15*g*(2*f + 3*g)^2*(f + g*x)^ 
(3/2))
 

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 48

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp 
[(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ 
a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

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 1245

Int[((d_) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[(d + e*x)^(m + p)*(f + g*x)^n*(a/d 
+ (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[c*d^2 - 
 b*d*e + a*e^2, 0] && GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]
 

Maple [A] (verified)

Time = 1.41 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {\sqrt {-36 x^{2}+30 x +36}\, \left (2 x -3\right ) \left (18 f x +53 g x +38 f +18 g \right )}{45 \sqrt {3 x +2}\, \left (g x +f \right )^{\frac {5}{2}} \left (2 f +3 g \right )^{2}}\)	\(58\)
gosper	\(\frac {\left (2 x -3\right ) \left (18 f x +53 g x +38 f +18 g \right ) \sqrt {-36 x^{2}+30 x +36}}{45 \left (g x +f \right )^{\frac {5}{2}} \left (4 f^{2}+12 f g +9 g^{2}\right ) \sqrt {3 x +2}}\)	\(66\)
orering	\(\frac {\left (2 x -3\right ) \left (18 f x +53 g x +38 f +18 g \right ) \sqrt {-36 x^{2}+30 x +36}}{45 \left (g x +f \right )^{\frac {5}{2}} \left (4 f^{2}+12 f g +9 g^{2}\right ) \sqrt {3 x +2}}\)	\(66\)

Input:

int(1/6*(3*x+2)^(1/2)*(-36*x^2+30*x+36)^(1/2)/(g*x+f)^(7/2),x,method=_RETU 
RNVERBOSE)
 

Output:

1/45/(3*x+2)^(1/2)*(-36*x^2+30*x+36)^(1/2)/(g*x+f)^(5/2)*(2*x-3)*(18*f*x+5 
3*g*x+38*f+18*g)/(2*f+3*g)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (77) = 154\).

Time = 0.09 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.06 \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{(f+g x)^{7/2}} \, dx=\frac {{\left (2 \, {\left (18 \, f + 53 \, g\right )} x^{2} + {\left (22 \, f - 123 \, g\right )} x - 114 \, f - 54 \, g\right )} \sqrt {g x + f} \sqrt {-36 \, x^{2} + 30 \, x + 36} \sqrt {3 \, x + 2}}{45 \, {\left (8 \, f^{5} + 24 \, f^{4} g + 18 \, f^{3} g^{2} + 3 \, {\left (4 \, f^{2} g^{3} + 12 \, f g^{4} + 9 \, g^{5}\right )} x^{4} + {\left (36 \, f^{3} g^{2} + 116 \, f^{2} g^{3} + 105 \, f g^{4} + 18 \, g^{5}\right )} x^{3} + 3 \, {\left (12 \, f^{4} g + 44 \, f^{3} g^{2} + 51 \, f^{2} g^{3} + 18 \, f g^{4}\right )} x^{2} + 3 \, {\left (4 \, f^{5} + 20 \, f^{4} g + 33 \, f^{3} g^{2} + 18 \, f^{2} g^{3}\right )} x\right )}} \] Input:

integrate(1/6*(2+3*x)^(1/2)*(-36*x^2+30*x+36)^(1/2)/(g*x+f)^(7/2),x, algor 
ithm="fricas")
 

Output:

1/45*(2*(18*f + 53*g)*x^2 + (22*f - 123*g)*x - 114*f - 54*g)*sqrt(g*x + f) 
*sqrt(-36*x^2 + 30*x + 36)*sqrt(3*x + 2)/(8*f^5 + 24*f^4*g + 18*f^3*g^2 + 
3*(4*f^2*g^3 + 12*f*g^4 + 9*g^5)*x^4 + (36*f^3*g^2 + 116*f^2*g^3 + 105*f*g 
^4 + 18*g^5)*x^3 + 3*(12*f^4*g + 44*f^3*g^2 + 51*f^2*g^3 + 18*f*g^4)*x^2 + 
 3*(4*f^5 + 20*f^4*g + 33*f^3*g^2 + 18*f^2*g^3)*x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{(f+g x)^{7/2}} \, dx=\text {Timed out} \] Input:

integrate(1/6*(2+3*x)**(1/2)*(-36*x**2+30*x+36)**(1/2)/(g*x+f)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{(f+g x)^{7/2}} \, dx=\int { \frac {\sqrt {-36 \, x^{2} + 30 \, x + 36} \sqrt {3 \, x + 2}}{6 \, {\left (g x + f\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate(1/6*(2+3*x)^(1/2)*(-36*x^2+30*x+36)^(1/2)/(g*x+f)^(7/2),x, algor 
ithm="maxima")
 

Output:

1/6*integrate(sqrt(-36*x^2 + 30*x + 36)*sqrt(3*x + 2)/(g*x + f)^(7/2), x)
 

Giac [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{(f+g x)^{7/2}} \, dx=\frac {2 \, \sqrt {6} {\left (\frac {{\left (18 \, \sqrt {2} f g^{2} + 53 \, \sqrt {2} g^{3}\right )} {\left (2 \, x - 3\right )}}{4 \, f^{2} g^{2} + 12 \, f g^{3} + 9 \, g^{4}} + \frac {65 \, {\left (2 \, \sqrt {2} f g^{2} + 3 \, \sqrt {2} g^{3}\right )}}{4 \, f^{2} g^{2} + 12 \, f g^{3} + 9 \, g^{4}}\right )} {\left (2 \, x - 3\right )} \sqrt {-2 \, x + 3}}{45 \, {\left (g {\left (2 \, x - 3\right )} + 2 \, f + 3 \, g\right )}^{\frac {5}{2}}} \] Input:

integrate(1/6*(2+3*x)^(1/2)*(-36*x^2+30*x+36)^(1/2)/(g*x+f)^(7/2),x, algor 
ithm="giac")
 

Output:

2/45*sqrt(6)*((18*sqrt(2)*f*g^2 + 53*sqrt(2)*g^3)*(2*x - 3)/(4*f^2*g^2 + 1 
2*f*g^3 + 9*g^4) + 65*(2*sqrt(2)*f*g^2 + 3*sqrt(2)*g^3)/(4*f^2*g^2 + 12*f* 
g^3 + 9*g^4))*(2*x - 3)*sqrt(-2*x + 3)/(g*(2*x - 3) + 2*f + 3*g)^(5/2)
 

Mupad [B] (verification not implemented)

Time = 11.71 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.77 \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{(f+g x)^{7/2}} \, dx=\frac {\sqrt {-36\,x^2+30\,x+36}\,\left (\frac {x\,\sqrt {3\,x+2}\,\left (22\,f-123\,g\right )}{135\,g^2\,{\left (2\,f+3\,g\right )}^2}-\frac {\sqrt {3\,x+2}\,\left (114\,f+54\,g\right )}{135\,g^2\,{\left (2\,f+3\,g\right )}^2}+\frac {x^2\,\sqrt {3\,x+2}\,\left (36\,f+106\,g\right )}{135\,g^2\,{\left (2\,f+3\,g\right )}^2}\right )}{x^3\,\sqrt {f+g\,x}+\frac {2\,f^2\,\sqrt {f+g\,x}}{3\,g^2}+\frac {2\,x^2\,\sqrt {f+g\,x}\,\left (3\,f+g\right )}{3\,g}+\frac {f\,x\,\sqrt {f+g\,x}\,\left (3\,f+4\,g\right )}{3\,g^2}} \] Input:

int(((3*x + 2)^(1/2)*(30*x - 36*x^2 + 36)^(1/2))/(6*(f + g*x)^(7/2)),x)
 

Output:

((30*x - 36*x^2 + 36)^(1/2)*((x*(3*x + 2)^(1/2)*(22*f - 123*g))/(135*g^2*( 
2*f + 3*g)^2) - ((3*x + 2)^(1/2)*(114*f + 54*g))/(135*g^2*(2*f + 3*g)^2) + 
 (x^2*(3*x + 2)^(1/2)*(36*f + 106*g))/(135*g^2*(2*f + 3*g)^2)))/(x^3*(f + 
g*x)^(1/2) + (2*f^2*(f + g*x)^(1/2))/(3*g^2) + (2*x^2*(f + g*x)^(1/2)*(3*f 
 + g))/(3*g) + (f*x*(f + g*x)^(1/2)*(3*f + 4*g))/(3*g^2))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{(f+g x)^{7/2}} \, dx=\frac {\sqrt {-2 x +3}\, \sqrt {6}\, \left (36 f \,x^{2}+106 g \,x^{2}+22 f x -123 g x -114 f -54 g \right )}{45 \sqrt {g x +f}\, \left (4 f^{2} g^{2} x^{2}+12 f \,g^{3} x^{2}+9 g^{4} x^{2}+8 f^{3} g x +24 f^{2} g^{2} x +18 f \,g^{3} x +4 f^{4}+12 f^{3} g +9 f^{2} g^{2}\right )} \] Input:

int(1/6*(2+3*x)^(1/2)*(-36*x^2+30*x+36)^(1/2)/(g*x+f)^(7/2),x)
 

Output:

(sqrt( - 2*x + 3)*sqrt(6)*(36*f*x**2 + 22*f*x - 114*f + 106*g*x**2 - 123*g 
*x - 54*g))/(45*sqrt(f + g*x)*(4*f**4 + 8*f**3*g*x + 12*f**3*g + 4*f**2*g* 
*2*x**2 + 24*f**2*g**2*x + 9*f**2*g**2 + 12*f*g**3*x**2 + 18*f*g**3*x + 9* 
g**4*x**2))