\(\int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{f+g x} \, dx\) [299]

Optimal result

Integrand size = 33, antiderivative size = 106 \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{f+g x} \, dx=-\frac {\sqrt {\frac {2}{3}} (3 f-2 g) \sqrt {3-2 x}}{g^2}-\frac {(3-2 x)^{3/2}}{\sqrt {6} g}+\frac {\sqrt {\frac {2}{3}} (3 f-2 g) \sqrt {2 f+3 g} \text {arctanh}\left (\frac {\sqrt {g} \sqrt {3-2 x}}{\sqrt {2 f+3 g}}\right )}{g^{5/2}} \] Output:

-1/3*6^(1/2)*(3*f-2*g)*(3-2*x)^(1/2)/g^2-1/6*(3-2*x)^(3/2)*6^(1/2)/g+1/3*6 
^(1/2)*(3*f-2*g)*(2*f+3*g)^(1/2)*arctanh(g^(1/2)*(3-2*x)^(1/2)/(2*f+3*g)^( 
1/2))/g^(5/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.23 \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{f+g x} \, dx=\frac {\frac {3 \sqrt {g} (-6 f+g+2 g x) \sqrt {6+5 x-6 x^2}}{\sqrt {2+3 x}}+\frac {6 \left (6 f^2+5 f g-6 g^2\right ) \text {arctanh}\left (\frac {\sqrt {2 f+3 g} \sqrt {6+5 x-6 x^2}}{\sqrt {g} (3-2 x) \sqrt {2+3 x}}\right )}{\sqrt {2 f+3 g}}}{3 \sqrt {6} g^{5/2}} \] Input:

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

Output:

((3*Sqrt[g]*(-6*f + g + 2*g*x)*Sqrt[6 + 5*x - 6*x^2])/Sqrt[2 + 3*x] + (6*( 
6*f^2 + 5*f*g - 6*g^2)*ArcTanh[(Sqrt[2*f + 3*g]*Sqrt[6 + 5*x - 6*x^2])/(Sq 
rt[g]*(3 - 2*x)*Sqrt[2 + 3*x])])/Sqrt[2*f + 3*g])/(3*Sqrt[6]*g^(5/2))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1245, 90, 27, 60, 73, 221}

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.

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

\(\Big \downarrow \) 1245

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

\(\displaystyle -\frac {(3 f-2 g) \left (\frac {2 \sqrt {3-2 x}}{g}-\frac {(2 f+3 g) \int \frac {1}{\frac {1}{2} (2 f+3 g)-\frac {1}{2} g (3-2 x)}d\sqrt {3-2 x}}{g}\right )}{\sqrt {6} g}-\frac {(3-2 x)^{3/2}}{\sqrt {6} g}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(3 f-2 g) \left (\frac {2 \sqrt {3-2 x}}{g}-\frac {2 \sqrt {2 f+3 g} \text {arctanh}\left (\frac {\sqrt {g} \sqrt {3-2 x}}{\sqrt {2 f+3 g}}\right )}{g^{3/2}}\right )}{\sqrt {6} g}-\frac {(3-2 x)^{3/2}}{\sqrt {6} g}\)

Input:

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

Output:

-((3 - 2*x)^(3/2)/(Sqrt[6]*g)) - ((3*f - 2*g)*((2*Sqrt[3 - 2*x])/g - (2*Sq 
rt[2*f + 3*g]*ArcTanh[(Sqrt[g]*Sqrt[3 - 2*x])/Sqrt[2*f + 3*g]])/g^(3/2)))/ 
(Sqrt[6]*g)
 

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 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 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 221

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

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 = 1.39 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.57

method	result	size
risch	\(\frac {\left (-2 g x +6 f -g \right ) \left (2 x -3\right ) \sqrt {\frac {-36 x^{2}+30 x +36}{3 x +2}}\, \sqrt {3 x +2}}{g^{2} \sqrt {-12 x +18}\, \sqrt {-36 x^{2}+30 x +36}}+\frac {\left (6 f^{2}+5 f g -6 g^{2}\right ) \sqrt {6}\, \operatorname {arctanh}\left (\frac {g \sqrt {-12 x +18}\, \sqrt {6}}{6 \sqrt {g \left (2 f +3 g \right )}}\right ) \sqrt {\frac {-36 x^{2}+30 x +36}{3 x +2}}\, \sqrt {3 x +2}}{3 g^{2} \sqrt {g \left (2 f +3 g \right )}\, \sqrt {-36 x^{2}+30 x +36}}\)	\(166\)
default	\(\frac {\sqrt {-6 x^{2}+5 x +6}\, \sqrt {6}\, \left (2 \sqrt {g \left (2 f +3 g \right )}\, \sqrt {3-2 x}\, g x +12 \,\operatorname {arctanh}\left (\frac {g \sqrt {-12 x +18}\, \sqrt {6}}{6 \sqrt {g \left (2 f +3 g \right )}}\right ) f^{2}+10 \,\operatorname {arctanh}\left (\frac {g \sqrt {-12 x +18}\, \sqrt {6}}{6 \sqrt {g \left (2 f +3 g \right )}}\right ) f g -12 \,\operatorname {arctanh}\left (\frac {g \sqrt {-12 x +18}\, \sqrt {6}}{6 \sqrt {g \left (2 f +3 g \right )}}\right ) g^{2}-6 \sqrt {g \left (2 f +3 g \right )}\, \sqrt {3-2 x}\, f +\sqrt {g \left (2 f +3 g \right )}\, \sqrt {3-2 x}\, g \right )}{6 \sqrt {3 x +2}\, \sqrt {3-2 x}\, \sqrt {g \left (2 f +3 g \right )}\, g^{2}}\)	\(199\)

Input:

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

Output:

(-2*g*x+6*f-g)*(2*x-3)/g^2/(-12*x+18)^(1/2)*((-36*x^2+30*x+36)/(3*x+2))^(1 
/2)*(3*x+2)^(1/2)/(-36*x^2+30*x+36)^(1/2)+1/3*(6*f^2+5*f*g-6*g^2)/g^2*6^(1 
/2)/(g*(2*f+3*g))^(1/2)*arctanh(1/6*g*(-12*x+18)^(1/2)*6^(1/2)/(g*(2*f+3*g 
))^(1/2))*((-36*x^2+30*x+36)/(3*x+2))^(1/2)*(3*x+2)^(1/2)/(-36*x^2+30*x+36 
)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.92 \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{f+g x} \, dx=\left [-\frac {3 \, \sqrt {\frac {2}{3}} {\left (3 \, {\left (3 \, f - 2 \, g\right )} x + 6 \, f - 4 \, g\right )} \sqrt {\frac {2 \, f + 3 \, g}{g}} \log \left (-\frac {6 \, g x^{2} + \sqrt {\frac {2}{3}} \sqrt {-36 \, x^{2} + 30 \, x + 36} g \sqrt {3 \, x + 2} \sqrt {\frac {2 \, f + 3 \, g}{g}} - 2 \, {\left (3 \, f + 7 \, g\right )} x - 4 \, f - 12 \, g}{3 \, g x^{2} + {\left (3 \, f + 2 \, g\right )} x + 2 \, f}\right ) - {\left (2 \, g x - 6 \, f + g\right )} \sqrt {-36 \, x^{2} + 30 \, x + 36} \sqrt {3 \, x + 2}}{6 \, {\left (3 \, g^{2} x + 2 \, g^{2}\right )}}, -\frac {6 \, \sqrt {\frac {2}{3}} {\left (3 \, {\left (3 \, f - 2 \, g\right )} x + 6 \, f - 4 \, g\right )} \sqrt {-\frac {2 \, f + 3 \, g}{g}} \arctan \left (\frac {\sqrt {\frac {2}{3}} \sqrt {-36 \, x^{2} + 30 \, x + 36} g \sqrt {3 \, x + 2} \sqrt {-\frac {2 \, f + 3 \, g}{g}}}{2 \, {\left (3 \, {\left (2 \, f + 3 \, g\right )} x + 4 \, f + 6 \, g\right )}}\right ) - {\left (2 \, g x - 6 \, f + g\right )} \sqrt {-36 \, x^{2} + 30 \, x + 36} \sqrt {3 \, x + 2}}{6 \, {\left (3 \, g^{2} x + 2 \, g^{2}\right )}}\right ] \] Input:

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

Output:

[-1/6*(3*sqrt(2/3)*(3*(3*f - 2*g)*x + 6*f - 4*g)*sqrt((2*f + 3*g)/g)*log(- 
(6*g*x^2 + sqrt(2/3)*sqrt(-36*x^2 + 30*x + 36)*g*sqrt(3*x + 2)*sqrt((2*f + 
 3*g)/g) - 2*(3*f + 7*g)*x - 4*f - 12*g)/(3*g*x^2 + (3*f + 2*g)*x + 2*f)) 
- (2*g*x - 6*f + g)*sqrt(-36*x^2 + 30*x + 36)*sqrt(3*x + 2))/(3*g^2*x + 2* 
g^2), -1/6*(6*sqrt(2/3)*(3*(3*f - 2*g)*x + 6*f - 4*g)*sqrt(-(2*f + 3*g)/g) 
*arctan(1/2*sqrt(2/3)*sqrt(-36*x^2 + 30*x + 36)*g*sqrt(3*x + 2)*sqrt(-(2*f 
 + 3*g)/g)/(3*(2*f + 3*g)*x + 4*f + 6*g)) - (2*g*x - 6*f + g)*sqrt(-36*x^2 
 + 30*x + 36)*sqrt(3*x + 2))/(3*g^2*x + 2*g^2)]
 

Sympy [F]

\[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{f+g x} \, dx=\frac {\sqrt {6} \int \frac {\sqrt {3 x + 2} \sqrt {- 6 x^{2} + 5 x + 6}}{f + g x}\, dx}{6} \] Input:

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

Output:

sqrt(6)*Integral(sqrt(3*x + 2)*sqrt(-6*x**2 + 5*x + 6)/(f + g*x), x)/6
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{f+g x} \, dx=-\frac {1}{6} \, \sqrt {6} {\left (\frac {2 \, {\left (6 \, f^{2} + 5 \, f g - 6 \, g^{2}\right )} \arctan \left (\frac {g \sqrt {-2 \, x + 3}}{\sqrt {-2 \, f g - 3 \, g^{2}}}\right )}{\sqrt {-2 \, f g - 3 \, g^{2}} g^{2}} + \frac {g^{2} {\left (-2 \, x + 3\right )}^{\frac {3}{2}} + 6 \, f g \sqrt {-2 \, x + 3} - 4 \, g^{2} \sqrt {-2 \, x + 3}}{g^{3}}\right )} \] Input:

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

Output:

-1/6*sqrt(6)*(2*(6*f^2 + 5*f*g - 6*g^2)*arctan(g*sqrt(-2*x + 3)/sqrt(-2*f* 
g - 3*g^2))/(sqrt(-2*f*g - 3*g^2)*g^2) + (g^2*(-2*x + 3)^(3/2) + 6*f*g*sqr 
t(-2*x + 3) - 4*g^2*sqrt(-2*x + 3))/g^3)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {2+3 x} \sqrt {1+\frac {5 x}{6}-x^2}}{f+g x} \, dx=\frac {\sqrt {6}\, \left (6 \sqrt {g}\, \sqrt {-2 f -3 g}\, \mathit {atan} \left (\frac {\sqrt {-2 x +3}\, g}{\sqrt {g}\, \sqrt {-2 f -3 g}}\right ) f -4 \sqrt {g}\, \sqrt {-2 f -3 g}\, \mathit {atan} \left (\frac {\sqrt {-2 x +3}\, g}{\sqrt {g}\, \sqrt {-2 f -3 g}}\right ) g -6 \sqrt {-2 x +3}\, f g +2 \sqrt {-2 x +3}\, g^{2} x +\sqrt {-2 x +3}\, g^{2}\right )}{6 g^{3}} \] Input:

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

Output:

(sqrt(6)*(6*sqrt(g)*sqrt( - 2*f - 3*g)*atan((sqrt( - 2*x + 3)*g)/(sqrt(g)* 
sqrt( - 2*f - 3*g)))*f - 4*sqrt(g)*sqrt( - 2*f - 3*g)*atan((sqrt( - 2*x + 
3)*g)/(sqrt(g)*sqrt( - 2*f - 3*g)))*g - 6*sqrt( - 2*x + 3)*f*g + 2*sqrt( - 
 2*x + 3)*g**2*x + sqrt( - 2*x + 3)*g**2))/(6*g**3)