\(\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx\) [226]

Optimal result

Integrand size = 28, antiderivative size = 137 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx=-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}+\frac {\sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c^2 d^{5/2} \sqrt {a+b x+c x^2}} \] Output:

-1/3*(c*x^2+b*x+a)^(1/2)/c/d/(2*c*d*x+b*d)^(3/2)+1/3*(-4*a*c+b^2)^(1/4)*(- 
c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*EllipticF((2*c*d*x+b*d)^(1/2)/(-4*a*c+ 
b^2)^(1/4)/d^(1/2),I)/c^2/d^(5/2)/(c*x^2+b*x+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx=-\frac {\sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{2},\frac {1}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{6 c d (d (b+2 c x))^{3/2} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[Sqrt[a + b*x + c*x^2]/(b*d + 2*c*d*x)^(5/2),x]
 

Output:

-1/6*(Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-3/4, -1/2, 1/4, (b + 2*c*x) 
^2/(b^2 - 4*a*c)])/(c*d*(d*(b + 2*c*x))^(3/2)*Sqrt[(c*(a + x*(b + c*x)))/( 
-b^2 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1108, 1115, 1113, 762}

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.

\(\displaystyle \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {\int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx}{6 c d^2}-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}\)

\(\Big \downarrow \) 1115

\(\displaystyle \frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{6 c d^2 \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}\)

\(\Big \downarrow \) 1113

\(\displaystyle \frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{3 c^2 d^3 \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {\sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c^2 d^{5/2} \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}\)

Input:

Int[Sqrt[a + b*x + c*x^2]/(b*d + 2*c*d*x)^(5/2),x]
 

Output:

-1/3*Sqrt[a + b*x + c*x^2]/(c*d*(b*d + 2*c*d*x)^(3/2)) + ((b^2 - 4*a*c)^(1 
/4)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sqrt[b*d 
 + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(3*c^2*d^(5/2)*Sqrt[a + b 
*x + c*x^2])
 

Defintions of rubi rules used

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 1108

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 1))), x] - Si 
mp[b*(p/(d*e*(m + 1)))   Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x 
], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 
3, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0] 
) && IntegerQ[2*p]
 

rule 1113

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_ 
Symbol] :> Simp[(4/e)*Sqrt[-c/(b^2 - 4*a*c)]   Subst[Int[1/Sqrt[Simp[1 - b^ 
2*(x^4/(d^2*(b^2 - 4*a*c))), x]], x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, 
 c, d, e}, x] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]
 

rule 1115

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sym 
bol] :> Simp[Sqrt[(-c)*((a + b*x + c*x^2)/(b^2 - 4*a*c))]/Sqrt[a + b*x + c* 
x^2]   Int[(d + e*x)^m/Sqrt[(-a)*(c/(b^2 - 4*a*c)) - b*c*(x/(b^2 - 4*a*c)) 
- c^2*(x^2/(b^2 - 4*a*c))], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c* 
d - b*e, 0] && EqQ[m^2, 1/4]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(318\) vs. \(2(115)=230\).

Time = 1.66 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.33

method	result	size
default	\(\frac {\left (2 \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, c x +\sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, b -2 c^{2} x^{2}-2 c b x -2 a c \right ) \sqrt {d \left (2 c x +b \right )}}{6 d^{3} \sqrt {c \,x^{2}+b x +a}\, \left (2 c x +b \right )^{2} c^{2}}\)	\(319\)
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{12 d^{3} c^{3} \left (x +\frac {b}{2 c}\right )^{2}}+\frac {\left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{3 d^{2} c \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(469\)

Input:

int((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*(2*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x+ 
b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^( 
1/2))^(1/2)*EllipticF(1/2*2^(1/2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2 
)^(1/2)+b))^(1/2),2^(1/2))*(-4*a*c+b^2)^(1/2)*c*x+(1/(-4*a*c+b^2)^(1/2)*(2 
*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*(( 
-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*2^(1/ 
2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2),2^(1/2))*(-4* 
a*c+b^2)^(1/2)*b-2*c^2*x^2-2*c*b*x-2*a*c)/d^3*(d*(2*c*x+b))^(1/2)/(c*x^2+b 
*x+a)^(1/2)/(2*c*x+b)^2/c^2
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx=-\frac {2 \, \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a} c^{2} - \sqrt {2} {\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )}{6 \, {\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\right )}} \] Input:

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^(5/2),x, algorithm="fricas")
 

Output:

-1/6*(2*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)*c^2 - sqrt(2)*(4*c^2*x^2 
 + 4*b*c*x + b^2)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/ 
2*(2*c*x + b)/c))/(4*c^5*d^3*x^2 + 4*b*c^4*d^3*x + b^2*c^3*d^3)
 

Sympy [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx=\int \frac {\sqrt {a + b x + c x^{2}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((c*x**2+b*x+a)**(1/2)/(2*c*d*x+b*d)**(5/2),x)
 

Output:

Integral(sqrt(a + b*x + c*x**2)/(d*(b + 2*c*x))**(5/2), x)
 

Maxima [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x + a}}{{\left (2 \, c d x + b d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^(5/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(c*x^2 + b*x + a)/(2*c*d*x + b*d)^(5/2), x)
 

Giac [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x + a}}{{\left (2 \, c d x + b d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^(5/2),x, algorithm="giac")
 

Output:

integrate(sqrt(c*x^2 + b*x + a)/(2*c*d*x + b*d)^(5/2), x)
 

Mupad [F(-1)]

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

int((a + b*x + c*x^2)^(1/2)/(b*d + 2*c*d*x)^(5/2),x)
 

Output:

int((a + b*x + c*x^2)^(1/2)/(b*d + 2*c*d*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx=\text {too large to display} \] Input:

int((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^(5/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(d)*( - 2*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a + 24*int((sqrt(b + 
 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(6*a**2*b**3*c + 36*a**2*b**2*c**2*x 
+ 72*a**2*b*c**3*x**2 + 48*a**2*c**4*x**3 - a*b**5 + 30*a*b**3*c**2*x**2 + 
 100*a*b**2*c**3*x**3 + 120*a*b*c**4*x**4 + 48*a*c**5*x**5 - b**6*x - 7*b* 
*5*c*x**2 - 18*b**4*c**2*x**3 - 20*b**3*c**3*x**4 - 8*b**2*c**4*x**5),x)*a 
**2*b**2*c**3 + 96*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(6*a* 
*2*b**3*c + 36*a**2*b**2*c**2*x + 72*a**2*b*c**3*x**2 + 48*a**2*c**4*x**3 
- a*b**5 + 30*a*b**3*c**2*x**2 + 100*a*b**2*c**3*x**3 + 120*a*b*c**4*x**4 
+ 48*a*c**5*x**5 - b**6*x - 7*b**5*c*x**2 - 18*b**4*c**2*x**3 - 20*b**3*c* 
*3*x**4 - 8*b**2*c**4*x**5),x)*a**2*b*c**4*x + 96*int((sqrt(b + 2*c*x)*sqr 
t(a + b*x + c*x**2)*x**2)/(6*a**2*b**3*c + 36*a**2*b**2*c**2*x + 72*a**2*b 
*c**3*x**2 + 48*a**2*c**4*x**3 - a*b**5 + 30*a*b**3*c**2*x**2 + 100*a*b**2 
*c**3*x**3 + 120*a*b*c**4*x**4 + 48*a*c**5*x**5 - b**6*x - 7*b**5*c*x**2 - 
 18*b**4*c**2*x**3 - 20*b**3*c**3*x**4 - 8*b**2*c**4*x**5),x)*a**2*c**5*x* 
*2 - 10*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(6*a**2*b**3*c + 
 36*a**2*b**2*c**2*x + 72*a**2*b*c**3*x**2 + 48*a**2*c**4*x**3 - a*b**5 + 
30*a*b**3*c**2*x**2 + 100*a*b**2*c**3*x**3 + 120*a*b*c**4*x**4 + 48*a*c**5 
*x**5 - b**6*x - 7*b**5*c*x**2 - 18*b**4*c**2*x**3 - 20*b**3*c**3*x**4 - 8 
*b**2*c**4*x**5),x)*a*b**4*c**2 - 40*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c 
*x**2)*x**2)/(6*a**2*b**3*c + 36*a**2*b**2*c**2*x + 72*a**2*b*c**3*x**2...