\(\int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx\) [262]

Optimal result

Integrand size = 28, antiderivative size = 144 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=\frac {4 \sqrt {a+b x+c x^2}}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}+\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c \left (b^2-4 a c\right )^{3/4} d^{5/2} \sqrt {a+b x+c x^2}} \] Output:

4/3*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(3/2)+2/3*(-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/(-4*a*c+b^2)^(3/4)/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 = 10.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.63 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=-\frac {2 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 c d (d (b+2 c x))^{3/2} \sqrt {a+x (b+c x)}} \] Input:

Integrate[1/((b*d + 2*c*d*x)^(5/2)*Sqrt[a + b*x + c*x^2]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 144, 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 = {1117, 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.

\(\displaystyle \int \frac {1}{\sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 1117

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

\(\Big \downarrow \) 1115

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

\(\Big \downarrow \) 1113

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

\(\Big \downarrow \) 762

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

Input:

Int[1/((b*d + 2*c*d*x)^(5/2)*Sqrt[a + b*x + c*x^2]),x]
 

Output:

(4*Sqrt[a + b*x + c*x^2])/(3*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)) + (2*S 
qrt[-((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*(b^2 - 4*a*c)^(3/4)*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 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]
 

rule 1117

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[-2*b*d*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m 
+ 1)*(b^2 - 4*a*c))), x] + Simp[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 - 4*a* 
c)))   Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] & 
& (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || IntegerQ[(m + 2*p + 3) 
/2])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(361\) vs. \(2(122)=244\).

Time = 4.27 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.51

method	result	size
default	\(-\frac {\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, \left (2 \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, c x +\sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, b +4 c^{2} x^{2}+4 c b x +4 a c \right )}{3 d^{3} \left (2 x^{3} c^{2}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right ) \left (4 a c -b^{2}\right ) c \left (2 c x +b \right )}\)	\(362\)
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{3 \left (4 a c -b^{2}\right ) d^{3} c^{2} \left (x +\frac {b}{2 c}\right )^{2}}-\frac {2 \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{3 \left (4 a c -b^{2}\right ) d^{2} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(490\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=\frac {4 \, \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a} c^{2} + \sqrt {2} {\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )}{3 \, {\left (4 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{3} x^{2} + 4 \, {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} d^{3} x + {\left (b^{4} c^{2} - 4 \, a b^{2} c^{3}\right )} d^{3}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}}{8 c^{4} x^{5}+20 b \,c^{3} x^{4}+8 a \,c^{3} x^{3}+18 b^{2} c^{2} x^{3}+12 a b \,c^{2} x^{2}+7 b^{3} c \,x^{2}+6 a \,b^{2} c x +b^{4} x +a \,b^{3}}d x \right )}{d^{3}} \] Input:

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

Output:

(sqrt(d)*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2))/(a*b**3 + 6*a*b**2*c 
*x + 12*a*b*c**2*x**2 + 8*a*c**3*x**3 + b**4*x + 7*b**3*c*x**2 + 18*b**2*c 
**2*x**3 + 20*b*c**3*x**4 + 8*c**4*x**5),x))/d**3