\(\int \frac {(b d+2 c d x)^{3/2}}{(a+b x+c x^2)^{3/2}} \, dx\) [277]

Optimal result

Integrand size = 28, antiderivative size = 125 \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}+\frac {4 \sqrt [4]{b^2-4 a c} d^{3/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{\sqrt {a+b x+c x^2}} \] Output:

-2*d*(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(1/2)+4*(-4*a*c+b^2)^(1/4)*d^(3/2)* 
(-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*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 = 10.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.70 \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 d \sqrt {d (b+2 c x)} \left (-1+2 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{\sqrt {a+x (b+c x)}} \] Input:

Integrate[(b*d + 2*c*d*x)^(3/2)/(a + b*x + c*x^2)^(3/2),x]
 

Output:

(2*d*Sqrt[d*(b + 2*c*x)]*(-1 + 2*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c) 
]*Hypergeometric2F1[1/4, 1/2, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/Sqrt[a + 
 x*(b + c*x)]
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 125, 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 = {1110, 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.

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

\(\Big \downarrow \) 1110

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

\(\Big \downarrow \) 1115

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

\(\Big \downarrow \) 1113

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

\(\Big \downarrow \) 762

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

Input:

Int[(b*d + 2*c*d*x)^(3/2)/(a + b*x + c*x^2)^(3/2),x]
 

Output:

(-2*d*Sqrt[b*d + 2*c*d*x])/Sqrt[a + b*x + c*x^2] + (4*(b^2 - 4*a*c)^(1/4)* 
d^(3/2)*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])/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 1110

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), 
x] - Simp[d*e*((m - 1)/(b*(p + 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] && N 
eQ[m + 2*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && 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 [A] (verified)

Time = 2.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.55

method	result	size
default	\(\frac {2 \sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, d \left (\sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right )-2 c x -b \right )}{2 x^{3} c^{2}+3 b c \,x^{2}+2 a c x +b^{2} x +a b}\)	\(194\)
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \sqrt {d \left (2 c x +b \right )}\, \left (-\frac {2 \left (2 c^{2} d x +d b c \right ) d}{c \sqrt {\left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right ) \left (2 c^{2} d x +d b c \right )}}+\frac {4 c \,d^{2} \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, d}\)	\(472\)

Input:

int((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.78 \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a} c d - \sqrt {2} {\left (c d x^{2} + b d x + a d\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right )}}{c^{2} x^{2} + b c x + a c} \] Input:

integrate((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {\left (d \left (b + 2 c x\right )\right )^{\frac {3}{2}}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((2*c*d*x+b*d)**(3/2)/(c*x**2+b*x+a)**(3/2),x)
 

Output:

Integral((d*(b + 2*c*x))**(3/2)/(a + b*x + c*x**2)**(3/2), x)
 

Maxima [F]

\[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((2*c*d*x + b*d)^(3/2)/(c*x^2 + b*x + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((2*c*d*x + b*d)^(3/2)/(c*x^2 + b*x + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \] Input:

int((b*d + 2*c*d*x)^(3/2)/(a + b*x + c*x^2)^(3/2),x)
 

Output:

int((b*d + 2*c*d*x)^(3/2)/(a + b*x + c*x^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\text {too large to display} \] Input:

int((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

(2*sqrt(d)*d*(sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**2 + 8*int((sqrt(b 
+ 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2*a**3*b*c + 4*a**3*c**2*x - a**2*b 
**3 + 2*a**2*b**2*c*x + 12*a**2*b*c**2*x**2 + 8*a**2*c**3*x**3 - 2*a*b**4* 
x - 4*a*b**3*c*x**2 + 4*a*b**2*c**2*x**3 + 10*a*b*c**3*x**4 + 4*a*c**4*x** 
5 - b**5*x**2 - 4*b**4*c*x**3 - 5*b**3*c**2*x**4 - 2*b**2*c**3*x**5),x)*a* 
*3*c**4 - 6*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2*a**3*b*c 
+ 4*a**3*c**2*x - a**2*b**3 + 2*a**2*b**2*c*x + 12*a**2*b*c**2*x**2 + 8*a* 
*2*c**3*x**3 - 2*a*b**4*x - 4*a*b**3*c*x**2 + 4*a*b**2*c**2*x**3 + 10*a*b* 
c**3*x**4 + 4*a*c**4*x**5 - b**5*x**2 - 4*b**4*c*x**3 - 5*b**3*c**2*x**4 - 
 2*b**2*c**3*x**5),x)*a**2*b**2*c**3 + 8*int((sqrt(b + 2*c*x)*sqrt(a + b*x 
 + c*x**2)*x**2)/(2*a**3*b*c + 4*a**3*c**2*x - a**2*b**3 + 2*a**2*b**2*c*x 
 + 12*a**2*b*c**2*x**2 + 8*a**2*c**3*x**3 - 2*a*b**4*x - 4*a*b**3*c*x**2 + 
 4*a*b**2*c**2*x**3 + 10*a*b*c**3*x**4 + 4*a*c**4*x**5 - b**5*x**2 - 4*b** 
4*c*x**3 - 5*b**3*c**2*x**4 - 2*b**2*c**3*x**5),x)*a**2*b*c**4*x + 8*int(( 
sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2*a**3*b*c + 4*a**3*c**2*x - 
 a**2*b**3 + 2*a**2*b**2*c*x + 12*a**2*b*c**2*x**2 + 8*a**2*c**3*x**3 - 2* 
a*b**4*x - 4*a*b**3*c*x**2 + 4*a*b**2*c**2*x**3 + 10*a*b*c**3*x**4 + 4*a*c 
**4*x**5 - b**5*x**2 - 4*b**4*c*x**3 - 5*b**3*c**2*x**4 - 2*b**2*c**3*x**5 
),x)*a**2*c**5*x**2 + int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2 
*a**3*b*c + 4*a**3*c**2*x - a**2*b**3 + 2*a**2*b**2*c*x + 12*a**2*b*c**...