\(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{21/2}} \, dx\) [252]

Optimal result

Integrand size = 28, antiderivative size = 305 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{21/2}} \, dx=-\frac {\sqrt {a+b x+c x^2}}{836 c^3 d^5 (b d+2 c d x)^{11/2}}+\frac {\sqrt {a+b x+c x^2}}{2926 c^3 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{7/2}}+\frac {5 \sqrt {a+b x+c x^2}}{8778 c^3 \left (b^2-4 a c\right )^2 d^9 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{114 c^2 d^3 (b d+2 c d x)^{15/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{19 c d (b d+2 c d x)^{19/2}}+\frac {5 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{17556 c^4 \left (b^2-4 a c\right )^{7/4} d^{21/2} \sqrt {a+b x+c x^2}} \] Output:

-1/836*(c*x^2+b*x+a)^(1/2)/c^3/d^5/(2*c*d*x+b*d)^(11/2)+1/2926*(c*x^2+b*x+ 
a)^(1/2)/c^3/(-4*a*c+b^2)/d^7/(2*c*d*x+b*d)^(7/2)+5/8778*(c*x^2+b*x+a)^(1/ 
2)/c^3/(-4*a*c+b^2)^2/d^9/(2*c*d*x+b*d)^(3/2)-1/114*(c*x^2+b*x+a)^(3/2)/c^ 
2/d^3/(2*c*d*x+b*d)^(15/2)-1/19*(c*x^2+b*x+a)^(5/2)/c/d/(2*c*d*x+b*d)^(19/ 
2)+5/17556*(-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/(-4*a*c+b^2)^(7/4)/d^(21/2)/(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 = 5.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.36 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{21/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2 \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {19}{4},-\frac {5}{2},-\frac {15}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{608 c^3 d^9 (b+2 c x)^8 (d (b+2 c x))^{3/2} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(21/2),x]
 

Output:

-1/608*((b^2 - 4*a*c)^2*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-19/4, -5/ 
2, -15/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c^3*d^9*(b + 2*c*x)^8*(d*(b + 2*c 
*x))^(3/2)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1108, 1108, 1108, 1117, 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.

\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{21/2}} \, dx\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {5 \int \frac {\left (c x^2+b x+a\right )^{3/2}}{(b d+2 c x d)^{17/2}}dx}{38 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{19 c d (b d+2 c d x)^{19/2}}\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {5 \left (\frac {\int \frac {\sqrt {c x^2+b x+a}}{(b d+2 c x d)^{13/2}}dx}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}\right )}{38 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{19 c d (b d+2 c d x)^{19/2}}\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {5 \left (\frac {\frac {\int \frac {1}{(b d+2 c x d)^{9/2} \sqrt {c x^2+b x+a}}dx}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}\right )}{38 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{19 c d (b d+2 c d x)^{19/2}}\)

\(\Big \downarrow \) 1117

\(\displaystyle \frac {5 \left (\frac {\frac {\frac {5 \int \frac {1}{(b d+2 c x d)^{5/2} \sqrt {c x^2+b x+a}}dx}{7 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}\right )}{38 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{19 c d (b d+2 c d x)^{19/2}}\)

\(\Big \downarrow \) 1117

\(\displaystyle \frac {5 \left (\frac {\frac {\frac {5 \left (\frac {\int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx}{3 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{7 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}\right )}{38 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{19 c d (b d+2 c d x)^{19/2}}\)

\(\Big \downarrow \) 1115

\(\displaystyle \frac {5 \left (\frac {\frac {\frac {5 \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{3 d^2 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{7 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}\right )}{38 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{19 c d (b d+2 c d x)^{19/2}}\)

\(\Big \downarrow \) 1113

\(\displaystyle \frac {5 \left (\frac {\frac {\frac {5 \left (\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{3 c d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{7 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}\right )}{38 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{19 c d (b d+2 c d x)^{19/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {5 \left (\frac {\frac {\frac {5 \left (\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c d^{5/2} \left (b^2-4 a c\right )^{3/4} \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{7 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}\right )}{38 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{19 c d (b d+2 c d x)^{19/2}}\)

Input:

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(21/2),x]
 

Output:

-1/19*(a + b*x + c*x^2)^(5/2)/(c*d*(b*d + 2*c*d*x)^(19/2)) + (5*(-1/15*(a 
+ b*x + c*x^2)^(3/2)/(c*d*(b*d + 2*c*d*x)^(15/2)) + (-1/11*Sqrt[a + b*x + 
c*x^2]/(c*d*(b*d + 2*c*d*x)^(11/2)) + ((4*Sqrt[a + b*x + c*x^2])/(7*(b^2 - 
 4*a*c)*d*(b*d + 2*c*d*x)^(7/2)) + (5*((4*Sqrt[a + b*x + c*x^2])/(3*(b^2 - 
 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)) + (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])/(3*c*(b^2 - 4*a*c)^(3/4)*d^(5/2)*Sqrt[a + b*x + c*x^2])))/(7*(b^2 
 - 4*a*c)*d^2))/(22*c*d^2))/(10*c*d^2)))/(38*c*d^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]
 

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. \(749\) vs. \(2(259)=518\).

Time = 10.96 (sec) , antiderivative size = 750, normalized size of antiderivative = 2.46

method	result	size
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{311296 c^{13} d^{11} \left (x +\frac {b}{2 c}\right )^{10}}-\frac {\left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{29184 c^{11} d^{11} \left (x +\frac {b}{2 c}\right )^{8}}-\frac {67 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{642048 c^{9} d^{11} \left (x +\frac {b}{2 c}\right )^{6}}-\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{46816 c^{7} \left (4 a c -b^{2}\right ) d^{11} \left (x +\frac {b}{2 c}\right )^{4}}+\frac {5 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{35112 c^{5} \left (4 a c -b^{2}\right )^{2} d^{11} \left (x +\frac {b}{2 c}\right )^{2}}+\frac {5 \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{17556 c^{3} \left (4 a c -b^{2}\right )^{2} d^{10} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(750\)
default	\(\text {Expression too large to display}\)	\(1843\)

Input:

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(21/2),x,method=_RETURNVERBOSE)
 

Output:

(d*(2*c*x+b)*(c*x^2+b*x+a))^(1/2)/(d*(2*c*x+b))^(1/2)/(c*x^2+b*x+a)^(1/2)* 
(-1/311296*(16*a^2*c^2-8*a*b^2*c+b^4)/c^13/d^11*(2*c^2*d*x^3+3*b*c*d*x^2+2 
*a*c*d*x+b^2*d*x+a*b*d)^(1/2)/(x+1/2*b/c)^10-1/29184/c^11*(4*a*c-b^2)/d^11 
*(2*c^2*d*x^3+3*b*c*d*x^2+2*a*c*d*x+b^2*d*x+a*b*d)^(1/2)/(x+1/2*b/c)^8-67/ 
642048/c^9/d^11*(2*c^2*d*x^3+3*b*c*d*x^2+2*a*c*d*x+b^2*d*x+a*b*d)^(1/2)/(x 
+1/2*b/c)^6-1/46816/c^7/(4*a*c-b^2)/d^11*(2*c^2*d*x^3+3*b*c*d*x^2+2*a*c*d* 
x+b^2*d*x+a*b*d)^(1/2)/(x+1/2*b/c)^4+5/35112/c^5/(4*a*c-b^2)^2/d^11*(2*c^2 
*d*x^3+3*b*c*d*x^2+2*a*c*d*x+b^2*d*x+a*b*d)^(1/2)/(x+1/2*b/c)^2+5/17556/c^ 
3/(4*a*c-b^2)^2/d^10*(1/2/c*(-b+(-4*a*c+b^2)^(1/2))+1/2*(b+(-4*a*c+b^2)^(1 
/2))/c)*((x+1/2*(b+(-4*a*c+b^2)^(1/2))/c)/(1/2/c*(-b+(-4*a*c+b^2)^(1/2))+1 
/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2)*((x+1/2*b/c)/(-1/2*(b+(-4*a*c+b^2)^(1/ 
2))/c+1/2*b/c))^(1/2)*((x-1/2/c*(-b+(-4*a*c+b^2)^(1/2)))/(-1/2*(b+(-4*a*c+ 
b^2)^(1/2))/c-1/2/c*(-b+(-4*a*c+b^2)^(1/2))))^(1/2)/(2*c^2*d*x^3+3*b*c*d*x 
^2+2*a*c*d*x+b^2*d*x+a*b*d)^(1/2)*EllipticF(((x+1/2*(b+(-4*a*c+b^2)^(1/2)) 
/c)/(1/2/c*(-b+(-4*a*c+b^2)^(1/2))+1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2),(( 
-1/2*(b+(-4*a*c+b^2)^(1/2))/c-1/2/c*(-b+(-4*a*c+b^2)^(1/2)))/(-1/2*(b+(-4* 
a*c+b^2)^(1/2))/c+1/2*b/c))^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 810 vs. \(2 (258) = 516\).

Time = 0.25 (sec) , antiderivative size = 810, normalized size of antiderivative = 2.66 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{21/2}} \, dx =\text {Too large to display} \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(21/2),x, algorithm="fricas")
 

Output:

1/35112*(5*sqrt(2)*(1024*c^10*x^10 + 5120*b*c^9*x^9 + 11520*b^2*c^8*x^8 + 
15360*b^3*c^7*x^7 + 13440*b^4*c^6*x^6 + 8064*b^5*c^5*x^5 + 3360*b^6*c^4*x^ 
4 + 960*b^7*c^3*x^3 + 180*b^8*c^2*x^2 + 20*b^9*c*x + b^10)*sqrt(c^2*d)*wei 
erstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c) + 2*(2560*c^10*x 
^8 + 10240*b*c^9*x^7 - 5*b^8*c^2 - 10*a*b^6*c^3 - 28*a^2*b^4*c^4 + 4928*a^ 
3*b^2*c^5 - 14784*a^4*c^6 + 128*(143*b^2*c^8 - 12*a*c^9)*x^6 + 128*(149*b^ 
3*c^7 - 36*a*b*c^8)*x^5 + 4*(2691*b^4*c^6 + 2312*a*b^2*c^7 - 7504*a^2*c^8) 
*x^4 + 8*(211*b^5*c^5 + 3272*a*b^3*c^6 - 7504*a^2*b*c^7)*x^3 - 2*(359*b^6* 
c^4 - 6840*a*b^4*c^5 + 7728*a^2*b^2*c^6 + 19712*a^3*c^7)*x^2 - 2*(45*b^7*c 
^3 + 88*a*b^5*c^4 - 7280*a^2*b^3*c^5 + 19712*a^3*b*c^6)*x)*sqrt(2*c*d*x + 
b*d)*sqrt(c*x^2 + b*x + a))/(1024*(b^4*c^15 - 8*a*b^2*c^16 + 16*a^2*c^17)* 
d^11*x^10 + 5120*(b^5*c^14 - 8*a*b^3*c^15 + 16*a^2*b*c^16)*d^11*x^9 + 1152 
0*(b^6*c^13 - 8*a*b^4*c^14 + 16*a^2*b^2*c^15)*d^11*x^8 + 15360*(b^7*c^12 - 
 8*a*b^5*c^13 + 16*a^2*b^3*c^14)*d^11*x^7 + 13440*(b^8*c^11 - 8*a*b^6*c^12 
 + 16*a^2*b^4*c^13)*d^11*x^6 + 8064*(b^9*c^10 - 8*a*b^7*c^11 + 16*a^2*b^5* 
c^12)*d^11*x^5 + 3360*(b^10*c^9 - 8*a*b^8*c^10 + 16*a^2*b^6*c^11)*d^11*x^4 
 + 960*(b^11*c^8 - 8*a*b^9*c^9 + 16*a^2*b^7*c^10)*d^11*x^3 + 180*(b^12*c^7 
 - 8*a*b^10*c^8 + 16*a^2*b^8*c^9)*d^11*x^2 + 20*(b^13*c^6 - 8*a*b^11*c^7 + 
 16*a^2*b^9*c^8)*d^11*x + (b^14*c^5 - 8*a*b^12*c^6 + 16*a^2*b^10*c^7)*d^11 
)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{21/2}} \, dx=\text {Timed out} \] Input:

integrate((c*x**2+b*x+a)**(5/2)/(2*c*d*x+b*d)**(21/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{21/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {21}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(21/2),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(21/2), x)
 

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{21/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {21}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(21/2),x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(21/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{21/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{21/2}} \,d x \] Input:

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(21/2),x)
 

Output:

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(21/2), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{21/2}} \, dx=\text {too large to display} \] Input:

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(21/2),x)
 

Output:

(sqrt(d)*( - 936*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**3*c**2 - 128*sq 
rt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*b**2*c - 2432*sqrt(b + 2*c*x)*sq 
rt(a + b*x + c*x**2)*a**2*b*c**2*x - 2432*sqrt(b + 2*c*x)*sqrt(a + b*x + c 
*x**2)*a**2*c**3*x**2 - 20*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**4 - 
 316*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**3*c*x - 2292*sqrt(b + 2*c 
*x)*sqrt(a + b*x + c*x**2)*a*b**2*c**2*x**2 - 3952*sqrt(b + 2*c*x)*sqrt(a 
+ b*x + c*x**2)*a*b*c**3*x**3 - 1976*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2 
)*a*c**4*x**4 + 10*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**5*x + 62*sqrt 
(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**4*c*x**2 + 104*sqrt(b + 2*c*x)*sqrt( 
a + b*x + c*x**2)*b**3*c**2*x**3 + 52*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x** 
2)*b**2*c**3*x**4 + 36480*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2 
)/(38*a**2*b**11*c + 836*a**2*b**10*c**2*x + 8360*a**2*b**9*c**3*x**2 + 50 
160*a**2*b**8*c**4*x**3 + 200640*a**2*b**7*c**5*x**4 + 561792*a**2*b**6*c* 
*6*x**5 + 1123584*a**2*b**5*c**7*x**6 + 1605120*a**2*b**4*c**8*x**7 + 1605 
120*a**2*b**3*c**9*x**8 + 1070080*a**2*b**2*c**10*x**9 + 428032*a**2*b*c** 
11*x**10 + 77824*a**2*c**12*x**11 - a*b**13 + 16*a*b**12*c*x + 654*a*b**11 
*c**2*x**2 + 7876*a*b**10*c**3*x**3 + 53240*a*b**9*c**4*x**4 + 236016*a*b* 
*8*c**5*x**5 + 732864*a*b**7*c**6*x**6 + 1643136*a*b**6*c**7*x**7 + 268646 
4*a*b**5*c**8*x**8 + 3182080*a*b**4*c**9*x**9 + 2663936*a*b**3*c**10*x**10 
 + 1496064*a*b**2*c**11*x**11 + 505856*a*b*c**12*x**12 + 77824*a*c**13*...