\(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^{17/2}} \, dx\) [239]

Optimal result

Integrand size = 28, antiderivative size = 268 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{17/2}} \, dx=-\frac {\sqrt {a+b x+c x^2}}{110 c^2 d^3 (b d+2 c d x)^{11/2}}+\frac {\sqrt {a+b x+c x^2}}{385 c^2 \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{7/2}}+\frac {\sqrt {a+b x+c x^2}}{231 c^2 \left (b^2-4 a c\right )^2 d^7 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{462 c^3 \left (b^2-4 a c\right )^{7/4} d^{17/2} \sqrt {a+b x+c x^2}} \] Output:

-1/110*(c*x^2+b*x+a)^(1/2)/c^2/d^3/(2*c*d*x+b*d)^(11/2)+1/385*(c*x^2+b*x+a 
)^(1/2)/c^2/(-4*a*c+b^2)/d^5/(2*c*d*x+b*d)^(7/2)+1/231*(c*x^2+b*x+a)^(1/2) 
/c^2/(-4*a*c+b^2)^2/d^7/(2*c*d*x+b*d)^(3/2)-1/15*(c*x^2+b*x+a)^(3/2)/c/d/( 
2*c*d*x+b*d)^(15/2)+1/462*(-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^3/(-4*a*c+b^2)^(7/4)/d 
^(17/2)/(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 = 5.03 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.40 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{17/2}} \, dx=\frac {\left (b^2-4 a c\right ) \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {15}{4},-\frac {3}{2},-\frac {11}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{120 c^2 d^9 (b+2 c x)^8 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(17/2),x]
 

Output:

((b^2 - 4*a*c)*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1 
[-15/4, -3/2, -11/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(120*c^2*d^9*(b + 2*c*x 
)^8*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.

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

\(\Big \downarrow \) 1108

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

\(\Big \downarrow \) 1108

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

\(\Big \downarrow \) 1117

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

\(\Big \downarrow \) 1117

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

\(\Big \downarrow \) 1115

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

\(\Big \downarrow \) 1113

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

\(\Big \downarrow \) 762

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

Input:

Int[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(17/2),x]
 

Output:

-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)
 

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. \(676\) vs. \(2(228)=456\).

Time = 8.53 (sec) , antiderivative size = 677, normalized size of antiderivative = 2.53

method	result	size
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{15360 c^{10} d^{9} \left (x +\frac {b}{2 c}\right )^{8}}-\frac {17 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{42240 c^{8} d^{9} \left (x +\frac {b}{2 c}\right )^{6}}-\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{6160 c^{6} \left (4 a c -b^{2}\right ) d^{9} \left (x +\frac {b}{2 c}\right )^{4}}+\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{924 c^{4} \left (4 a c -b^{2}\right )^{2} d^{9} \left (x +\frac {b}{2 c}\right )^{2}}+\frac {\left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{462 c^{2} \left (4 a c -b^{2}\right )^{2} d^{8} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(677\)
default	\(\text {Expression too large to display}\)	\(1431\)

Input:

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(17/2),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)* 
(-1/15360*(4*a*c-b^2)/c^10/d^9*(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)^8-17/42240/c^8/d^9*(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)^6-1/6160/c^6/(4*a*c-b^2)/d^9*(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+1/924/c^4/( 
4*a*c-b^2)^2/d^9*(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+1/462/c^2/(4*a*c-b^2)^2/d^8*(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. 620 vs. \(2 (227) = 454\).

Time = 0.15 (sec) , antiderivative size = 620, normalized size of antiderivative = 2.31 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{17/2}} \, dx=\frac {5 \, \sqrt {2} {\left (256 \, c^{8} x^{8} + 1024 \, b c^{7} x^{7} + 1792 \, b^{2} c^{6} x^{6} + 1792 \, b^{3} c^{5} x^{5} + 1120 \, b^{4} c^{4} x^{4} + 448 \, b^{5} c^{3} x^{3} + 112 \, b^{6} c^{2} x^{2} + 16 \, b^{7} c x + b^{8}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + 2 \, {\left (640 \, c^{8} x^{6} + 1920 \, b c^{7} x^{5} - 5 \, b^{6} c^{2} - 10 \, a b^{4} c^{3} + 896 \, a^{2} b^{2} c^{4} - 2464 \, a^{3} c^{5} + 192 \, {\left (13 \, b^{2} c^{6} - 2 \, a c^{7}\right )} x^{4} + 256 \, {\left (7 \, b^{3} c^{5} - 3 \, a b c^{6}\right )} x^{3} + 2 \, {\left (253 \, b^{4} c^{4} + 664 \, a b^{2} c^{5} - 1904 \, a^{2} c^{6}\right )} x^{2} - 2 \, {\left (35 \, b^{5} c^{3} - 856 \, a b^{3} c^{4} + 1904 \, a^{2} b c^{5}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{4620 \, {\left (256 \, {\left (b^{4} c^{12} - 8 \, a b^{2} c^{13} + 16 \, a^{2} c^{14}\right )} d^{9} x^{8} + 1024 \, {\left (b^{5} c^{11} - 8 \, a b^{3} c^{12} + 16 \, a^{2} b c^{13}\right )} d^{9} x^{7} + 1792 \, {\left (b^{6} c^{10} - 8 \, a b^{4} c^{11} + 16 \, a^{2} b^{2} c^{12}\right )} d^{9} x^{6} + 1792 \, {\left (b^{7} c^{9} - 8 \, a b^{5} c^{10} + 16 \, a^{2} b^{3} c^{11}\right )} d^{9} x^{5} + 1120 \, {\left (b^{8} c^{8} - 8 \, a b^{6} c^{9} + 16 \, a^{2} b^{4} c^{10}\right )} d^{9} x^{4} + 448 \, {\left (b^{9} c^{7} - 8 \, a b^{7} c^{8} + 16 \, a^{2} b^{5} c^{9}\right )} d^{9} x^{3} + 112 \, {\left (b^{10} c^{6} - 8 \, a b^{8} c^{7} + 16 \, a^{2} b^{6} c^{8}\right )} d^{9} x^{2} + 16 \, {\left (b^{11} c^{5} - 8 \, a b^{9} c^{6} + 16 \, a^{2} b^{7} c^{7}\right )} d^{9} x + {\left (b^{12} c^{4} - 8 \, a b^{10} c^{5} + 16 \, a^{2} b^{8} c^{6}\right )} d^{9}\right )}} \] Input:

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(17/2),x, algorithm="fricas")
 

Output:

1/4620*(5*sqrt(2)*(256*c^8*x^8 + 1024*b*c^7*x^7 + 1792*b^2*c^6*x^6 + 1792* 
b^3*c^5*x^5 + 1120*b^4*c^4*x^4 + 448*b^5*c^3*x^3 + 112*b^6*c^2*x^2 + 16*b^ 
7*c*x + b^8)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2* 
c*x + b)/c) + 2*(640*c^8*x^6 + 1920*b*c^7*x^5 - 5*b^6*c^2 - 10*a*b^4*c^3 + 
 896*a^2*b^2*c^4 - 2464*a^3*c^5 + 192*(13*b^2*c^6 - 2*a*c^7)*x^4 + 256*(7* 
b^3*c^5 - 3*a*b*c^6)*x^3 + 2*(253*b^4*c^4 + 664*a*b^2*c^5 - 1904*a^2*c^6)* 
x^2 - 2*(35*b^5*c^3 - 856*a*b^3*c^4 + 1904*a^2*b*c^5)*x)*sqrt(2*c*d*x + b* 
d)*sqrt(c*x^2 + b*x + a))/(256*(b^4*c^12 - 8*a*b^2*c^13 + 16*a^2*c^14)*d^9 
*x^8 + 1024*(b^5*c^11 - 8*a*b^3*c^12 + 16*a^2*b*c^13)*d^9*x^7 + 1792*(b^6* 
c^10 - 8*a*b^4*c^11 + 16*a^2*b^2*c^12)*d^9*x^6 + 1792*(b^7*c^9 - 8*a*b^5*c 
^10 + 16*a^2*b^3*c^11)*d^9*x^5 + 1120*(b^8*c^8 - 8*a*b^6*c^9 + 16*a^2*b^4* 
c^10)*d^9*x^4 + 448*(b^9*c^7 - 8*a*b^7*c^8 + 16*a^2*b^5*c^9)*d^9*x^3 + 112 
*(b^10*c^6 - 8*a*b^8*c^7 + 16*a^2*b^6*c^8)*d^9*x^2 + 16*(b^11*c^5 - 8*a*b^ 
9*c^6 + 16*a^2*b^7*c^7)*d^9*x + (b^12*c^4 - 8*a*b^10*c^5 + 16*a^2*b^8*c^6) 
*d^9)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{17/2}} \, dx=\text {Timed out} \] Input:

integrate((c*x**2+b*x+a)**(3/2)/(2*c*d*x+b*d)**(17/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{17/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {17}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(17/2),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(17/2), x)
 

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{17/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {17}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(17/2),x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(17/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{17/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{17/2}} \,d x \] Input:

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(17/2),x)
 

Output:

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(17/2), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{17/2}} \, dx=\text {too large to display} \] Input:

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(17/2),x)
 

Output:

(sqrt(d)*( - 36*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*c - 4*sqrt(b + 
 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**2 - 60*sqrt(b + 2*c*x)*sqrt(a + b*x + 
c*x**2)*a*b*c*x - 60*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*c**2*x**2 + 
2*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**3*x + 2*sqrt(b + 2*c*x)*sqrt(a 
 + b*x + c*x**2)*b**2*c*x**2 - 1440*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c* 
x**2)*x**2)/(30*a**2*b**9*c + 540*a**2*b**8*c**2*x + 4320*a**2*b**7*c**3*x 
**2 + 20160*a**2*b**6*c**4*x**3 + 60480*a**2*b**5*c**5*x**4 + 120960*a**2* 
b**4*c**6*x**5 + 161280*a**2*b**3*c**7*x**6 + 138240*a**2*b**2*c**8*x**7 + 
 69120*a**2*b*c**9*x**8 + 15360*a**2*c**10*x**9 - a*b**11 + 12*a*b**10*c*x 
 + 426*a*b**9*c**2*x**2 + 4188*a*b**8*c**3*x**3 + 22464*a*b**7*c**4*x**4 + 
 76608*a*b**6*c**5*x**5 + 176064*a*b**5*c**6*x**6 + 277632*a*b**4*c**7*x** 
7 + 297216*a*b**3*c**8*x**8 + 206848*a*b**2*c**9*x**9 + 84480*a*b*c**10*x* 
*10 + 15360*a*c**11*x**11 - b**12*x - 19*b**11*c*x**2 - 162*b**10*c**2*x** 
3 - 816*b**9*c**3*x**4 - 2688*b**8*c**4*x**5 - 6048*b**7*c**5*x**6 - 9408* 
b**6*c**6*x**7 - 9984*b**5*c**7*x**8 - 6912*b**4*c**8*x**9 - 2816*b**3*c** 
9*x**10 - 512*b**2*c**10*x**11),x)*a**3*b**8*c**4 - 23040*int((sqrt(b + 2* 
c*x)*sqrt(a + b*x + c*x**2)*x**2)/(30*a**2*b**9*c + 540*a**2*b**8*c**2*x + 
 4320*a**2*b**7*c**3*x**2 + 20160*a**2*b**6*c**4*x**3 + 60480*a**2*b**5*c* 
*5*x**4 + 120960*a**2*b**4*c**6*x**5 + 161280*a**2*b**3*c**7*x**6 + 138240 
*a**2*b**2*c**8*x**7 + 69120*a**2*b*c**9*x**8 + 15360*a**2*c**10*x**9 -...