\(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^{3/2}} \, dx\) [242]

Optimal result

Integrand size = 28, antiderivative size = 271 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{3/2}} \, dx=\frac {3 (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{10 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{c d \sqrt {b d+2 c d x}}-\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{10 c^3 d^{3/2} \sqrt {a+b x+c x^2}}+\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{10 c^3 d^{3/2} \sqrt {a+b x+c x^2}} \] Output:

3/10*(2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a)^(1/2)/c^2/d^3-(c*x^2+b*x+a)^(3/2)/c 
/d/(2*c*d*x+b*d)^(1/2)-3/10*(-4*a*c+b^2)^(7/4)*(-c*(c*x^2+b*x+a)/(-4*a*c+b 
^2))^(1/2)*EllipticE((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)/c^3 
/d^(3/2)/(c*x^2+b*x+a)^(1/2)+3/10*(-4*a*c+b^2)^(7/4)*(-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^3/d^(3/2)/(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 = 99, normalized size of antiderivative = 0.37 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{3/2}} \, dx=\frac {\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{4},\frac {3}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{8 c^2 d \sqrt {d (b+2 c x)} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1108, 1109, 1115, 1114, 836, 27, 762, 1389, 327}

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.

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

\(\Big \downarrow \) 1108

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

\(\Big \downarrow \) 1109

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

\(\Big \downarrow \) 1115

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

\(\Big \downarrow \) 1114

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

\(\Big \downarrow \) 836

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 762

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

\(\Big \downarrow \) 1389

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

\(\Big \downarrow \) 327

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

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 836

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, 
Simp[-q^(-1)   Int[1/Sqrt[a + b*x^4], x], x] + Simp[1/q   Int[(1 + q*x^2)/S 
qrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]
 

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 1109

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 + 2*p + 1))), x 
] - Simp[d*p*((b^2 - 4*a*c)/(b*e*(m + 2*p + 1)))   Int[(d + e*x)^m*(a + b*x 
 + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[2*c*d - b* 
e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1 
)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p])) && RationalQ[m] && IntegerQ[2*p]
 

rule 1114

Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symb 
ol] :> Simp[(4/e)*Sqrt[-c/(b^2 - 4*a*c)]   Subst[Int[x^2/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 1389

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Simp[d/Sq 
rt[a]   Int[Sqrt[1 + e*(x^2/d)]/Sqrt[1 - e*(x^2/d)], x], x] /; FreeQ[{a, c, 
 d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && NegQ[c/a] && GtQ[a, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(503\) vs. \(2(229)=458\).

Time = 3.58 (sec) , antiderivative size = 504, normalized size of antiderivative = 1.86

method	result	size
default	\(\frac {\sqrt {c \,x^{2}+b x +a}\, \sqrt {d \left (2 c x +b \right )}\, \left (48 \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticE}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) a^{2} c^{2}-24 \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticE}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) a \,b^{2} c +3 \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticE}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) b^{4}+4 c^{4} x^{4}+8 b \,c^{3} x^{3}-16 a \,c^{3} x^{2}+10 b^{2} c^{2} x^{2}-16 a b \,c^{2} x +6 b^{3} c x -20 a^{2} c^{2}+6 c a \,b^{2}\right )}{20 d^{2} \left (2 x^{3} c^{2}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right ) c^{3}}\)	\(504\)
elliptic	\(\text {Expression too large to display}\)	\(1229\)
risch	\(\text {Expression too large to display}\)	\(2891\)

Input:

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

Output:

1/20*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)*(48*(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)*EllipticE(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))*a^2*c 
^2-24*(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)*EllipticE(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))*a*b^2*c+3*(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)*EllipticE(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))*b^4+4*c^4*x^4+8*b* 
c^3*x^3-16*a*c^3*x^2+10*b^2*c^2*x^2-16*a*b*c^2*x+6*b^3*c*x-20*a^2*c^2+6*c* 
a*b^2)/d^2/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)/c^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.55 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{3/2}} \, dx=\frac {3 \, \sqrt {2} {\left (b^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x\right )} \sqrt {c^{2} d} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + {\left (2 \, c^{3} x^{2} + 2 \, b c^{2} x + 3 \, b^{2} c - 10 \, a c^{2}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{10 \, {\left (2 \, c^{4} d^{2} x + b c^{3} d^{2}\right )}} \] Input:

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

Output:

1/10*(3*sqrt(2)*(b^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*x)*sqrt(c^2*d)*weiers 
trassZeta((b^2 - 4*a*c)/c^2, 0, weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 
1/2*(2*c*x + b)/c)) + (2*c^3*x^2 + 2*b*c^2*x + 3*b^2*c - 10*a*c^2)*sqrt(2* 
c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(2*c^4*d^2*x + b*c^3*d^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(d)*( - 20*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 + 4*sqrt(b + 2*c*x)*sqrt(a + b*x + c 
*x**2)*a*b*c*x + 4*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 + 96*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2 
)*x**2)/(2*a**2*b**2*c + 8*a**2*b*c**2*x + 8*a**2*c**3*x**2 - a*b**4 - 2*a 
*b**3*c*x + 6*a*b**2*c**2*x**2 + 16*a*b*c**3*x**3 + 8*a*c**4*x**4 - b**5*x 
 - 5*b**4*c*x**2 - 8*b**3*c**2*x**3 - 4*b**2*c**3*x**4),x)*a**3*b*c**4 + 1 
92*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2*a**2*b**2*c + 8*a* 
*2*b*c**2*x + 8*a**2*c**3*x**2 - a*b**4 - 2*a*b**3*c*x + 6*a*b**2*c**2*x** 
2 + 16*a*b*c**3*x**3 + 8*a*c**4*x**4 - b**5*x - 5*b**4*c*x**2 - 8*b**3*c** 
2*x**3 - 4*b**2*c**3*x**4),x)*a**3*c**5*x - 96*int((sqrt(b + 2*c*x)*sqrt(a 
 + b*x + c*x**2)*x**2)/(2*a**2*b**2*c + 8*a**2*b*c**2*x + 8*a**2*c**3*x**2 
 - a*b**4 - 2*a*b**3*c*x + 6*a*b**2*c**2*x**2 + 16*a*b*c**3*x**3 + 8*a*c** 
4*x**4 - b**5*x - 5*b**4*c*x**2 - 8*b**3*c**2*x**3 - 4*b**2*c**3*x**4),x)* 
a**2*b**3*c**3 - 192*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2* 
a**2*b**2*c + 8*a**2*b*c**2*x + 8*a**2*c**3*x**2 - a*b**4 - 2*a*b**3*c*x + 
 6*a*b**2*c**2*x**2 + 16*a*b*c**3*x**3 + 8*a*c**4*x**4 - b**5*x - 5*b**4*c 
*x**2 - 8*b**3*c**2*x**3 - 4*b**2*c**3*x**4),x)*a**2*b**2*c**4*x + 30*int( 
(sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2*a**2*b**2*c + 8*a**2*b...