3.14.33 \(\int (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2} \, dx\) [1333]

3.14.33.1 Optimal result

Integrand size = 28, antiderivative size = 279 \[ \int (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2} \, dx=-\frac {2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{45 c}+\frac {(b d+2 c d x)^{7/2} \sqrt {a+b x+c x^2}}{9 c d}-\frac {\left (b^2-4 a c\right )^{11/4} d^{5/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{15 c^2 \sqrt {a+b x+c x^2}}+\frac {\left (b^2-4 a c\right )^{11/4} d^{5/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{15 c^2 \sqrt {a+b x+c x^2}} \]

output

-2/45*(-4*a*c+b^2)*d*(2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a)^(1/2)/c+1/9*(2*c*d* 
x+b*d)^(7/2)*(c*x^2+b*x+a)^(1/2)/c/d-1/15*(-4*a*c+b^2)^(11/4)*d^(5/2)*Elli 
pticE((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)*(-c*(c*x^2+b*x+a)/ 
(-4*a*c+b^2))^(1/2)/c^2/(c*x^2+b*x+a)^(1/2)+1/15*(-4*a*c+b^2)^(11/4)*d^(5/ 
2)*EllipticF((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)*(-c*(c*x^2+ 
b*x+a)/(-4*a*c+b^2))^(1/2)/c^2/(c*x^2+b*x+a)^(1/2)
 

3.14.33.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.39 \[ \int (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2} \, dx=\frac {1}{18} d (d (b+2 c x))^{3/2} \sqrt {a+x (b+c x)} \left (8 (a+x (b+c x))+\frac {\left (b^2-4 a c\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}}\right ) \]

input

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

output

(d*(d*(b + 2*c*x))^(3/2)*Sqrt[a + x*(b + c*x)]*(8*(a + x*(b + c*x)) + ((b^ 
2 - 4*a*c)*Hypergeometric2F1[-1/2, 3/4, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]) 
/(c*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])))/18
 

3.14.33.3 Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.89, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1109, 1116, 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.

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

\(\Big \downarrow \) 1109

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{7/2}}{9 c d}-\frac {\left (b^2-4 a c\right ) \int \frac {(b d+2 c x d)^{5/2}}{\sqrt {c x^2+b x+a}}dx}{18 c}\)

\(\Big \downarrow \) 1116

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{7/2}}{9 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {3}{5} d^2 \left (b^2-4 a c\right ) \int \frac {\sqrt {b d+2 c x d}}{\sqrt {c x^2+b x+a}}dx+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )}{18 c}\)

\(\Big \downarrow \) 1115

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{7/2}}{9 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {3 d^2 \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {\sqrt {b d+2 c x d}}{\sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{5 \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )}{18 c}\)

\(\Big \downarrow \) 1114

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{7/2}}{9 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {b d+2 c x d}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )}{18 c}\)

\(\Big \downarrow \) 836

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{7/2}}{9 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{d \sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )}{18 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{7/2}}{9 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )}{18 c}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{7/2}}{9 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )}{18 c}\)

\(\Big \downarrow \) 1389

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{7/2}}{9 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {\sqrt {\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}+1}}{\sqrt {1-\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )}{18 c}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{7/2}}{9 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d^{3/2} \left (b^2-4 a c\right )^{3/4} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )}{18 c}\)

input

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

output

((b*d + 2*c*d*x)^(7/2)*Sqrt[a + b*x + c*x^2])/(9*c*d) - ((b^2 - 4*a*c)*((4 
*d*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/5 + (6*(b^2 - 4*a*c)*d*Sqr 
t[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*((b^2 - 4*a*c)^(3/4)*d^(3/2)*Ell 
ipticE[ArcSin[Sqrt[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*Sqrt[a + b*x + c*x^2])))/(18*c)
 

3.14.33.3.1 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 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 1116

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

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]
 

3.14.33.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(702\) vs. \(2(235)=470\).

Time = 3.11 (sec) , antiderivative size = 703, normalized size of antiderivative = 2.52

method	result	size
default	\(-\frac {\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, d^{2} \left (-160 c^{6} x^{6}-480 b \,c^{5} x^{5}+192 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a^{3} c^{3}-144 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a^{2} b^{2} c^{2}+36 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a \,b^{4} c -3 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) b^{6}-224 a \,c^{5} x^{4}-544 b^{2} c^{4} x^{4}-448 a b \,c^{4} x^{3}-288 x^{3} b^{3} c^{3}-64 a^{2} c^{4} x^{2}-304 a \,b^{2} c^{3} x^{2}-70 x^{2} b^{4} c^{2}-64 a^{2} b \,c^{3} x -80 x a \,b^{3} c^{2}-6 x \,b^{5} c -16 a^{2} b^{2} c^{2}-6 a \,b^{4} c \right )}{90 c^{2} \left (2 c^{2} x^{3}+3 c b \,x^{2}+2 a c x +b^{2} x +a b \right )}\)	\(703\)
elliptic	\(\text {Expression too large to display}\)	\(1669\)
risch	\(\text {Expression too large to display}\)	\(1786\)

input

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

output

-1/90*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d^2*(-160*c^6*x^6-480*b*c^5* 
x^5+192*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b 
)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1 
/2))^(1/2)*EllipticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)) 
^(1/2)*2^(1/2),2^(1/2))*a^3*c^3-144*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+ 
b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c 
+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticE(1/2*((b+2*c*x+(-4*a*c+b^2 
)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*a^2*b^2*c^2+36*((b+2*c 
*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^ 
(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*Elli 
pticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2), 
2^(1/2))*a*b^4*c-3*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2) 
*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4* 
a*c+b^2)^(1/2))^(1/2)*EllipticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+ 
b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*b^6-224*a*c^5*x^4-544*b^2*c^4*x^4-448*a 
*b*c^4*x^3-288*x^3*b^3*c^3-64*a^2*c^4*x^2-304*a*b^2*c^3*x^2-70*x^2*b^4*c^2 
-64*a^2*b*c^3*x-80*x*a*b^3*c^2-6*x*b^5*c-16*a^2*b^2*c^2-6*a*b^4*c)/c^2/(2* 
c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)
 

3.14.33.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.59 \[ \int (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2} \, dx=\frac {3 \, \sqrt {2} {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c^{2} d} d^{2} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + {\left (40 \, c^{4} d^{2} x^{3} + 60 \, b c^{3} d^{2} x^{2} + 2 \, {\left (13 \, b^{2} c^{2} + 8 \, a c^{3}\right )} d^{2} x + {\left (3 \, b^{3} c + 8 \, a b c^{2}\right )} d^{2}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{45 \, c^{2}} \]

input

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

output

1/45*(3*sqrt(2)*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c^2*d)*d^2*weierstrass 
Zeta((b^2 - 4*a*c)/c^2, 0, weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*( 
2*c*x + b)/c)) + (40*c^4*d^2*x^3 + 60*b*c^3*d^2*x^2 + 2*(13*b^2*c^2 + 8*a* 
c^3)*d^2*x + (3*b^3*c + 8*a*b*c^2)*d^2)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b 
*x + a))/c^2
 

3.14.33.6 Sympy [F]

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

input

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

output

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

3.14.33.7 Maxima [F]

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

input

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

output

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

3.14.33.8 Giac [F]

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

input

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

output

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

3.14.33.9 Mupad [F(-1)]

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

input

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

output

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