\(\int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx\) [263]

Optimal result

Integrand size = 28, antiderivative size = 188 \[ \int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx=\frac {4 \sqrt {a+b x+c x^2}}{7 \left (b^2-4 a c\right ) d (b d+2 c d x)^{7/2}}+\frac {20 \sqrt {a+b x+c x^2}}{21 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{3/2}}+\frac {10 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{21 c \left (b^2-4 a c\right )^{7/4} d^{9/2} \sqrt {a+b x+c x^2}} \] Output:

4/7*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(7/2)+20/21*(c*x^2+b* 
x+a)^(1/2)/(-4*a*c+b^2)^2/d^3/(2*c*d*x+b*d)^(3/2)+10/21*(-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*a*c+b^2)^(7/4)/d^(9/2)/(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 = 10.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx=-\frac {2 \sqrt {d (b+2 c x)} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {1}{2},-\frac {3}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{7 c d^5 (b+2 c x)^4 \sqrt {a+x (b+c x)}} \] Input:

Integrate[1/((b*d + 2*c*d*x)^(9/2)*Sqrt[a + b*x + c*x^2]),x]
 

Output:

(-2*Sqrt[d*(b + 2*c*x)]*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hyperge 
ometric2F1[-7/4, 1/2, -3/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(7*c*d^5*(b + 2* 
c*x)^4*Sqrt[a + x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {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.

\(\displaystyle \int \frac {1}{\sqrt {a+b x+c x^2} (b d+2 c d x)^{9/2}} \, dx\)

\(\Big \downarrow \) 1117

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

\(\Big \downarrow \) 1117

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

\(\Big \downarrow \) 1115

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

\(\Big \downarrow \) 1113

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

\(\Big \downarrow \) 762

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

Input:

Int[1/((b*d + 2*c*d*x)^(9/2)*Sqrt[a + b*x + c*x^2]),x]
 

Output:

(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*S 
qrt[-((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)
 

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 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. \(555\) vs. \(2(160)=320\).

Time = 8.11 (sec) , antiderivative size = 556, normalized size of antiderivative = 2.96

method	result	size
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{28 \left (4 a c -b^{2}\right ) d^{5} c^{4} \left (x +\frac {b}{2 c}\right )^{4}}+\frac {5 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{21 \left (4 a c -b^{2}\right )^{2} d^{5} c^{2} \left (x +\frac {b}{2 c}\right )^{2}}+\frac {10 \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{21 \left (4 a c -b^{2}\right )^{2} d^{4} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(556\)
default	\(\frac {\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, \left (40 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) c^{3} x^{3}+60 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) b \,c^{2} x^{2}+30 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) b^{2} c x +5 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) b^{3}+80 c^{4} x^{4}+160 b \,c^{3} x^{3}+32 a \,c^{3} x^{2}+112 b^{2} c^{2} x^{2}+32 a b \,c^{2} x +32 b^{3} c x -48 a^{2} c^{2}+32 c a \,b^{2}\right )}{21 d^{5} \left (2 x^{3} c^{2}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right ) \left (4 a c -b^{2}\right )^{2} \left (2 c x +b \right )^{3} c}\)	\(691\)

Input:

int(1/(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^(1/2),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)* 
(-1/28/(4*a*c-b^2)/d^5/c^4*(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+5/21/(4*a*c-b^2)^2/d^5/c^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)/(x+1/2*b/c)^2+10/21/(4*a*c-b^2)^2/d^4*(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 [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.63 \[ \int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx=\frac {5 \, \sqrt {2} {\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + 16 \, {\left (5 \, c^{4} x^{2} + 5 \, b c^{3} x + 2 \, b^{2} c^{2} - 3 \, a c^{3}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{21 \, {\left (16 \, {\left (b^{4} c^{6} - 8 \, a b^{2} c^{7} + 16 \, a^{2} c^{8}\right )} d^{5} x^{4} + 32 \, {\left (b^{5} c^{5} - 8 \, a b^{3} c^{6} + 16 \, a^{2} b c^{7}\right )} d^{5} x^{3} + 24 \, {\left (b^{6} c^{4} - 8 \, a b^{4} c^{5} + 16 \, a^{2} b^{2} c^{6}\right )} d^{5} x^{2} + 8 \, {\left (b^{7} c^{3} - 8 \, a b^{5} c^{4} + 16 \, a^{2} b^{3} c^{5}\right )} d^{5} x + {\left (b^{8} c^{2} - 8 \, a b^{6} c^{3} + 16 \, a^{2} b^{4} c^{4}\right )} d^{5}\right )}} \] Input:

integrate(1/(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

1/21*(5*sqrt(2)*(16*c^4*x^4 + 32*b*c^3*x^3 + 24*b^2*c^2*x^2 + 8*b^3*c*x + 
b^4)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b) 
/c) + 16*(5*c^4*x^2 + 5*b*c^3*x + 2*b^2*c^2 - 3*a*c^3)*sqrt(2*c*d*x + b*d) 
*sqrt(c*x^2 + b*x + a))/(16*(b^4*c^6 - 8*a*b^2*c^7 + 16*a^2*c^8)*d^5*x^4 + 
 32*(b^5*c^5 - 8*a*b^3*c^6 + 16*a^2*b*c^7)*d^5*x^3 + 24*(b^6*c^4 - 8*a*b^4 
*c^5 + 16*a^2*b^2*c^6)*d^5*x^2 + 8*(b^7*c^3 - 8*a*b^5*c^4 + 16*a^2*b^3*c^5 
)*d^5*x + (b^8*c^2 - 8*a*b^6*c^3 + 16*a^2*b^4*c^4)*d^5)
 

Sympy [F]

\[ \int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx=\int \frac {1}{\left (d \left (b + 2 c x\right )\right )^{\frac {9}{2}} \sqrt {a + b x + c x^{2}}}\, dx \] Input:

integrate(1/(2*c*d*x+b*d)**(9/2)/(c*x**2+b*x+a)**(1/2),x)
 

Output:

Integral(1/((d*(b + 2*c*x))**(9/2)*sqrt(a + b*x + c*x**2)), x)
 

Maxima [F]

\[ \int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx=\int { \frac {1}{{\left (2 \, c d x + b d\right )}^{\frac {9}{2}} \sqrt {c x^{2} + b x + a}} \,d x } \] Input:

integrate(1/(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/((2*c*d*x + b*d)^(9/2)*sqrt(c*x^2 + b*x + a)), x)
 

Giac [F]

\[ \int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx=\int { \frac {1}{{\left (2 \, c d x + b d\right )}^{\frac {9}{2}} \sqrt {c x^{2} + b x + a}} \,d x } \] Input:

integrate(1/(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/((2*c*d*x + b*d)^(9/2)*sqrt(c*x^2 + b*x + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx=\int \frac {1}{{\left (b\,d+2\,c\,d\,x\right )}^{9/2}\,\sqrt {c\,x^2+b\,x+a}} \,d x \] Input:

int(1/((b*d + 2*c*d*x)^(9/2)*(a + b*x + c*x^2)^(1/2)),x)
 

Output:

int(1/((b*d + 2*c*d*x)^(9/2)*(a + b*x + c*x^2)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}}{32 c^{6} x^{7}+112 b \,c^{5} x^{6}+32 a \,c^{5} x^{5}+160 b^{2} c^{4} x^{5}+80 a b \,c^{4} x^{4}+120 b^{3} c^{3} x^{4}+80 a \,b^{2} c^{3} x^{3}+50 b^{4} c^{2} x^{3}+40 a \,b^{3} c^{2} x^{2}+11 b^{5} c \,x^{2}+10 a \,b^{4} c x +b^{6} x +a \,b^{5}}d x \right )}{d^{5}} \] Input:

int(1/(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(d)*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2))/(a*b**5 + 10*a*b**4* 
c*x + 40*a*b**3*c**2*x**2 + 80*a*b**2*c**3*x**3 + 80*a*b*c**4*x**4 + 32*a* 
c**5*x**5 + b**6*x + 11*b**5*c*x**2 + 50*b**4*c**2*x**3 + 120*b**3*c**3*x* 
*4 + 160*b**2*c**4*x**5 + 112*b*c**5*x**6 + 32*c**6*x**7),x))/d**5