Integrand size = 28, antiderivative size = 258 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}+\frac {56}{5} c d^3 (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}+\frac {84 \left (b^2-4 a c\right )^{7/4} d^{9/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{5 \sqrt {a+b x+c x^2}}-\frac {84 \left (b^2-4 a c\right )^{7/4} d^{9/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{5 \sqrt {a+b x+c x^2}} \]
-2*d*(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(1/2)+56/5*c*d^3*(2*c*d*x+b*d)^(3/2 )*(c*x^2+b*x+a)^(1/2)+84/5*(-4*a*c+b^2)^(7/4)*d^(9/2)*EllipticE((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*x^2+b*x+a)^(1/2)-84/5*(-4*a*c+b^2)^(7/4)*d^(9/2)*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*x^2+b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.47 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {8 d^3 (d (b+2 c x))^{3/2} \left (-2 \left (2 b^2+b c x+c \left (-7 a+c x^2\right )\right )+7 \left (b^2-4 a c\right ) \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{5 \sqrt {a+x (b+c x)}} \]
(-8*d^3*(d*(b + 2*c*x))^(3/2)*(-2*(2*b^2 + b*c*x + c*(-7*a + c*x^2)) + 7*( b^2 - 4*a*c)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[ 3/4, 3/2, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(5*Sqrt[a + x*(b + c*x)])
Time = 0.55 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.90, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1110, 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.
\(\displaystyle \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1110 |
\(\displaystyle 14 c d^2 \int \frac {(b d+2 c x d)^{5/2}}{\sqrt {c x^2+b x+a}}dx-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 1116 |
\(\displaystyle 14 c d^2 \left (\frac {3}{5} d^2 \left (b^2-4 a c\right ) \int \frac {\sqrt {b d+2 c x d}}{\sqrt {c x^2+b x+a}}dx+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 1115 |
\(\displaystyle 14 c d^2 \left (\frac {3 d^2 \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {\sqrt {b d+2 c x d}}{\sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{5 \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 1114 |
\(\displaystyle 14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {b d+2 c x d}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 836 |
\(\displaystyle 14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{d \sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle 14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 1389 |
\(\displaystyle 14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {\sqrt {\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}+1}}{\sqrt {1-\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle 14 c d^2 \left (\frac {6 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d^{3/2} \left (b^2-4 a c\right )^{3/4} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}\right )-\frac {2 d (b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}}\) |
(-2*d*(b*d + 2*c*d*x)^(7/2))/Sqrt[a + b*x + c*x^2] + 14*c*d^2*((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*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*((b^2 - 4*a*c)^(3/4)*d^(3/2)*EllipticE[Ar cSin[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]))
3.14.84.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy mbol] :> Simp[d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), x] - Simp[d*e*((m - 1)/(b*(p + 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] && N eQ[m + 2*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
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]
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]
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])
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]
Leaf count of result is larger than twice the leaf count of optimal. \(497\) vs. \(2(216)=432\).
Time = 4.91 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.93
method | result | size |
default | \(-\frac {2 d^{4} \left (336 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a^{2} c^{2}-168 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a \,b^{2} c +21 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) b^{4}-32 c^{4} x^{4}-64 b \,c^{3} x^{3}-112 x^{2} c^{3} a -20 b^{2} c^{2} x^{2}-112 a b \,c^{2} x +12 b^{3} c x -28 a \,b^{2} c +5 b^{4}\right ) \sqrt {c \,x^{2}+b x +a}\, \sqrt {d \left (2 c x +b \right )}}{5 \left (2 c^{2} x^{3}+3 c b \,x^{2}+2 a c x +b^{2} x +a b \right )}\) | \(498\) |
elliptic | \(\text {Expression too large to display}\) | \(1242\) |
risch | \(\text {Expression too large to display}\) | \(2873\) |
-2/5*d^4*(336*((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*c^2-168*((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*b^2*c+21*((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)*El lipticE(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^4-32*c^4*x^4-64*b*c^3*x^3-112*x^2*c^3*a-20*b^2*c^2*x^2-112*a* b*c^2*x+12*b^3*c*x-28*a*b^2*c+5*b^4)*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/ 2)/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.77 \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (42 \, \sqrt {2} {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} x^{2} + {\left (b^{3} - 4 \, a b c\right )} d^{4} x + {\left (a b^{2} - 4 \, a^{2} c\right )} d^{4}\right )} \sqrt {c^{2} d} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) - {\left (16 \, c^{3} d^{4} x^{3} + 24 \, b c^{2} d^{4} x^{2} - 2 \, {\left (b^{2} c - 28 \, a c^{2}\right )} d^{4} x - {\left (5 \, b^{3} - 28 \, a b c\right )} d^{4}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}\right )}}{5 \, {\left (c x^{2} + b x + a\right )}} \]
-2/5*(42*sqrt(2)*((b^2*c - 4*a*c^2)*d^4*x^2 + (b^3 - 4*a*b*c)*d^4*x + (a*b ^2 - 4*a^2*c)*d^4)*sqrt(c^2*d)*weierstrassZeta((b^2 - 4*a*c)/c^2, 0, weier strassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c)) - (16*c^3*d^4*x^3 + 24*b*c^2*d^4*x^2 - 2*(b^2*c - 28*a*c^2)*d^4*x - (5*b^3 - 28*a*b*c)*d^4) *sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(c*x^2 + b*x + a)
Timed out. \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {9}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {9}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^{9/2}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]