Integrand size = 28, antiderivative size = 258 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{17/2}} \, dx=-\frac {\sqrt {a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}+\frac {\sqrt {a+b x+c x^2}}{462 c^3 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{924 c^4 \left (b^2-4 a c\right )^{3/4} d^{17/2} \sqrt {a+b x+c x^2}} \]
-1/66*(c*x^2+b*x+a)^(3/2)/c^2/d^3/(2*c*d*x+b*d)^(11/2)-1/15*(c*x^2+b*x+a)^ (5/2)/c/d/(2*c*d*x+b*d)^(15/2)-1/308*(c*x^2+b*x+a)^(1/2)/c^3/d^5/(2*c*d*x+ b*d)^(7/2)+1/462*(c*x^2+b*x+a)^(1/2)/c^3/(-4*a*c+b^2)/d^7/(2*c*d*x+b*d)^(3 /2)+1/924*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^4/(-4*a*c+b^2)^(3/4)/d^(17/2)/(c*x^2+b *x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.26 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{17/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2 \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {15}{4},-\frac {5}{2},-\frac {11}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{480 c^3 d^9 (b+2 c x)^8 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]
-1/480*((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeo metric2F1[-15/4, -5/2, -11/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c^3*d^9*(b + 2*c*x)^8*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])
Time = 0.51 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1108, 1108, 1108, 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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{17/2}} \, dx\) |
\(\Big \downarrow \) 1108 |
\(\displaystyle \frac {\int \frac {\left (c x^2+b x+a\right )^{3/2}}{(b d+2 c x d)^{13/2}}dx}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}\) |
\(\Big \downarrow \) 1108 |
\(\displaystyle \frac {\frac {3 \int \frac {\sqrt {c x^2+b x+a}}{(b d+2 c x d)^{9/2}}dx}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{11 c d (b d+2 c d x)^{11/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}\) |
\(\Big \downarrow \) 1108 |
\(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{(b d+2 c x d)^{5/2} \sqrt {c x^2+b x+a}}dx}{14 c d^2}-\frac {\sqrt {a+b x+c x^2}}{7 c d (b d+2 c d x)^{7/2}}\right )}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{11 c d (b d+2 c d x)^{11/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}\) |
\(\Big \downarrow \) 1117 |
\(\displaystyle \frac {\frac {3 \left (\frac {\frac {\int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx}{3 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}}{14 c d^2}-\frac {\sqrt {a+b x+c x^2}}{7 c d (b d+2 c d x)^{7/2}}\right )}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{11 c d (b d+2 c d x)^{11/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}\) |
\(\Big \downarrow \) 1115 |
\(\displaystyle \frac {\frac {3 \left (\frac {\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{3 d^2 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}}{14 c d^2}-\frac {\sqrt {a+b x+c x^2}}{7 c d (b d+2 c d x)^{7/2}}\right )}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{11 c d (b d+2 c d x)^{11/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}\) |
\(\Big \downarrow \) 1113 |
\(\displaystyle \frac {\frac {3 \left (\frac {\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{3 c d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}}{14 c d^2}-\frac {\sqrt {a+b x+c x^2}}{7 c d (b d+2 c d x)^{7/2}}\right )}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{11 c d (b d+2 c d x)^{11/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {\frac {3 \left (\frac {\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c d^{5/2} \left (b^2-4 a c\right )^{3/4} \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}}{14 c d^2}-\frac {\sqrt {a+b x+c x^2}}{7 c d (b d+2 c d x)^{7/2}}\right )}{22 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{11 c d (b d+2 c d x)^{11/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}\) |
-1/15*(a + b*x + c*x^2)^(5/2)/(c*d*(b*d + 2*c*d*x)^(15/2)) + (-1/11*(a + b *x + c*x^2)^(3/2)/(c*d*(b*d + 2*c*d*x)^(11/2)) + (3*(-1/7*Sqrt[a + b*x + c *x^2]/(c*d*(b*d + 2*c*d*x)^(7/2)) + ((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]))/(14*c*d^2) ))/(22*c*d^2))/(6*c*d^2)
3.14.55.3.1 Defintions of rubi rules used
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[((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]
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]
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*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])
Leaf count of result is larger than twice the leaf count of optimal. \(683\) vs. \(2(218)=436\).
Time = 6.95 (sec) , antiderivative size = 684, normalized size of antiderivative = 2.65
method | result | size |
elliptic | \(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{61440 c^{11} d^{9} \left (x +\frac {b}{2 c}\right )^{8}}-\frac {\left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{5280 c^{9} d^{9} \left (x +\frac {b}{2 c}\right )^{6}}-\frac {69 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{98560 c^{7} d^{9} \left (x +\frac {b}{2 c}\right )^{4}}-\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{1848 c^{5} \left (4 a c -b^{2}\right ) d^{9} \left (x +\frac {b}{2 c}\right )^{2}}-\frac {\left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, F\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{924 c^{3} \left (4 a c -b^{2}\right ) d^{8} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\) | \(684\) |
default | \(\text {Expression too large to display}\) | \(1431\) |
(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)* (-1/61440*(16*a^2*c^2-8*a*b^2*c+b^4)/c^11/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/c*b)^8-1/5280/c^9*(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/c*b)^6-69/98560/ c^7/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/c*b )^4-1/1848/c^5/(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/c*b)^2-1/924/c^3/(4*a*c-b^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/c*b)/(-1/2*(b+(-4*a*c+b^2)^(1/2))/c+1/2/c*b))^(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)*Elliptic F(((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/c*b))^(1/2)))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.21 (sec) , antiderivative size = 521, normalized size of antiderivative = 2.02 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{17/2}} \, dx=\frac {5 \, \sqrt {2} {\left (256 \, c^{8} x^{8} + 1024 \, b c^{7} x^{7} + 1792 \, b^{2} c^{6} x^{6} + 1792 \, b^{3} c^{5} x^{5} + 1120 \, b^{4} c^{4} x^{4} + 448 \, b^{5} c^{3} x^{3} + 112 \, b^{6} c^{2} x^{2} + 16 \, b^{7} c x + b^{8}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + 2 \, {\left (640 \, c^{8} x^{6} + 1920 \, b c^{7} x^{5} - 5 \, b^{6} c^{2} - 10 \, a b^{4} c^{3} - 28 \, a^{2} b^{2} c^{4} + 1232 \, a^{3} c^{5} + 12 \, {\left (131 \, b^{2} c^{6} + 276 \, a c^{7}\right )} x^{4} - 8 \, {\left (7 \, b^{3} c^{5} - 828 \, a b c^{6}\right )} x^{3} - 2 \, {\left (209 \, b^{4} c^{4} - 1588 \, a b^{2} c^{5} - 1792 \, a^{2} c^{6}\right )} x^{2} - 2 \, {\left (35 \, b^{5} c^{3} + 68 \, a b^{3} c^{4} - 1792 \, a^{2} b c^{5}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{9240 \, {\left (256 \, {\left (b^{2} c^{13} - 4 \, a c^{14}\right )} d^{9} x^{8} + 1024 \, {\left (b^{3} c^{12} - 4 \, a b c^{13}\right )} d^{9} x^{7} + 1792 \, {\left (b^{4} c^{11} - 4 \, a b^{2} c^{12}\right )} d^{9} x^{6} + 1792 \, {\left (b^{5} c^{10} - 4 \, a b^{3} c^{11}\right )} d^{9} x^{5} + 1120 \, {\left (b^{6} c^{9} - 4 \, a b^{4} c^{10}\right )} d^{9} x^{4} + 448 \, {\left (b^{7} c^{8} - 4 \, a b^{5} c^{9}\right )} d^{9} x^{3} + 112 \, {\left (b^{8} c^{7} - 4 \, a b^{6} c^{8}\right )} d^{9} x^{2} + 16 \, {\left (b^{9} c^{6} - 4 \, a b^{7} c^{7}\right )} d^{9} x + {\left (b^{10} c^{5} - 4 \, a b^{8} c^{6}\right )} d^{9}\right )}} \]
1/9240*(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 - 28*a^2*b^2*c^4 + 1232*a^3*c^5 + 12*(131*b^2*c^6 + 276*a*c^7)*x^4 - 8*(7*b ^3*c^5 - 828*a*b*c^6)*x^3 - 2*(209*b^4*c^4 - 1588*a*b^2*c^5 - 1792*a^2*c^6 )*x^2 - 2*(35*b^5*c^3 + 68*a*b^3*c^4 - 1792*a^2*b*c^5)*x)*sqrt(2*c*d*x + b *d)*sqrt(c*x^2 + b*x + a))/(256*(b^2*c^13 - 4*a*c^14)*d^9*x^8 + 1024*(b^3* c^12 - 4*a*b*c^13)*d^9*x^7 + 1792*(b^4*c^11 - 4*a*b^2*c^12)*d^9*x^6 + 1792 *(b^5*c^10 - 4*a*b^3*c^11)*d^9*x^5 + 1120*(b^6*c^9 - 4*a*b^4*c^10)*d^9*x^4 + 448*(b^7*c^8 - 4*a*b^5*c^9)*d^9*x^3 + 112*(b^8*c^7 - 4*a*b^6*c^8)*d^9*x ^2 + 16*(b^9*c^6 - 4*a*b^7*c^7)*d^9*x + (b^10*c^5 - 4*a*b^8*c^6)*d^9)
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{17/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{17/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {17}{2}}} \,d x } \]
\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{17/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {17}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{17/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{17/2}} \,d x \]