Integrand size = 26, antiderivative size = 147 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^3} \, dx=-\frac {5 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}{64 c^3 d^3}+\frac {5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}+\frac {5 \left (b^2-4 a c\right )^{3/2} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{128 c^{7/2} d^3} \]
5/48*(c*x^2+b*x+a)^(3/2)/c^2/d^3-1/4*(c*x^2+b*x+a)^(5/2)/c/d^3/(2*c*x+b)^2 +5/128*(-4*a*c+b^2)^(3/2)*arctan(2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2 )^(1/2))/c^(7/2)/d^3-5/64*(-4*a*c+b^2)*(c*x^2+b*x+a)^(1/2)/c^3/d^3
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^3} \, dx=\frac {2 (a+x (b+c x))^{7/2} \operatorname {Hypergeometric2F1}\left (2,\frac {7}{2},\frac {9}{2},\frac {4 c (a+x (b+c x))}{-b^2+4 a c}\right )}{7 \left (b^2-4 a c\right )^2 d^3} \]
(2*(a + x*(b + c*x))^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(7*(b^2 - 4*a*c)^2*d^3)
Time = 0.31 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1108, 27, 1109, 1109, 1112, 218}
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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^3} \, dx\) |
\(\Big \downarrow \) 1108 |
\(\displaystyle \frac {5 \int \frac {\left (c x^2+b x+a\right )^{3/2}}{d (b+2 c x)}dx}{8 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5 \int \frac {\left (c x^2+b x+a\right )^{3/2}}{b+2 c x}dx}{8 c d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}\) |
\(\Big \downarrow \) 1109 |
\(\displaystyle \frac {5 \left (\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac {\left (b^2-4 a c\right ) \int \frac {\sqrt {c x^2+b x+a}}{b+2 c x}dx}{4 c}\right )}{8 c d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}\) |
\(\Big \downarrow \) 1109 |
\(\displaystyle \frac {5 \left (\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac {\left (b^2-4 a c\right ) \left (\frac {\sqrt {a+b x+c x^2}}{2 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{(b+2 c x) \sqrt {c x^2+b x+a}}dx}{4 c}\right )}{4 c}\right )}{8 c d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}\) |
\(\Big \downarrow \) 1112 |
\(\displaystyle \frac {5 \left (\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac {\left (b^2-4 a c\right ) \left (\frac {\sqrt {a+b x+c x^2}}{2 c}-\left (b^2-4 a c\right ) \int \frac {1}{8 \left (c x^2+b x+a\right ) c^2+2 \left (b^2-4 a c\right ) c}d\sqrt {c x^2+b x+a}\right )}{4 c}\right )}{8 c d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac {\left (b^2-4 a c\right ) \left (\frac {\sqrt {a+b x+c x^2}}{2 c}-\frac {\sqrt {b^2-4 a c} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{4 c^{3/2}}\right )}{4 c}\right )}{8 c d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}\) |
-1/4*(a + b*x + c*x^2)^(5/2)/(c*d^3*(b + 2*c*x)^2) + (5*((a + b*x + c*x^2) ^(3/2)/(6*c) - ((b^2 - 4*a*c)*(Sqrt[a + b*x + c*x^2]/(2*c) - (Sqrt[b^2 - 4 *a*c]*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(4*c^(3 /2))))/(4*c)))/(8*c*d^3)
3.13.24.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[((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]
Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symb ol] :> Simp[4*c Subst[Int[1/(b^2*e - 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Time = 2.76 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.18
method | result | size |
pseudoelliptic | \(-\frac {5 \left (\left (2 c x +b \right )^{2} \left (-\frac {b^{2}}{4}+a c \right )^{2} \operatorname {arctanh}\left (\frac {2 c \sqrt {c \,x^{2}+b x +a}}{\sqrt {4 c^{2} a -b^{2} c}}\right )+\frac {2 \sqrt {4 c^{2} a -b^{2} c}\, \left (-\frac {2 c^{4} x^{4}}{3}-\frac {14 x^{2} \left (\frac {2 b x}{7}+a \right ) c^{3}}{3}+\left (\frac {1}{6} b^{2} x^{2}-\frac {14}{3} a b x +a^{2}\right ) c^{2}-\frac {5 \left (-\frac {b x}{2}+a \right ) b^{2} c}{3}+\frac {5 b^{4}}{16}\right ) \sqrt {c \,x^{2}+b x +a}}{5}\right )}{8 \sqrt {4 c^{2} a -b^{2} c}\, \left (2 c x +b \right )^{2} c^{3} d^{3}}\) | \(174\) |
default | \(\frac {-\frac {2 c \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {7}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2}}+\frac {10 c^{2} \left (\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{5}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{3}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{2 c \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 c}\right )}{4 c}\right )}{4 a c -b^{2}}}{8 d^{3} c^{3}}\) | \(318\) |
risch | \(\frac {\left (2 c^{2} x^{2}+2 b c x +14 a c -3 b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{48 c^{3} d^{3}}+\frac {-\frac {\left (24 a^{2} c^{2}-12 a \,b^{2} c +\frac {3}{2} b^{4}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{c \sqrt {\frac {4 a c -b^{2}}{c}}}+\frac {\left (-24 a^{2} b^{2} c^{2}+6 a \,b^{4} c -\frac {1}{2} b^{6}+32 c^{3} a^{3}\right ) \left (-\frac {2 c \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2}}+\frac {4 c^{2} \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{\left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{8 c^{3}}}{32 c^{3} d^{3}}\) | \(374\) |
-5/8*((2*c*x+b)^2*(-1/4*b^2+a*c)^2*arctanh(2*c*(c*x^2+b*x+a)^(1/2)/(4*a*c^ 2-b^2*c)^(1/2))+2/5*(4*a*c^2-b^2*c)^(1/2)*(-2/3*c^4*x^4-14/3*x^2*(2/7*b*x+ a)*c^3+(1/6*b^2*x^2-14/3*a*b*x+a^2)*c^2-5/3*(-1/2*b*x+a)*b^2*c+5/16*b^4)*( c*x^2+b*x+a)^(1/2))/(4*a*c^2-b^2*c)^(1/2)/(2*c*x+b)^2/c^3/d^3
Time = 0.45 (sec) , antiderivative size = 480, normalized size of antiderivative = 3.27 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^3} \, dx=\left [-\frac {15 \, {\left (b^{4} - 4 \, a b^{2} c + 4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 4 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x\right )} \sqrt {-\frac {b^{2} - 4 \, a c}{c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {c x^{2} + b x + a} c \sqrt {-\frac {b^{2} - 4 \, a c}{c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) - 4 \, {\left (32 \, c^{4} x^{4} + 64 \, b c^{3} x^{3} - 15 \, b^{4} + 80 \, a b^{2} c - 48 \, a^{2} c^{2} - 8 \, {\left (b^{2} c^{2} - 28 \, a c^{3}\right )} x^{2} - 8 \, {\left (5 \, b^{3} c - 28 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, {\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\right )}}, -\frac {15 \, {\left (b^{4} - 4 \, a b^{2} c + 4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 4 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x\right )} \sqrt {\frac {b^{2} - 4 \, a c}{c}} \arctan \left (\frac {\sqrt {\frac {b^{2} - 4 \, a c}{c}}}{2 \, \sqrt {c x^{2} + b x + a}}\right ) - 2 \, {\left (32 \, c^{4} x^{4} + 64 \, b c^{3} x^{3} - 15 \, b^{4} + 80 \, a b^{2} c - 48 \, a^{2} c^{2} - 8 \, {\left (b^{2} c^{2} - 28 \, a c^{3}\right )} x^{2} - 8 \, {\left (5 \, b^{3} c - 28 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, {\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\right )}}\right ] \]
[-1/768*(15*(b^4 - 4*a*b^2*c + 4*(b^2*c^2 - 4*a*c^3)*x^2 + 4*(b^3*c - 4*a* b*c^2)*x)*sqrt(-(b^2 - 4*a*c)/c)*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(c*x^2 + b*x + a)*c*sqrt(-(b^2 - 4*a*c)/c))/(4*c^2*x^2 + 4*b*c*x + b^2)) - 4*(32*c^4*x^4 + 64*b*c^3*x^3 - 15*b^4 + 80*a*b^2*c - 48*a^2*c^2 - 8*(b^2*c^2 - 28*a*c^3)*x^2 - 8*(5*b^3*c - 28*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(4*c^5*d^3*x^2 + 4*b*c^4*d^3*x + b^2*c^3*d^3), -1/384*(15*(b^4 - 4*a *b^2*c + 4*(b^2*c^2 - 4*a*c^3)*x^2 + 4*(b^3*c - 4*a*b*c^2)*x)*sqrt((b^2 - 4*a*c)/c)*arctan(1/2*sqrt((b^2 - 4*a*c)/c)/sqrt(c*x^2 + b*x + a)) - 2*(32* c^4*x^4 + 64*b*c^3*x^3 - 15*b^4 + 80*a*b^2*c - 48*a^2*c^2 - 8*(b^2*c^2 - 2 8*a*c^3)*x^2 - 8*(5*b^3*c - 28*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(4*c^5*d ^3*x^2 + 4*b*c^4*d^3*x + b^2*c^3*d^3)]
\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^3} \, dx=\frac {\int \frac {a^{2} \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {b^{2} x^{2} \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {c^{2} x^{4} \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {2 a b x \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {2 a c x^{2} \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {2 b c x^{3} \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx}{d^{3}} \]
(Integral(a**2*sqrt(a + b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(b**2*x**2*sqrt(a + b*x + c*x**2)/(b**3 + 6*b **2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(c**2*x**4*sqrt(a + b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Int egral(2*a*b*x*sqrt(a + b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(2*a*c*x**2*sqrt(a + b*x + c*x**2)/(b**3 + 6*b **2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(2*b*c*x**3*sqrt(a + b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x))/d** 3
Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Leaf count of result is larger than twice the leaf count of optimal. 510 vs. \(2 (123) = 246\).
Time = 0.31 (sec) , antiderivative size = 510, normalized size of antiderivative = 3.47 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^3} \, dx=\frac {1}{48} \, \sqrt {c x^{2} + b x + a} {\left (2 \, x {\left (\frac {x}{c d^{3}} + \frac {b}{c^{2} d^{3}}\right )} - \frac {3 \, b^{2} c^{5} d^{9} - 14 \, a c^{6} d^{9}}{c^{8} d^{12}}\right )} + \frac {5 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \arctan \left (-\frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} c + b \sqrt {c}}{\sqrt {b^{2} c - 4 \, a c^{2}}}\right )}{64 \, \sqrt {b^{2} c - 4 \, a c^{2}} c^{3} d^{3}} + \frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} b^{4} c - 16 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} a b^{2} c^{2} + 32 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} a^{2} c^{3} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} b^{5} \sqrt {c} - 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} a b^{3} c^{\frac {3}{2}} + 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} a^{2} b c^{\frac {5}{2}} + {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b^{6} - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} a b^{4} c + 32 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} a^{3} c^{3} + a b^{5} \sqrt {c} - 8 \, a^{2} b^{3} c^{\frac {3}{2}} + 16 \, a^{3} b c^{\frac {5}{2}}}{64 \, {\left (2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} c + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b \sqrt {c} + b^{2} - 2 \, a c\right )}^{2} c^{3} d^{3}} \]
1/48*sqrt(c*x^2 + b*x + a)*(2*x*(x/(c*d^3) + b/(c^2*d^3)) - (3*b^2*c^5*d^9 - 14*a*c^6*d^9)/(c^8*d^12)) + 5/64*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*arctan( -(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c + b*sqrt(c))/sqrt(b^2*c - 4*a*c^ 2))/(sqrt(b^2*c - 4*a*c^2)*c^3*d^3) + 1/64*(2*(sqrt(c)*x - sqrt(c*x^2 + b* x + a))^3*b^4*c - 16*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b^2*c^2 + 32* (sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^2*c^3 + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b^5*sqrt(c) - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b^3 *c^(3/2) + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^2*b*c^(5/2) + (sqrt( c)*x - sqrt(c*x^2 + b*x + a))*b^6 - 6*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))* a*b^4*c + 32*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^3*c^3 + a*b^5*sqrt(c) - 8*a^2*b^3*c^(3/2) + 16*a^3*b*c^(5/2))/((2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b*sqrt(c) + b^2 - 2*a*c)^ 2*c^3*d^3)
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^3} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^3} \,d x \]