Integrand size = 28, antiderivative size = 357 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{15/2}} \, dx=-\frac {\sqrt {a+b x+c x^2}}{156 c^3 d^5 (b d+2 c d x)^{5/2}}+\frac {\sqrt {a+b x+c x^2}}{78 c^3 \left (b^2-4 a c\right ) d^7 \sqrt {b d+2 c d x}}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{234 c^2 d^3 (b d+2 c d x)^{9/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}-\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{156 c^4 \sqrt [4]{b^2-4 a c} d^{15/2} \sqrt {a+b x+c x^2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{156 c^4 \sqrt [4]{b^2-4 a c} d^{15/2} \sqrt {a+b x+c x^2}} \]
-5/234*(c*x^2+b*x+a)^(3/2)/c^2/d^3/(2*c*d*x+b*d)^(9/2)-1/13*(c*x^2+b*x+a)^ (5/2)/c/d/(2*c*d*x+b*d)^(13/2)-1/156*(c*x^2+b*x+a)^(1/2)/c^3/d^5/(2*c*d*x+ b*d)^(5/2)+1/78*(c*x^2+b*x+a)^(1/2)/c^3/(-4*a*c+b^2)/d^7/(2*c*d*x+b*d)^(1/ 2)-1/156*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^4/(-4*a*c+b^2)^(1/4)/d^(15/2)/(c*x^2+b* x+a)^(1/2)+1/156*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)^(1/4)/d^(15/2)/( c*x^2+b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.07 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.31 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{15/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2 \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {13}{4},-\frac {5}{2},-\frac {9}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{416 c^3 d^8 (b+2 c x)^7 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]
-1/416*((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeo metric2F1[-13/4, -5/2, -9/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c^3*d^8*(b + 2 *c*x)^7*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])
Time = 0.71 (sec) , antiderivative size = 352, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {1108, 1108, 1108, 1117, 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.
| \(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{15/2}} \, dx\) |
| \(\Big \downarrow \) 1108 |
| \(\displaystyle \frac {5 \int \frac {\left (c x^2+b x+a\right )^{3/2}}{(b d+2 c x d)^{11/2}}dx}{26 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}\) |
| \(\Big \downarrow \) 1108 |
| \(\displaystyle \frac {5 \left (\frac {\int \frac {\sqrt {c x^2+b x+a}}{(b d+2 c x d)^{7/2}}dx}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{9 c d (b d+2 c d x)^{9/2}}\right )}{26 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}\) |
| \(\Big \downarrow \) 1108 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\int \frac {1}{(b d+2 c x d)^{3/2} \sqrt {c x^2+b x+a}}dx}{10 c d^2}-\frac {\sqrt {a+b x+c x^2}}{5 c d (b d+2 c d x)^{5/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{9 c d (b d+2 c d x)^{9/2}}\right )}{26 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}\) |
| \(\Big \downarrow \) 1117 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\frac {4 \sqrt {a+b x+c x^2}}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {\int \frac {\sqrt {b d+2 c x d}}{\sqrt {c x^2+b x+a}}dx}{d^2 \left (b^2-4 a c\right )}}{10 c d^2}-\frac {\sqrt {a+b x+c x^2}}{5 c d (b d+2 c d x)^{5/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{9 c d (b d+2 c d x)^{9/2}}\right )}{26 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}\) |
| \(\Big \downarrow \) 1115 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\frac {4 \sqrt {a+b x+c x^2}}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {\sqrt {b d+2 c x d}}{\sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{d^2 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{10 c d^2}-\frac {\sqrt {a+b x+c x^2}}{5 c d (b d+2 c d x)^{5/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{9 c d (b d+2 c d x)^{9/2}}\right )}{26 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}\) |
| \(\Big \downarrow \) 1114 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\frac {4 \sqrt {a+b x+c x^2}}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {b d+2 c x d}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{c d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{10 c d^2}-\frac {\sqrt {a+b x+c x^2}}{5 c d (b d+2 c d x)^{5/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{9 c d (b d+2 c d x)^{9/2}}\right )}{26 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}\) |
| \(\Big \downarrow \) 836 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\frac {4 \sqrt {a+b x+c x^2}}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{d \sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{c d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{10 c d^2}-\frac {\sqrt {a+b x+c x^2}}{5 c d (b d+2 c d x)^{5/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{9 c d (b d+2 c d x)^{9/2}}\right )}{26 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\frac {4 \sqrt {a+b x+c x^2}}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{c d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{10 c d^2}-\frac {\sqrt {a+b x+c x^2}}{5 c d (b d+2 c d x)^{5/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{9 c d (b d+2 c d x)^{9/2}}\right )}{26 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}\) |
| \(\Big \downarrow \) 762 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\frac {4 \sqrt {a+b x+c x^2}}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{c d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{10 c d^2}-\frac {\sqrt {a+b x+c x^2}}{5 c d (b d+2 c d x)^{5/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{9 c d (b d+2 c d x)^{9/2}}\right )}{26 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}\) |
| \(\Big \downarrow \) 1389 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\frac {4 \sqrt {a+b x+c x^2}}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {\sqrt {\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}+1}}{\sqrt {1-\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{c d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{10 c d^2}-\frac {\sqrt {a+b x+c x^2}}{5 c d (b d+2 c d x)^{5/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{9 c d (b d+2 c d x)^{9/2}}\right )}{26 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}\) |
| \(\Big \downarrow \) 327 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\frac {4 \sqrt {a+b x+c x^2}}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d^{3/2} \left (b^2-4 a c\right )^{3/4} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{c d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{10 c d^2}-\frac {\sqrt {a+b x+c x^2}}{5 c d (b d+2 c d x)^{5/2}}}{6 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{9 c d (b d+2 c d x)^{9/2}}\right )}{26 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{13 c d (b d+2 c d x)^{13/2}}\) |
-1/13*(a + b*x + c*x^2)^(5/2)/(c*d*(b*d + 2*c*d*x)^(13/2)) + (5*(-1/9*(a + b*x + c*x^2)^(3/2)/(c*d*(b*d + 2*c*d*x)^(9/2)) + (-1/5*Sqrt[a + b*x + c*x ^2]/(c*d*(b*d + 2*c*d*x)^(5/2)) + ((4*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c )*d*Sqrt[b*d + 2*c*d*x]) - (2*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))] *((b^2 - 4*a*c)^(3/4)*d^(3/2)*EllipticE[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[ArcSi n[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1]))/(c*(b^2 - 4*a* c)*d^3*Sqrt[a + b*x + c*x^2]))/(10*c*d^2))/(6*c*d^2)))/(26*c*d^2)
3.14.62.3.1 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_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[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*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])
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. \(1233\) vs. \(2(301)=602\).
Time = 5.69 (sec) , antiderivative size = 1234, normalized size of antiderivative = 3.46
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(1234\) |
| default | \(\text {Expression too large to display}\) | \(2125\) |
(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)* (-1/26624*(16*a^2*c^2-8*a*b^2*c+b^4)/c^10/d^8*(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)^7-7/14976/c^8*(4*a*c-b^2)/d^8*(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)^5-31/14976 /c^6/d^8*(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)^3-1/156*(2*c^2*d*x^2+2*b*c*d*x+2*a*c*d)/c^4/(4*a*c-b^2)/d^8/((x+1/2/c*b )*(2*c^2*d*x^2+2*b*c*d*x+2*a*c*d))^(1/2)+1/156/c^3*b/(4*a*c-b^2)/d^7*(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)*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/c*b) )^(1/2))+1/78/c^2/(4*a*c-b^2)/d^7*(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...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.18 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{15/2}} \, dx=\frac {3 \, \sqrt {2} {\left (128 \, c^{7} x^{7} + 448 \, b c^{6} x^{6} + 672 \, b^{2} c^{5} x^{5} + 560 \, b^{3} c^{4} x^{4} + 280 \, b^{4} c^{3} x^{3} + 84 \, b^{5} c^{2} x^{2} + 14 \, b^{6} c x + b^{7}\right )} \sqrt {c^{2} d} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + {\left (384 \, c^{7} x^{6} + 1152 \, b c^{6} x^{5} + 3 \, b^{6} c + 2 \, a b^{4} c^{2} + 4 \, a^{2} b^{2} c^{3} + 144 \, a^{3} c^{4} + 4 \, {\left (329 \, b^{2} c^{5} + 124 \, a c^{6}\right )} x^{4} + 8 \, {\left (89 \, b^{3} c^{4} + 124 \, a b c^{5}\right )} x^{3} + 2 \, {\left (101 \, b^{4} c^{3} + 260 \, a b^{2} c^{4} + 224 \, a^{2} c^{5}\right )} x^{2} + 2 \, {\left (19 \, b^{5} c^{2} + 12 \, a b^{3} c^{3} + 224 \, a^{2} b c^{4}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{468 \, {\left (128 \, {\left (b^{2} c^{11} - 4 \, a c^{12}\right )} d^{8} x^{7} + 448 \, {\left (b^{3} c^{10} - 4 \, a b c^{11}\right )} d^{8} x^{6} + 672 \, {\left (b^{4} c^{9} - 4 \, a b^{2} c^{10}\right )} d^{8} x^{5} + 560 \, {\left (b^{5} c^{8} - 4 \, a b^{3} c^{9}\right )} d^{8} x^{4} + 280 \, {\left (b^{6} c^{7} - 4 \, a b^{4} c^{8}\right )} d^{8} x^{3} + 84 \, {\left (b^{7} c^{6} - 4 \, a b^{5} c^{7}\right )} d^{8} x^{2} + 14 \, {\left (b^{8} c^{5} - 4 \, a b^{6} c^{6}\right )} d^{8} x + {\left (b^{9} c^{4} - 4 \, a b^{7} c^{5}\right )} d^{8}\right )}} \]
1/468*(3*sqrt(2)*(128*c^7*x^7 + 448*b*c^6*x^6 + 672*b^2*c^5*x^5 + 560*b^3* c^4*x^4 + 280*b^4*c^3*x^3 + 84*b^5*c^2*x^2 + 14*b^6*c*x + b^7)*sqrt(c^2*d) *weierstrassZeta((b^2 - 4*a*c)/c^2, 0, weierstrassPInverse((b^2 - 4*a*c)/c ^2, 0, 1/2*(2*c*x + b)/c)) + (384*c^7*x^6 + 1152*b*c^6*x^5 + 3*b^6*c + 2*a *b^4*c^2 + 4*a^2*b^2*c^3 + 144*a^3*c^4 + 4*(329*b^2*c^5 + 124*a*c^6)*x^4 + 8*(89*b^3*c^4 + 124*a*b*c^5)*x^3 + 2*(101*b^4*c^3 + 260*a*b^2*c^4 + 224*a ^2*c^5)*x^2 + 2*(19*b^5*c^2 + 12*a*b^3*c^3 + 224*a^2*b*c^4)*x)*sqrt(2*c*d* x + b*d)*sqrt(c*x^2 + b*x + a))/(128*(b^2*c^11 - 4*a*c^12)*d^8*x^7 + 448*( b^3*c^10 - 4*a*b*c^11)*d^8*x^6 + 672*(b^4*c^9 - 4*a*b^2*c^10)*d^8*x^5 + 56 0*(b^5*c^8 - 4*a*b^3*c^9)*d^8*x^4 + 280*(b^6*c^7 - 4*a*b^4*c^8)*d^8*x^3 + 84*(b^7*c^6 - 4*a*b^5*c^7)*d^8*x^2 + 14*(b^8*c^5 - 4*a*b^6*c^6)*d^8*x + (b ^9*c^4 - 4*a*b^7*c^5)*d^8)
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{15/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{15/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {15}{2}}} \,d x } \]
\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{15/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {15}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{15/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{15/2}} \,d x \]