Integrand size = 28, antiderivative size = 184 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}-\frac {40 c \sqrt {a+b x+c x^2}}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac {20 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 \left (b^2-4 a c\right )^{7/4} d^{5/2} \sqrt {a+b x+c x^2}} \]
-2/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^(1/2)-40/3*c*(c*x^2+b* x+a)^(1/2)/(-4*a*c+b^2)^2/d/(2*c*d*x+b*d)^(3/2)-20/3*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)/(-4*a*c+b^2)^(7/4)/d^(5/2)/(c*x^2+b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {8 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 \left (b^2-4 a c\right ) d (d (b+2 c x))^{3/2} \sqrt {a+x (b+c x)}} \]
(8*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[-3/4, 3/2, 1/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(3*(b^2 - 4*a*c)*d*(d*(b + 2*c*x))^(3/ 2)*Sqrt[a + x*(b + c*x)])
Time = 0.43 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1111, 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.
| \(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1111 |
| \(\displaystyle -\frac {10 c \int \frac {1}{(b d+2 c x d)^{5/2} \sqrt {c x^2+b x+a}}dx}{b^2-4 a c}-\frac {2}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 1117 |
| \(\displaystyle -\frac {10 c \left (\frac {\int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx}{3 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{b^2-4 a c}-\frac {2}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 1115 |
| \(\displaystyle -\frac {10 c \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{3 d^2 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{b^2-4 a c}-\frac {2}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 1113 |
| \(\displaystyle -\frac {10 c \left (\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{3 c d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{b^2-4 a c}-\frac {2}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}}\) |
| \(\Big \downarrow \) 762 |
| \(\displaystyle -\frac {10 c \left (\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c d^{5/2} \left (b^2-4 a c\right )^{3/4} \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{b^2-4 a c}-\frac {2}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}}\) |
-2/((b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2]) - (10*c*( (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])))/(b^2 - 4*a*c)
3.14.83.3.1 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[2*c*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)* (b^2 - 4*a*c))), x] - Simp[2*c*e*((m + 2*p + 3)/(e*(p + 1)*(b^2 - 4*a*c))) 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] && LtQ[p, -1] && !G tQ[m, 1] && RationalQ[m] && 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. \(364\) vs. \(2(158)=316\).
Time = 4.39 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.98
| method | result | size |
| default | \(-\frac {2 \sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, \left (10 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, F\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, c x +5 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, F\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, b +20 c^{2} x^{2}+20 b c x +8 a c +3 b^{2}\right )}{3 d^{3} \left (2 c^{2} x^{3}+3 c b \,x^{2}+2 a c x +b^{2} x +a b \right ) \left (4 a c -b^{2}\right )^{2} \left (2 c x +b \right )}\) | \(365\) |
| elliptic | \(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {4 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{3 \left (4 a c -b^{2}\right )^{2} d^{3} c \left (x +\frac {b}{2 c}\right )^{2}}-\frac {2 \left (2 c^{2} d x +b c d \right )}{\left (4 a c -b^{2}\right )^{2} d^{3} c \sqrt {\left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right ) \left (2 c^{2} d x +b c d \right )}}-\frac {20 c \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, F\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{3 \left (4 a c -b^{2}\right )^{2} d^{2} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\) | \(553\) |
-2/3*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*(10*((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)*EllipticF(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))*(-4*a*c+b ^2)^(1/2)*c*x+5*((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)*EllipticF(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))*(-4*a*c+b^2)^(1/2)*b+20*c^2*x^2+20*b*c*x+8 *a*c+3*b^2)/d^3/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)/(4*a*c-b^2)^2/(2*c *x+b)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.15 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.79 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (5 \, \sqrt {2} {\left (4 \, c^{3} x^{4} + 8 \, b c^{2} x^{3} + a b^{2} + {\left (5 \, b^{2} c + 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} + 4 \, a b c\right )} x\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + {\left (20 \, c^{3} x^{2} + 20 \, b c^{2} x + 3 \, b^{2} c + 8 \, a c^{2}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left (4 \, {\left (b^{4} c^{4} - 8 \, a b^{2} c^{5} + 16 \, a^{2} c^{6}\right )} d^{3} x^{4} + 8 \, {\left (b^{5} c^{3} - 8 \, a b^{3} c^{4} + 16 \, a^{2} b c^{5}\right )} d^{3} x^{3} + {\left (5 \, b^{6} c^{2} - 36 \, a b^{4} c^{3} + 48 \, a^{2} b^{2} c^{4} + 64 \, a^{3} c^{5}\right )} d^{3} x^{2} + {\left (b^{7} c - 4 \, a b^{5} c^{2} - 16 \, a^{2} b^{3} c^{3} + 64 \, a^{3} b c^{4}\right )} d^{3} x + {\left (a b^{6} c - 8 \, a^{2} b^{4} c^{2} + 16 \, a^{3} b^{2} c^{3}\right )} d^{3}\right )}} \]
-2/3*(5*sqrt(2)*(4*c^3*x^4 + 8*b*c^2*x^3 + a*b^2 + (5*b^2*c + 4*a*c^2)*x^2 + (b^3 + 4*a*b*c)*x)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0 , 1/2*(2*c*x + b)/c) + (20*c^3*x^2 + 20*b*c^2*x + 3*b^2*c + 8*a*c^2)*sqrt( 2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(4*(b^4*c^4 - 8*a*b^2*c^5 + 16*a^2*c ^6)*d^3*x^4 + 8*(b^5*c^3 - 8*a*b^3*c^4 + 16*a^2*b*c^5)*d^3*x^3 + (5*b^6*c^ 2 - 36*a*b^4*c^3 + 48*a^2*b^2*c^4 + 64*a^3*c^5)*d^3*x^2 + (b^7*c - 4*a*b^5 *c^2 - 16*a^2*b^3*c^3 + 64*a^3*b*c^4)*d^3*x + (a*b^6*c - 8*a^2*b^4*c^2 + 1 6*a^3*b^2*c^3)*d^3)
\[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (d \left (b + 2 c x\right )\right )^{\frac {5}{2}} \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (2 \, c d x + b d\right )}^{\frac {5}{2}} {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (2 \, c d x + b d\right )}^{\frac {5}{2}} {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]