Integrand size = 28, antiderivative size = 211 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{9/2}} \, dx=\frac {5 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{84 c^3 d^5}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{42 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{7 c d (b d+2 c d x)^{7/2}}-\frac {5 \left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{84 c^4 d^{9/2} \sqrt {a+b x+c x^2}} \]
-5/42*(c*x^2+b*x+a)^(3/2)/c^2/d^3/(2*c*d*x+b*d)^(3/2)-1/7*(c*x^2+b*x+a)^(5 /2)/c/d/(2*c*d*x+b*d)^(7/2)+5/84*(2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^(1/2)/c ^3/d^5-5/84*(-4*a*c+b^2)^(5/4)*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/d^(9/2)/(c*x^2+ b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.06 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.52 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{9/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2 \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {7}{4},-\frac {3}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{224 c^3 d^5 (b+2 c x)^4 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]
-1/224*((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeo metric2F1[-5/2, -7/4, -3/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c^3*d^5*(b + 2* c*x)^4*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])
Time = 0.43 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1108, 1108, 1109, 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.
| \(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{9/2}} \, dx\) |
| \(\Big \downarrow \) 1108 |
| \(\displaystyle \frac {5 \int \frac {\left (c x^2+b x+a\right )^{3/2}}{(b d+2 c x d)^{5/2}}dx}{14 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{7 c d (b d+2 c d x)^{7/2}}\) |
| \(\Big \downarrow \) 1108 |
| \(\displaystyle \frac {5 \left (\frac {\int \frac {\sqrt {c x^2+b x+a}}{\sqrt {b d+2 c x d}}dx}{2 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}\right )}{14 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{7 c d (b d+2 c d x)^{7/2}}\) |
| \(\Big \downarrow \) 1109 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}}{3 c d}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx}{6 c}}{2 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}\right )}{14 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{7 c d (b d+2 c d x)^{7/2}}\) |
| \(\Big \downarrow \) 1115 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}}{3 c d}-\frac {\left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{6 c \sqrt {a+b x+c x^2}}}{2 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}\right )}{14 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{7 c d (b d+2 c d x)^{7/2}}\) |
| \(\Big \downarrow \) 1113 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}}{3 c d}-\frac {\left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{3 c^2 d \sqrt {a+b x+c x^2}}}{2 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}\right )}{14 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{7 c d (b d+2 c d x)^{7/2}}\) |
| \(\Big \downarrow \) 762 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}}{3 c d}-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c^2 \sqrt {d} \sqrt {a+b x+c x^2}}}{2 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}\right )}{14 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{7 c d (b d+2 c d x)^{7/2}}\) |
-1/7*(a + b*x + c*x^2)^(5/2)/(c*d*(b*d + 2*c*d*x)^(7/2)) + (5*(-1/3*(a + b *x + c*x^2)^(3/2)/(c*d*(b*d + 2*c*d*x)^(3/2)) + ((Sqrt[b*d + 2*c*d*x]*Sqrt [a + b*x + c*x^2])/(3*c*d) - ((b^2 - 4*a*c)^(5/4)*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^2*Sqrt[d]*Sqrt[a + b*x + c*x^2]))/(2*c*d^2)))/(1 4*c*d^2)
3.14.53.3.1 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[((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/(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]
Leaf count of result is larger than twice the leaf count of optimal. \(645\) vs. \(2(177)=354\).
Time = 8.21 (sec) , antiderivative size = 646, normalized size of antiderivative = 3.06
| method | result | size |
| elliptic | \(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{1792 c^{7} d^{5} \left (x +\frac {b}{2 c}\right )^{4}}-\frac {\left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{84 c^{5} d^{5} \left (x +\frac {b}{2 c}\right )^{2}}+\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{48 d^{5} c^{3}}+\frac {2 \left (\frac {6 a c -b^{2}}{32 d^{4} c^{3}}-\frac {4 a c -b^{2}}{84 c^{3} d^{4}}-\frac {a c d +\frac {1}{2} b^{2} d}{48 d^{5} c^{3}}\right ) \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, F\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\) | \(646\) |
| default | \(\text {Expression too large to display}\) | \(1310\) |
| risch | \(\text {Expression too large to display}\) | \(1829\) |
(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)* (-1/1792*(16*a^2*c^2-8*a*b^2*c+b^4)/c^7/d^5*(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)^4-1/84*(4*a*c-b^2)/c^5/d^5*(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+1/48/d^5/c^3* (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)+2*(1/32*(6*a*c-b^2 )/d^4/c^3-1/84*(4*a*c-b^2)/c^3/d^4-1/48/d^5/c^3*(a*c*d+1/2*b^2*d))*(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)))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.42 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{9/2}} \, dx=-\frac {5 \, \sqrt {2} {\left (b^{6} - 4 \, a b^{4} c + 16 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{4} + 32 \, {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{3} + 24 \, {\left (b^{4} c^{2} - 4 \, a b^{2} c^{3}\right )} x^{2} + 8 \, {\left (b^{5} c - 4 \, a b^{3} c^{2}\right )} x\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) - 2 \, {\left (28 \, c^{6} x^{4} + 56 \, b c^{5} x^{3} + 5 \, b^{4} c^{2} - 10 \, a b^{2} c^{3} - 12 \, a^{2} c^{4} + 2 \, {\left (29 \, b^{2} c^{4} - 32 \, a c^{5}\right )} x^{2} + 2 \, {\left (15 \, b^{3} c^{3} - 32 \, a b c^{4}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{168 \, {\left (16 \, c^{9} d^{5} x^{4} + 32 \, b c^{8} d^{5} x^{3} + 24 \, b^{2} c^{7} d^{5} x^{2} + 8 \, b^{3} c^{6} d^{5} x + b^{4} c^{5} d^{5}\right )}} \]
-1/168*(5*sqrt(2)*(b^6 - 4*a*b^4*c + 16*(b^2*c^4 - 4*a*c^5)*x^4 + 32*(b^3* c^3 - 4*a*b*c^4)*x^3 + 24*(b^4*c^2 - 4*a*b^2*c^3)*x^2 + 8*(b^5*c - 4*a*b^3 *c^2)*x)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c) - 2*(28*c^6*x^4 + 56*b*c^5*x^3 + 5*b^4*c^2 - 10*a*b^2*c^3 - 12*a^2 *c^4 + 2*(29*b^2*c^4 - 32*a*c^5)*x^2 + 2*(15*b^3*c^3 - 32*a*b*c^4)*x)*sqrt (2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(16*c^9*d^5*x^4 + 32*b*c^8*d^5*x^3 + 24*b^2*c^7*d^5*x^2 + 8*b^3*c^6*d^5*x + b^4*c^5*d^5)
\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{9/2}} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {9}{2}}}\, dx \]
\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{9/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {9}{2}}} \,d x } \]
\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{9/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{9/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{9/2}} \,d x \]