Integrand size = 28, antiderivative size = 274 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2} \, dx=-\frac {\left (b^2-4 a c\right )^3 d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{462 c^3}+\frac {\left (b^2-4 a c\right )^2 (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}}{308 c^3 d}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{66 c^2 d}+\frac {(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{5/2}}{15 c d}-\frac {\left (b^2-4 a c\right )^{17/4} d^{3/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{924 c^4 \sqrt {a+b x+c x^2}} \]
-1/66*(-4*a*c+b^2)*(2*c*d*x+b*d)^(5/2)*(c*x^2+b*x+a)^(3/2)/c^2/d+1/15*(2*c *d*x+b*d)^(5/2)*(c*x^2+b*x+a)^(5/2)/c/d+1/308*(-4*a*c+b^2)^2*(2*c*d*x+b*d) ^(5/2)*(c*x^2+b*x+a)^(1/2)/c^3/d-1/462*(-4*a*c+b^2)^3*d*(2*c*d*x+b*d)^(1/2 )*(c*x^2+b*x+a)^(1/2)/c^3-1/924*(-4*a*c+b^2)^(17/4)*d^(3/2)*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/(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 = 117, normalized size of antiderivative = 0.43 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {2}{15} d \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \left (2 (a+x (b+c x))^3+\frac {\left (b^2-4 a c\right )^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1}{4},\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{64 c^3 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}}\right ) \]
(2*d*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*(2*(a + x*(b + c*x))^3 + (( b^2 - 4*a*c)^3*Hypergeometric2F1[-5/2, 1/4, 5/4, (b + 2*c*x)^2/(b^2 - 4*a* c)])/(64*c^3*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])))/15
Time = 0.54 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1109, 1109, 1109, 1116, 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.
| \(\displaystyle \int \left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{3/2} \, dx\) |
| \(\Big \downarrow \) 1109 |
| \(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{5/2}}{15 c d}-\frac {\left (b^2-4 a c\right ) \int (b d+2 c x d)^{3/2} \left (c x^2+b x+a\right )^{3/2}dx}{6 c}\) |
| \(\Big \downarrow \) 1109 |
| \(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{5/2}}{15 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{11 c d}-\frac {3 \left (b^2-4 a c\right ) \int (b d+2 c x d)^{3/2} \sqrt {c x^2+b x+a}dx}{22 c}\right )}{6 c}\) |
| \(\Big \downarrow \) 1109 |
| \(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{5/2}}{15 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{11 c d}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}}{7 c d}-\frac {\left (b^2-4 a c\right ) \int \frac {(b d+2 c x d)^{3/2}}{\sqrt {c x^2+b x+a}}dx}{14 c}\right )}{22 c}\right )}{6 c}\) |
| \(\Big \downarrow \) 1116 |
| \(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{5/2}}{15 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{11 c d}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}}{7 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {1}{3} d^2 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )}{14 c}\right )}{22 c}\right )}{6 c}\) |
| \(\Big \downarrow \) 1115 |
| \(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{5/2}}{15 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{11 c d}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}}{7 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {d^2 \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{3 \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )}{14 c}\right )}{22 c}\right )}{6 c}\) |
| \(\Big \downarrow \) 1113 |
| \(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{5/2}}{15 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{11 c d}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}}{7 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {2 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{3 c \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )}{14 c}\right )}{22 c}\right )}{6 c}\) |
| \(\Big \downarrow \) 762 |
| \(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{5/2}}{15 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{11 c d}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}}{7 c d}-\frac {\left (b^2-4 a c\right ) \left (\frac {2 d^{3/2} \left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )}{14 c}\right )}{22 c}\right )}{6 c}\) |
((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^(5/2))/(15*c*d) - ((b^2 - 4*a*c)* (((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^(3/2))/(11*c*d) - (3*(b^2 - 4*a* c)*(((b*d + 2*c*d*x)^(5/2)*Sqrt[a + b*x + c*x^2])/(7*c*d) - ((b^2 - 4*a*c) *((4*d*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/3 + (2*(b^2 - 4*a*c)^(5/ 4)*d^(3/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[S qrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(3*c*Sqrt[a + b*x + c*x^2])))/(14*c)))/(22*c)))/(6*c)
3.14.50.3.1 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 + 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]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))) 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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || OddQ[m])
Leaf count of result is larger than twice the leaf count of optimal. \(641\) vs. \(2(234)=468\).
Time = 3.48 (sec) , antiderivative size = 642, normalized size of antiderivative = 2.34
| method | result | size |
| risch | \(\frac {\left (1232 c^{6} x^{6}+3696 b \,c^{5} x^{5}+3584 a \,c^{5} x^{4}+3724 b^{2} c^{4} x^{4}+7168 a b \,c^{4} x^{3}+1288 x^{3} b^{3} c^{3}+3312 a^{2} c^{4} x^{2}+3720 a \,b^{2} c^{3} x^{2}+18 x^{2} b^{4} c^{2}+3312 a^{2} b \,c^{3} x +136 x a \,b^{3} c^{2}-10 x \,b^{5} c +640 c^{3} a^{3}+348 a^{2} b^{2} c^{2}-70 a \,b^{4} c +5 b^{6}\right ) \left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, d^{2}}{4620 c^{3} \sqrt {d \left (2 c x +b \right )}}-\frac {\left (256 a^{4} c^{4}-256 a^{3} b^{2} c^{3}+96 a^{2} b^{4} c^{2}-16 a \,b^{6} c +b^{8}\right ) \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, F\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right ) d^{2} \sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}}{924 c^{3} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}\, \sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\) | \(642\) |
| default | \(\text {Expression too large to display}\) | \(1055\) |
| elliptic | \(\text {Expression too large to display}\) | \(5587\) |
1/4620/c^3*(1232*c^6*x^6+3696*b*c^5*x^5+3584*a*c^5*x^4+3724*b^2*c^4*x^4+71 68*a*b*c^4*x^3+1288*b^3*c^3*x^3+3312*a^2*c^4*x^2+3720*a*b^2*c^3*x^2+18*b^4 *c^2*x^2+3312*a^2*b*c^3*x+136*a*b^3*c^2*x-10*b^5*c*x+640*a^3*c^3+348*a^2*b ^2*c^2-70*a*b^4*c+5*b^6)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)*d^2/(d*(2*c*x+b))^( 1/2)-1/924/c^3*(256*a^4*c^4-256*a^3*b^2*c^3+96*a^2*b^4*c^2-16*a*b^6*c+b^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)*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))*d^2*(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)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2} \, dx=-\frac {5 \, \sqrt {2} {\left (b^{8} - 16 \, a b^{6} c + 96 \, a^{2} b^{4} c^{2} - 256 \, a^{3} b^{2} c^{3} + 256 \, a^{4} c^{4}\right )} \sqrt {c^{2} d} d {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) - 2 \, {\left (1232 \, c^{8} d x^{6} + 3696 \, b c^{7} d x^{5} + 28 \, {\left (133 \, b^{2} c^{6} + 128 \, a c^{7}\right )} d x^{4} + 56 \, {\left (23 \, b^{3} c^{5} + 128 \, a b c^{6}\right )} d x^{3} + 6 \, {\left (3 \, b^{4} c^{4} + 620 \, a b^{2} c^{5} + 552 \, a^{2} c^{6}\right )} d x^{2} - 2 \, {\left (5 \, b^{5} c^{3} - 68 \, a b^{3} c^{4} - 1656 \, a^{2} b c^{5}\right )} d x + {\left (5 \, b^{6} c^{2} - 70 \, a b^{4} c^{3} + 348 \, a^{2} b^{2} c^{4} + 640 \, a^{3} c^{5}\right )} d\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{9240 \, c^{5}} \]
-1/9240*(5*sqrt(2)*(b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*sqrt(c^2*d)*d*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*( 2*c*x + b)/c) - 2*(1232*c^8*d*x^6 + 3696*b*c^7*d*x^5 + 28*(133*b^2*c^6 + 1 28*a*c^7)*d*x^4 + 56*(23*b^3*c^5 + 128*a*b*c^6)*d*x^3 + 6*(3*b^4*c^4 + 620 *a*b^2*c^5 + 552*a^2*c^6)*d*x^2 - 2*(5*b^5*c^3 - 68*a*b^3*c^4 - 1656*a^2*b *c^5)*d*x + (5*b^6*c^2 - 70*a*b^4*c^3 + 348*a^2*b^2*c^4 + 640*a^3*c^5)*d)* sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/c^5
\[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2} \, dx=\int \left (d \left (b + 2 c x\right )\right )^{\frac {3}{2}} \left (a + b x + c x^{2}\right )^{\frac {5}{2}}\, dx \]
\[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2} \, dx=\int { {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} \,d x } \]
\[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2} \, dx=\int { {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2} \, dx=\int {\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,{\left (c\,x^2+b\,x+a\right )}^{5/2} \,d x \]