Integrand size = 28, antiderivative size = 219 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 d (b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}}+\frac {12 \left (b^2-4 a c\right )^{3/4} d^{5/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\sqrt {a+b x+c x^2}}-\frac {12 \left (b^2-4 a c\right )^{3/4} d^{5/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{\sqrt {a+b x+c x^2}} \]
-2*d*(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^(1/2)+12*(-4*a*c+b^2)^(3/4)*d^(5/2) *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*x^2+b*x+a)^(1/2)-12*(-4*a*c+b^2)^(3/4)*d^(5/2) *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*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 = 88, normalized size of antiderivative = 0.40 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {4 d (d (b+2 c x))^{3/2} \left (-1+2 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{\sqrt {a+x (b+c x)}} \]
(-4*d*(d*(b + 2*c*x))^(3/2)*(-1 + 2*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a *c)]*Hypergeometric2F1[3/4, 3/2, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/Sqrt[ a + x*(b + c*x)]
Time = 0.47 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.83, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1110, 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.
| \(\displaystyle \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1110 |
| \(\displaystyle 6 c d^2 \int \frac {\sqrt {b d+2 c x d}}{\sqrt {c x^2+b x+a}}dx-\frac {2 d (b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}}\) |
| \(\Big \downarrow \) 1115 |
| \(\displaystyle \frac {6 c d^2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {\sqrt {b d+2 c x d}}{\sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{\sqrt {a+b x+c x^2}}-\frac {2 d (b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}}\) |
| \(\Big \downarrow \) 1114 |
| \(\displaystyle \frac {12 d \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {b d+2 c x d}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{\sqrt {a+b x+c x^2}}-\frac {2 d (b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}}\) |
| \(\Big \downarrow \) 836 |
| \(\displaystyle \frac {12 d \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{d \sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{\sqrt {a+b x+c x^2}}-\frac {2 d (b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {12 d \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{\sqrt {a+b x+c x^2}}-\frac {2 d (b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}}\) |
| \(\Big \downarrow \) 762 |
| \(\displaystyle \frac {12 d \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{\sqrt {a+b x+c x^2}}-\frac {2 d (b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}}\) |
| \(\Big \downarrow \) 1389 |
| \(\displaystyle \frac {12 d \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {\sqrt {\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}+1}}{\sqrt {1-\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{\sqrt {a+b x+c x^2}}-\frac {2 d (b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}}\) |
| \(\Big \downarrow \) 327 |
| \(\displaystyle \frac {12 d \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d^{3/2} \left (b^2-4 a c\right )^{3/4} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{\sqrt {a+b x+c x^2}}-\frac {2 d (b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}}\) |
(-2*d*(b*d + 2*c*d*x)^(3/2))/Sqrt[a + b*x + c*x^2] + (12*d*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*((b^2 - 4*a*c)^(3/4)*d^(3/2)*EllipticE[ArcSi n[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[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sq rt[d])], -1]))/Sqrt[a + b*x + c*x^2]
3.14.85.3.1 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_Sy mbol] :> Simp[d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), x] - Simp[d*e*((m - 1)/(b*(p + 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] && N eQ[m + 2*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && 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_)^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]
Time = 2.91 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.47
| method | result | size |
| default | \(\frac {2 \left (12 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a c -3 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) b^{2}-4 c^{2} x^{2}-4 b c x -b^{2}\right ) d^{2} \sqrt {c \,x^{2}+b x +a}\, \sqrt {d \left (2 c x +b \right )}}{2 c^{2} x^{3}+3 c b \,x^{2}+2 a c x +b^{2} x +a b}\) | \(323\) |
| elliptic | \(\text {Expression too large to display}\) | \(1014\) |
2*(12*((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)*EllipticE(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))*a*c-3*((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)*EllipticE(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))*b^2-4*c^2*x^2-4*b*c*x-b^2)*d^2 *(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2* x+a*b)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.56 \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (6 \, \sqrt {2} {\left (c d^{2} x^{2} + b d^{2} x + a d^{2}\right )} \sqrt {c^{2} d} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}\right )}}{c x^{2} + b x + a} \]
-2*(6*sqrt(2)*(c*d^2*x^2 + b*d^2*x + a*d^2)*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)) + (2*c*d^2*x + b*d^2)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/( c*x^2 + b*x + a)
\[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {\left (d \left (b + 2 c x\right )\right )^{\frac {5}{2}}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {5}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {5}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]