Integrand size = 25, antiderivative size = 183 \[ \int \frac {(2-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{x^{5/2}} \, dx=-\frac {34 \sqrt {x} (2+3 x)}{3 \sqrt {2+5 x+3 x^2}}+\frac {2 (2-x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}}-\frac {2 (2+3 x) \left (2+5 x+3 x^2\right )^{3/2}}{3 x^{3/2}}+\frac {34 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {2+5 x+3 x^2}}-\frac {14 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}} \]
-2/3*(2+3*x)*(3*x^2+5*x+2)^(3/2)/x^(3/2)-34/3*(2+3*x)*x^(1/2)/(3*x^2+5*x+2 )^(1/2)+34/3*(1+x)^(3/2)*(1/(1+x))^(1/2)*EllipticE(x^(1/2)/(1+x)^(1/2),1/2 *I*2^(1/2))*2^(1/2)*((2+3*x)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)-14*(1+x)^(3/ 2)*(1/(1+x))^(1/2)*EllipticF(x^(1/2)/(1+x)^(1/2),1/2*I*2^(1/2))*2^(1/2)*(( 2+3*x)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)+2*(2-x)*(3*x^2+5*x+2)^(1/2)/x^(1/2 )
Result contains complex when optimal does not.
Time = 21.17 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.89 \[ \int \frac {(2-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{x^{5/2}} \, dx=\frac {-34 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{5/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-2 \left (8+74 x+195 x^2+219 x^3+117 x^4+27 x^5+4 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{5/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )\right )}{3 x^{3/2} \sqrt {2+5 x+3 x^2}} \]
((-34*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(5/2)*EllipticE[I*ArcSin h[Sqrt[2/3]/Sqrt[x]], 3/2] - 2*(8 + 74*x + 195*x^2 + 219*x^3 + 117*x^4 + 2 7*x^5 + (4*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(5/2)*EllipticF[I*A rcSinh[Sqrt[2/3]/Sqrt[x]], 3/2]))/(3*x^(3/2)*Sqrt[2 + 5*x + 3*x^2])
Time = 0.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {1230, 27, 1230, 27, 1240, 1503, 1413, 1456}
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.
| \(\displaystyle \int \frac {(2-5 x) \left (3 x^2+5 x+2\right )^{3/2}}{x^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1230 |
| \(\displaystyle -\frac {2}{5} \int \frac {5 (3 x+2) \sqrt {3 x^2+5 x+2}}{2 x^{3/2}}dx-\frac {2 (3 x+2) \left (3 x^2+5 x+2\right )^{3/2}}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\int \frac {(3 x+2) \sqrt {3 x^2+5 x+2}}{x^{3/2}}dx-\frac {2 (3 x+2) \left (3 x^2+5 x+2\right )^{3/2}}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 1230 |
| \(\displaystyle \frac {2}{3} \int -\frac {3 (17 x+14)}{2 \sqrt {x} \sqrt {3 x^2+5 x+2}}dx+\frac {2 (2-x) \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {2 (3 x+2) \left (3 x^2+5 x+2\right )^{3/2}}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\int \frac {17 x+14}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx+\frac {2 (2-x) \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {2 (3 x+2) \left (3 x^2+5 x+2\right )^{3/2}}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 1240 |
| \(\displaystyle -2 \int \frac {17 x+14}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {2 (2-x) \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {2 (3 x+2) \left (3 x^2+5 x+2\right )^{3/2}}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle -2 \left (14 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+17 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )+\frac {2 (2-x) \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {2 (3 x+2) \left (3 x^2+5 x+2\right )^{3/2}}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 1413 |
| \(\displaystyle -2 \left (17 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {7 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}\right )+\frac {2 (2-x) \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {2 (3 x+2) \left (3 x^2+5 x+2\right )^{3/2}}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 1456 |
| \(\displaystyle -2 \left (\frac {7 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+17 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )+\frac {2 (2-x) \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {2 (3 x+2) \left (3 x^2+5 x+2\right )^{3/2}}{3 x^{3/2}}\) |
(2*(2 - x)*Sqrt[2 + 5*x + 3*x^2])/Sqrt[x] - (2*(2 + 3*x)*(2 + 5*x + 3*x^2) ^(3/2))/(3*x^(3/2)) - 2*(17*((Sqrt[x]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/ 2])/(3*Sqrt[2 + 5*x + 3*x^2])) + (7*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x) ]*EllipticF[ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2])
3.11.49.3.1 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[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + b*x + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Simp[p/(e^2*(m + 1)*(m + 2*p + 2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (LtQ[m, - 1] || EqQ[p, 1] || (IntegerQ[p] && !RationalQ[m])) && NeQ[m, -1] && !ILtQ [m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Simp[2 Subst[Int[(f + g*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, b, c, f, g}, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b ^2 - 4*a*c, 2]}, Simp[(2*a + (b - q)*x^2)*(Sqrt[(2*a + (b + q)*x^2)/(2*a + (b - q)*x^2)]/(2*a*Rt[(b - q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*EllipticF [ArcTan[Rt[(b - q)/(2*a), 2]*x], -2*(q/(b - q))], x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[x*((b - q + 2*c*x^2)/(2*c*Sqrt[a + b*x^2 + c*x^4 ])), x] - Simp[Rt[(b - q)/(2*a), 2]*(2*a + (b - q)*x^2)*(Sqrt[(2*a + (b + q )*x^2)/(2*a + (b - q)*x^2)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan [Rt[(b - q)/(2*a), 2]*x], -2*(q/(b - q))], x] /; PosQ[(b - q)/a]] /; FreeQ[ {a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[d Int[1/Sqrt[a + b*x^2 + c*x^4] , x], x] + Simp[e Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b + q) /a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]
Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.68
| method | result | size |
| default | \(\frac {9 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x -17 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x -162 x^{5}-702 x^{4}-1008 x^{3}-660 x^{2}-240 x -48}{9 \sqrt {3 x^{2}+5 x +2}\, x^{\frac {3}{2}}}\) | \(125\) |
| risch | \(-\frac {2 \left (27 x^{5}+117 x^{4}+168 x^{3}+110 x^{2}+40 x +8\right )}{3 x^{\frac {3}{2}} \sqrt {3 x^{2}+5 x +2}}-\frac {\left (\frac {14 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {17 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right ) \sqrt {x \left (3 x^{2}+5 x +2\right )}}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(203\) |
| elliptic | \(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {8 \sqrt {3 x^{3}+5 x^{2}+2 x}}{3 x^{2}}-\frac {20 \left (3 x^{2}+5 x +2\right )}{3 \sqrt {x \left (3 x^{2}+5 x +2\right )}}-6 x \sqrt {3 x^{3}+5 x^{2}+2 x}-16 \sqrt {3 x^{3}+5 x^{2}+2 x}-\frac {14 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {17 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(243\) |
1/9*(9*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticF(1/2*(6*x+4 )^(1/2),I*2^(1/2))*x-17*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*Ell ipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))*x-162*x^5-702*x^4-1008*x^3-660*x^2-240 *x-48)/(3*x^2+5*x+2)^(1/2)/x^(3/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.38 \[ \int \frac {(2-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{x^{5/2}} \, dx=-\frac {2 \, {\left (41 \, \sqrt {3} x^{2} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 153 \, \sqrt {3} x^{2} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) + 9 \, {\left (9 \, x^{3} + 24 \, x^{2} + 10 \, x + 4\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}\right )}}{27 \, x^{2}} \]
-2/27*(41*sqrt(3)*x^2*weierstrassPInverse(28/27, 80/729, x + 5/9) - 153*sq rt(3)*x^2*weierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729 , x + 5/9)) + 9*(9*x^3 + 24*x^2 + 10*x + 4)*sqrt(3*x^2 + 5*x + 2)*sqrt(x)) /x^2
\[ \int \frac {(2-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{x^{5/2}} \, dx=- \int \left (- \frac {4 \sqrt {3 x^{2} + 5 x + 2}}{x^{\frac {5}{2}}}\right )\, dx - \int \frac {19 \sqrt {3 x^{2} + 5 x + 2}}{\sqrt {x}}\, dx - \int 15 \sqrt {x} \sqrt {3 x^{2} + 5 x + 2}\, dx \]
-Integral(-4*sqrt(3*x**2 + 5*x + 2)/x**(5/2), x) - Integral(19*sqrt(3*x**2 + 5*x + 2)/sqrt(x), x) - Integral(15*sqrt(x)*sqrt(3*x**2 + 5*x + 2), x)
\[ \int \frac {(2-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{x^{5/2}} \, dx=\int { -\frac {{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} {\left (5 \, x - 2\right )}}{x^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(2-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{x^{5/2}} \, dx=\int { -\frac {{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} {\left (5 \, x - 2\right )}}{x^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(2-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{x^{5/2}} \, dx=\int -\frac {\left (5\,x-2\right )\,{\left (3\,x^2+5\,x+2\right )}^{3/2}}{x^{5/2}} \,d x \]