Integrand size = 25, antiderivative size = 159 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}} \, dx=\frac {88 \sqrt {x} (2+3 x)}{27 \sqrt {2+5 x+3 x^2}}+\frac {2}{9} (1-9 x) \sqrt {x} \sqrt {2+5 x+3 x^2}-\frac {88 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{27 \sqrt {2+5 x+3 x^2}}+\frac {34 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{9 \sqrt {2+5 x+3 x^2}} \]
88/27*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)-88/27*(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)+34/9*(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)+2/9*(1-9*x)*x^(1/2)*(3*x^2+5*x+2)^(1/2)
Result contains complex when optimal does not.
Time = 20.44 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.99 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}} \, dx=\frac {2 \left (88+226 x+93 x^2-126 x^3-81 x^4\right )+88 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+14 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{27 \sqrt {x} \sqrt {2+5 x+3 x^2}} \]
(2*(88 + 226*x + 93*x^2 - 126*x^3 - 81*x^4) + (88*I)*Sqrt[2]*Sqrt[1 + x^(- 1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + ( 14*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[S qrt[2/3]/Sqrt[x]], 3/2])/(27*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])
Time = 0.32 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1231, 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.
| \(\displaystyle \int \frac {(2-5 x) \sqrt {3 x^2+5 x+2}}{\sqrt {x}} \, dx\) |
| \(\Big \downarrow \) 1231 |
| \(\displaystyle \frac {2}{9} (1-9 x) \sqrt {x} \sqrt {3 x^2+5 x+2}-\frac {2}{45} \int -\frac {5 (22 x+17)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{9} \int \frac {22 x+17}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx+\frac {2}{9} \sqrt {x} \sqrt {3 x^2+5 x+2} (1-9 x)\) |
| \(\Big \downarrow \) 1240 |
| \(\displaystyle \frac {4}{9} \int \frac {22 x+17}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {2}{9} \sqrt {x} \sqrt {3 x^2+5 x+2} (1-9 x)\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle \frac {4}{9} \left (17 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+22 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )+\frac {2}{9} \sqrt {x} \sqrt {3 x^2+5 x+2} (1-9 x)\) |
| \(\Big \downarrow \) 1413 |
| \(\displaystyle \frac {4}{9} \left (22 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {17 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}\right )+\frac {2}{9} \sqrt {x} \sqrt {3 x^2+5 x+2} (1-9 x)\) |
| \(\Big \downarrow \) 1456 |
| \(\displaystyle \frac {4}{9} \left (\frac {17 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}+22 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )+\frac {2}{9} \sqrt {x} \sqrt {3 x^2+5 x+2} (1-9 x)\) |
(2*(1 - 9*x)*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])/9 + (4*(22*((Sqrt[x]*(2 + 3*x) )/(3*Sqrt[2 + 5*x + 3*x^2]) - (Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*Ell ipticE[ArcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2])) + (17*(1 + x)*Sq rt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(Sqrt[2]*Sqrt[2 + 5*x + 3*x^2])))/9
3.11.38.3.1 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)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ (c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)) Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c ^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x ] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] || !R ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) && !ILtQ[m + 2*p, 0] && (Integer Q[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 = 0.25 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.74
| method | result | size |
| default | \(-\frac {2 \left (15 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-22 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+243 x^{4}+378 x^{3}+117 x^{2}-18 x \right )}{81 \sqrt {3 x^{2}+5 x +2}\, \sqrt {x}}\) | \(117\) |
| risch | \(-\frac {2 \left (-1+9 x \right ) \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{9}-\frac {\left (-\frac {34 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{27 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {44 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{27 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right ) \sqrt {x \left (3 x^{2}+5 x +2\right )}}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(183\) |
| elliptic | \(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (-2 x \sqrt {3 x^{3}+5 x^{2}+2 x}+\frac {2 \sqrt {3 x^{3}+5 x^{2}+2 x}}{9}+\frac {34 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{27 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {44 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{27 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(196\) |
-2/81/(3*x^2+5*x+2)^(1/2)/x^(1/2)*(15*(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))-22*(6*x+4)^(1/2)*(3+3*x) ^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))+243*x^4+3 78*x^3+117*x^2-18*x)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.30 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}} \, dx=-\frac {2}{9} \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (9 \, x - 1\right )} \sqrt {x} + \frac {172}{243} \, \sqrt {3} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - \frac {88}{27} \, \sqrt {3} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) \]
-2/9*sqrt(3*x^2 + 5*x + 2)*(9*x - 1)*sqrt(x) + 172/243*sqrt(3)*weierstrass PInverse(28/27, 80/729, x + 5/9) - 88/27*sqrt(3)*weierstrassZeta(28/27, 80 /729, weierstrassPInverse(28/27, 80/729, x + 5/9))
\[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}} \, dx=- \int \left (- \frac {2 \sqrt {3 x^{2} + 5 x + 2}}{\sqrt {x}}\right )\, dx - \int 5 \sqrt {x} \sqrt {3 x^{2} + 5 x + 2}\, dx \]
-Integral(-2*sqrt(3*x**2 + 5*x + 2)/sqrt(x), x) - Integral(5*sqrt(x)*sqrt( 3*x**2 + 5*x + 2), x)
\[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}} \, dx=\int { -\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (5 \, x - 2\right )}}{\sqrt {x}} \,d x } \]
\[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}} \, dx=\int { -\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (5 \, x - 2\right )}}{\sqrt {x}} \,d x } \]
Timed out. \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}} \, dx=-\int \frac {\left (5\,x-2\right )\,\sqrt {3\,x^2+5\,x+2}}{\sqrt {x}} \,d x \]