Integrand size = 25, antiderivative size = 154 \[ \int \frac {(2-5 x) \sqrt {x}}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {136 \sqrt {x} (2+3 x)}{27 \sqrt {2+5 x+3 x^2}}-\frac {10}{9} \sqrt {x} \sqrt {2+5 x+3 x^2}-\frac {136 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{27 \sqrt {2+5 x+3 x^2}}+\frac {10 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{9 \sqrt {2+5 x+3 x^2}} \]
136/27*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)-136/27*(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)+10/9*(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)-10/9*x^(1/2)*(3*x^2+5*x+2)^(1/2)
Result contains complex when optimal does not.
Time = 21.15 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97 \[ \int \frac {(2-5 x) \sqrt {x}}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {272+620 x+258 x^2-90 x^3+136 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-106 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{27 \sqrt {x} \sqrt {2+5 x+3 x^2}} \]
(272 + 620*x + 258*x^2 - 90*x^3 + (136*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (106*I)*Sqrt [2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/S qrt[x]], 3/2])/(27*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])
Time = 0.31 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1236, 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.
\(\displaystyle \int \frac {(2-5 x) \sqrt {x}}{\sqrt {3 x^2+5 x+2}} \, dx\) |
\(\Big \downarrow \) 1236 |
\(\displaystyle \frac {2}{9} \int \frac {34 x+5}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {10}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1240 |
\(\displaystyle \frac {4}{9} \int \frac {34 x+5}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}-\frac {10}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1503 |
\(\displaystyle \frac {4}{9} \left (5 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+34 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )-\frac {10}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1413 |
\(\displaystyle \frac {4}{9} \left (34 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {5 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}\right )-\frac {10}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1456 |
\(\displaystyle \frac {4}{9} \left (\frac {5 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}+34 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )-\frac {10}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\) |
(-10*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])/9 + (4*(34*((Sqrt[x]*(2 + 3*x))/(3*Sqr t[2 + 5*x + 3*x^2]) - (Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[A rcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2])) + (5*(1 + x)*Sqrt[(2 + 3 *x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(Sqrt[2]*Sqrt[2 + 5*x + 3*x ^2])))/9
3.11.59.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[1/(c*(m + 2*p + 2)) Int[(d + e*x)^(m - 1 )*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m *(c*e*f + c*d*g - b*e*g) + e*(p + 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[ {a, b, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && EqQ[f, 0])
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 = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.73
method | result | size |
default | \(-\frac {2 \left (87 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-34 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+135 x^{3}+225 x^{2}+90 x \right )}{81 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(112\) |
elliptic | \(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {10 \sqrt {3 x^{3}+5 x^{2}+2 x}}{9}+\frac {10 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{27 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {68 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{27 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(177\) |
risch | \(-\frac {10 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{9}-\frac {\left (-\frac {10 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{27 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {68 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{27 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right ) \sqrt {x \left (3 x^{2}+5 x +2\right )}}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(178\) |
-2/81/x^(1/2)/(3*x^2+5*x+2)^(1/2)*(87*(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))-34*(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))+135*x^3+2 25*x^2+90*x)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.28 \[ \int \frac {(2-5 x) \sqrt {x}}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {500}{243} \, \sqrt {3} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - \frac {136}{27} \, \sqrt {3} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - \frac {10}{9} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x} \]
-500/243*sqrt(3)*weierstrassPInverse(28/27, 80/729, x + 5/9) - 136/27*sqrt (3)*weierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729, x + 5/9)) - 10/9*sqrt(3*x^2 + 5*x + 2)*sqrt(x)
\[ \int \frac {(2-5 x) \sqrt {x}}{\sqrt {2+5 x+3 x^2}} \, dx=- \int \left (- \frac {2 \sqrt {x}}{\sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {5 x^{\frac {3}{2}}}{\sqrt {3 x^{2} + 5 x + 2}}\, dx \]
-Integral(-2*sqrt(x)/sqrt(3*x**2 + 5*x + 2), x) - Integral(5*x**(3/2)/sqrt (3*x**2 + 5*x + 2), x)
\[ \int \frac {(2-5 x) \sqrt {x}}{\sqrt {2+5 x+3 x^2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} \sqrt {x}}{\sqrt {3 \, x^{2} + 5 \, x + 2}} \,d x } \]
\[ \int \frac {(2-5 x) \sqrt {x}}{\sqrt {2+5 x+3 x^2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} \sqrt {x}}{\sqrt {3 \, x^{2} + 5 \, x + 2}} \,d x } \]
Timed out. \[ \int \frac {(2-5 x) \sqrt {x}}{\sqrt {2+5 x+3 x^2}} \, dx=-\int \frac {\sqrt {x}\,\left (5\,x-2\right )}{\sqrt {3\,x^2+5\,x+2}} \,d x \]