Integrand size = 25, antiderivative size = 179 \[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=-\frac {2 \sqrt {x} (30+37 x)}{3 \left (2+5 x+3 x^2\right )^{3/2}}-\frac {198 \sqrt {x} (2+3 x)}{\sqrt {2+5 x+3 x^2}}+\frac {2 \sqrt {x} (250+297 x)}{\sqrt {2+5 x+3 x^2}}+\frac {198 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}}-\frac {245 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}} \]
-2/3*(30+37*x)*x^(1/2)/(3*x^2+5*x+2)^(3/2)-198*(2+3*x)*x^(1/2)/(3*x^2+5*x+ 2)^(1/2)+2*(250+297*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)+198*(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)-245*(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)
Result contains complex when optimal does not.
Time = 21.24 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.92 \[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=-\frac {2 \left (1188+4470 x+5494 x^2+2205 x^3\right )}{3 \sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}}-\frac {198 i \sqrt {2+\frac {2}{x}} \sqrt {3+\frac {2}{x}} x E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )}{\sqrt {2+5 x+3 x^2}}-\frac {47 i \sqrt {2+\frac {2}{x}} \sqrt {3+\frac {2}{x}} x \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{\sqrt {2+5 x+3 x^2}} \]
(-2*(1188 + 4470*x + 5494*x^2 + 2205*x^3))/(3*Sqrt[x]*(2 + 5*x + 3*x^2)^(3 /2)) - ((198*I)*Sqrt[2 + 2/x]*Sqrt[3 + 2/x]*x*EllipticE[I*ArcSinh[Sqrt[2/3 ]/Sqrt[x]], 3/2])/Sqrt[2 + 5*x + 3*x^2] - ((47*I)*Sqrt[2 + 2/x]*Sqrt[3 + 2 /x]*x*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/Sqrt[2 + 5*x + 3*x^2]
Time = 0.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1234, 27, 1235, 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.
| \(\displaystyle \int \frac {(2-5 x) \sqrt {x}}{\left (3 x^2+5 x+2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1234 |
| \(\displaystyle -\frac {2}{3} \int -\frac {3 (10-37 x)}{2 \sqrt {x} \left (3 x^2+5 x+2\right )^{3/2}}dx-\frac {2 \sqrt {x} (37 x+30)}{3 \left (3 x^2+5 x+2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {10-37 x}{\sqrt {x} \left (3 x^2+5 x+2\right )^{3/2}}dx-\frac {2 \sqrt {x} (37 x+30)}{3 \left (3 x^2+5 x+2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle -\int \frac {297 x+245}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {2 \sqrt {x} (37 x+30)}{3 \left (3 x^2+5 x+2\right )^{3/2}}+\frac {2 \sqrt {x} (297 x+250)}{\sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1240 |
| \(\displaystyle -2 \int \frac {297 x+245}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}-\frac {2 \sqrt {x} (37 x+30)}{3 \left (3 x^2+5 x+2\right )^{3/2}}+\frac {2 \sqrt {x} (297 x+250)}{\sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle -2 \left (245 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+297 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )-\frac {2 \sqrt {x} (37 x+30)}{3 \left (3 x^2+5 x+2\right )^{3/2}}+\frac {2 \sqrt {x} (297 x+250)}{\sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1413 |
| \(\displaystyle -2 \left (297 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {245 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}\right )-\frac {2 \sqrt {x} (37 x+30)}{3 \left (3 x^2+5 x+2\right )^{3/2}}+\frac {2 \sqrt {x} (297 x+250)}{\sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1456 |
| \(\displaystyle -2 \left (\frac {245 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}+297 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )-\frac {2 \sqrt {x} (37 x+30)}{3 \left (3 x^2+5 x+2\right )^{3/2}}+\frac {2 \sqrt {x} (297 x+250)}{\sqrt {3 x^2+5 x+2}}\) |
(-2*Sqrt[x]*(30 + 37*x))/(3*(2 + 5*x + 3*x^2)^(3/2)) + (2*Sqrt[x]*(250 + 2 97*x))/Sqrt[2 + 5*x + 3*x^2] - 2*(297*((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[Sqr t[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2])) + (245*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(Sqrt[2]*Sqrt[2 + 5*x + 3*x^2]))
3.11.78.3.1 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*(a + b*x + c*x^2)^(p + 1)*( (f*b - 2*a*g + (2*c*f - b*g)*x)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g *(2*a*e*m + b*d*(2*p + 3)) - f*(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)* (m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1 ] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 *a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m *(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && (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 = 0.22 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.20
| method | result | size |
| elliptic | \(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (\frac {\left (-\frac {20}{9}-\frac {74 x}{27}\right ) \sqrt {3 x^{3}+5 x^{2}+2 x}}{\left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right )^{2}}-\frac {2 x \left (-\frac {250}{3}-99 x \right ) \sqrt {3}}{\sqrt {x \left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right )}}-\frac {245 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {99 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{\sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(214\) |
| default | \(\frac {\left (156 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x^{2}-297 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x^{2}+260 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x -495 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x +104 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-198 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+5346 x^{4}+13410 x^{3}+10990 x^{2}+2940 x \right ) \sqrt {3 x^{2}+5 x +2}}{3 \sqrt {x}\, \left (2+3 x \right )^{2} \left (1+x \right )^{2}}\) | \(297\) |
(x*(3*x^2+5*x+2))^(1/2)/x^(1/2)/(3*x^2+5*x+2)^(1/2)*((-20/9-74/27*x)*(3*x^ 3+5*x^2+2*x)^(1/2)/(x^2+5/3*x+2/3)^2-2*x*(-250/3-99*x)*3^(1/2)/(x*(x^2+5/3 *x+2/3))^(1/2)-245/3*(6*x+4)^(1/2)*(3+3*x)^(1/2)*(-6*x)^(1/2)/(3*x^3+5*x^2 +2*x)^(1/2)*EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))-99*(6*x+4)^(1/2)*(3+3*x )^(1/2)*(-6*x)^(1/2)/(3*x^3+5*x^2+2*x)^(1/2)*(1/3*EllipticE(1/2*(6*x+4)^(1 /2),I*2^(1/2))-EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.68 \[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (80 \, \sqrt {3} {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 297 \, \sqrt {3} {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - {\left (2673 \, x^{3} + 6705 \, x^{2} + 5495 \, x + 1470\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}\right )}}{3 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \]
-2/3*(80*sqrt(3)*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*weierstrassPInverse( 28/27, 80/729, x + 5/9) - 297*sqrt(3)*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4) *weierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729, x + 5/9 )) - (2673*x^3 + 6705*x^2 + 5495*x + 1470)*sqrt(3*x^2 + 5*x + 2)*sqrt(x))/ (9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)
\[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=- \int \left (- \frac {2 \sqrt {x}}{9 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 30 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 37 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 20 x \sqrt {3 x^{2} + 5 x + 2} + 4 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {5 x^{\frac {3}{2}}}{9 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 30 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 37 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 20 x \sqrt {3 x^{2} + 5 x + 2} + 4 \sqrt {3 x^{2} + 5 x + 2}}\, dx \]
-Integral(-2*sqrt(x)/(9*x**4*sqrt(3*x**2 + 5*x + 2) + 30*x**3*sqrt(3*x**2 + 5*x + 2) + 37*x**2*sqrt(3*x**2 + 5*x + 2) + 20*x*sqrt(3*x**2 + 5*x + 2) + 4*sqrt(3*x**2 + 5*x + 2)), x) - Integral(5*x**(3/2)/(9*x**4*sqrt(3*x**2 + 5*x + 2) + 30*x**3*sqrt(3*x**2 + 5*x + 2) + 37*x**2*sqrt(3*x**2 + 5*x + 2) + 20*x*sqrt(3*x**2 + 5*x + 2) + 4*sqrt(3*x**2 + 5*x + 2)), x)
\[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} \sqrt {x}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} \sqrt {x}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=-\int \frac {\sqrt {x}\,\left (5\,x-2\right )}{{\left (3\,x^2+5\,x+2\right )}^{5/2}} \,d x \]