Integrand size = 25, antiderivative size = 159 \[ \int \frac {(2-5 x) x^{3/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {200 \sqrt {x} (2+3 x)}{9 \sqrt {2+5 x+3 x^2}}+\frac {2 \sqrt {x} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}+\frac {200 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{9 \sqrt {2+5 x+3 x^2}}-\frac {74 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{3 \sqrt {2+5 x+3 x^2}} \]
-200/9*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)+2/3*(74+95*x)*x^(1/2)/(3*x^2+5* x+2)^(1/2)+200/9*(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)-74/3*(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 = 21.15 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.91 \[ \int \frac {(2-5 x) x^{3/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {-400-556 x-30 x^2-200 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-22 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{9 \sqrt {x} \sqrt {2+5 x+3 x^2}} \]
(-400 - 556*x - 30*x^2 - (200*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^ (3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (22*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3 /2])/(9*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 = {1233, 25, 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.
| \(\displaystyle \int \frac {(2-5 x) x^{3/2}}{\left (3 x^2+5 x+2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1233 |
| \(\displaystyle \frac {2}{3} \int -\frac {50 x+37}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx+\frac {2 \sqrt {x} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \sqrt {x} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {2}{3} \int \frac {50 x+37}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx\) |
| \(\Big \downarrow \) 1240 |
| \(\displaystyle \frac {2 \sqrt {x} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {4}{3} \int \frac {50 x+37}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle \frac {2 \sqrt {x} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {4}{3} \left (37 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+50 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\) |
| \(\Big \downarrow \) 1413 |
| \(\displaystyle \frac {2 \sqrt {x} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {4}{3} \left (50 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {37 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}\right )\) |
| \(\Big \downarrow \) 1456 |
| \(\displaystyle \frac {2 \sqrt {x} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {4}{3} \left (\frac {37 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}+50 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )\) |
(2*Sqrt[x]*(74 + 95*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (4*(50*((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])) + (37*(1 + x )*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(Sqrt[2]*Sqrt[ 2 + 5*x + 3*x^2])))/3
3.11.66.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2) ^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c *(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f *(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | | !ILtQ[m + 2*p + 3, 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 = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.67
| method | result | size |
| default | \(\frac {\frac {26 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{9}-\frac {100 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{27}+\frac {190 x^{2}}{3}+\frac {148 x}{3}}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(107\) |
| elliptic | \(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {2 x \left (-\frac {74}{9}-\frac {95 x}{9}\right ) \sqrt {3}}{\sqrt {x \left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right )}}-\frac {74 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{9 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {100 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{9 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(182\) |
2/27*(39*(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))-50*(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))+855*x^2+666*x)/x^(1/2)/(3*x^2+5*x+2)^( 1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.52 \[ \int \frac {(2-5 x) x^{3/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (166 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 900 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - 27 \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (95 \, x + 74\right )} \sqrt {x}\right )}}{81 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \]
-2/81*(166*sqrt(3)*(3*x^2 + 5*x + 2)*weierstrassPInverse(28/27, 80/729, x + 5/9) - 900*sqrt(3)*(3*x^2 + 5*x + 2)*weierstrassZeta(28/27, 80/729, weie rstrassPInverse(28/27, 80/729, x + 5/9)) - 27*sqrt(3*x^2 + 5*x + 2)*(95*x + 74)*sqrt(x))/(3*x^2 + 5*x + 2)
\[ \int \frac {(2-5 x) x^{3/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=- \int \left (- \frac {2 x^{\frac {3}{2}}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {5 x^{\frac {5}{2}}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\, dx \]
-Integral(-2*x**(3/2)/(3*x**2*sqrt(3*x**2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5 *x + 2) + 2*sqrt(3*x**2 + 5*x + 2)), x) - Integral(5*x**(5/2)/(3*x**2*sqrt (3*x**2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5*x + 2) + 2*sqrt(3*x**2 + 5*x + 2) ), x)
\[ \int \frac {(2-5 x) x^{3/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} x^{\frac {3}{2}}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(2-5 x) x^{3/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} x^{\frac {3}{2}}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(2-5 x) x^{3/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\int \frac {x^{3/2}\,\left (5\,x-2\right )}{{\left (3\,x^2+5\,x+2\right )}^{3/2}} \,d x \]