Integrand size = 25, antiderivative size = 172 \[ \int \frac {2-5 x}{x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {39 \sqrt {x} (2+3 x)}{\sqrt {2+5 x+3 x^2}}+\frac {2 (38+45 x)}{\sqrt {x} \sqrt {2+5 x+3 x^2}}-\frac {39 \sqrt {2+5 x+3 x^2}}{\sqrt {x}}-\frac {39 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}}+\frac {45 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}} \]
2*(38+45*x)/x^(1/2)/(3*x^2+5*x+2)^(1/2)+39*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^( 1/2)-39*(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)+45*(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)-39*(3*x^2+5*x+2)^(1/2)/x^(1/2)
Result contains complex when optimal does not.
Time = 21.16 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.80 \[ \int \frac {2-5 x}{x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {76+90 x+39 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+6 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \]
(76 + 90*x + (39*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*Ellipti cE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + (6*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqr t[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(Sqrt[x]* Sqrt[2 + 5*x + 3*x^2])
Time = 0.38 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {1235, 27, 1237, 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.
| \(\displaystyle \int \frac {2-5 x}{x^{3/2} \left (3 x^2+5 x+2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle \frac {2 (45 x+38)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}-\int -\frac {3 (15 x+13)}{x^{3/2} \sqrt {3 x^2+5 x+2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 3 \int \frac {15 x+13}{x^{3/2} \sqrt {3 x^2+5 x+2}}dx+\frac {2 (45 x+38)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1237 |
| \(\displaystyle 3 \left (-\int -\frac {3 (13 x+10)}{2 \sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {13 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {2 (45 x+38)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 3 \left (\frac {3}{2} \int \frac {13 x+10}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {13 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {2 (45 x+38)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1240 |
| \(\displaystyle 3 \left (3 \int \frac {13 x+10}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}-\frac {13 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {2 (45 x+38)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle 3 \left (3 \left (10 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+13 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )-\frac {13 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {2 (45 x+38)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1413 |
| \(\displaystyle 3 \left (3 \left (13 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {5 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}\right )-\frac {13 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {2 (45 x+38)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1456 |
| \(\displaystyle 3 \left (3 \left (\frac {5 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+13 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )-\frac {13 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {2 (45 x+38)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}\) |
(2*(38 + 45*x))/(Sqrt[x]*Sqrt[2 + 5*x + 3*x^2]) + 3*((-13*Sqrt[2 + 5*x + 3 *x^2])/Sqrt[x] + 3*(13*((Sqrt[x]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (S qrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/( 3*Sqrt[2 + 5*x + 3*x^2])) + (5*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*Ell ipticF[ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2]))
3.11.69.3.1 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)*(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[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) *(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ (c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -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 = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.63
| method | result | size |
| default | \(-\frac {9 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-13 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+234 x^{2}+210 x +4}{2 \sqrt {3 x^{2}+5 x +2}\, \sqrt {x}}\) | \(108\) |
| elliptic | \(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {3 x^{2}+5 x +2}{\sqrt {x \left (3 x^{2}+5 x +2\right )}}-\frac {2 x \left (\frac {50}{3}+19 x \right ) \sqrt {3}}{\sqrt {x \left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right )}}+\frac {15 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{\sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {39 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{2 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(208\) |
-1/2*(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))-13*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*Elli pticE(1/2*(6*x+4)^(1/2),I*2^(1/2))+234*x^2+210*x+4)/(3*x^2+5*x+2)^(1/2)/x^ (1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.58 \[ \int \frac {2-5 x}{x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {25 \, \sqrt {3} {\left (3 \, x^{3} + 5 \, x^{2} + 2 \, x\right )} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 117 \, \sqrt {3} {\left (3 \, x^{3} + 5 \, x^{2} + 2 \, x\right )} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - 3 \, {\left (117 \, x^{2} + 105 \, x + 2\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}}{3 \, {\left (3 \, x^{3} + 5 \, x^{2} + 2 \, x\right )}} \]
1/3*(25*sqrt(3)*(3*x^3 + 5*x^2 + 2*x)*weierstrassPInverse(28/27, 80/729, x + 5/9) - 117*sqrt(3)*(3*x^3 + 5*x^2 + 2*x)*weierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729, x + 5/9)) - 3*(117*x^2 + 105*x + 2)*sq rt(3*x^2 + 5*x + 2)*sqrt(x))/(3*x^3 + 5*x^2 + 2*x)
\[ \int \frac {2-5 x}{x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=- \int \frac {5}{3 x^{\frac {5}{2}} \sqrt {3 x^{2} + 5 x + 2} + 5 x^{\frac {3}{2}} \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {x} \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {2}{3 x^{\frac {7}{2}} \sqrt {3 x^{2} + 5 x + 2} + 5 x^{\frac {5}{2}} \sqrt {3 x^{2} + 5 x + 2} + 2 x^{\frac {3}{2}} \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx \]
-Integral(5/(3*x**(5/2)*sqrt(3*x**2 + 5*x + 2) + 5*x**(3/2)*sqrt(3*x**2 + 5*x + 2) + 2*sqrt(x)*sqrt(3*x**2 + 5*x + 2)), x) - Integral(-2/(3*x**(7/2) *sqrt(3*x**2 + 5*x + 2) + 5*x**(5/2)*sqrt(3*x**2 + 5*x + 2) + 2*x**(3/2)*s qrt(3*x**2 + 5*x + 2)), x)
\[ \int \frac {2-5 x}{x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {5 \, x - 2}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} x^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {2-5 x}{x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {5 \, x - 2}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} x^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {2-5 x}{x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int -\frac {5\,x-2}{x^{3/2}\,{\left (3\,x^2+5\,x+2\right )}^{3/2}} \,d x \]