Integrand size = 22, antiderivative size = 40 \[ \int \frac {\sqrt {5-2 x}}{\sqrt {2+3 x+x^2}} \, dx=\frac {6 \sqrt {-2-3 x-x^2} E\left (\arcsin \left (\sqrt {2+x}\right )|\frac {2}{9}\right )}{\sqrt {2+3 x+x^2}} \] Output:
6*(-x^2-3*x-2)^(1/2)*EllipticE((2+x)^(1/2),1/3*2^(1/2))/(x^2+3*x+2)^(1/2)
Time = 30.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.78 \[ \int \frac {\sqrt {5-2 x}}{\sqrt {2+3 x+x^2}} \, dx=\frac {6 \sqrt {1+x} \sqrt {2+x} \left (E\left (\arcsin \left (\frac {\sqrt {5-2 x}}{\sqrt {7}}\right )|\frac {7}{9}\right )-\operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {5-2 x}}{\sqrt {7}}\right ),\frac {7}{9}\right )\right )}{\sqrt {2+3 x+x^2}} \] Input:
Integrate[Sqrt[5 - 2*x]/Sqrt[2 + 3*x + x^2],x]
Output:
(6*Sqrt[1 + x]*Sqrt[2 + x]*(EllipticE[ArcSin[Sqrt[5 - 2*x]/Sqrt[7]], 7/9] - EllipticF[ArcSin[Sqrt[5 - 2*x]/Sqrt[7]], 7/9]))/Sqrt[2 + 3*x + x^2]
Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1172, 27, 327}
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.
\(\displaystyle \int \frac {\sqrt {5-2 x}}{\sqrt {x^2+3 x+2}} \, dx\) |
\(\Big \downarrow \) 1172 |
\(\displaystyle \frac {6 \sqrt {-x^2-3 x-2} \int \frac {\sqrt {9-2 (x+2)}}{3 \sqrt {-x-1}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \sqrt {-x^2-3 x-2} \int \frac {\sqrt {9-2 (x+2)}}{\sqrt {-x-1}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {6 \sqrt {-x^2-3 x-2} E\left (\arcsin \left (\sqrt {x+2}\right )|\frac {2}{9}\right )}{\sqrt {x^2+3 x+2}}\) |
Input:
Int[Sqrt[5 - 2*x]/Sqrt[2 + 3*x + x^2],x]
Output:
(6*Sqrt[-2 - 3*x - x^2]*EllipticE[ArcSin[Sqrt[2 + x]], 2/9])/Sqrt[2 + 3*x + x^2]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ (Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) )], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sy mbol] :> Simp[2*Rt[b^2 - 4*a*c, 2]*(d + e*x)^m*(Sqrt[(-c)*((a + b*x + c*x^2 )/(b^2 - 4*a*c))]/(c*Sqrt[a + b*x + c*x^2]*(2*c*((d + e*x)/(2*c*d - b*e - e *Rt[b^2 - 4*a*c, 2])))^m)) Subst[Int[(1 + 2*e*Rt[b^2 - 4*a*c, 2]*(x^2/(2* c*d - b*e - e*Rt[b^2 - 4*a*c, 2])))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^ 2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], x] /; FreeQ[{a, b, c, d, e }, x] && EqQ[m^2, 1/4]
Time = 1.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.48
method | result | size |
default | \(\frac {\left (-\operatorname {EllipticF}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )+\operatorname {EllipticE}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )\right ) \sqrt {2 x +4}\, \sqrt {14 x +14}}{\sqrt {x^{2}+3 x +2}}\) | \(59\) |
elliptic | \(\frac {\sqrt {-\left (-5+2 x \right ) \left (x^{2}+3 x +2\right )}\, \left (-\frac {5 \sqrt {5-2 x}\, \sqrt {14 x +14}\, \sqrt {2 x +4}\, \operatorname {EllipticF}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )}{7 \sqrt {-2 x^{3}-x^{2}+11 x +10}}+\frac {2 \sqrt {5-2 x}\, \sqrt {14 x +14}\, \sqrt {2 x +4}\, \left (\frac {7 \operatorname {EllipticE}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )}{2}-\operatorname {EllipticF}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )\right )}{7 \sqrt {-2 x^{3}-x^{2}+11 x +10}}\right )}{\sqrt {5-2 x}\, \sqrt {x^{2}+3 x +2}}\) | \(167\) |
Input:
int((5-2*x)^(1/2)/(x^2+3*x+2)^(1/2),x,method=_RETURNVERBOSE)
Output:
(-EllipticF(1/3*(5-2*x)^(1/2),3/7*7^(1/2))+EllipticE(1/3*(5-2*x)^(1/2),3/7 *7^(1/2)))*(2*x+4)^(1/2)*(14*x+14)^(1/2)/(x^2+3*x+2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {5-2 x}}{\sqrt {2+3 x+x^2}} \, dx=-\frac {16}{3} \, \sqrt {-2} {\rm weierstrassPInverse}\left (\frac {67}{3}, \frac {440}{27}, x + \frac {1}{6}\right ) - 2 \, \sqrt {-2} {\rm weierstrassZeta}\left (\frac {67}{3}, \frac {440}{27}, {\rm weierstrassPInverse}\left (\frac {67}{3}, \frac {440}{27}, x + \frac {1}{6}\right )\right ) \] Input:
integrate((5-2*x)^(1/2)/(x^2+3*x+2)^(1/2),x, algorithm="fricas")
Output:
-16/3*sqrt(-2)*weierstrassPInverse(67/3, 440/27, x + 1/6) - 2*sqrt(-2)*wei erstrassZeta(67/3, 440/27, weierstrassPInverse(67/3, 440/27, x + 1/6))
\[ \int \frac {\sqrt {5-2 x}}{\sqrt {2+3 x+x^2}} \, dx=\int \frac {\sqrt {5 - 2 x}}{\sqrt {\left (x + 1\right ) \left (x + 2\right )}}\, dx \] Input:
integrate((5-2*x)**(1/2)/(x**2+3*x+2)**(1/2),x)
Output:
Integral(sqrt(5 - 2*x)/sqrt((x + 1)*(x + 2)), x)
\[ \int \frac {\sqrt {5-2 x}}{\sqrt {2+3 x+x^2}} \, dx=\int { \frac {\sqrt {-2 \, x + 5}}{\sqrt {x^{2} + 3 \, x + 2}} \,d x } \] Input:
integrate((5-2*x)^(1/2)/(x^2+3*x+2)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(-2*x + 5)/sqrt(x^2 + 3*x + 2), x)
\[ \int \frac {\sqrt {5-2 x}}{\sqrt {2+3 x+x^2}} \, dx=\int { \frac {\sqrt {-2 \, x + 5}}{\sqrt {x^{2} + 3 \, x + 2}} \,d x } \] Input:
integrate((5-2*x)^(1/2)/(x^2+3*x+2)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(-2*x + 5)/sqrt(x^2 + 3*x + 2), x)
Timed out. \[ \int \frac {\sqrt {5-2 x}}{\sqrt {2+3 x+x^2}} \, dx=\int \frac {\sqrt {5-2\,x}}{\sqrt {x^2+3\,x+2}} \,d x \] Input:
int((5 - 2*x)^(1/2)/(3*x + x^2 + 2)^(1/2),x)
Output:
int((5 - 2*x)^(1/2)/(3*x + x^2 + 2)^(1/2), x)
\[ \int \frac {\sqrt {5-2 x}}{\sqrt {2+3 x+x^2}} \, dx=\int \frac {\sqrt {-2 x +5}\, \sqrt {x^{2}+3 x +2}}{x^{2}+3 x +2}d x \] Input:
int((5-2*x)^(1/2)/(x^2+3*x+2)^(1/2),x)
Output:
int((sqrt( - 2*x + 5)*sqrt(x**2 + 3*x + 2))/(x**2 + 3*x + 2),x)