Integrand size = 25, antiderivative size = 173 \[ \int \frac {(2-5 x) x^{3/2}}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {412 \sqrt {x} (2+3 x)}{81 \sqrt {2+5 x+3 x^2}}+\frac {52}{27} \sqrt {x} \sqrt {2+5 x+3 x^2}-\frac {2}{3} x^{3/2} \sqrt {2+5 x+3 x^2}+\frac {412 \sqrt {2} \sqrt {2+5 x+3 x^2} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{81 \sqrt {1+x} \sqrt {2+3 x}}-\frac {52 \sqrt {2} \sqrt {1+x} \sqrt {2+3 x} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{27 \sqrt {2+5 x+3 x^2}} \] Output:
-412/81*x^(1/2)*(2+3*x)/(3*x^2+5*x+2)^(1/2)+52/27*x^(1/2)*(3*x^2+5*x+2)^(1 /2)-2/3*x^(3/2)*(3*x^2+5*x+2)^(1/2)+412/81*2^(1/2)*(3*x^2+5*x+2)^(1/2)*Ell ipticE(x^(1/2)/(1+x)^(1/2),1/2*I*2^(1/2))/(1+x)^(1/2)/(2+3*x)^(1/2)-52/27* 2^(1/2)*(1+x)^(1/2)*(2+3*x)^(1/2)*InverseJacobiAM(arctan(x^(1/2)),1/2*I*2^ (1/2))/(3*x^2+5*x+2)^(1/2)
Result contains complex when optimal does not.
Time = 21.21 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.91 \[ \int \frac {(2-5 x) x^{3/2}}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {-2 \left (412+874 x+282 x^2-99 x^3+81 x^4\right )-412 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+256 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{81 \sqrt {x} \sqrt {2+5 x+3 x^2}} \] Input:
Integrate[((2 - 5*x)*x^(3/2))/Sqrt[2 + 5*x + 3*x^2],x]
Output:
(-2*(412 + 874*x + 282*x^2 - 99*x^3 + 81*x^4) - (412*I)*Sqrt[2]*Sqrt[1 + x ^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + (256*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSi nh[Sqrt[2/3]/Sqrt[x]], 3/2])/(81*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])
Time = 0.37 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {1236, 27, 1236, 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.
\(\displaystyle \int \frac {(2-5 x) x^{3/2}}{\sqrt {3 x^2+5 x+2}} \, dx\) |
\(\Big \downarrow \) 1236 |
\(\displaystyle \frac {2}{15} \int \frac {5 \sqrt {x} (13 x+3)}{\sqrt {3 x^2+5 x+2}}dx-\frac {2}{3} x^{3/2} \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{3} \int \frac {\sqrt {x} (13 x+3)}{\sqrt {3 x^2+5 x+2}}dx-\frac {2}{3} x^{3/2} \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1236 |
\(\displaystyle \frac {2}{3} \left (\frac {2}{9} \int -\frac {103 x+26}{2 \sqrt {x} \sqrt {3 x^2+5 x+2}}dx+\frac {26}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\right )-\frac {2}{3} x^{3/2} \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{3} \left (\frac {26}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}-\frac {1}{9} \int \frac {103 x+26}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx\right )-\frac {2}{3} x^{3/2} \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1240 |
\(\displaystyle \frac {2}{3} \left (\frac {26}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}-\frac {2}{9} \int \frac {103 x+26}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )-\frac {2}{3} x^{3/2} \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1503 |
\(\displaystyle \frac {2}{3} \left (\frac {26}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}-\frac {2}{9} \left (26 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+103 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\right )-\frac {2}{3} x^{3/2} \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1413 |
\(\displaystyle \frac {2}{3} \left (\frac {26}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}-\frac {2}{9} \left (103 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {13 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}\right )\right )-\frac {2}{3} x^{3/2} \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1456 |
\(\displaystyle \frac {2}{3} \left (\frac {26}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}-\frac {2}{9} \left (\frac {13 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+103 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )\right )-\frac {2}{3} x^{3/2} \sqrt {3 x^2+5 x+2}\) |
Input:
Int[((2 - 5*x)*x^(3/2))/Sqrt[2 + 5*x + 3*x^2],x]
Output:
(-2*x^(3/2)*Sqrt[2 + 5*x + 3*x^2])/3 + (2*((26*Sqrt[x]*Sqrt[2 + 5*x + 3*x^ 2])/9 - (2*(103*((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])) + (13*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF [ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2]))/9))/3
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[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 = 1.15 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.68
method | result | size |
default | \(\frac {\frac {154 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{81}-\frac {206 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{243}-2 x^{4}+\frac {22 x^{3}}{9}+\frac {224 x^{2}}{27}+\frac {104 x}{27}}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(117\) |
risch | \(-\frac {2 \left (-26+9 x \right ) \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{27}-\frac {\left (\frac {52 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{81 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {206 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {\operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-\operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{81 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right ) \sqrt {\left (3 x^{2}+5 x +2\right ) x}}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(183\) |
elliptic | \(\frac {\sqrt {\left (3 x^{2}+5 x +2\right ) x}\, \left (-\frac {2 x \sqrt {3 x^{3}+5 x^{2}+2 x}}{3}+\frac {52 \sqrt {3 x^{3}+5 x^{2}+2 x}}{27}-\frac {52 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{81 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {206 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {\operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-\operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{81 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(196\) |
Input:
int((2-5*x)*x^(3/2)/(3*x^2+5*x+2)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/243/x^(1/2)/(3*x^2+5*x+2)^(1/2)*(231*(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))-103*(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))-243*x^4 +297*x^3+1008*x^2+468*x)
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.28 \[ \int \frac {(2-5 x) x^{3/2}}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {2}{27} \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (9 \, x - 26\right )} \sqrt {x} + \frac {1124}{729} \, \sqrt {3} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) + \frac {412}{81} \, \sqrt {3} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) \] Input:
integrate((2-5*x)*x^(3/2)/(3*x^2+5*x+2)^(1/2),x, algorithm="fricas")
Output:
-2/27*sqrt(3*x^2 + 5*x + 2)*(9*x - 26)*sqrt(x) + 1124/729*sqrt(3)*weierstr assPInverse(28/27, 80/729, x + 5/9) + 412/81*sqrt(3)*weierstrassZeta(28/27 , 80/729, weierstrassPInverse(28/27, 80/729, x + 5/9))
\[ \int \frac {(2-5 x) x^{3/2}}{\sqrt {2+5 x+3 x^2}} \, dx=- \int \left (- \frac {2 x^{\frac {3}{2}}}{\sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {5 x^{\frac {5}{2}}}{\sqrt {3 x^{2} + 5 x + 2}}\, dx \] Input:
integrate((2-5*x)*x**(3/2)/(3*x**2+5*x+2)**(1/2),x)
Output:
-Integral(-2*x**(3/2)/sqrt(3*x**2 + 5*x + 2), x) - Integral(5*x**(5/2)/sqr t(3*x**2 + 5*x + 2), x)
\[ \int \frac {(2-5 x) x^{3/2}}{\sqrt {2+5 x+3 x^2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} x^{\frac {3}{2}}}{\sqrt {3 \, x^{2} + 5 \, x + 2}} \,d x } \] Input:
integrate((2-5*x)*x^(3/2)/(3*x^2+5*x+2)^(1/2),x, algorithm="maxima")
Output:
-integrate((5*x - 2)*x^(3/2)/sqrt(3*x^2 + 5*x + 2), x)
\[ \int \frac {(2-5 x) x^{3/2}}{\sqrt {2+5 x+3 x^2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} x^{\frac {3}{2}}}{\sqrt {3 \, x^{2} + 5 \, x + 2}} \,d x } \] Input:
integrate((2-5*x)*x^(3/2)/(3*x^2+5*x+2)^(1/2),x, algorithm="giac")
Output:
integrate(-(5*x - 2)*x^(3/2)/sqrt(3*x^2 + 5*x + 2), x)
Timed out. \[ \int \frac {(2-5 x) x^{3/2}}{\sqrt {2+5 x+3 x^2}} \, dx=-\int \frac {x^{3/2}\,\left (5\,x-2\right )}{\sqrt {3\,x^2+5\,x+2}} \,d x \] Input:
int(-(x^(3/2)*(5*x - 2))/(5*x + 3*x^2 + 2)^(1/2),x)
Output:
-int((x^(3/2)*(5*x - 2))/(5*x + 3*x^2 + 2)^(1/2), x)
\[ \int \frac {(2-5 x) x^{3/2}}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {2 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}\, x}{3}+\frac {2 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{5}+\frac {103 \left (\int \frac {\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}\, x}{3 x^{2}+5 x +2}d x \right )}{15}-\frac {2 \left (\int \frac {\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{3 x^{3}+5 x^{2}+2 x}d x \right )}{5} \] Input:
int((2-5*x)*x^(3/2)/(3*x^2+5*x+2)^(1/2),x)
Output:
( - 10*sqrt(x)*sqrt(3*x**2 + 5*x + 2)*x + 6*sqrt(x)*sqrt(3*x**2 + 5*x + 2) + 103*int((sqrt(x)*sqrt(3*x**2 + 5*x + 2)*x)/(3*x**2 + 5*x + 2),x) - 6*in t((sqrt(x)*sqrt(3*x**2 + 5*x + 2))/(3*x**3 + 5*x**2 + 2*x),x))/15