Integrand size = 25, antiderivative size = 201 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{9/2}} \, dx=\frac {62 \sqrt {x} (2+3 x)}{21 \sqrt {2+5 x+3 x^2}}-\frac {4 (1-3 x) \sqrt {2+5 x+3 x^2}}{7 x^{7/2}}+\frac {43 \sqrt {2+5 x+3 x^2}}{21 x^{3/2}}-\frac {62 \sqrt {2+5 x+3 x^2}}{21 \sqrt {x}}-\frac {62 \sqrt {2} \sqrt {2+5 x+3 x^2} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{21 \sqrt {1+x} \sqrt {2+3 x}}+\frac {43 \sqrt {1+x} \sqrt {2+3 x} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{7 \sqrt {2} \sqrt {2+5 x+3 x^2}} \] Output:
62/21*x^(1/2)*(2+3*x)/(3*x^2+5*x+2)^(1/2)-4/7*(1-3*x)*(3*x^2+5*x+2)^(1/2)/ x^(7/2)+43/21*(3*x^2+5*x+2)^(1/2)/x^(3/2)-62/21*(3*x^2+5*x+2)^(1/2)/x^(1/2 )-62/21*2^(1/2)*(3*x^2+5*x+2)^(1/2)*EllipticE(x^(1/2)/(1+x)^(1/2),1/2*I*2^ (1/2))/(1+x)^(1/2)/(2+3*x)^(1/2)+43/14*2^(1/2)*(1+x)^(1/2)*(2+3*x)^(1/2)*I nverseJacobiAM(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 = 21.20 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.77 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{9/2}} \, dx=\frac {-48+24 x+460 x^2+646 x^3+258 x^4+124 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{9/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+5 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{9/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{42 x^{7/2} \sqrt {2+5 x+3 x^2}} \] Input:
Integrate[((2 - 5*x)*Sqrt[2 + 5*x + 3*x^2])/x^(9/2),x]
Output:
(-48 + 24*x + 460*x^2 + 646*x^3 + 258*x^4 + (124*I)*Sqrt[2]*Sqrt[1 + x^(-1 )]*Sqrt[3 + 2/x]*x^(9/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + (5 *I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(9/2)*EllipticF[I*ArcSinh[Sqr t[2/3]/Sqrt[x]], 3/2])/(42*x^(7/2)*Sqrt[2 + 5*x + 3*x^2])
Time = 0.41 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1229, 27, 1237, 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.
\(\displaystyle \int \frac {(2-5 x) \sqrt {3 x^2+5 x+2}}{x^{9/2}} \, dx\) |
\(\Big \downarrow \) 1229 |
\(\displaystyle -\frac {1}{35} \int \frac {5 (51 x+43)}{x^{5/2} \sqrt {3 x^2+5 x+2}}dx-\frac {4 \sqrt {3 x^2+5 x+2} (1-3 x)}{7 x^{7/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{7} \int \frac {51 x+43}{x^{5/2} \sqrt {3 x^2+5 x+2}}dx-\frac {4 \sqrt {3 x^2+5 x+2} (1-3 x)}{7 x^{7/2}}\) |
\(\Big \downarrow \) 1237 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \int \frac {129 x+124}{2 x^{3/2} \sqrt {3 x^2+5 x+2}}dx+\frac {43 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )-\frac {4 (1-3 x) \sqrt {3 x^2+5 x+2}}{7 x^{7/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{6} \int \frac {129 x+124}{x^{3/2} \sqrt {3 x^2+5 x+2}}dx+\frac {43 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )-\frac {4 (1-3 x) \sqrt {3 x^2+5 x+2}}{7 x^{7/2}}\) |
\(\Big \downarrow \) 1237 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{6} \left (-\int -\frac {3 (62 x+43)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {124 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {43 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )-\frac {4 (1-3 x) \sqrt {3 x^2+5 x+2}}{7 x^{7/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{6} \left (3 \int \frac {62 x+43}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {124 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {43 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )-\frac {4 (1-3 x) \sqrt {3 x^2+5 x+2}}{7 x^{7/2}}\) |
\(\Big \downarrow \) 1240 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{6} \left (6 \int \frac {62 x+43}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}-\frac {124 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {43 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )-\frac {4 (1-3 x) \sqrt {3 x^2+5 x+2}}{7 x^{7/2}}\) |
\(\Big \downarrow \) 1503 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{6} \left (6 \left (43 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+62 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )-\frac {124 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {43 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )-\frac {4 (1-3 x) \sqrt {3 x^2+5 x+2}}{7 x^{7/2}}\) |
\(\Big \downarrow \) 1413 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{6} \left (6 \left (62 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {43 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}\right )-\frac {124 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {43 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )-\frac {4 (1-3 x) \sqrt {3 x^2+5 x+2}}{7 x^{7/2}}\) |
\(\Big \downarrow \) 1456 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{6} \left (6 \left (\frac {43 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}+62 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )-\frac {124 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )+\frac {43 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )-\frac {4 (1-3 x) \sqrt {3 x^2+5 x+2}}{7 x^{7/2}}\) |
Input:
Int[((2 - 5*x)*Sqrt[2 + 5*x + 3*x^2])/x^(9/2),x]
Output:
(-4*(1 - 3*x)*Sqrt[2 + 5*x + 3*x^2])/(7*x^(7/2)) + ((43*Sqrt[2 + 5*x + 3*x ^2])/(3*x^(3/2)) + ((-124*Sqrt[2 + 5*x + 3*x^2])/Sqrt[x] + 6*(62*((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])) + (43* (1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(Sqrt[2] *Sqrt[2 + 5*x + 3*x^2])))/6)/7
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))*((a + b*x + c*x^2 )^p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)))*((d*g - e*f*(m + 2))*(c* d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x), x] - Simp[p/(e^2*(m + 1 )*(m + 2)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2 )^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1) - c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c *(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1) - b*(d*g*( m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g }, x] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] && !ILtQ[m + 2*p + 3, 0]
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 = 0.95 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.64
method | result | size |
default | \(\frac {62 \operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, x^{3}-57 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x^{3}-1116 x^{5}-1086 x^{4}+1194 x^{3}+1380 x^{2}+72 x -144}{126 \sqrt {3 x^{2}+5 x +2}\, x^{\frac {7}{2}}}\) | \(129\) |
elliptic | \(\frac {\sqrt {\left (3 x^{2}+5 x +2\right ) x}\, \left (-\frac {4 \sqrt {3 x^{3}+5 x^{2}+2 x}}{7 x^{4}}+\frac {12 \sqrt {3 x^{3}+5 x^{2}+2 x}}{7 x^{3}}+\frac {43 \sqrt {3 x^{3}+5 x^{2}+2 x}}{21 x^{2}}-\frac {62 \left (3 x^{2}+5 x +2\right )}{21 \sqrt {\left (3 x^{2}+5 x +2\right ) x}}+\frac {43 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{42 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {31 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {\operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-\operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{21 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(248\) |
Input:
int((2-5*x)*(3*x^2+5*x+2)^(1/2)/x^(9/2),x,method=_RETURNVERBOSE)
Output:
1/126*(62*EllipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))*(6*x+4)^(1/2)*(3+3*x)^(1/ 2)*6^(1/2)*(-x)^(1/2)*x^3-57*(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))*x^3-1116*x^5-1086*x^4+1194*x^3+13 80*x^2+72*x-144)/(3*x^2+5*x+2)^(1/2)/x^(7/2)
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.34 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{9/2}} \, dx=\frac {77 \, \sqrt {3} x^{4} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 558 \, \sqrt {3} x^{4} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - 9 \, {\left (62 \, x^{3} - 43 \, x^{2} - 36 \, x + 12\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}}{189 \, x^{4}} \] Input:
integrate((2-5*x)*(3*x^2+5*x+2)^(1/2)/x^(9/2),x, algorithm="fricas")
Output:
1/189*(77*sqrt(3)*x^4*weierstrassPInverse(28/27, 80/729, x + 5/9) - 558*sq rt(3)*x^4*weierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729 , x + 5/9)) - 9*(62*x^3 - 43*x^2 - 36*x + 12)*sqrt(3*x^2 + 5*x + 2)*sqrt(x ))/x^4
\[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{9/2}} \, dx=- \int \left (- \frac {2 \sqrt {3 x^{2} + 5 x + 2}}{x^{\frac {9}{2}}}\right )\, dx - \int \frac {5 \sqrt {3 x^{2} + 5 x + 2}}{x^{\frac {7}{2}}}\, dx \] Input:
integrate((2-5*x)*(3*x**2+5*x+2)**(1/2)/x**(9/2),x)
Output:
-Integral(-2*sqrt(3*x**2 + 5*x + 2)/x**(9/2), x) - Integral(5*sqrt(3*x**2 + 5*x + 2)/x**(7/2), x)
\[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{9/2}} \, dx=\int { -\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (5 \, x - 2\right )}}{x^{\frac {9}{2}}} \,d x } \] Input:
integrate((2-5*x)*(3*x^2+5*x+2)^(1/2)/x^(9/2),x, algorithm="maxima")
Output:
-integrate(sqrt(3*x^2 + 5*x + 2)*(5*x - 2)/x^(9/2), x)
\[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{9/2}} \, dx=\int { -\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (5 \, x - 2\right )}}{x^{\frac {9}{2}}} \,d x } \] Input:
integrate((2-5*x)*(3*x^2+5*x+2)^(1/2)/x^(9/2),x, algorithm="giac")
Output:
integrate(-sqrt(3*x^2 + 5*x + 2)*(5*x - 2)/x^(9/2), x)
Timed out. \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{9/2}} \, dx=\int -\frac {\left (5\,x-2\right )\,\sqrt {3\,x^2+5\,x+2}}{x^{9/2}} \,d x \] Input:
int(-((5*x - 2)*(5*x + 3*x^2 + 2)^(1/2))/x^(9/2),x)
Output:
int(-((5*x - 2)*(5*x + 3*x^2 + 2)^(1/2))/x^(9/2), x)
\[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{9/2}} \, dx=\frac {70 \sqrt {3 x^{2}+5 x +2}\, x -12 \sqrt {3 x^{2}+5 x +2}+170 \sqrt {x}\, \left (\int \frac {\sqrt {3 x^{2}+5 x +2}}{3 \sqrt {x}\, x^{5}+5 \sqrt {x}\, x^{4}+2 \sqrt {x}\, x^{3}}d x \right ) x^{3}+85 \sqrt {x}\, \left (\int \frac {\sqrt {3 x^{2}+5 x +2}}{3 \sqrt {x}\, x^{4}+5 \sqrt {x}\, x^{3}+2 \sqrt {x}\, x^{2}}d x \right ) x^{3}+126 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{3 x^{5}+5 x^{4}+2 x^{3}}d x \right ) x^{3}}{21 \sqrt {x}\, x^{3}} \] Input:
int((2-5*x)*(3*x^2+5*x+2)^(1/2)/x^(9/2),x)
Output:
(70*sqrt(3*x**2 + 5*x + 2)*x - 12*sqrt(3*x**2 + 5*x + 2) + 170*sqrt(x)*int (sqrt(3*x**2 + 5*x + 2)/(3*sqrt(x)*x**5 + 5*sqrt(x)*x**4 + 2*sqrt(x)*x**3) ,x)*x**3 + 85*sqrt(x)*int(sqrt(3*x**2 + 5*x + 2)/(3*sqrt(x)*x**4 + 5*sqrt( x)*x**3 + 2*sqrt(x)*x**2),x)*x**3 + 126*sqrt(x)*int((sqrt(x)*sqrt(3*x**2 + 5*x + 2))/(3*x**5 + 5*x**4 + 2*x**3),x)*x**3)/(21*sqrt(x)*x**3)