Integrand size = 25, antiderivative size = 180 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{7/2}} \, dx=-\frac {139 \sqrt {x} (2+3 x)}{15 \sqrt {2+5 x+3 x^2}}-\frac {4 (3-10 x) \sqrt {2+5 x+3 x^2}}{15 x^{5/2}}+\frac {139 \sqrt {2+5 x+3 x^2}}{15 \sqrt {x}}+\frac {139 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{15 \sqrt {2+5 x+3 x^2}}-\frac {11 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}} \]
-139/15*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)+139/15*(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)-11*(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)-4/15*(3-10*x)*(3*x^2+5*x+2)^(1/2)/x^(5/2)+139/15*(3*x^2+5*x+2)^(1/2)/x ^(1/2)
Result contains complex when optimal does not.
Time = 21.15 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.85 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{7/2}} \, dx=\frac {4 \left (-6+5 x+41 x^2+30 x^3\right )-139 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{7/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-26 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{7/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{15 x^{5/2} \sqrt {2+5 x+3 x^2}} \]
(4*(-6 + 5*x + 41*x^2 + 30*x^3) - (139*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(7/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (26*I)*Sqrt[ 2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(7/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sq rt[x]], 3/2])/(15*x^(5/2)*Sqrt[2 + 5*x + 3*x^2])
Time = 0.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1229, 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.
| \(\displaystyle \int \frac {(2-5 x) \sqrt {3 x^2+5 x+2}}{x^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1229 |
| \(\displaystyle -\frac {1}{15} \int \frac {165 x+139}{x^{3/2} \sqrt {3 x^2+5 x+2}}dx-\frac {4 \sqrt {3 x^2+5 x+2} (3-10 x)}{15 x^{5/2}}\) |
| \(\Big \downarrow \) 1237 |
| \(\displaystyle \frac {1}{15} \left (\int -\frac {3 (139 x+110)}{2 \sqrt {x} \sqrt {3 x^2+5 x+2}}dx+\frac {139 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )-\frac {4 (3-10 x) \sqrt {3 x^2+5 x+2}}{15 x^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{15} \left (\frac {139 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {3}{2} \int \frac {139 x+110}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx\right )-\frac {4 (3-10 x) \sqrt {3 x^2+5 x+2}}{15 x^{5/2}}\) |
| \(\Big \downarrow \) 1240 |
| \(\displaystyle \frac {1}{15} \left (\frac {139 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-3 \int \frac {139 x+110}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )-\frac {4 (3-10 x) \sqrt {3 x^2+5 x+2}}{15 x^{5/2}}\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle \frac {1}{15} \left (\frac {139 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-3 \left (110 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+139 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\right )-\frac {4 (3-10 x) \sqrt {3 x^2+5 x+2}}{15 x^{5/2}}\) |
| \(\Big \downarrow \) 1413 |
| \(\displaystyle \frac {1}{15} \left (\frac {139 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-3 \left (139 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {55 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}\right )\right )-\frac {4 (3-10 x) \sqrt {3 x^2+5 x+2}}{15 x^{5/2}}\) |
| \(\Big \downarrow \) 1456 |
| \(\displaystyle \frac {1}{15} \left (\frac {139 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-3 \left (\frac {55 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+139 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )\right )-\frac {4 (3-10 x) \sqrt {3 x^2+5 x+2}}{15 x^{5/2}}\) |
(-4*(3 - 10*x)*Sqrt[2 + 5*x + 3*x^2])/(15*x^(5/2)) + ((139*Sqrt[2 + 5*x + 3*x^2])/Sqrt[x] - 3*(139*((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])) + (55*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]* EllipticF[ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2]))/15
3.11.41.3.1 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))*((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 = 0.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.69
| method | result | size |
| default | \(\frac {87 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x^{2}-139 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x^{2}+2502 x^{4}+4890 x^{3}+2652 x^{2}+120 x -144}{90 \sqrt {3 x^{2}+5 x +2}\, x^{\frac {5}{2}}}\) | \(124\) |
| risch | \(\frac {417 x^{4}+815 x^{3}+442 x^{2}+20 x -24}{15 x^{\frac {5}{2}} \sqrt {3 x^{2}+5 x +2}}-\frac {\left (\frac {11 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {139 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{30 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right ) \sqrt {x \left (3 x^{2}+5 x +2\right )}}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(198\) |
| elliptic | \(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {4 \sqrt {3 x^{3}+5 x^{2}+2 x}}{5 x^{3}}+\frac {8 \sqrt {3 x^{3}+5 x^{2}+2 x}}{3 x^{2}}+\frac {\frac {139}{5} x^{2}+\frac {139}{3} x +\frac {278}{15}}{\sqrt {x \left (3 x^{2}+5 x +2\right )}}-\frac {11 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {139 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{30 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(227\) |
1/90*(87*(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^2-139*(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))*x^2+2502*x^4+4890*x^3+2652*x^2+12 0*x-144)/(3*x^2+5*x+2)^(1/2)/x^(5/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.36 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{7/2}} \, dx=-\frac {295 \, \sqrt {3} x^{3} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 1251 \, \sqrt {3} x^{3} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - 9 \, {\left (139 \, x^{2} + 40 \, x - 12\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}}{135 \, x^{3}} \]
-1/135*(295*sqrt(3)*x^3*weierstrassPInverse(28/27, 80/729, x + 5/9) - 1251 *sqrt(3)*x^3*weierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/ 729, x + 5/9)) - 9*(139*x^2 + 40*x - 12)*sqrt(3*x^2 + 5*x + 2)*sqrt(x))/x^ 3
\[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{7/2}} \, dx=- \int \left (- \frac {2 \sqrt {3 x^{2} + 5 x + 2}}{x^{\frac {7}{2}}}\right )\, dx - \int \frac {5 \sqrt {3 x^{2} + 5 x + 2}}{x^{\frac {5}{2}}}\, dx \]
-Integral(-2*sqrt(3*x**2 + 5*x + 2)/x**(7/2), x) - Integral(5*sqrt(3*x**2 + 5*x + 2)/x**(5/2), x)
\[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{7/2}} \, dx=\int { -\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (5 \, x - 2\right )}}{x^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{7/2}} \, dx=\int { -\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (5 \, x - 2\right )}}{x^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{7/2}} \, dx=\int -\frac {\left (5\,x-2\right )\,\sqrt {3\,x^2+5\,x+2}}{x^{7/2}} \,d x \]