Integrand size = 29, antiderivative size = 170 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{7/2}} \, dx=\frac {49 \sqrt {2+5 x+3 x^2}}{125 \sqrt {3+2 x}}+\frac {(32+43 x) \sqrt {2+5 x+3 x^2}}{25 (3+2 x)^{5/2}}-\frac {49 \sqrt {3} \sqrt {-2-5 x-3 x^2} E\left (\arcsin \left (\sqrt {3} \sqrt {1+x}\right )|-\frac {2}{3}\right )}{250 \sqrt {2+5 x+3 x^2}}+\frac {9 \sqrt {3} \sqrt {-2-5 x-3 x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {1+x}\right ),-\frac {2}{3}\right )}{50 \sqrt {2+5 x+3 x^2}} \]
-49/250*EllipticE(3^(1/2)*(1+x)^(1/2),1/3*I*6^(1/2))*(-3*x^2-5*x-2)^(1/2)* 3^(1/2)/(3*x^2+5*x+2)^(1/2)+9/50*EllipticF(3^(1/2)*(1+x)^(1/2),1/3*I*6^(1/ 2))*(-3*x^2-5*x-2)^(1/2)*3^(1/2)/(3*x^2+5*x+2)^(1/2)+1/25*(32+43*x)*(3*x^2 +5*x+2)^(1/2)/(3+2*x)^(5/2)+49/125*(3*x^2+5*x+2)^(1/2)/(3+2*x)^(1/2)
Time = 31.26 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.07 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{7/2}} \, dx=\frac {640+2460 x+3110 x^2+1290 x^3-49 \sqrt {5} \sqrt {\frac {1+x}{3+2 x}} (3+2 x)^{7/2} \sqrt {\frac {2+3 x}{3+2 x}} E\left (\arcsin \left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {3+2 x}}\right )|\frac {3}{5}\right )+22 \sqrt {5} \sqrt {\frac {1+x}{3+2 x}} (3+2 x)^{7/2} \sqrt {\frac {2+3 x}{3+2 x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {3+2 x}}\right ),\frac {3}{5}\right )}{250 (3+2 x)^{5/2} \sqrt {2+5 x+3 x^2}} \]
(640 + 2460*x + 3110*x^2 + 1290*x^3 - 49*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*( 3 + 2*x)^(7/2)*Sqrt[(2 + 3*x)/(3 + 2*x)]*EllipticE[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5] + 22*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^(7/2)*Sqrt[( 2 + 3*x)/(3 + 2*x)]*EllipticF[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5])/(250* (3 + 2*x)^(5/2)*Sqrt[2 + 5*x + 3*x^2])
Time = 0.34 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {1229, 27, 1237, 27, 1269, 1172, 27, 321, 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.
| \(\displaystyle \int \frac {(5-x) \sqrt {3 x^2+5 x+2}}{(2 x+3)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1229 |
| \(\displaystyle \frac {(43 x+32) \sqrt {3 x^2+5 x+2}}{25 (2 x+3)^{5/2}}-\frac {1}{150} \int -\frac {3 (27 x+16)}{(2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{50} \int \frac {27 x+16}{(2 x+3)^{3/2} \sqrt {3 x^2+5 x+2}}dx+\frac {\sqrt {3 x^2+5 x+2} (43 x+32)}{25 (2 x+3)^{5/2}}\) |
| \(\Big \downarrow \) 1237 |
| \(\displaystyle \frac {1}{50} \left (\frac {98 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {2}{5} \int \frac {3 (49 x+51)}{2 \sqrt {2 x+3} \sqrt {3 x^2+5 x+2}}dx\right )+\frac {\sqrt {3 x^2+5 x+2} (43 x+32)}{25 (2 x+3)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{50} \left (\frac {98 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \int \frac {49 x+51}{\sqrt {2 x+3} \sqrt {3 x^2+5 x+2}}dx\right )+\frac {\sqrt {3 x^2+5 x+2} (43 x+32)}{25 (2 x+3)^{5/2}}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {1}{50} \left (\frac {98 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \left (\frac {49}{2} \int \frac {\sqrt {2 x+3}}{\sqrt {3 x^2+5 x+2}}dx-\frac {45}{2} \int \frac {1}{\sqrt {2 x+3} \sqrt {3 x^2+5 x+2}}dx\right )\right )+\frac {\sqrt {3 x^2+5 x+2} (43 x+32)}{25 (2 x+3)^{5/2}}\) |
| \(\Big \downarrow \) 1172 |
| \(\displaystyle \frac {1}{50} \left (\frac {98 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \left (\frac {49 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {3} \sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}-\frac {15 \sqrt {3} \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {3}}{\sqrt {1-3 (x+1)} \sqrt {6 (x+1)+3}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3 x^2+5 x+2}}\right )\right )+\frac {\sqrt {3 x^2+5 x+2} (43 x+32)}{25 (2 x+3)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{50} \left (\frac {98 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \left (\frac {49 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{3 \sqrt {3 x^2+5 x+2}}-\frac {45 \sqrt {-3 x^2-5 x-2} \int \frac {1}{\sqrt {1-3 (x+1)} \sqrt {6 (x+1)+3}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3 x^2+5 x+2}}\right )\right )+\frac {\sqrt {3 x^2+5 x+2} (43 x+32)}{25 (2 x+3)^{5/2}}\) |
| \(\Big \downarrow \) 321 |
| \(\displaystyle \frac {1}{50} \left (\frac {98 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \left (\frac {49 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{3 \sqrt {3 x^2+5 x+2}}-\frac {15 \sqrt {3} \sqrt {-3 x^2-5 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right ),-\frac {2}{3}\right )}{\sqrt {3 x^2+5 x+2}}\right )\right )+\frac {\sqrt {3 x^2+5 x+2} (43 x+32)}{25 (2 x+3)^{5/2}}\) |
| \(\Big \downarrow \) 327 |
| \(\displaystyle \frac {1}{50} \left (\frac {98 \sqrt {3 x^2+5 x+2}}{5 \sqrt {2 x+3}}-\frac {3}{5} \left (\frac {49 \sqrt {-3 x^2-5 x-2} E\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}-\frac {15 \sqrt {3} \sqrt {-3 x^2-5 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right ),-\frac {2}{3}\right )}{\sqrt {3 x^2+5 x+2}}\right )\right )+\frac {\sqrt {3 x^2+5 x+2} (43 x+32)}{25 (2 x+3)^{5/2}}\) |
((32 + 43*x)*Sqrt[2 + 5*x + 3*x^2])/(25*(3 + 2*x)^(5/2)) + ((98*Sqrt[2 + 5 *x + 3*x^2])/(5*Sqrt[3 + 2*x]) - (3*((49*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(Sqrt[3]*Sqrt[2 + 5*x + 3*x^2]) - (15* Sqrt[3]*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3 ])/Sqrt[2 + 5*x + 3*x^2]))/5)/50
3.26.82.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[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] && !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
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]
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[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Time = 0.34 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.45
| method | result | size |
| elliptic | \(\frac {\sqrt {\left (3+2 x \right ) \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {13 \sqrt {6 x^{3}+19 x^{2}+19 x +6}}{80 \left (x +\frac {3}{2}\right )^{3}}+\frac {43 \sqrt {6 x^{3}+19 x^{2}+19 x +6}}{200 \left (x +\frac {3}{2}\right )^{2}}+\frac {\frac {147}{125} x^{2}+\frac {49}{25} x +\frac {98}{125}}{\sqrt {\left (x +\frac {3}{2}\right ) \left (6 x^{2}+10 x +4\right )}}+\frac {51 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {45+30 x}\, F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )}{1250 \sqrt {6 x^{3}+19 x^{2}+19 x +6}}+\frac {49 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {45+30 x}\, \left (\frac {E\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )}{3}-F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )\right )}{1250 \sqrt {6 x^{3}+19 x^{2}+19 x +6}}\right )}{\sqrt {3+2 x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(246\) |
| default | \(\frac {24 \sqrt {15}\, F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right ) x^{2} \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {3+2 x}+196 \sqrt {15}\, E\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right ) x^{2} \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {3+2 x}+72 F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right ) \sqrt {15}\, x \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {3+2 x}+588 \sqrt {15}\, E\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right ) x \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {3+2 x}+54 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {15}\, \sqrt {3+2 x}\, F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )+441 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {15}\, \sqrt {3+2 x}\, E\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )+17640 x^{4}+101670 x^{3}+186300 x^{2}+138330 x +36060}{3750 \sqrt {3 x^{2}+5 x +2}\, \left (3+2 x \right )^{\frac {5}{2}}}\) | \(296\) |
((3+2*x)*(3*x^2+5*x+2))^(1/2)/(3+2*x)^(1/2)/(3*x^2+5*x+2)^(1/2)*(-13/80*(6 *x^3+19*x^2+19*x+6)^(1/2)/(x+3/2)^3+43/200*(6*x^3+19*x^2+19*x+6)^(1/2)/(x+ 3/2)^2+49/250*(6*x^2+10*x+4)/((x+3/2)*(6*x^2+10*x+4))^(1/2)+51/1250*(-20-3 0*x)^(1/2)*(3+3*x)^(1/2)*(45+30*x)^(1/2)/(6*x^3+19*x^2+19*x+6)^(1/2)*Ellip ticF(1/5*(-20-30*x)^(1/2),1/2*10^(1/2))+49/1250*(-20-30*x)^(1/2)*(3+3*x)^( 1/2)*(45+30*x)^(1/2)/(6*x^3+19*x^2+19*x+6)^(1/2)*(1/3*EllipticE(1/5*(-20-3 0*x)^(1/2),1/2*10^(1/2))-EllipticF(1/5*(-20-30*x)^(1/2),1/2*10^(1/2))))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.62 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{7/2}} \, dx=\frac {13 \, \sqrt {6} {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )} {\rm weierstrassPInverse}\left (\frac {19}{27}, -\frac {28}{729}, x + \frac {19}{18}\right ) + 882 \, \sqrt {6} {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )} {\rm weierstrassZeta}\left (\frac {19}{27}, -\frac {28}{729}, {\rm weierstrassPInverse}\left (\frac {19}{27}, -\frac {28}{729}, x + \frac {19}{18}\right )\right ) + 36 \, {\left (196 \, x^{2} + 803 \, x + 601\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {2 \, x + 3}}{4500 \, {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )}} \]
1/4500*(13*sqrt(6)*(8*x^3 + 36*x^2 + 54*x + 27)*weierstrassPInverse(19/27, -28/729, x + 19/18) + 882*sqrt(6)*(8*x^3 + 36*x^2 + 54*x + 27)*weierstras sZeta(19/27, -28/729, weierstrassPInverse(19/27, -28/729, x + 19/18)) + 36 *(196*x^2 + 803*x + 601)*sqrt(3*x^2 + 5*x + 2)*sqrt(2*x + 3))/(8*x^3 + 36* x^2 + 54*x + 27)
\[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{7/2}} \, dx=- \int \left (- \frac {5 \sqrt {3 x^{2} + 5 x + 2}}{8 x^{3} \sqrt {2 x + 3} + 36 x^{2} \sqrt {2 x + 3} + 54 x \sqrt {2 x + 3} + 27 \sqrt {2 x + 3}}\right )\, dx - \int \frac {x \sqrt {3 x^{2} + 5 x + 2}}{8 x^{3} \sqrt {2 x + 3} + 36 x^{2} \sqrt {2 x + 3} + 54 x \sqrt {2 x + 3} + 27 \sqrt {2 x + 3}}\, dx \]
-Integral(-5*sqrt(3*x**2 + 5*x + 2)/(8*x**3*sqrt(2*x + 3) + 36*x**2*sqrt(2 *x + 3) + 54*x*sqrt(2*x + 3) + 27*sqrt(2*x + 3)), x) - Integral(x*sqrt(3*x **2 + 5*x + 2)/(8*x**3*sqrt(2*x + 3) + 36*x**2*sqrt(2*x + 3) + 54*x*sqrt(2 *x + 3) + 27*sqrt(2*x + 3)), x)
\[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{7/2}} \, dx=\int { -\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (x - 5\right )}}{{\left (2 \, x + 3\right )}^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{7/2}} \, dx=\int { -\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (x - 5\right )}}{{\left (2 \, x + 3\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{7/2}} \, dx=-\int \frac {\left (x-5\right )\,\sqrt {3\,x^2+5\,x+2}}{{\left (2\,x+3\right )}^{7/2}} \,d x \]