Integrand size = 31, antiderivative size = 320 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2 a^5 A x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {2 a^4 (5 A b+a B) x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {10 a^3 b (2 A b+a B) x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {20 a^2 b^2 (A b+a B) x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {10 a b^3 (A b+2 a B) x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {2 b^4 (A b+5 a B) x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (a+b x)}+\frac {2 b^5 B x^{15/2} \sqrt {a^2+2 a b x+b^2 x^2}}{15 (a+b x)} \]
2/3*a^5*A*x^(3/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/5*a^4*(5*A*b+B*a)*x^(5/2)*(( b*x+a)^2)^(1/2)/(b*x+a)+10/7*a^3*b*(2*A*b+B*a)*x^(7/2)*((b*x+a)^2)^(1/2)/( b*x+a)+20/9*a^2*b^2*(A*b+B*a)*x^(9/2)*((b*x+a)^2)^(1/2)/(b*x+a)+10/11*a*b^ 3*(A*b+2*B*a)*x^(11/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/13*b^4*(A*b+5*B*a)*x^(1 3/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/15*b^5*B*x^(15/2)*((b*x+a)^2)^(1/2)/(b*x+ a)
Time = 0.07 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.40 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2 x^{3/2} \sqrt {(a+b x)^2} \left (3003 a^5 (5 A+3 B x)+6435 a^4 b x (7 A+5 B x)+7150 a^3 b^2 x^2 (9 A+7 B x)+4550 a^2 b^3 x^3 (11 A+9 B x)+1575 a b^4 x^4 (13 A+11 B x)+231 b^5 x^5 (15 A+13 B x)\right )}{45045 (a+b x)} \]
(2*x^(3/2)*Sqrt[(a + b*x)^2]*(3003*a^5*(5*A + 3*B*x) + 6435*a^4*b*x*(7*A + 5*B*x) + 7150*a^3*b^2*x^2*(9*A + 7*B*x) + 4550*a^2*b^3*x^3*(11*A + 9*B*x) + 1575*a*b^4*x^4*(13*A + 11*B*x) + 231*b^5*x^5*(15*A + 13*B*x)))/(45045*( a + b*x))
Time = 0.30 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.50, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1187, 27, 85, 2009}
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.
| \(\displaystyle \int \sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x) \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^5 \sqrt {x} (a+b x)^5 (A+B x)dx}{b^5 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \sqrt {x} (a+b x)^5 (A+B x)dx}{a+b x}\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^5 B x^{13/2}+b^4 (A b+5 a B) x^{11/2}+5 a b^3 (A b+2 a B) x^{9/2}+10 a^2 b^2 (A b+a B) x^{7/2}+5 a^3 b (2 A b+a B) x^{5/2}+a^4 (5 A b+a B) x^{3/2}+a^5 A \sqrt {x}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {2}{3} a^5 A x^{3/2}+\frac {2}{5} a^4 x^{5/2} (a B+5 A b)+\frac {10}{7} a^3 b x^{7/2} (a B+2 A b)+\frac {20}{9} a^2 b^2 x^{9/2} (a B+A b)+\frac {2}{13} b^4 x^{13/2} (5 a B+A b)+\frac {10}{11} a b^3 x^{11/2} (2 a B+A b)+\frac {2}{15} b^5 B x^{15/2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*a^5*A*x^(3/2))/3 + (2*a^4*(5*A*b + a*B) *x^(5/2))/5 + (10*a^3*b*(2*A*b + a*B)*x^(7/2))/7 + (20*a^2*b^2*(A*b + a*B) *x^(9/2))/9 + (10*a*b^3*(A*b + 2*a*B)*x^(11/2))/11 + (2*b^4*(A*b + 5*a*B)* x^(13/2))/13 + (2*b^5*B*x^(15/2))/15))/(a + b*x)
3.9.3.3.1 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_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ IntPart[p]*(b/2 + c*x)^(2*FracPart[p])) Int[(d + e*x)^m*(f + g*x)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 0.36 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.44
| method | result | size |
| gosper | \(\frac {2 x^{\frac {3}{2}} \left (3003 B \,b^{5} x^{6}+3465 A \,b^{5} x^{5}+17325 B a \,b^{4} x^{5}+20475 A a \,b^{4} x^{4}+40950 B \,a^{2} b^{3} x^{4}+50050 A \,a^{2} b^{3} x^{3}+50050 B \,a^{3} b^{2} x^{3}+64350 A \,a^{3} b^{2} x^{2}+32175 B \,a^{4} b \,x^{2}+45045 A \,a^{4} b x +9009 a^{5} B x +15015 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{45045 \left (b x +a \right )^{5}}\) | \(140\) |
| default | \(\frac {2 x^{\frac {3}{2}} \left (3003 B \,b^{5} x^{6}+3465 A \,b^{5} x^{5}+17325 B a \,b^{4} x^{5}+20475 A a \,b^{4} x^{4}+40950 B \,a^{2} b^{3} x^{4}+50050 A \,a^{2} b^{3} x^{3}+50050 B \,a^{3} b^{2} x^{3}+64350 A \,a^{3} b^{2} x^{2}+32175 B \,a^{4} b \,x^{2}+45045 A \,a^{4} b x +9009 a^{5} B x +15015 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{45045 \left (b x +a \right )^{5}}\) | \(140\) |
| risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, x^{\frac {3}{2}} \left (3003 B \,b^{5} x^{6}+3465 A \,b^{5} x^{5}+17325 B a \,b^{4} x^{5}+20475 A a \,b^{4} x^{4}+40950 B \,a^{2} b^{3} x^{4}+50050 A \,a^{2} b^{3} x^{3}+50050 B \,a^{3} b^{2} x^{3}+64350 A \,a^{3} b^{2} x^{2}+32175 B \,a^{4} b \,x^{2}+45045 A \,a^{4} b x +9009 a^{5} B x +15015 A \,a^{5}\right )}{45045 \left (b x +a \right )}\) | \(140\) |
2/45045*x^(3/2)*(3003*B*b^5*x^6+3465*A*b^5*x^5+17325*B*a*b^4*x^5+20475*A*a *b^4*x^4+40950*B*a^2*b^3*x^4+50050*A*a^2*b^3*x^3+50050*B*a^3*b^2*x^3+64350 *A*a^3*b^2*x^2+32175*B*a^4*b*x^2+45045*A*a^4*b*x+9009*B*a^5*x+15015*A*a^5) *((b*x+a)^2)^(5/2)/(b*x+a)^5
Time = 0.31 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.38 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2}{45045} \, {\left (3003 \, B b^{5} x^{7} + 15015 \, A a^{5} x + 3465 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{6} + 20475 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{5} + 50050 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{4} + 32175 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{3} + 9009 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2}\right )} \sqrt {x} \]
2/45045*(3003*B*b^5*x^7 + 15015*A*a^5*x + 3465*(5*B*a*b^4 + A*b^5)*x^6 + 2 0475*(2*B*a^2*b^3 + A*a*b^4)*x^5 + 50050*(B*a^3*b^2 + A*a^2*b^3)*x^4 + 321 75*(B*a^4*b + 2*A*a^3*b^2)*x^3 + 9009*(B*a^5 + 5*A*a^4*b)*x^2)*sqrt(x)
\[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\int \sqrt {x} \left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.75 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2}{45045} \, {\left (315 \, {\left (11 \, b^{5} x^{2} + 13 \, a b^{4} x\right )} x^{\frac {9}{2}} + 1820 \, {\left (9 \, a b^{4} x^{2} + 11 \, a^{2} b^{3} x\right )} x^{\frac {7}{2}} + 4290 \, {\left (7 \, a^{2} b^{3} x^{2} + 9 \, a^{3} b^{2} x\right )} x^{\frac {5}{2}} + 5148 \, {\left (5 \, a^{3} b^{2} x^{2} + 7 \, a^{4} b x\right )} x^{\frac {3}{2}} + 3003 \, {\left (3 \, a^{4} b x^{2} + 5 \, a^{5} x\right )} \sqrt {x}\right )} A + \frac {2}{45045} \, {\left (231 \, {\left (13 \, b^{5} x^{2} + 15 \, a b^{4} x\right )} x^{\frac {11}{2}} + 1260 \, {\left (11 \, a b^{4} x^{2} + 13 \, a^{2} b^{3} x\right )} x^{\frac {9}{2}} + 2730 \, {\left (9 \, a^{2} b^{3} x^{2} + 11 \, a^{3} b^{2} x\right )} x^{\frac {7}{2}} + 2860 \, {\left (7 \, a^{3} b^{2} x^{2} + 9 \, a^{4} b x\right )} x^{\frac {5}{2}} + 1287 \, {\left (5 \, a^{4} b x^{2} + 7 \, a^{5} x\right )} x^{\frac {3}{2}}\right )} B \]
2/45045*(315*(11*b^5*x^2 + 13*a*b^4*x)*x^(9/2) + 1820*(9*a*b^4*x^2 + 11*a^ 2*b^3*x)*x^(7/2) + 4290*(7*a^2*b^3*x^2 + 9*a^3*b^2*x)*x^(5/2) + 5148*(5*a^ 3*b^2*x^2 + 7*a^4*b*x)*x^(3/2) + 3003*(3*a^4*b*x^2 + 5*a^5*x)*sqrt(x))*A + 2/45045*(231*(13*b^5*x^2 + 15*a*b^4*x)*x^(11/2) + 1260*(11*a*b^4*x^2 + 13 *a^2*b^3*x)*x^(9/2) + 2730*(9*a^2*b^3*x^2 + 11*a^3*b^2*x)*x^(7/2) + 2860*( 7*a^3*b^2*x^2 + 9*a^4*b*x)*x^(5/2) + 1287*(5*a^4*b*x^2 + 7*a^5*x)*x^(3/2)) *B
Time = 0.27 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.62 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2}{15} \, B b^{5} x^{\frac {15}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{13} \, B a b^{4} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{13} \, A b^{5} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{11} \, B a^{2} b^{3} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{11} \, A a b^{4} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{9} \, B a^{3} b^{2} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{9} \, A a^{2} b^{3} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, B a^{4} b x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{7} \, A a^{3} b^{2} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, B a^{5} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a^{4} b x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, A a^{5} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) \]
2/15*B*b^5*x^(15/2)*sgn(b*x + a) + 10/13*B*a*b^4*x^(13/2)*sgn(b*x + a) + 2 /13*A*b^5*x^(13/2)*sgn(b*x + a) + 20/11*B*a^2*b^3*x^(11/2)*sgn(b*x + a) + 10/11*A*a*b^4*x^(11/2)*sgn(b*x + a) + 20/9*B*a^3*b^2*x^(9/2)*sgn(b*x + a) + 20/9*A*a^2*b^3*x^(9/2)*sgn(b*x + a) + 10/7*B*a^4*b*x^(7/2)*sgn(b*x + a) + 20/7*A*a^3*b^2*x^(7/2)*sgn(b*x + a) + 2/5*B*a^5*x^(5/2)*sgn(b*x + a) + 2 *A*a^4*b*x^(5/2)*sgn(b*x + a) + 2/3*A*a^5*x^(3/2)*sgn(b*x + a)
Timed out. \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\int \sqrt {x}\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2} \,d x \]