Integrand size = 31, antiderivative size = 320 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2 a^5 A x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {2 a^4 (5 A b+a B) x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {10 a^3 b (2 A b+a B) x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {20 a^2 b^2 (A b+a B) x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {10 a b^3 (A b+2 a B) x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (a+b x)}+\frac {2 b^4 (A b+5 a B) x^{15/2} \sqrt {a^2+2 a b x+b^2 x^2}}{15 (a+b x)}+\frac {2 b^5 B x^{17/2} \sqrt {a^2+2 a b x+b^2 x^2}}{17 (a+b x)} \]
2/5*a^5*A*x^(5/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/7*a^4*(5*A*b+B*a)*x^(7/2)*(( b*x+a)^2)^(1/2)/(b*x+a)+10/9*a^3*b*(2*A*b+B*a)*x^(9/2)*((b*x+a)^2)^(1/2)/( b*x+a)+20/11*a^2*b^2*(A*b+B*a)*x^(11/2)*((b*x+a)^2)^(1/2)/(b*x+a)+10/13*a* b^3*(A*b+2*B*a)*x^(13/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/15*b^4*(A*b+5*B*a)*x^ (15/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/17*b^5*B*x^(17/2)*((b*x+a)^2)^(1/2)/(b* x+a)
Time = 0.07 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.40 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2 x^{5/2} \sqrt {(a+b x)^2} \left (21879 a^5 (7 A+5 B x)+60775 a^4 b x (9 A+7 B x)+77350 a^3 b^2 x^2 (11 A+9 B x)+53550 a^2 b^3 x^3 (13 A+11 B x)+19635 a b^4 x^4 (15 A+13 B x)+3003 b^5 x^5 (17 A+15 B x)\right )}{765765 (a+b x)} \]
(2*x^(5/2)*Sqrt[(a + b*x)^2]*(21879*a^5*(7*A + 5*B*x) + 60775*a^4*b*x*(9*A + 7*B*x) + 77350*a^3*b^2*x^2*(11*A + 9*B*x) + 53550*a^2*b^3*x^3*(13*A + 1 1*B*x) + 19635*a*b^4*x^4*(15*A + 13*B*x) + 3003*b^5*x^5*(17*A + 15*B*x)))/ (765765*(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.
| \(\displaystyle \int x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x) \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^5 x^{3/2} (a+b x)^5 (A+B x)dx}{b^5 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^{3/2} (a+b x)^5 (A+B x)dx}{a+b x}\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^5 B x^{15/2}+b^4 (A b+5 a B) x^{13/2}+5 a b^3 (A b+2 a B) x^{11/2}+10 a^2 b^2 (A b+a B) x^{9/2}+5 a^3 b (2 A b+a B) x^{7/2}+a^4 (5 A b+a B) x^{5/2}+a^5 A x^{3/2}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {2}{5} a^5 A x^{5/2}+\frac {2}{7} a^4 x^{7/2} (a B+5 A b)+\frac {10}{9} a^3 b x^{9/2} (a B+2 A b)+\frac {20}{11} a^2 b^2 x^{11/2} (a B+A b)+\frac {2}{15} b^4 x^{15/2} (5 a B+A b)+\frac {10}{13} a b^3 x^{13/2} (2 a B+A b)+\frac {2}{17} b^5 B x^{17/2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*a^5*A*x^(5/2))/5 + (2*a^4*(5*A*b + a*B) *x^(7/2))/7 + (10*a^3*b*(2*A*b + a*B)*x^(9/2))/9 + (20*a^2*b^2*(A*b + a*B) *x^(11/2))/11 + (10*a*b^3*(A*b + 2*a*B)*x^(13/2))/13 + (2*b^4*(A*b + 5*a*B )*x^(15/2))/15 + (2*b^5*B*x^(17/2))/17))/(a + b*x)
3.9.2.3.1 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 = 0.14 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.44
| method | result | size |
| gosper | \(\frac {2 x^{\frac {5}{2}} \left (45045 B \,b^{5} x^{6}+51051 A \,b^{5} x^{5}+255255 B a \,b^{4} x^{5}+294525 A a \,b^{4} x^{4}+589050 B \,a^{2} b^{3} x^{4}+696150 A \,a^{2} b^{3} x^{3}+696150 B \,a^{3} b^{2} x^{3}+850850 A \,a^{3} b^{2} x^{2}+425425 B \,a^{4} b \,x^{2}+546975 A \,a^{4} b x +109395 a^{5} B x +153153 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{765765 \left (b x +a \right )^{5}}\) | \(140\) |
| default | \(\frac {2 x^{\frac {5}{2}} \left (45045 B \,b^{5} x^{6}+51051 A \,b^{5} x^{5}+255255 B a \,b^{4} x^{5}+294525 A a \,b^{4} x^{4}+589050 B \,a^{2} b^{3} x^{4}+696150 A \,a^{2} b^{3} x^{3}+696150 B \,a^{3} b^{2} x^{3}+850850 A \,a^{3} b^{2} x^{2}+425425 B \,a^{4} b \,x^{2}+546975 A \,a^{4} b x +109395 a^{5} B x +153153 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{765765 \left (b x +a \right )^{5}}\) | \(140\) |
| risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, x^{\frac {5}{2}} \left (45045 B \,b^{5} x^{6}+51051 A \,b^{5} x^{5}+255255 B a \,b^{4} x^{5}+294525 A a \,b^{4} x^{4}+589050 B \,a^{2} b^{3} x^{4}+696150 A \,a^{2} b^{3} x^{3}+696150 B \,a^{3} b^{2} x^{3}+850850 A \,a^{3} b^{2} x^{2}+425425 B \,a^{4} b \,x^{2}+546975 A \,a^{4} b x +109395 a^{5} B x +153153 A \,a^{5}\right )}{765765 \left (b x +a \right )}\) | \(140\) |
2/765765*x^(5/2)*(45045*B*b^5*x^6+51051*A*b^5*x^5+255255*B*a*b^4*x^5+29452 5*A*a*b^4*x^4+589050*B*a^2*b^3*x^4+696150*A*a^2*b^3*x^3+696150*B*a^3*b^2*x ^3+850850*A*a^3*b^2*x^2+425425*B*a^4*b*x^2+546975*A*a^4*b*x+109395*B*a^5*x +153153*A*a^5)*((b*x+a)^2)^(5/2)/(b*x+a)^5
Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.39 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2}{765765} \, {\left (45045 \, B b^{5} x^{8} + 153153 \, A a^{5} x^{2} + 51051 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{7} + 294525 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{6} + 696150 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{5} + 425425 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} + 109395 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{3}\right )} \sqrt {x} \]
2/765765*(45045*B*b^5*x^8 + 153153*A*a^5*x^2 + 51051*(5*B*a*b^4 + A*b^5)*x ^7 + 294525*(2*B*a^2*b^3 + A*a*b^4)*x^6 + 696150*(B*a^3*b^2 + A*a^2*b^3)*x ^5 + 425425*(B*a^4*b + 2*A*a^3*b^2)*x^4 + 109395*(B*a^5 + 5*A*a^4*b)*x^3)* sqrt(x)
Timed out. \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.75 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2}{45045} \, {\left (231 \, {\left (13 \, b^{5} x^{2} + 15 \, a b^{4} x\right )} x^{\frac {11}{2}} + 1260 \, {\left (11 \, a b^{4} x^{2} + 13 \, a^{2} b^{3} x\right )} x^{\frac {9}{2}} + 2730 \, {\left (9 \, a^{2} b^{3} x^{2} + 11 \, a^{3} b^{2} x\right )} x^{\frac {7}{2}} + 2860 \, {\left (7 \, a^{3} b^{2} x^{2} + 9 \, a^{4} b x\right )} x^{\frac {5}{2}} + 1287 \, {\left (5 \, a^{4} b x^{2} + 7 \, a^{5} x\right )} x^{\frac {3}{2}}\right )} A + \frac {2}{765765} \, {\left (3003 \, {\left (15 \, b^{5} x^{2} + 17 \, a b^{4} x\right )} x^{\frac {13}{2}} + 15708 \, {\left (13 \, a b^{4} x^{2} + 15 \, a^{2} b^{3} x\right )} x^{\frac {11}{2}} + 32130 \, {\left (11 \, a^{2} b^{3} x^{2} + 13 \, a^{3} b^{2} x\right )} x^{\frac {9}{2}} + 30940 \, {\left (9 \, a^{3} b^{2} x^{2} + 11 \, a^{4} b x\right )} x^{\frac {7}{2}} + 12155 \, {\left (7 \, a^{4} b x^{2} + 9 \, a^{5} x\right )} x^{\frac {5}{2}}\right )} B \]
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))* A + 2/765765*(3003*(15*b^5*x^2 + 17*a*b^4*x)*x^(13/2) + 15708*(13*a*b^4*x^ 2 + 15*a^2*b^3*x)*x^(11/2) + 32130*(11*a^2*b^3*x^2 + 13*a^3*b^2*x)*x^(9/2) + 30940*(9*a^3*b^2*x^2 + 11*a^4*b*x)*x^(7/2) + 12155*(7*a^4*b*x^2 + 9*a^5 *x)*x^(5/2))*B
Time = 0.27 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.62 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {2}{17} \, B b^{5} x^{\frac {17}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, B a b^{4} x^{\frac {15}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{15} \, A b^{5} x^{\frac {15}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{13} \, B a^{2} b^{3} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{13} \, A a b^{4} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{11} \, B a^{3} b^{2} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{11} \, A a^{2} b^{3} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{9} \, B a^{4} b x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{9} \, A a^{3} b^{2} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{7} \, B a^{5} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, A a^{4} b x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, A a^{5} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) \]
2/17*B*b^5*x^(17/2)*sgn(b*x + a) + 2/3*B*a*b^4*x^(15/2)*sgn(b*x + a) + 2/1 5*A*b^5*x^(15/2)*sgn(b*x + a) + 20/13*B*a^2*b^3*x^(13/2)*sgn(b*x + a) + 10 /13*A*a*b^4*x^(13/2)*sgn(b*x + a) + 20/11*B*a^3*b^2*x^(11/2)*sgn(b*x + a) + 20/11*A*a^2*b^3*x^(11/2)*sgn(b*x + a) + 10/9*B*a^4*b*x^(9/2)*sgn(b*x + a ) + 20/9*A*a^3*b^2*x^(9/2)*sgn(b*x + a) + 2/7*B*a^5*x^(7/2)*sgn(b*x + a) + 10/7*A*a^4*b*x^(7/2)*sgn(b*x + a) + 2/5*A*a^5*x^(5/2)*sgn(b*x + a)
Timed out. \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\int x^{3/2}\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2} \,d x \]