Integrand size = 31, antiderivative size = 316 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {x}} \, dx=\frac {2 a^5 A \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 a^4 (5 A b+a B) x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {2 a^3 b (2 A b+a B) x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {20 a^2 b^2 (A b+a B) x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {10 a b^3 (A b+2 a B) x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {2 b^4 (A b+5 a B) x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {2 b^5 B x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (a+b x)} \]
2/3*a^4*(5*A*b+B*a)*x^(3/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2*a^3*b*(2*A*b+B*a)* x^(5/2)*((b*x+a)^2)^(1/2)/(b*x+a)+20/7*a^2*b^2*(A*b+B*a)*x^(7/2)*((b*x+a)^ 2)^(1/2)/(b*x+a)+10/9*a*b^3*(A*b+2*B*a)*x^(9/2)*((b*x+a)^2)^(1/2)/(b*x+a)+ 2/11*b^4*(A*b+5*B*a)*x^(11/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/13*b^5*B*x^(13/2 )*((b*x+a)^2)^(1/2)/(b*x+a)+2*a^5*A*x^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.40 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {x}} \, dx=\frac {2 \sqrt {x} \sqrt {(a+b x)^2} \left (3003 a^5 (3 A+B x)+3003 a^4 b x (5 A+3 B x)+2574 a^3 b^2 x^2 (7 A+5 B x)+1430 a^2 b^3 x^3 (9 A+7 B x)+455 a b^4 x^4 (11 A+9 B x)+63 b^5 x^5 (13 A+11 B x)\right )}{9009 (a+b x)} \]
(2*Sqrt[x]*Sqrt[(a + b*x)^2]*(3003*a^5*(3*A + B*x) + 3003*a^4*b*x*(5*A + 3 *B*x) + 2574*a^3*b^2*x^2*(7*A + 5*B*x) + 1430*a^2*b^3*x^3*(9*A + 7*B*x) + 455*a*b^4*x^4*(11*A + 9*B*x) + 63*b^5*x^5*(13*A + 11*B*x)))/(9009*(a + b*x ))
Time = 0.29 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.49, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x)}{\sqrt {x}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5 (A+B x)}{\sqrt {x}}dx}{b^5 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5 (A+B x)}{\sqrt {x}}dx}{a+b x}\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^5 B x^{11/2}+b^4 (A b+5 a B) x^{9/2}+5 a b^3 (A b+2 a B) x^{7/2}+10 a^2 b^2 (A b+a B) x^{5/2}+5 a^3 b (2 A b+a B) x^{3/2}+a^4 (5 A b+a B) \sqrt {x}+\frac {a^5 A}{\sqrt {x}}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (2 a^5 A \sqrt {x}+\frac {2}{3} a^4 x^{3/2} (a B+5 A b)+2 a^3 b x^{5/2} (a B+2 A b)+\frac {20}{7} a^2 b^2 x^{7/2} (a B+A b)+\frac {2}{11} b^4 x^{11/2} (5 a B+A b)+\frac {10}{9} a b^3 x^{9/2} (2 a B+A b)+\frac {2}{13} b^5 B x^{13/2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(2*a^5*A*Sqrt[x] + (2*a^4*(5*A*b + a*B)*x^( 3/2))/3 + 2*a^3*b*(2*A*b + a*B)*x^(5/2) + (20*a^2*b^2*(A*b + a*B)*x^(7/2)) /7 + (10*a*b^3*(A*b + 2*a*B)*x^(9/2))/9 + (2*b^4*(A*b + 5*a*B)*x^(11/2))/1 1 + (2*b^5*B*x^(13/2))/13))/(a + b*x)
3.9.4.3.1 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 | \(\frac {2 \sqrt {x}\, \left (693 B \,b^{5} x^{6}+819 A \,b^{5} x^{5}+4095 B a \,b^{4} x^{5}+5005 A a \,b^{4} x^{4}+10010 B \,a^{2} b^{3} x^{4}+12870 A \,a^{2} b^{3} x^{3}+12870 B \,a^{3} b^{2} x^{3}+18018 A \,a^{3} b^{2} x^{2}+9009 B \,a^{4} b \,x^{2}+15015 A \,a^{4} b x +3003 a^{5} B x +9009 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{9009 \left (b x +a \right )^{5}}\) | \(140\) |
| default | \(\frac {2 \sqrt {x}\, \left (693 B \,b^{5} x^{6}+819 A \,b^{5} x^{5}+4095 B a \,b^{4} x^{5}+5005 A a \,b^{4} x^{4}+10010 B \,a^{2} b^{3} x^{4}+12870 A \,a^{2} b^{3} x^{3}+12870 B \,a^{3} b^{2} x^{3}+18018 A \,a^{3} b^{2} x^{2}+9009 B \,a^{4} b \,x^{2}+15015 A \,a^{4} b x +3003 a^{5} B x +9009 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{9009 \left (b x +a \right )^{5}}\) | \(140\) |
| risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (693 B \,b^{5} x^{6}+819 A \,b^{5} x^{5}+4095 B a \,b^{4} x^{5}+5005 A a \,b^{4} x^{4}+10010 B \,a^{2} b^{3} x^{4}+12870 A \,a^{2} b^{3} x^{3}+12870 B \,a^{3} b^{2} x^{3}+18018 A \,a^{3} b^{2} x^{2}+9009 B \,a^{4} b \,x^{2}+15015 A \,a^{4} b x +3003 a^{5} B x +9009 A \,a^{5}\right ) \sqrt {x}}{9009 \left (b x +a \right )}\) | \(140\) |
2/9009*x^(1/2)*(693*B*b^5*x^6+819*A*b^5*x^5+4095*B*a*b^4*x^5+5005*A*a*b^4* x^4+10010*B*a^2*b^3*x^4+12870*A*a^2*b^3*x^3+12870*B*a^3*b^2*x^3+18018*A*a^ 3*b^2*x^2+9009*B*a^4*b*x^2+15015*A*a^4*b*x+3003*B*a^5*x+9009*A*a^5)*((b*x+ a)^2)^(5/2)/(b*x+a)^5
Time = 0.31 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.38 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {x}} \, dx=\frac {2}{9009} \, {\left (693 \, B b^{5} x^{6} + 9009 \, A a^{5} + 819 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 5005 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 12870 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 9009 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 3003 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x\right )} \sqrt {x} \]
2/9009*(693*B*b^5*x^6 + 9009*A*a^5 + 819*(5*B*a*b^4 + A*b^5)*x^5 + 5005*(2 *B*a^2*b^3 + A*a*b^4)*x^4 + 12870*(B*a^3*b^2 + A*a^2*b^3)*x^3 + 9009*(B*a^ 4*b + 2*A*a^3*b^2)*x^2 + 3003*(B*a^5 + 5*A*a^4*b)*x)*sqrt(x)
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {x}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{\sqrt {x}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.76 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {x}} \, dx=\frac {2}{3465} \, {\left (35 \, {\left (9 \, b^{5} x^{2} + 11 \, a b^{4} x\right )} x^{\frac {7}{2}} + 220 \, {\left (7 \, a b^{4} x^{2} + 9 \, a^{2} b^{3} x\right )} x^{\frac {5}{2}} + 594 \, {\left (5 \, a^{2} b^{3} x^{2} + 7 \, a^{3} b^{2} x\right )} x^{\frac {3}{2}} + 924 \, {\left (3 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x\right )} \sqrt {x} + \frac {1155 \, {\left (a^{4} b x^{2} + 3 \, a^{5} x\right )}}{\sqrt {x}}\right )} A + \frac {2}{45045} \, {\left (315 \, {\left (11 \, b^{5} x^{2} + 13 \, a b^{4} x\right )} x^{\frac {9}{2}} + 1820 \, {\left (9 \, a b^{4} x^{2} + 11 \, a^{2} b^{3} x\right )} x^{\frac {7}{2}} + 4290 \, {\left (7 \, a^{2} b^{3} x^{2} + 9 \, a^{3} b^{2} x\right )} x^{\frac {5}{2}} + 5148 \, {\left (5 \, a^{3} b^{2} x^{2} + 7 \, a^{4} b x\right )} x^{\frac {3}{2}} + 3003 \, {\left (3 \, a^{4} b x^{2} + 5 \, a^{5} x\right )} \sqrt {x}\right )} B \]
2/3465*(35*(9*b^5*x^2 + 11*a*b^4*x)*x^(7/2) + 220*(7*a*b^4*x^2 + 9*a^2*b^3 *x)*x^(5/2) + 594*(5*a^2*b^3*x^2 + 7*a^3*b^2*x)*x^(3/2) + 924*(3*a^3*b^2*x ^2 + 5*a^4*b*x)*sqrt(x) + 1155*(a^4*b*x^2 + 3*a^5*x)/sqrt(x))*A + 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))*B
Time = 0.27 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {x}} \, dx=\frac {2}{13} \, B b^{5} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{11} \, B a b^{4} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{11} \, A b^{5} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{9} \, B a^{2} b^{3} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{9} \, A a b^{4} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{7} \, B a^{3} b^{2} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{7} \, A a^{2} b^{3} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a^{4} b x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + 4 \, A a^{3} b^{2} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, B a^{5} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, A a^{4} b x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a^{5} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) \]
2/13*B*b^5*x^(13/2)*sgn(b*x + a) + 10/11*B*a*b^4*x^(11/2)*sgn(b*x + a) + 2 /11*A*b^5*x^(11/2)*sgn(b*x + a) + 20/9*B*a^2*b^3*x^(9/2)*sgn(b*x + a) + 10 /9*A*a*b^4*x^(9/2)*sgn(b*x + a) + 20/7*B*a^3*b^2*x^(7/2)*sgn(b*x + a) + 20 /7*A*a^2*b^3*x^(7/2)*sgn(b*x + a) + 2*B*a^4*b*x^(5/2)*sgn(b*x + a) + 4*A*a ^3*b^2*x^(5/2)*sgn(b*x + a) + 2/3*B*a^5*x^(3/2)*sgn(b*x + a) + 10/3*A*a^4* b*x^(3/2)*sgn(b*x + a) + 2*A*a^5*sqrt(x)*sgn(b*x + a)
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {x}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{\sqrt {x}} \,d x \]