Integrand size = 31, antiderivative size = 314 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{5/2}} \, dx=-\frac {2 a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}-\frac {2 a^4 (5 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {10 a^3 b (2 A b+a B) \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {20 a^2 b^2 (A b+a B) x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {2 a b^3 (A b+2 a B) x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 b^4 (A b+5 a B) x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {2 b^5 B x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \]
-2/3*a^5*A*((b*x+a)^2)^(1/2)/x^(3/2)/(b*x+a)+20/3*a^2*b^2*(A*b+B*a)*x^(3/2 )*((b*x+a)^2)^(1/2)/(b*x+a)+2*a*b^3*(A*b+2*B*a)*x^(5/2)*((b*x+a)^2)^(1/2)/ (b*x+a)+2/7*b^4*(A*b+5*B*a)*x^(7/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/9*b^5*B*x^ (9/2)*((b*x+a)^2)^(1/2)/(b*x+a)-2*a^4*(5*A*b+B*a)*((b*x+a)^2)^(1/2)/(b*x+a )/x^(1/2)+10*a^3*b*(2*A*b+B*a)*x^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.07 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.39 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{5/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (315 a^4 b x (A-B x)-210 a^3 b^2 x^2 (3 A+B x)+21 a^5 (A+3 B x)-42 a^2 b^3 x^3 (5 A+3 B x)-9 a b^4 x^4 (7 A+5 B x)-b^5 x^5 (9 A+7 B x)\right )}{63 x^{3/2} (a+b x)} \]
(-2*Sqrt[(a + b*x)^2]*(315*a^4*b*x*(A - B*x) - 210*a^3*b^2*x^2*(3*A + B*x) + 21*a^5*(A + 3*B*x) - 42*a^2*b^3*x^3*(5*A + 3*B*x) - 9*a*b^4*x^4*(7*A + 5*B*x) - b^5*x^5*(9*A + 7*B*x)))/(63*x^(3/2)*(a + b*x))
Time = 0.30 (sec) , antiderivative size = 153, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x)}{x^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5 (A+B x)}{x^{5/2}}dx}{b^5 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5 (A+B x)}{x^{5/2}}dx}{a+b x}\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^5}{x^{5/2}}+\frac {(5 A b+a B) a^4}{x^{3/2}}+\frac {5 b (2 A b+a B) a^3}{\sqrt {x}}+10 b^2 (A b+a B) \sqrt {x} a^2+5 b^3 (A b+2 a B) x^{3/2} a+b^5 B x^{7/2}+b^4 (A b+5 a B) x^{5/2}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 a^5 A}{3 x^{3/2}}-\frac {2 a^4 (a B+5 A b)}{\sqrt {x}}+10 a^3 b \sqrt {x} (a B+2 A b)+\frac {20}{3} a^2 b^2 x^{3/2} (a B+A b)+\frac {2}{7} b^4 x^{7/2} (5 a B+A b)+2 a b^3 x^{5/2} (2 a B+A b)+\frac {2}{9} b^5 B x^{9/2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*a^5*A)/(3*x^(3/2)) - (2*a^4*(5*A*b + a *B))/Sqrt[x] + 10*a^3*b*(2*A*b + a*B)*Sqrt[x] + (20*a^2*b^2*(A*b + a*B)*x^ (3/2))/3 + 2*a*b^3*(A*b + 2*a*B)*x^(5/2) + (2*b^4*(A*b + 5*a*B)*x^(7/2))/7 + (2*b^5*B*x^(9/2))/9))/(a + b*x)
3.9.6.3.1 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 = 0.45
| method | result | size |
| gosper | \(-\frac {2 \left (-7 B \,b^{5} x^{6}-9 A \,b^{5} x^{5}-45 B a \,b^{4} x^{5}-63 A a \,b^{4} x^{4}-126 B \,a^{2} b^{3} x^{4}-210 A \,a^{2} b^{3} x^{3}-210 B \,a^{3} b^{2} x^{3}-630 A \,a^{3} b^{2} x^{2}-315 B \,a^{4} b \,x^{2}+315 A \,a^{4} b x +63 a^{5} B x +21 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{63 x^{\frac {3}{2}} \left (b x +a \right )^{5}}\) | \(140\) |
| default | \(-\frac {2 \left (-7 B \,b^{5} x^{6}-9 A \,b^{5} x^{5}-45 B a \,b^{4} x^{5}-63 A a \,b^{4} x^{4}-126 B \,a^{2} b^{3} x^{4}-210 A \,a^{2} b^{3} x^{3}-210 B \,a^{3} b^{2} x^{3}-630 A \,a^{3} b^{2} x^{2}-315 B \,a^{4} b \,x^{2}+315 A \,a^{4} b x +63 a^{5} B x +21 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{63 x^{\frac {3}{2}} \left (b x +a \right )^{5}}\) | \(140\) |
| risch | \(-\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (-7 B \,b^{5} x^{6}-9 A \,b^{5} x^{5}-45 B a \,b^{4} x^{5}-63 A a \,b^{4} x^{4}-126 B \,a^{2} b^{3} x^{4}-210 A \,a^{2} b^{3} x^{3}-210 B \,a^{3} b^{2} x^{3}-630 A \,a^{3} b^{2} x^{2}-315 B \,a^{4} b \,x^{2}+315 A \,a^{4} b x +63 a^{5} B x +21 A \,a^{5}\right )}{63 \left (b x +a \right ) x^{\frac {3}{2}}}\) | \(140\) |
-2/63*(-7*B*b^5*x^6-9*A*b^5*x^5-45*B*a*b^4*x^5-63*A*a*b^4*x^4-126*B*a^2*b^ 3*x^4-210*A*a^2*b^3*x^3-210*B*a^3*b^2*x^3-630*A*a^3*b^2*x^2-315*B*a^4*b*x^ 2+315*A*a^4*b*x+63*B*a^5*x+21*A*a^5)*((b*x+a)^2)^(5/2)/x^(3/2)/(b*x+a)^5
Time = 0.32 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.38 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{5/2}} \, dx=\frac {2 \, {\left (7 \, B b^{5} x^{6} - 21 \, A a^{5} + 9 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 63 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 210 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 315 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 63 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x\right )}}{63 \, x^{\frac {3}{2}}} \]
2/63*(7*B*b^5*x^6 - 21*A*a^5 + 9*(5*B*a*b^4 + A*b^5)*x^5 + 63*(2*B*a^2*b^3 + A*a*b^4)*x^4 + 210*(B*a^3*b^2 + A*a^2*b^3)*x^3 + 315*(B*a^4*b + 2*A*a^3 *b^2)*x^2 - 63*(B*a^5 + 5*A*a^4*b)*x)/x^(3/2)
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{\frac {5}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.75 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{5/2}} \, dx=\frac {2}{105} \, {\left (3 \, {\left (5 \, b^{5} x^{2} + 7 \, a b^{4} x\right )} x^{\frac {3}{2}} + 28 \, {\left (3 \, a b^{4} x^{2} + 5 \, a^{2} b^{3} x\right )} \sqrt {x} + \frac {210 \, {\left (a^{2} b^{3} x^{2} + 3 \, a^{3} b^{2} x\right )}}{\sqrt {x}} + \frac {420 \, {\left (a^{3} b^{2} x^{2} - a^{4} b x\right )}}{x^{\frac {3}{2}}} - \frac {35 \, {\left (3 \, a^{4} b x^{2} + a^{5} x\right )}}{x^{\frac {5}{2}}}\right )} A + \frac {2}{315} \, {\left (5 \, {\left (7 \, b^{5} x^{2} + 9 \, a b^{4} x\right )} x^{\frac {5}{2}} + 36 \, {\left (5 \, a b^{4} x^{2} + 7 \, a^{2} b^{3} x\right )} x^{\frac {3}{2}} + 126 \, {\left (3 \, a^{2} b^{3} x^{2} + 5 \, a^{3} b^{2} x\right )} \sqrt {x} + \frac {420 \, {\left (a^{3} b^{2} x^{2} + 3 \, a^{4} b x\right )}}{\sqrt {x}} + \frac {315 \, {\left (a^{4} b x^{2} - a^{5} x\right )}}{x^{\frac {3}{2}}}\right )} B \]
2/105*(3*(5*b^5*x^2 + 7*a*b^4*x)*x^(3/2) + 28*(3*a*b^4*x^2 + 5*a^2*b^3*x)* sqrt(x) + 210*(a^2*b^3*x^2 + 3*a^3*b^2*x)/sqrt(x) + 420*(a^3*b^2*x^2 - a^4 *b*x)/x^(3/2) - 35*(3*a^4*b*x^2 + a^5*x)/x^(5/2))*A + 2/315*(5*(7*b^5*x^2 + 9*a*b^4*x)*x^(5/2) + 36*(5*a*b^4*x^2 + 7*a^2*b^3*x)*x^(3/2) + 126*(3*a^2 *b^3*x^2 + 5*a^3*b^2*x)*sqrt(x) + 420*(a^3*b^2*x^2 + 3*a^4*b*x)/sqrt(x) + 315*(a^4*b*x^2 - a^5*x)/x^(3/2))*B
Time = 0.26 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{5/2}} \, dx=\frac {2}{9} \, B b^{5} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, B a b^{4} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{7} \, A b^{5} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + 4 \, B a^{2} b^{3} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a b^{4} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{3} \, B a^{3} b^{2} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{3} \, A a^{2} b^{3} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 10 \, B a^{4} b \sqrt {x} \mathrm {sgn}\left (b x + a\right ) + 20 \, A a^{3} b^{2} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - \frac {2 \, {\left (3 \, B a^{5} x \mathrm {sgn}\left (b x + a\right ) + 15 \, A a^{4} b x \mathrm {sgn}\left (b x + a\right ) + A a^{5} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, x^{\frac {3}{2}}} \]
2/9*B*b^5*x^(9/2)*sgn(b*x + a) + 10/7*B*a*b^4*x^(7/2)*sgn(b*x + a) + 2/7*A *b^5*x^(7/2)*sgn(b*x + a) + 4*B*a^2*b^3*x^(5/2)*sgn(b*x + a) + 2*A*a*b^4*x ^(5/2)*sgn(b*x + a) + 20/3*B*a^3*b^2*x^(3/2)*sgn(b*x + a) + 20/3*A*a^2*b^3 *x^(3/2)*sgn(b*x + a) + 10*B*a^4*b*sqrt(x)*sgn(b*x + a) + 20*A*a^3*b^2*sqr t(x)*sgn(b*x + a) - 2/3*(3*B*a^5*x*sgn(b*x + a) + 15*A*a^4*b*x*sgn(b*x + a ) + A*a^5*sgn(b*x + a))/x^(3/2)
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{5/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{x^{5/2}} \,d x \]