Integrand size = 29, antiderivative size = 306 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {a^5 A x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a^4 (5 A b+a B) x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}+\frac {5 a^3 b (2 A b+a B) x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {5 a^2 b^2 (A b+a B) x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {5 a b^3 (A b+2 a B) x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {b^4 (A b+5 a B) x^{10} \sqrt {a^2+2 a b x+b^2 x^2}}{10 (a+b x)}+\frac {b^5 B x^{11} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)} \]
1/5*a^5*A*x^5*((b*x+a)^2)^(1/2)/(b*x+a)+1/6*a^4*(5*A*b+B*a)*x^6*((b*x+a)^2 )^(1/2)/(b*x+a)+5/7*a^3*b*(2*A*b+B*a)*x^7*((b*x+a)^2)^(1/2)/(b*x+a)+5/4*a^ 2*b^2*(A*b+B*a)*x^8*((b*x+a)^2)^(1/2)/(b*x+a)+5/9*a*b^3*(A*b+2*B*a)*x^9*(( b*x+a)^2)^(1/2)/(b*x+a)+1/10*b^4*(A*b+5*B*a)*x^10*((b*x+a)^2)^(1/2)/(b*x+a )+1/11*b^5*B*x^11*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 1.03 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.41 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {x^5 \sqrt {(a+b x)^2} \left (462 a^5 (6 A+5 B x)+1650 a^4 b x (7 A+6 B x)+2475 a^3 b^2 x^2 (8 A+7 B x)+1925 a^2 b^3 x^3 (9 A+8 B x)+770 a b^4 x^4 (10 A+9 B x)+126 b^5 x^5 (11 A+10 B x)\right )}{13860 (a+b x)} \]
(x^5*Sqrt[(a + b*x)^2]*(462*a^5*(6*A + 5*B*x) + 1650*a^4*b*x*(7*A + 6*B*x) + 2475*a^3*b^2*x^2*(8*A + 7*B*x) + 1925*a^2*b^3*x^3*(9*A + 8*B*x) + 770*a *b^4*x^4*(10*A + 9*B*x) + 126*b^5*x^5*(11*A + 10*B*x)))/(13860*(a + b*x))
Time = 0.31 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.47, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, 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.
| \(\displaystyle \int x^4 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x) \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^5 x^4 (a+b x)^5 (A+B x)dx}{b^5 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^4 (a+b x)^5 (A+B x)dx}{a+b x}\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^5 B x^{10}+b^4 (A b+5 a B) x^9+5 a b^3 (A b+2 a B) x^8+10 a^2 b^2 (A b+a B) x^7+5 a^3 b (2 A b+a B) x^6+a^4 (5 A b+a B) x^5+a^5 A x^4\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{5} a^5 A x^5+\frac {1}{6} a^4 x^6 (a B+5 A b)+\frac {5}{7} a^3 b x^7 (a B+2 A b)+\frac {5}{4} a^2 b^2 x^8 (a B+A b)+\frac {1}{10} b^4 x^{10} (5 a B+A b)+\frac {5}{9} a b^3 x^9 (2 a B+A b)+\frac {1}{11} b^5 B x^{11}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((a^5*A*x^5)/5 + (a^4*(5*A*b + a*B)*x^6)/6 + (5*a^3*b*(2*A*b + a*B)*x^7)/7 + (5*a^2*b^2*(A*b + a*B)*x^8)/4 + (5*a*b^3 *(A*b + 2*a*B)*x^9)/9 + (b^4*(A*b + 5*a*B)*x^10)/10 + (b^5*B*x^11)/11))/(a + b*x)
3.7.85.3.1 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 = 0.39 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.46
| method | result | size |
| gosper | \(\frac {x^{5} \left (1260 B \,b^{5} x^{6}+1386 A \,b^{5} x^{5}+6930 B a \,b^{4} x^{5}+7700 A a \,b^{4} x^{4}+15400 B \,a^{2} b^{3} x^{4}+17325 A \,a^{2} b^{3} x^{3}+17325 B \,a^{3} b^{2} x^{3}+19800 A \,a^{3} b^{2} x^{2}+9900 B \,a^{4} b \,x^{2}+11550 A \,a^{4} b x +2310 a^{5} B x +2772 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{13860 \left (b x +a \right )^{5}}\) | \(140\) |
| default | \(\frac {x^{5} \left (1260 B \,b^{5} x^{6}+1386 A \,b^{5} x^{5}+6930 B a \,b^{4} x^{5}+7700 A a \,b^{4} x^{4}+15400 B \,a^{2} b^{3} x^{4}+17325 A \,a^{2} b^{3} x^{3}+17325 B \,a^{3} b^{2} x^{3}+19800 A \,a^{3} b^{2} x^{2}+9900 B \,a^{4} b \,x^{2}+11550 A \,a^{4} b x +2310 a^{5} B x +2772 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{13860 \left (b x +a \right )^{5}}\) | \(140\) |
| risch | \(\frac {b^{5} B \,x^{11} \sqrt {\left (b x +a \right )^{2}}}{11 b x +11 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A \,b^{5}+5 B a \,b^{4}\right ) x^{10}}{10 b x +10 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (5 A a \,b^{4}+10 B \,a^{2} b^{3}\right ) x^{9}}{9 b x +9 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (10 A \,a^{2} b^{3}+10 B \,a^{3} b^{2}\right ) x^{8}}{8 b x +8 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (10 A \,a^{3} b^{2}+5 B \,a^{4} b \right ) x^{7}}{7 b x +7 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (5 A \,a^{4} b +a^{5} B \right ) x^{6}}{6 b x +6 a}+\frac {a^{5} A \,x^{5} \sqrt {\left (b x +a \right )^{2}}}{5 b x +5 a}\) | \(236\) |
1/13860*x^5*(1260*B*b^5*x^6+1386*A*b^5*x^5+6930*B*a*b^4*x^5+7700*A*a*b^4*x ^4+15400*B*a^2*b^3*x^4+17325*A*a^2*b^3*x^3+17325*B*a^3*b^2*x^3+19800*A*a^3 *b^2*x^2+9900*B*a^4*b*x^2+11550*A*a^4*b*x+2310*B*a^5*x+2772*A*a^5)*((b*x+a )^2)^(5/2)/(b*x+a)^5
Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.39 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{11} \, B b^{5} x^{11} + \frac {1}{5} \, A a^{5} x^{5} + \frac {1}{10} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{10} + \frac {5}{9} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{9} + \frac {5}{4} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{8} + \frac {5}{7} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{7} + \frac {1}{6} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{6} \]
1/11*B*b^5*x^11 + 1/5*A*a^5*x^5 + 1/10*(5*B*a*b^4 + A*b^5)*x^10 + 5/9*(2*B *a^2*b^3 + A*a*b^4)*x^9 + 5/4*(B*a^3*b^2 + A*a^2*b^3)*x^8 + 5/7*(B*a^4*b + 2*A*a^3*b^2)*x^7 + 1/6*(B*a^5 + 5*A*a^4*b)*x^6
Leaf count of result is larger than twice the leaf count of optimal. 15400 vs. \(2 (233) = 466\).
Time = 1.05 (sec) , antiderivative size = 15400, normalized size of antiderivative = 50.33 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\text {Too large to display} \]
Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(B*b**4*x**10/11 + x**9*(A*b** 6 + 45*B*a*b**5/11)/(10*b**2) + x**8*(6*A*a*b**5 + 155*B*a**2*b**4/11 - 19 *a*(A*b**6 + 45*B*a*b**5/11)/(10*b))/(9*b**2) + x**7*(15*A*a**2*b**4 + 20* B*a**3*b**3 - 9*a**2*(A*b**6 + 45*B*a*b**5/11)/(10*b**2) - 17*a*(6*A*a*b** 5 + 155*B*a**2*b**4/11 - 19*a*(A*b**6 + 45*B*a*b**5/11)/(10*b))/(9*b))/(8* b**2) + x**6*(20*A*a**3*b**3 + 15*B*a**4*b**2 - 8*a**2*(6*A*a*b**5 + 155*B *a**2*b**4/11 - 19*a*(A*b**6 + 45*B*a*b**5/11)/(10*b))/(9*b**2) - 15*a*(15 *A*a**2*b**4 + 20*B*a**3*b**3 - 9*a**2*(A*b**6 + 45*B*a*b**5/11)/(10*b**2) - 17*a*(6*A*a*b**5 + 155*B*a**2*b**4/11 - 19*a*(A*b**6 + 45*B*a*b**5/11)/ (10*b))/(9*b))/(8*b))/(7*b**2) + x**5*(15*A*a**4*b**2 + 6*B*a**5*b - 7*a** 2*(15*A*a**2*b**4 + 20*B*a**3*b**3 - 9*a**2*(A*b**6 + 45*B*a*b**5/11)/(10* b**2) - 17*a*(6*A*a*b**5 + 155*B*a**2*b**4/11 - 19*a*(A*b**6 + 45*B*a*b**5 /11)/(10*b))/(9*b))/(8*b**2) - 13*a*(20*A*a**3*b**3 + 15*B*a**4*b**2 - 8*a **2*(6*A*a*b**5 + 155*B*a**2*b**4/11 - 19*a*(A*b**6 + 45*B*a*b**5/11)/(10* b))/(9*b**2) - 15*a*(15*A*a**2*b**4 + 20*B*a**3*b**3 - 9*a**2*(A*b**6 + 45 *B*a*b**5/11)/(10*b**2) - 17*a*(6*A*a*b**5 + 155*B*a**2*b**4/11 - 19*a*(A* b**6 + 45*B*a*b**5/11)/(10*b))/(9*b))/(8*b))/(7*b))/(6*b**2) + x**4*(6*A*a **5*b + B*a**6 - 6*a**2*(20*A*a**3*b**3 + 15*B*a**4*b**2 - 8*a**2*(6*A*a*b **5 + 155*B*a**2*b**4/11 - 19*a*(A*b**6 + 45*B*a*b**5/11)/(10*b))/(9*b**2) - 15*a*(15*A*a**2*b**4 + 20*B*a**3*b**3 - 9*a**2*(A*b**6 + 45*B*a*b**5...
Time = 0.20 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.18 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B x^{4}}{11 \, b^{2}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a x^{3}}{22 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A x^{3}}{10 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{5} x}{6 \, b^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{4} x}{6 \, b^{4}} + \frac {31 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{2} x^{2}}{198 \, b^{4}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a x^{2}}{90 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{6}}{6 \, b^{6}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{5}}{6 \, b^{5}} - \frac {65 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{3} x}{396 \, b^{5}} + \frac {29 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a^{2} x}{180 \, b^{4}} + \frac {461 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{4}}{2772 \, b^{6}} - \frac {209 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a^{3}}{1260 \, b^{5}} \]
1/11*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*x^4/b^2 - 3/22*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a*x^3/b^3 + 1/10*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*x^3/b^2 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^5*x/b^5 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a^4*x/b^4 + 31/198*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^2*x^ 2/b^4 - 13/90*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a*x^2/b^3 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^6/b^6 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a^ 5/b^5 - 65/396*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^3*x/b^5 + 29/180*(b^2*x ^2 + 2*a*b*x + a^2)^(7/2)*A*a^2*x/b^4 + 461/2772*(b^2*x^2 + 2*a*b*x + a^2) ^(7/2)*B*a^4/b^6 - 209/1260*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a^3/b^5
Time = 0.27 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.73 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{11} \, B b^{5} x^{11} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a b^{4} x^{10} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{10} \, A b^{5} x^{10} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{9} \, B a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{9} \, A a b^{4} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, B a^{3} b^{2} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, A a^{2} b^{3} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{7} \, B a^{4} b x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, A a^{3} b^{2} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, B a^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{6} \, A a^{4} b x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A a^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (5 \, B a^{11} - 11 \, A a^{10} b\right )} \mathrm {sgn}\left (b x + a\right )}{13860 \, b^{6}} \]
1/11*B*b^5*x^11*sgn(b*x + a) + 1/2*B*a*b^4*x^10*sgn(b*x + a) + 1/10*A*b^5* x^10*sgn(b*x + a) + 10/9*B*a^2*b^3*x^9*sgn(b*x + a) + 5/9*A*a*b^4*x^9*sgn( b*x + a) + 5/4*B*a^3*b^2*x^8*sgn(b*x + a) + 5/4*A*a^2*b^3*x^8*sgn(b*x + a) + 5/7*B*a^4*b*x^7*sgn(b*x + a) + 10/7*A*a^3*b^2*x^7*sgn(b*x + a) + 1/6*B* a^5*x^6*sgn(b*x + a) + 5/6*A*a^4*b*x^6*sgn(b*x + a) + 1/5*A*a^5*x^5*sgn(b* x + a) - 1/13860*(5*B*a^11 - 11*A*a^10*b)*sgn(b*x + a)/b^6
Timed out. \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\int x^4\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2} \,d x \]