Integrand size = 29, antiderivative size = 210 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{12}} \, dx=-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{11 x^{11} (a+b x)}-\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{10 x^{10} (a+b x)}-\frac {a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^9 (a+b x)}-\frac {b^2 (A b+3 a B) \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)} \]
-1/11*a^3*A*((b*x+a)^2)^(1/2)/x^11/(b*x+a)-1/10*a^2*(3*A*b+B*a)*((b*x+a)^2 )^(1/2)/x^10/(b*x+a)-1/3*a*b*(A*b+B*a)*((b*x+a)^2)^(1/2)/x^9/(b*x+a)-1/8*b ^2*(A*b+3*B*a)*((b*x+a)^2)^(1/2)/x^8/(b*x+a)-1/7*b^3*B*((b*x+a)^2)^(1/2)/x ^7/(b*x+a)
Time = 0.84 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.41 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{12}} \, dx=-\frac {\sqrt {(a+b x)^2} \left (165 b^3 x^3 (7 A+8 B x)+385 a b^2 x^2 (8 A+9 B x)+308 a^2 b x (9 A+10 B x)+84 a^3 (10 A+11 B x)\right )}{9240 x^{11} (a+b x)} \]
-1/9240*(Sqrt[(a + b*x)^2]*(165*b^3*x^3*(7*A + 8*B*x) + 385*a*b^2*x^2*(8*A + 9*B*x) + 308*a^2*b*x*(9*A + 10*B*x) + 84*a^3*(10*A + 11*B*x)))/(x^11*(a + b*x))
Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.49, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^{12}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{x^{12}}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{x^{12}}dx}{a+b x}\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^3}{x^{12}}+\frac {(3 A b+a B) a^2}{x^{11}}+\frac {3 b (A b+a B) a}{x^{10}}+\frac {b^3 B}{x^8}+\frac {b^2 (A b+3 a B)}{x^9}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^3 A}{11 x^{11}}-\frac {a^2 (a B+3 A b)}{10 x^{10}}-\frac {b^2 (3 a B+A b)}{8 x^8}-\frac {a b (a B+A b)}{3 x^9}-\frac {b^3 B}{7 x^7}\right )}{a+b x}\) |
((-1/11*(a^3*A)/x^11 - (a^2*(3*A*b + a*B))/(10*x^10) - (a*b*(A*b + a*B))/( 3*x^9) - (b^2*(A*b + 3*a*B))/(8*x^8) - (b^3*B)/(7*x^7))*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x)
3.7.82.3.1 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 = 1.85 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.43
| method | result | size |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {x^{4} B \,b^{3}}{7}+\left (-\frac {1}{8} A \,b^{3}-\frac {3}{8} B a \,b^{2}\right ) x^{3}+\left (-\frac {1}{3} A a \,b^{2}-\frac {1}{3} B b \,a^{2}\right ) x^{2}+\left (-\frac {3}{10} A \,a^{2} b -\frac {1}{10} B \,a^{3}\right ) x -\frac {A \,a^{3}}{11}\right )}{\left (b x +a \right ) x^{11}}\) | \(90\) |
| gosper | \(-\frac {\left (1320 x^{4} B \,b^{3}+1155 A \,b^{3} x^{3}+3465 B a \,b^{2} x^{3}+3080 A a \,b^{2} x^{2}+3080 B \,a^{2} b \,x^{2}+2772 A \,a^{2} b x +924 a^{3} B x +840 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{9240 x^{11} \left (b x +a \right )^{3}}\) | \(92\) |
| default | \(-\frac {\left (1320 x^{4} B \,b^{3}+1155 A \,b^{3} x^{3}+3465 B a \,b^{2} x^{3}+3080 A a \,b^{2} x^{2}+3080 B \,a^{2} b \,x^{2}+2772 A \,a^{2} b x +924 a^{3} B x +840 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{9240 x^{11} \left (b x +a \right )^{3}}\) | \(92\) |
((b*x+a)^2)^(1/2)/(b*x+a)*(-1/7*x^4*B*b^3+(-1/8*A*b^3-3/8*B*a*b^2)*x^3+(-1 /3*A*a*b^2-1/3*B*b*a^2)*x^2+(-3/10*A*a^2*b-1/10*B*a^3)*x-1/11*A*a^3)/x^11
Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.35 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{12}} \, dx=-\frac {1320 \, B b^{3} x^{4} + 840 \, A a^{3} + 1155 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 3080 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 924 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{9240 \, x^{11}} \]
-1/9240*(1320*B*b^3*x^4 + 840*A*a^3 + 1155*(3*B*a*b^2 + A*b^3)*x^3 + 3080* (B*a^2*b + A*a*b^2)*x^2 + 924*(B*a^3 + 3*A*a^2*b)*x)/x^11
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{12}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{12}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 675 vs. \(2 (145) = 290\).
Time = 0.23 (sec) , antiderivative size = 675, normalized size of antiderivative = 3.21 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{12}} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{10}}{4 \, a^{10}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{11}}{4 \, a^{11}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{9}}{4 \, a^{9} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{10}}{4 \, a^{10} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{8}}{4 \, a^{10} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{9}}{4 \, a^{11} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{7}}{4 \, a^{9} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{8}}{4 \, a^{10} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{6}}{4 \, a^{8} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{7}}{4 \, a^{9} x^{4}} + \frac {209 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{5}}{840 \, a^{7} x^{5}} - \frac {329 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{6}}{1320 \, a^{8} x^{5}} - \frac {41 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{4}}{168 \, a^{6} x^{6}} + \frac {65 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{5}}{264 \, a^{7} x^{6}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{56 \, a^{5} x^{7}} - \frac {21 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{88 \, a^{6} x^{7}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{24 \, a^{4} x^{8}} + \frac {59 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{264 \, a^{5} x^{8}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{6 \, a^{3} x^{9}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{66 \, a^{4} x^{9}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{10 \, a^{2} x^{10}} + \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{110 \, a^{3} x^{10}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{11 \, a^{2} x^{11}} \]
1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^10/a^10 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^11/a^11 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^9/(a^9*x ) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^10/(a^10*x) - 1/4*(b^2*x^2 + 2 *a*b*x + a^2)^(5/2)*B*b^8/(a^10*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2) *A*b^9/(a^11*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^7/(a^9*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^8/(a^10*x^3) - 1/4*(b^2*x^2 + 2*a* b*x + a^2)^(5/2)*B*b^6/(a^8*x^4) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b ^7/(a^9*x^4) + 209/840*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^5/(a^7*x^5) - 3 29/1320*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^6/(a^8*x^5) - 41/168*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^4/(a^6*x^6) + 65/264*(b^2*x^2 + 2*a*b*x + a^2)^ (5/2)*A*b^5/(a^7*x^6) + 13/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^3/(a^5*x ^7) - 21/88*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4/(a^6*x^7) - 5/24*(b^2*x^ 2 + 2*a*b*x + a^2)^(5/2)*B*b^2/(a^4*x^8) + 59/264*(b^2*x^2 + 2*a*b*x + a^2 )^(5/2)*A*b^3/(a^5*x^8) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b/(a^3*x^9 ) - 13/66*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^9) - 1/10*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B/(a^2*x^10) + 17/110*(b^2*x^2 + 2*a*b*x + a^2)^(5/ 2)*A*b/(a^3*x^10) - 1/11*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A/(a^2*x^11)
Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{12}} \, dx=-\frac {{\left (11 \, B a b^{10} - 7 \, A b^{11}\right )} \mathrm {sgn}\left (b x + a\right )}{9240 \, a^{8}} - \frac {1320 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 3465 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 1155 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 3080 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 3080 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 924 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 2772 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 840 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{9240 \, x^{11}} \]
-1/9240*(11*B*a*b^10 - 7*A*b^11)*sgn(b*x + a)/a^8 - 1/9240*(1320*B*b^3*x^4 *sgn(b*x + a) + 3465*B*a*b^2*x^3*sgn(b*x + a) + 1155*A*b^3*x^3*sgn(b*x + a ) + 3080*B*a^2*b*x^2*sgn(b*x + a) + 3080*A*a*b^2*x^2*sgn(b*x + a) + 924*B* a^3*x*sgn(b*x + a) + 2772*A*a^2*b*x*sgn(b*x + a) + 840*A*a^3*sgn(b*x + a)) /x^11
Time = 10.02 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{12}} \, dx=-\frac {\left (\frac {B\,a^3}{10}+\frac {3\,A\,b\,a^2}{10}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^{10}\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^3}{8}+\frac {3\,B\,a\,b^2}{8}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^8\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{11\,x^{11}\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {a\,b\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,x^9\,\left (a+b\,x\right )} \]
- (((B*a^3)/10 + (3*A*a^2*b)/10)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^10*(a + b*x)) - (((A*b^3)/8 + (3*B*a*b^2)/8)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/( x^8*(a + b*x)) - (A*a^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(11*x^11*(a + b*x )) - (B*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(7*x^7*(a + b*x)) - (a*b*(A*b + B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(3*x^9*(a + b*x))