Integrand size = 29, antiderivative size = 293 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=-\frac {a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {a^4 (5 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {5 a^3 b (2 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {5 a^2 b^2 (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{x^2 (a+b x)}-\frac {5 a b^3 (A b+2 a B) \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^5 B x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b^4 (A b+5 a B) \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \]
-1/5*a^5*A*((b*x+a)^2)^(1/2)/x^5/(b*x+a)-1/4*a^4*(5*A*b+B*a)*((b*x+a)^2)^( 1/2)/x^4/(b*x+a)-5/3*a^3*b*(2*A*b+B*a)*((b*x+a)^2)^(1/2)/x^3/(b*x+a)-5*a^2 *b^2*(A*b+B*a)*((b*x+a)^2)^(1/2)/x^2/(b*x+a)-5*a*b^3*(A*b+2*B*a)*((b*x+a)^ 2)^(1/2)/x/(b*x+a)+b^5*B*x*((b*x+a)^2)^(1/2)/(b*x+a)+b^4*(A*b+5*B*a)*ln(x) *((b*x+a)^2)^(1/2)/(b*x+a)
Time = 1.04 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.43 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=-\frac {\sqrt {(a+b x)^2} \left (300 a A b^4 x^4-60 b^5 B x^6+300 a^2 b^3 x^3 (A+2 B x)+100 a^3 b^2 x^2 (2 A+3 B x)+25 a^4 b x (3 A+4 B x)+3 a^5 (4 A+5 B x)-60 b^4 (A b+5 a B) x^5 \log (x)\right )}{60 x^5 (a+b x)} \]
-1/60*(Sqrt[(a + b*x)^2]*(300*a*A*b^4*x^4 - 60*b^5*B*x^6 + 300*a^2*b^3*x^3 *(A + 2*B*x) + 100*a^3*b^2*x^2*(2*A + 3*B*x) + 25*a^4*b*x*(3*A + 4*B*x) + 3*a^5*(4*A + 5*B*x) - 60*b^4*(A*b + 5*a*B)*x^5*Log[x]))/(x^5*(a + b*x))
Time = 0.29 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.45, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x)}{x^6} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5 (A+B x)}{x^6}dx}{b^5 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5 (A+B x)}{x^6}dx}{a+b x}\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^5}{x^6}+\frac {(5 A b+a B) a^4}{x^5}+\frac {5 b (2 A b+a B) a^3}{x^4}+\frac {10 b^2 (A b+a B) a^2}{x^3}+\frac {5 b^3 (A b+2 a B) a}{x^2}+b^5 B+\frac {b^4 (A b+5 a B)}{x}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^5 A}{5 x^5}-\frac {a^4 (a B+5 A b)}{4 x^4}-\frac {5 a^3 b (a B+2 A b)}{3 x^3}-\frac {5 a^2 b^2 (a B+A b)}{x^2}+b^4 \log (x) (5 a B+A b)-\frac {5 a b^3 (2 a B+A b)}{x}+b^5 B x\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/5*(a^5*A)/x^5 - (a^4*(5*A*b + a*B))/(4* x^4) - (5*a^3*b*(2*A*b + a*B))/(3*x^3) - (5*a^2*b^2*(A*b + a*B))/x^2 - (5* a*b^3*(A*b + 2*a*B))/x + b^5*B*x + b^4*(A*b + 5*a*B)*Log[x]))/(a + b*x)
3.7.95.3.1 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 = 0.54 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.49
| method | result | size |
| default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (60 A \ln \left (x \right ) x^{5} b^{5}+300 B \ln \left (x \right ) x^{5} a \,b^{4}+60 B \,b^{5} x^{6}-300 A a \,b^{4} x^{4}-600 B \,a^{2} b^{3} x^{4}-300 A \,a^{2} b^{3} x^{3}-300 B \,a^{3} b^{2} x^{3}-200 A \,a^{3} b^{2} x^{2}-100 B \,a^{4} b \,x^{2}-75 A \,a^{4} b x -15 a^{5} B x -12 A \,a^{5}\right )}{60 \left (b x +a \right )^{5} x^{5}}\) | \(144\) |
| risch | \(\frac {b^{5} B x \sqrt {\left (b x +a \right )^{2}}}{b x +a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-5 A a \,b^{4}-10 B \,a^{2} b^{3}\right ) x^{4}+\left (-5 A \,a^{2} b^{3}-5 B \,a^{3} b^{2}\right ) x^{3}+\left (-\frac {10}{3} A \,a^{3} b^{2}-\frac {5}{3} B \,a^{4} b \right ) x^{2}+\left (-\frac {5}{4} A \,a^{4} b -\frac {1}{4} a^{5} B \right ) x -\frac {A \,a^{5}}{5}\right )}{\left (b x +a \right ) x^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A \,b^{5}+5 B a \,b^{4}\right ) \ln \left (x \right )}{b x +a}\) | \(164\) |
1/60*((b*x+a)^2)^(5/2)*(60*A*ln(x)*x^5*b^5+300*B*ln(x)*x^5*a*b^4+60*B*b^5* x^6-300*A*a*b^4*x^4-600*B*a^2*b^3*x^4-300*A*a^2*b^3*x^3-300*B*a^3*b^2*x^3- 200*A*a^3*b^2*x^2-100*B*a^4*b*x^2-75*A*a^4*b*x-15*a^5*B*x-12*A*a^5)/(b*x+a )^5/x^5
Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.41 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=\frac {60 \, B b^{5} x^{6} + 60 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} \log \left (x\right ) - 12 \, A a^{5} - 300 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} - 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} - 100 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 15 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{60 \, x^{5}} \]
1/60*(60*B*b^5*x^6 + 60*(5*B*a*b^4 + A*b^5)*x^5*log(x) - 12*A*a^5 - 300*(2 *B*a^2*b^3 + A*a*b^4)*x^4 - 300*(B*a^3*b^2 + A*a^2*b^3)*x^3 - 100*(B*a^4*b + 2*A*a^3*b^2)*x^2 - 15*(B*a^5 + 5*A*a^4*b)*x)/x^5
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{6}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 673 vs. \(2 (210) = 420\).
Time = 0.20 (sec) , antiderivative size = 673, normalized size of antiderivative = 2.30 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=5 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} B a b^{4} \log \left (2 \, b^{2} x + 2 \, a b\right ) + \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A b^{5} \log \left (2 \, b^{2} x + 2 \, a b\right ) - 5 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B a b^{4} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{5} x}{2 \, a} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{6} x}{2 \, a^{2}} + \frac {15}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{4} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{5}}{2 \, a} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{5} x}{4 \, a^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{6} x}{4 \, a^{4}} + \frac {35 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{4}}{12 \, a^{2}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{5}}{12 \, a^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{4}}{3 \, a^{4}} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{5}}{15 \, a^{5}} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{3 \, a^{3} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{3 \, a^{4} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b^{2}}{3 \, a^{4} x^{2}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{3}}{15 \, a^{5} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b}{12 \, a^{3} x^{3}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{2}}{60 \, a^{4} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B}{4 \, a^{2} x^{4}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b}{20 \, a^{3} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A}{5 \, a^{2} x^{5}} \]
5*(-1)^(2*b^2*x + 2*a*b)*B*a*b^4*log(2*b^2*x + 2*a*b) + (-1)^(2*b^2*x + 2* a*b)*A*b^5*log(2*b^2*x + 2*a*b) - 5*(-1)^(2*a*b*x + 2*a^2)*B*a*b^4*log(2*a *b*x/abs(x) + 2*a^2/abs(x)) - (-1)^(2*a*b*x + 2*a^2)*A*b^5*log(2*a*b*x/abs (x) + 2*a^2/abs(x)) + 5/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^5*x/a + 1/2*sq rt(b^2*x^2 + 2*a*b*x + a^2)*A*b^6*x/a^2 + 15/2*sqrt(b^2*x^2 + 2*a*b*x + a^ 2)*B*b^4 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^5/a + 5/4*(b^2*x^2 + 2*a* b*x + a^2)^(3/2)*B*b^5*x/a^3 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^6*x /a^4 + 35/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^4/a^2 + 7/12*(b^2*x^2 + 2 *a*b*x + a^2)^(3/2)*A*b^5/a^3 + 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^4/ a^4 - 2/15*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^5/a^5 - 2/3*(b^2*x^2 + 2*a* b*x + a^2)^(5/2)*B*b^3/(a^3*x) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4 /(a^4*x) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b^2/(a^4*x^2) + 2/15*(b^2 *x^2 + 2*a*b*x + a^2)^(7/2)*A*b^3/(a^5*x^2) + 1/12*(b^2*x^2 + 2*a*b*x + a^ 2)^(7/2)*B*b/(a^3*x^3) - 11/60*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^2/(a^4* x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B/(a^2*x^4) + 3/20*(b^2*x^2 + 2 *a*b*x + a^2)^(7/2)*A*b/(a^3*x^4) - 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A/ (a^2*x^5)
Time = 0.27 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.64 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=B b^{5} x \mathrm {sgn}\left (b x + a\right ) + {\left (5 \, B a b^{4} \mathrm {sgn}\left (b x + a\right ) + A b^{5} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) - \frac {12 \, A a^{5} \mathrm {sgn}\left (b x + a\right ) + 300 \, {\left (2 \, B a^{2} b^{3} \mathrm {sgn}\left (b x + a\right ) + A a b^{4} \mathrm {sgn}\left (b x + a\right )\right )} x^{4} + 300 \, {\left (B a^{3} b^{2} \mathrm {sgn}\left (b x + a\right ) + A a^{2} b^{3} \mathrm {sgn}\left (b x + a\right )\right )} x^{3} + 100 \, {\left (B a^{4} b \mathrm {sgn}\left (b x + a\right ) + 2 \, A a^{3} b^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 15 \, {\left (B a^{5} \mathrm {sgn}\left (b x + a\right ) + 5 \, A a^{4} b \mathrm {sgn}\left (b x + a\right )\right )} x}{60 \, x^{5}} \]
B*b^5*x*sgn(b*x + a) + (5*B*a*b^4*sgn(b*x + a) + A*b^5*sgn(b*x + a))*log(a bs(x)) - 1/60*(12*A*a^5*sgn(b*x + a) + 300*(2*B*a^2*b^3*sgn(b*x + a) + A*a *b^4*sgn(b*x + a))*x^4 + 300*(B*a^3*b^2*sgn(b*x + a) + A*a^2*b^3*sgn(b*x + a))*x^3 + 100*(B*a^4*b*sgn(b*x + a) + 2*A*a^3*b^2*sgn(b*x + a))*x^2 + 15* (B*a^5*sgn(b*x + a) + 5*A*a^4*b*sgn(b*x + a))*x)/x^5
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{x^6} \,d x \]