Integrand size = 29, antiderivative size = 297 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=-\frac {a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {a^4 (5 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {5 a^3 b (2 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {5 a b^3 (A b+2 a B) x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b^4 (A b+5 a B) x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {b^5 B x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {10 a^2 b^2 (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \]
-1/3*a^5*A*((b*x+a)^2)^(1/2)/x^3/(b*x+a)-1/2*a^4*(5*A*b+B*a)*((b*x+a)^2)^( 1/2)/x^2/(b*x+a)-5*a^3*b*(2*A*b+B*a)*((b*x+a)^2)^(1/2)/x/(b*x+a)+5*a*b^3*( A*b+2*B*a)*x*((b*x+a)^2)^(1/2)/(b*x+a)+1/2*b^4*(A*b+5*B*a)*x^2*((b*x+a)^2) ^(1/2)/(b*x+a)+1/3*b^5*B*x^3*((b*x+a)^2)^(1/2)/(b*x+a)+10*a^2*b^2*(A*b+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 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=\frac {\sqrt {(a+b x)^2} \left (-60 a^3 A b^2 x^2+60 a^2 b^3 B x^4+15 a b^4 x^4 (2 A+B x)-15 a^4 b x (A+2 B x)+b^5 x^5 (3 A+2 B x)-a^5 (2 A+3 B x)+60 a^2 b^2 (A b+a B) x^3 \log (x)\right )}{6 x^3 (a+b x)} \]
(Sqrt[(a + b*x)^2]*(-60*a^3*A*b^2*x^2 + 60*a^2*b^3*B*x^4 + 15*a*b^4*x^4*(2 *A + B*x) - 15*a^4*b*x*(A + 2*B*x) + b^5*x^5*(3*A + 2*B*x) - a^5*(2*A + 3* B*x) + 60*a^2*b^2*(A*b + a*B)*x^3*Log[x]))/(6*x^3*(a + b*x))
Time = 0.30 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.46, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x)}{x^4} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5 (A+B x)}{x^4}dx}{b^5 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5 (A+B x)}{x^4}dx}{a+b x}\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^5}{x^4}+\frac {(5 A b+a B) a^4}{x^3}+\frac {5 b (2 A b+a B) a^3}{x^2}+\frac {10 b^2 (A b+a B) a^2}{x}+5 b^3 (A b+2 a B) a+b^5 B x^2+b^4 (A b+5 a B) x\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^5 A}{3 x^3}-\frac {a^4 (a B+5 A b)}{2 x^2}-\frac {5 a^3 b (a B+2 A b)}{x}+10 a^2 b^2 \log (x) (a B+A b)+\frac {1}{2} b^4 x^2 (5 a B+A b)+5 a b^3 x (2 a B+A b)+\frac {1}{3} b^5 B x^3\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/3*(a^5*A)/x^3 - (a^4*(5*A*b + a*B))/(2* x^2) - (5*a^3*b*(2*A*b + a*B))/x + 5*a*b^3*(A*b + 2*a*B)*x + (b^4*(A*b + 5 *a*B)*x^2)/2 + (b^5*B*x^3)/3 + 10*a^2*b^2*(A*b + a*B)*Log[x]))/(a + b*x)
3.7.93.3.1 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 = 0.31 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.48
| method | result | size |
| default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (2 B \,b^{5} x^{6}+3 A \,b^{5} x^{5}+15 B a \,b^{4} x^{5}+60 A \ln \left (x \right ) x^{3} a^{2} b^{3}+30 A a \,b^{4} x^{4}+60 B \ln \left (x \right ) x^{3} a^{3} b^{2}+60 B \,a^{2} b^{3} x^{4}-60 A \,a^{3} b^{2} x^{2}-30 B \,a^{4} b \,x^{2}-15 A \,a^{4} b x -3 a^{5} B x -2 A \,a^{5}\right )}{6 \left (b x +a \right )^{5} x^{3}}\) | \(144\) |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{3} \left (\frac {1}{3} B \,b^{2} x^{3}+\frac {1}{2} A \,b^{2} x^{2}+\frac {5}{2} B a b \,x^{2}+5 a A b x +10 a^{2} B x \right )}{b x +a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-10 A \,a^{3} b^{2}-5 B \,a^{4} b \right ) x^{2}+\left (-\frac {5}{2} A \,a^{4} b -\frac {1}{2} a^{5} B \right ) x -\frac {A \,a^{5}}{3}\right )}{\left (b x +a \right ) x^{3}}+\frac {10 \sqrt {\left (b x +a \right )^{2}}\, \left (A \,a^{2} b^{3}+B \,a^{3} b^{2}\right ) \ln \left (x \right )}{b x +a}\) | \(163\) |
1/6*((b*x+a)^2)^(5/2)*(2*B*b^5*x^6+3*A*b^5*x^5+15*B*a*b^4*x^5+60*A*ln(x)*x ^3*a^2*b^3+30*A*a*b^4*x^4+60*B*ln(x)*x^3*a^3*b^2+60*B*a^2*b^3*x^4-60*A*a^3 *b^2*x^2-30*B*a^4*b*x^2-15*A*a^4*b*x-3*a^5*B*x-2*A*a^5)/(b*x+a)^5/x^3
Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.41 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=\frac {2 \, B b^{5} x^{6} - 2 \, A a^{5} + 3 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 30 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 60 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} \log \left (x\right ) - 30 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 3 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{6 \, x^{3}} \]
1/6*(2*B*b^5*x^6 - 2*A*a^5 + 3*(5*B*a*b^4 + A*b^5)*x^5 + 30*(2*B*a^2*b^3 + A*a*b^4)*x^4 + 60*(B*a^3*b^2 + A*a^2*b^3)*x^3*log(x) - 30*(B*a^4*b + 2*A* a^3*b^2)*x^2 - 3*(B*a^5 + 5*A*a^4*b)*x)/x^3
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{4}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 557 vs. \(2 (212) = 424\).
Time = 0.21 (sec) , antiderivative size = 557, normalized size of antiderivative = 1.88 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=10 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} B a^{3} b^{2} \log \left (2 \, b^{2} x + 2 \, a b\right ) + 10 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A a^{2} b^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) - 10 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B a^{3} b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - 10 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A a^{2} b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + 5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a b^{3} x + 5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{4} x + 15 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2} b^{2} + 15 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a b^{3} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{3} x}{2 \, a} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{4} x}{2 \, a^{2}} + \frac {35}{6} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{2} + \frac {35 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{3}}{6 \, a} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{2 \, a^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{6 \, a^{3}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{2 \, a x} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{6 \, a^{2} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B}{2 \, a^{2} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b}{6 \, a^{3} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A}{3 \, a^{2} x^{3}} \]
10*(-1)^(2*b^2*x + 2*a*b)*B*a^3*b^2*log(2*b^2*x + 2*a*b) + 10*(-1)^(2*b^2* x + 2*a*b)*A*a^2*b^3*log(2*b^2*x + 2*a*b) - 10*(-1)^(2*a*b*x + 2*a^2)*B*a^ 3*b^2*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) - 10*(-1)^(2*a*b*x + 2*a^2)*A*a^2 *b^3*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) + 5*sqrt(b^2*x^2 + 2*a*b*x + a^2)* B*a*b^3*x + 5*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^4*x + 15*sqrt(b^2*x^2 + 2* a*b*x + a^2)*B*a^2*b^2 + 15*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a*b^3 + 5/2*(b ^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^3*x/a + 5/2*(b^2*x^2 + 2*a*b*x + a^2)^(3 /2)*A*b^4*x/a^2 + 35/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^2 + 35/6*(b^2*x ^2 + 2*a*b*x + a^2)^(3/2)*A*b^3/a + 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B* b^2/a^2 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/a^3 - 3/2*(b^2*x^2 + 2 *a*b*x + a^2)^(5/2)*B*b/(a*x) - 11/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2 /(a^2*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B/(a^2*x^2) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b/(a^3*x^2) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(7/2) *A/(a^2*x^3)
Time = 0.26 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.64 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=\frac {1}{3} \, B b^{5} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, B a b^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A b^{5} x^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, B a^{2} b^{3} x \mathrm {sgn}\left (b x + a\right ) + 5 \, A a b^{4} x \mathrm {sgn}\left (b x + a\right ) + 10 \, {\left (B a^{3} b^{2} \mathrm {sgn}\left (b x + a\right ) + A a^{2} b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) - \frac {2 \, A a^{5} \mathrm {sgn}\left (b x + a\right ) + 30 \, {\left (B a^{4} b \mathrm {sgn}\left (b x + a\right ) + 2 \, A a^{3} b^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 3 \, {\left (B a^{5} \mathrm {sgn}\left (b x + a\right ) + 5 \, A a^{4} b \mathrm {sgn}\left (b x + a\right )\right )} x}{6 \, x^{3}} \]
1/3*B*b^5*x^3*sgn(b*x + a) + 5/2*B*a*b^4*x^2*sgn(b*x + a) + 1/2*A*b^5*x^2* sgn(b*x + a) + 10*B*a^2*b^3*x*sgn(b*x + a) + 5*A*a*b^4*x*sgn(b*x + a) + 10 *(B*a^3*b^2*sgn(b*x + a) + A*a^2*b^3*sgn(b*x + a))*log(abs(x)) - 1/6*(2*A* a^5*sgn(b*x + a) + 30*(B*a^4*b*sgn(b*x + a) + 2*A*a^3*b^2*sgn(b*x + a))*x^ 2 + 3*(B*a^5*sgn(b*x + a) + 5*A*a^4*b*sgn(b*x + a))*x)/x^3
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^4} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{x^4} \,d x \]