Integrand size = 29, antiderivative size = 75 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^6} \, dx=\frac {(A b-a B) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 a^2 x^4}-\frac {A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 a^2 x^5} \] Output:
1/4*(A*b-B*a)*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/a^2/x^4-1/5*A*(b^2*x^2+2 *a*b*x+a^2)^(5/2)/a^2/x^5
Time = 0.86 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.57 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^6} \, dx=-\frac {(a+b x)^3 \sqrt {(a+b x)^2} (4 a A-A b x+5 a B x)}{20 a^2 x^5} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^6,x]
Output:
-1/20*((a + b*x)^3*Sqrt[(a + b*x)^2]*(4*a*A - A*b*x + 5*a*B*x))/(a^2*x^5)
Time = 0.37 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1186, 1102, 27, 48}
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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^6} \, dx\) |
\(\Big \downarrow \) 1186 |
\(\displaystyle -\frac {(A b-a B) \int \frac {\left (a^2+2 b x a+b^2 x^2\right )^{3/2}}{x^5}dx}{a}-\frac {A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 a^2 x^5}\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (A b-a B) \int \frac {b^3 (a+b x)^3}{x^5}dx}{a b^3 (a+b x)}-\frac {A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 a^2 x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (A b-a B) \int \frac {(a+b x)^3}{x^5}dx}{a (a+b x)}-\frac {A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 a^2 x^5}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (A b-a B)}{4 a^2 x^4}-\frac {A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 a^2 x^5}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^6,x]
Output:
((A*b - a*B)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*a^2*x^4) - (A*( a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(5*a^2*x^5)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b*e)^2)), x] + Simp[(2*c*f - b*g)/ (2*c*d - b*e) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[ {a, b, c, d, e, f, g, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]
Time = 1.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.20
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-B \,b^{3} x^{4}+\left (-\frac {1}{2} A \,b^{3}-\frac {3}{2} B a \,b^{2}\right ) x^{3}+\left (-A a \,b^{2}-B \,a^{2} b \right ) x^{2}+\left (-\frac {3}{4} A \,a^{2} b -\frac {1}{4} B \,a^{3}\right ) x -\frac {a^{3} A}{5}\right )}{\left (b x +a \right ) x^{5}}\) | \(90\) |
gosper | \(-\frac {\left (20 B \,b^{3} x^{4}+10 A \,b^{3} x^{3}+30 B a \,b^{2} x^{3}+20 A a \,b^{2} x^{2}+20 B \,a^{2} b \,x^{2}+15 A \,a^{2} b x +5 B \,a^{3} x +4 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 x^{5} \left (b x +a \right )^{3}}\) | \(92\) |
default | \(-\frac {\left (20 B \,b^{3} x^{4}+10 A \,b^{3} x^{3}+30 B a \,b^{2} x^{3}+20 A a \,b^{2} x^{2}+20 B \,a^{2} b \,x^{2}+15 A \,a^{2} b x +5 B \,a^{3} x +4 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 x^{5} \left (b x +a \right )^{3}}\) | \(92\) |
orering | \(-\frac {\left (20 B \,b^{3} x^{4}+10 A \,b^{3} x^{3}+30 B a \,b^{2} x^{3}+20 A a \,b^{2} x^{2}+20 B \,a^{2} b \,x^{2}+15 A \,a^{2} b x +5 B \,a^{3} x +4 a^{3} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{20 x^{5} \left (b x +a \right )^{3}}\) | \(101\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^6,x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)*(-B*b^3*x^4+(-1/2*A*b^3-3/2*B*a*b^2)*x^3+(-A*a*b ^2-B*a^2*b)*x^2+(-3/4*A*a^2*b-1/4*B*a^3)*x-1/5*a^3*A)/x^5
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^6} \, dx=-\frac {20 \, B b^{3} x^{4} + 4 \, A a^{3} + 10 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 20 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{20 \, x^{5}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^6,x, algorithm="fricas")
Output:
-1/20*(20*B*b^3*x^4 + 4*A*a^3 + 10*(3*B*a*b^2 + A*b^3)*x^3 + 20*(B*a^2*b + A*a*b^2)*x^2 + 5*(B*a^3 + 3*A*a^2*b)*x)/x^5
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^6} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{6}}\, dx \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/x**6,x)
Output:
Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**6, x)
Leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (67) = 134\).
Time = 0.05 (sec) , antiderivative size = 315, normalized size of antiderivative = 4.20 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^6} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{4}}{4 \, a^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{5}}{4 \, a^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{3}}{4 \, a^{3} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{4}}{4 \, a^{4} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{4 \, a^{4} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{4 \, a^{5} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{4 \, a^{3} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{4 \, a^{4} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{4 \, a^{2} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{4 \, a^{3} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{5 \, a^{2} x^{5}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^6,x, algorithm="maxima")
Output:
1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^4/a^4 - 1/4*(b^2*x^2 + 2*a*b*x + a ^2)^(3/2)*A*b^5/a^5 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^3/(a^3*x) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^4/(a^4*x) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^2/(a^4*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/ (a^5*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b/(a^3*x^3) - 1/4*(b^2*x ^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^ (5/2)*B/(a^2*x^4) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^4) - 1/ 5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A/(a^2*x^5)
Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (67) = 134\).
Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.99 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^6} \, dx=-\frac {{\left (5 \, B a b^{4} - A b^{5}\right )} \mathrm {sgn}\left (b x + a\right )}{20 \, a^{2}} - \frac {20 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 30 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 20 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 20 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 15 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 4 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{20 \, x^{5}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^6,x, algorithm="giac")
Output:
-1/20*(5*B*a*b^4 - A*b^5)*sgn(b*x + a)/a^2 - 1/20*(20*B*b^3*x^4*sgn(b*x + a) + 30*B*a*b^2*x^3*sgn(b*x + a) + 10*A*b^3*x^3*sgn(b*x + a) + 20*B*a^2*b* x^2*sgn(b*x + a) + 20*A*a*b^2*x^2*sgn(b*x + a) + 5*B*a^3*x*sgn(b*x + a) + 15*A*a^2*b*x*sgn(b*x + a) + 4*A*a^3*sgn(b*x + a))/x^5
Time = 10.77 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.61 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^6} \, dx=-\frac {\left (\frac {B\,a^3}{4}+\frac {3\,A\,b\,a^2}{4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^4\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^3}{2}+\frac {3\,B\,a\,b^2}{2}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^2\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,x^5\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x\,\left (a+b\,x\right )}-\frac {a\,b\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^3\,\left (a+b\,x\right )} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^6,x)
Output:
- (((B*a^3)/4 + (3*A*a^2*b)/4)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^4*(a + b*x)) - (((A*b^3)/2 + (3*B*a*b^2)/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^2 *(a + b*x)) - (A*a^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(5*x^5*(a + b*x)) - (B*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x*(a + b*x)) - (a*b*(A*b + B*a)*( a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^3*(a + b*x))
Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.61 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^6} \, dx=\frac {-5 b^{4} x^{4}-10 a \,b^{3} x^{3}-10 a^{2} b^{2} x^{2}-5 a^{3} b x -a^{4}}{5 x^{5}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^6,x)
Output:
( - a**4 - 5*a**3*b*x - 10*a**2*b**2*x**2 - 10*a*b**3*x**3 - 5*b**4*x**4)/ (5*x**5)