Integrand size = 29, antiderivative size = 157 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^8} \, dx=\frac {b^2 (A b-a B) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 a^4 x^4}-\frac {A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{7 a^2 x^7}+\frac {(9 A b-7 a B) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{42 a^3 x^6}-\frac {b (51 A b-49 a B) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{210 a^4 x^5} \] Output:
1/4*b^2*(A*b-B*a)*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/a^4/x^4-1/7*A*(b^2*x ^2+2*a*b*x+a^2)^(5/2)/a^2/x^7+1/42*(9*A*b-7*B*a)*(b^2*x^2+2*a*b*x+a^2)^(5/ 2)/a^3/x^6-1/210*b*(51*A*b-49*B*a)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/a^4/x^5
Time = 1.01 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.55 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^8} \, dx=-\frac {\sqrt {(a+b x)^2} \left (35 b^3 x^3 (3 A+4 B x)+63 a b^2 x^2 (4 A+5 B x)+42 a^2 b x (5 A+6 B x)+10 a^3 (6 A+7 B x)\right )}{420 x^7 (a+b x)} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^8,x]
Output:
-1/420*(Sqrt[(a + b*x)^2]*(35*b^3*x^3*(3*A + 4*B*x) + 63*a*b^2*x^2*(4*A + 5*B*x) + 42*a^2*b*x*(5*A + 6*B*x) + 10*a^3*(6*A + 7*B*x)))/(x^7*(a + b*x))
Time = 0.43 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.66, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^8} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{x^8}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{x^8}dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^3}{x^8}+\frac {(3 A b+a B) a^2}{x^7}+\frac {3 b (A b+a B) a}{x^6}+\frac {b^3 B}{x^4}+\frac {b^2 (A b+3 a B)}{x^5}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^3 A}{7 x^7}-\frac {a^2 (a B+3 A b)}{6 x^6}-\frac {b^2 (3 a B+A b)}{4 x^4}-\frac {3 a b (a B+A b)}{5 x^5}-\frac {b^3 B}{3 x^3}\right )}{a+b x}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^8,x]
Output:
((-1/7*(a^3*A)/x^7 - (a^2*(3*A*b + a*B))/(6*x^6) - (3*a*b*(A*b + a*B))/(5* x^5) - (b^2*(A*b + 3*a*B))/(4*x^4) - (b^3*B)/(3*x^3))*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x)
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 = 1.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.57
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{3} x^{4}}{3}+\left (-\frac {1}{4} A \,b^{3}-\frac {3}{4} B a \,b^{2}\right ) x^{3}+\left (-\frac {3}{5} A a \,b^{2}-\frac {3}{5} B \,a^{2} b \right ) x^{2}+\left (-\frac {1}{2} A \,a^{2} b -\frac {1}{6} B \,a^{3}\right ) x -\frac {a^{3} A}{7}\right )}{\left (b x +a \right ) x^{7}}\) | \(90\) |
gosper | \(-\frac {\left (140 B \,b^{3} x^{4}+105 A \,b^{3} x^{3}+315 B a \,b^{2} x^{3}+252 A a \,b^{2} x^{2}+252 B \,a^{2} b \,x^{2}+210 A \,a^{2} b x +70 B \,a^{3} x +60 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{420 x^{7} \left (b x +a \right )^{3}}\) | \(92\) |
default | \(-\frac {\left (140 B \,b^{3} x^{4}+105 A \,b^{3} x^{3}+315 B a \,b^{2} x^{3}+252 A a \,b^{2} x^{2}+252 B \,a^{2} b \,x^{2}+210 A \,a^{2} b x +70 B \,a^{3} x +60 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{420 x^{7} \left (b x +a \right )^{3}}\) | \(92\) |
orering | \(-\frac {\left (140 B \,b^{3} x^{4}+105 A \,b^{3} x^{3}+315 B a \,b^{2} x^{3}+252 A a \,b^{2} x^{2}+252 B \,a^{2} b \,x^{2}+210 A \,a^{2} b x +70 B \,a^{3} x +60 a^{3} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{420 x^{7} \left (b x +a \right )^{3}}\) | \(101\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^8,x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)*(-1/3*B*b^3*x^4+(-1/4*A*b^3-3/4*B*a*b^2)*x^3+(-3 /5*A*a*b^2-3/5*B*a^2*b)*x^2+(-1/2*A*a^2*b-1/6*B*a^3)*x-1/7*a^3*A)/x^7
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.46 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^8} \, dx=-\frac {140 \, B b^{3} x^{4} + 60 \, A a^{3} + 105 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 252 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 70 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{420 \, x^{7}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^8,x, algorithm="fricas")
Output:
-1/420*(140*B*b^3*x^4 + 60*A*a^3 + 105*(3*B*a*b^2 + A*b^3)*x^3 + 252*(B*a^ 2*b + A*a*b^2)*x^2 + 70*(B*a^3 + 3*A*a^2*b)*x)/x^7
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^8} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{8}}\, dx \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/x**8,x)
Output:
Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**8, x)
Leaf count of result is larger than twice the leaf count of optimal. 435 vs. \(2 (141) = 282\).
Time = 0.04 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.77 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^8} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{6}}{4 \, a^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{7}}{4 \, a^{7}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{5}}{4 \, a^{5} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{6}}{4 \, a^{6} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{4}}{4 \, a^{6} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{5}}{4 \, a^{7} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{4 \, a^{5} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{4 \, a^{6} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{4 \, a^{4} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{4 \, a^{5} x^{4}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{30 \, a^{3} x^{5}} - \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{70 \, a^{4} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{6 \, a^{2} x^{6}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{14 \, a^{3} x^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{7 \, a^{2} x^{7}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^8,x, algorithm="maxima")
Output:
1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^6/a^6 - 1/4*(b^2*x^2 + 2*a*b*x + a ^2)^(3/2)*A*b^7/a^7 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^5/(a^5*x) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^6/(a^6*x) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^4/(a^6*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^5/ (a^7*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^3/(a^5*x^3) - 1/4*(b^2 *x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4/(a^6*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2 )^(5/2)*B*b^2/(a^4*x^4) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/(a^5*x ^4) + 7/30*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b/(a^3*x^5) - 17/70*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^5) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/ 2)*B/(a^2*x^6) + 3/14*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^6) - 1/7* (b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A/(a^2*x^7)
Time = 0.24 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^8} \, dx=-\frac {{\left (7 \, B a b^{6} - 3 \, A b^{7}\right )} \mathrm {sgn}\left (b x + a\right )}{420 \, a^{4}} - \frac {140 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 315 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 105 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 252 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 252 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 70 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 210 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 60 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{420 \, x^{7}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^8,x, algorithm="giac")
Output:
-1/420*(7*B*a*b^6 - 3*A*b^7)*sgn(b*x + a)/a^4 - 1/420*(140*B*b^3*x^4*sgn(b *x + a) + 315*B*a*b^2*x^3*sgn(b*x + a) + 105*A*b^3*x^3*sgn(b*x + a) + 252* B*a^2*b*x^2*sgn(b*x + a) + 252*A*a*b^2*x^2*sgn(b*x + a) + 70*B*a^3*x*sgn(b *x + a) + 210*A*a^2*b*x*sgn(b*x + a) + 60*A*a^3*sgn(b*x + a))/x^7
Time = 10.84 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.25 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^8} \, dx=-\frac {\left (\frac {B\,a^3}{6}+\frac {A\,b\,a^2}{2}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^6\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^3}{4}+\frac {3\,B\,a\,b^2}{4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^4\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,x^3\,\left (a+b\,x\right )}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,x^5\,\left (a+b\,x\right )} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^8,x)
Output:
- (((B*a^3)/6 + (A*a^2*b)/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^6*(a + b* x)) - (((A*b^3)/4 + (3*B*a*b^2)/4)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^4*( a + b*x)) - (A*a^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(7*x^7*(a + b*x)) - (B *b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(3*x^3*(a + b*x)) - (3*a*b*(A*b + B* a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(5*x^5*(a + b*x))
Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.29 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^8} \, dx=\frac {-35 b^{4} x^{4}-105 a \,b^{3} x^{3}-126 a^{2} b^{2} x^{2}-70 a^{3} b x -15 a^{4}}{105 x^{7}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^8,x)
Output:
( - 15*a**4 - 70*a**3*b*x - 126*a**2*b**2*x**2 - 105*a*b**3*x**3 - 35*b**4 *x**4)/(105*x**7)