Integrand size = 29, antiderivative size = 115 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=-\frac {b (A b-a B) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 a^3 x^4}-\frac {A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 a^2 x^6}+\frac {(7 A b-6 a B) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{30 a^3 x^5} \] Output:
-1/4*b*(A*b-B*a)*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/a^3/x^4-1/6*A*(b^2*x^ 2+2*a*b*x+a^2)^(5/2)/a^2/x^6+1/30*(7*A*b-6*B*a)*(b^2*x^2+2*a*b*x+a^2)^(5/2 )/a^3/x^5
Time = 0.89 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.76 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=-\frac {\sqrt {(a+b x)^2} \left (10 b^3 x^3 (2 A+3 B x)+15 a b^2 x^2 (3 A+4 B x)+9 a^2 b x (4 A+5 B x)+2 a^3 (5 A+6 B x)\right )}{60 x^6 (a+b x)} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^7,x]
Output:
-1/60*(Sqrt[(a + b*x)^2]*(10*b^3*x^3*(2*A + 3*B*x) + 15*a*b^2*x^2*(3*A + 4 *B*x) + 9*a^2*b*x*(4*A + 5*B*x) + 2*a^3*(5*A + 6*B*x)))/(x^6*(a + b*x))
Time = 0.43 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.90, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^7} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{x^7}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{x^7}dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^3}{x^7}+\frac {(3 A b+a B) a^2}{x^6}+\frac {3 b (A b+a B) a}{x^5}+\frac {b^3 B}{x^3}+\frac {b^2 (A b+3 a B)}{x^4}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^3 A}{6 x^6}-\frac {a^2 (a B+3 A b)}{5 x^5}-\frac {b^2 (3 a B+A b)}{3 x^3}-\frac {3 a b (a B+A b)}{4 x^4}-\frac {b^3 B}{2 x^2}\right )}{a+b x}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^7,x]
Output:
((-1/6*(a^3*A)/x^6 - (a^2*(3*A*b + a*B))/(5*x^5) - (3*a*b*(A*b + a*B))/(4* x^4) - (b^2*(A*b + 3*a*B))/(3*x^3) - (b^3*B)/(2*x^2))*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 = 1.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{3} x^{4}}{2}+\left (-\frac {1}{3} A \,b^{3}-B a \,b^{2}\right ) x^{3}+\left (-\frac {3}{4} A a \,b^{2}-\frac {3}{4} B \,a^{2} b \right ) x^{2}+\left (-\frac {3}{5} A \,a^{2} b -\frac {1}{5} B \,a^{3}\right ) x -\frac {a^{3} A}{6}\right )}{\left (b x +a \right ) x^{6}}\) | \(90\) |
gosper | \(-\frac {\left (30 B \,b^{3} x^{4}+20 A \,b^{3} x^{3}+60 B a \,b^{2} x^{3}+45 A a \,b^{2} x^{2}+45 B \,a^{2} b \,x^{2}+36 A \,a^{2} b x +12 B \,a^{3} x +10 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 x^{6} \left (b x +a \right )^{3}}\) | \(92\) |
default | \(-\frac {\left (30 B \,b^{3} x^{4}+20 A \,b^{3} x^{3}+60 B a \,b^{2} x^{3}+45 A a \,b^{2} x^{2}+45 B \,a^{2} b \,x^{2}+36 A \,a^{2} b x +12 B \,a^{3} x +10 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 x^{6} \left (b x +a \right )^{3}}\) | \(92\) |
orering | \(-\frac {\left (30 B \,b^{3} x^{4}+20 A \,b^{3} x^{3}+60 B a \,b^{2} x^{3}+45 A a \,b^{2} x^{2}+45 B \,a^{2} b \,x^{2}+36 A \,a^{2} b x +12 B \,a^{3} x +10 a^{3} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{60 x^{6} \left (b x +a \right )^{3}}\) | \(101\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^7,x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)*(-1/2*B*b^3*x^4+(-1/3*A*b^3-B*a*b^2)*x^3+(-3/4*A *a*b^2-3/4*B*a^2*b)*x^2+(-3/5*A*a^2*b-1/5*B*a^3)*x-1/6*a^3*A)/x^6
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.63 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=-\frac {30 \, B b^{3} x^{4} + 10 \, A a^{3} + 20 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 45 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 12 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{60 \, x^{6}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^7,x, algorithm="fricas")
Output:
-1/60*(30*B*b^3*x^4 + 10*A*a^3 + 20*(3*B*a*b^2 + A*b^3)*x^3 + 45*(B*a^2*b + A*a*b^2)*x^2 + 12*(B*a^3 + 3*A*a^2*b)*x)/x^6
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{7}}\, dx \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/x**7,x)
Output:
Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**7, x)
Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (103) = 206\).
Time = 0.04 (sec) , antiderivative size = 375, normalized size of antiderivative = 3.26 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{5}}{4 \, a^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{6}}{4 \, a^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{4}}{4 \, a^{4} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{5}}{4 \, a^{5} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{4 \, a^{5} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{4 \, a^{6} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{4 \, a^{4} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{4 \, a^{5} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{4 \, a^{3} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{4 \, a^{4} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{5 \, a^{2} x^{5}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{30 \, a^{3} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{6 \, a^{2} x^{6}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^7,x, algorithm="maxima")
Output:
-1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^5/a^5 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^6/a^6 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^4/(a^4*x) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^5/(a^5*x) + 1/4*(b^2*x^2 + 2*a*b* x + a^2)^(5/2)*B*b^3/(a^5*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4 /(a^6*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^2/(a^4*x^3) + 1/4*(b^ 2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/(a^5*x^3) + 1/4*(b^2*x^2 + 2*a*b*x + a^ 2)^(5/2)*B*b/(a^3*x^4) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^ 4) - 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B/(a^2*x^5) + 7/30*(b^2*x^2 + 2*a *b*x + a^2)^(5/2)*A*b/(a^3*x^5) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A/(a ^2*x^6)
Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.30 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=\frac {{\left (3 \, B a b^{5} - A b^{6}\right )} \mathrm {sgn}\left (b x + a\right )}{60 \, a^{3}} - \frac {30 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 60 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 20 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 45 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 45 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 12 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 36 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 10 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{60 \, x^{6}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^7,x, algorithm="giac")
Output:
1/60*(3*B*a*b^5 - A*b^6)*sgn(b*x + a)/a^3 - 1/60*(30*B*b^3*x^4*sgn(b*x + a ) + 60*B*a*b^2*x^3*sgn(b*x + a) + 20*A*b^3*x^3*sgn(b*x + a) + 45*B*a^2*b*x ^2*sgn(b*x + a) + 45*A*a*b^2*x^2*sgn(b*x + a) + 12*B*a^3*x*sgn(b*x + a) + 36*A*a^2*b*x*sgn(b*x + a) + 10*A*a^3*sgn(b*x + a))/x^6
Time = 10.85 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.70 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=-\frac {\left (\frac {B\,a^3}{5}+\frac {3\,A\,b\,a^2}{5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^5\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^3}{3}+B\,a\,b^2\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^3\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,x^6\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,x^2\,\left (a+b\,x\right )}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^4\,\left (a+b\,x\right )} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^7,x)
Output:
- (((B*a^3)/5 + (3*A*a^2*b)/5)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^5*(a + b*x)) - (((A*b^3)/3 + B*a*b^2)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^3*(a + b*x)) - (A*a^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6*x^6*(a + b*x)) - (B*b^3 *(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(2*x^2*(a + b*x)) - (3*a*b*(A*b + B*a)*( a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*x^4*(a + b*x))
Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.40 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=\frac {-15 b^{4} x^{4}-40 a \,b^{3} x^{3}-45 a^{2} b^{2} x^{2}-24 a^{3} b x -5 a^{4}}{30 x^{6}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^7,x)
Output:
( - 5*a**4 - 24*a**3*b*x - 45*a**2*b**2*x**2 - 40*a*b**3*x**3 - 15*b**4*x* *4)/(30*x**6)