Integrand size = 29, antiderivative size = 199 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^4} \, dx=-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {3 a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^3 B x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b^2 (A b+3 a B) \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \]
-1/3*a^3*A*((b*x+a)^2)^(1/2)/x^3/(b*x+a)-1/2*a^2*(3*A*b+B*a)*((b*x+a)^2)^( 1/2)/x^2/(b*x+a)-3*a*b*(A*b+B*a)*((b*x+a)^2)^(1/2)/x/(b*x+a)+b^3*B*x*((b*x +a)^2)^(1/2)/(b*x+a)+b^2*(A*b+3*B*a)*ln(x)*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 1.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.44 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^4} \, dx=-\frac {\sqrt {(a+b x)^2} \left (18 a A b^2 x^2-6 b^3 B x^4+9 a^2 b x (A+2 B x)+a^3 (2 A+3 B x)-6 b^2 (A b+3 a B) x^3 \log (x)\right )}{6 x^3 (a+b x)} \]
-1/6*(Sqrt[(a + b*x)^2]*(18*a*A*b^2*x^2 - 6*b^3*B*x^4 + 9*a^2*b*x*(A + 2*B *x) + a^3*(2*A + 3*B*x) - 6*b^2*(A*b + 3*a*B)*x^3*Log[x]))/(x^3*(a + b*x))
Time = 0.25 (sec) , antiderivative size = 92, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^4} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{x^4}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{x^4}dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^3}{x^4}+\frac {(3 A b+a B) a^2}{x^3}+\frac {3 b (A b+a B) a}{x^2}+b^3 B+\frac {b^2 (A b+3 a B)}{x}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^3 A}{3 x^3}-\frac {a^2 (a B+3 A b)}{2 x^2}+b^2 \log (x) (3 a B+A b)-\frac {3 a b (a B+A b)}{x}+b^3 B x\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/3*(a^3*A)/x^3 - (a^2*(3*A*b + a*B))/(2* x^2) - (3*a*b*(A*b + a*B))/x + b^3*B*x + b^2*(A*b + 3*a*B)*Log[x]))/(a + b *x)
3.7.74.3.1 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 = 0.51 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.48
method | result | size |
default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (6 A \ln \left (x \right ) x^{3} b^{3}+18 B \ln \left (x \right ) x^{3} a \,b^{2}+6 x^{4} B \,b^{3}-18 A a \,b^{2} x^{2}-18 B \,a^{2} b \,x^{2}-9 A \,a^{2} b x -3 a^{3} B x -2 A \,a^{3}\right )}{6 \left (b x +a \right )^{3} x^{3}}\) | \(96\) |
risch | \(\frac {b^{3} B x \sqrt {\left (b x +a \right )^{2}}}{b x +a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-3 A a \,b^{2}-3 B b \,a^{2}\right ) x^{2}+\left (-\frac {3}{2} A \,a^{2} b -\frac {1}{2} B \,a^{3}\right ) x -\frac {A \,a^{3}}{3}\right )}{\left (b x +a \right ) x^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A \,b^{3}+3 B a \,b^{2}\right ) \ln \left (x \right )}{b x +a}\) | \(118\) |
1/6*((b*x+a)^2)^(3/2)*(6*A*ln(x)*x^3*b^3+18*B*ln(x)*x^3*a*b^2+6*x^4*B*b^3- 18*A*a*b^2*x^2-18*B*a^2*b*x^2-9*A*a^2*b*x-3*a^3*B*x-2*A*a^3)/(b*x+a)^3/x^3
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.38 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^4} \, dx=\frac {6 \, B b^{3} x^{4} + 6 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} \log \left (x\right ) - 2 \, A a^{3} - 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} - 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{6 \, x^{3}} \]
1/6*(6*B*b^3*x^4 + 6*(3*B*a*b^2 + A*b^3)*x^3*log(x) - 2*A*a^3 - 18*(B*a^2* b + A*a*b^2)*x^2 - 3*(B*a^3 + 3*A*a^2*b)*x)/x^3
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^4} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{4}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 443 vs. \(2 (140) = 280\).
Time = 0.21 (sec) , antiderivative size = 443, normalized size of antiderivative = 2.23 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^4} \, dx=3 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} B a b^{2} \log \left (2 \, b^{2} x + 2 \, a b\right ) + \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A b^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) - 3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B a b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{3} x}{2 \, a} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{4} x}{2 \, a^{2}} + \frac {9}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{2} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{3}}{2 \, a} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{2}}{2 \, a^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{3}}{6 \, a^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b}{2 \, a x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{2}}{2 \, a^{2} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{2 \, a^{2} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{6 \, a^{3} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{3 \, a^{2} x^{3}} \]
3*(-1)^(2*b^2*x + 2*a*b)*B*a*b^2*log(2*b^2*x + 2*a*b) + (-1)^(2*b^2*x + 2* a*b)*A*b^3*log(2*b^2*x + 2*a*b) - 3*(-1)^(2*a*b*x + 2*a^2)*B*a*b^2*log(2*a *b*x/abs(x) + 2*a^2/abs(x)) - (-1)^(2*a*b*x + 2*a^2)*A*b^3*log(2*a*b*x/abs (x) + 2*a^2/abs(x)) + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^3*x/a + 1/2*sq rt(b^2*x^2 + 2*a*b*x + a^2)*A*b^4*x/a^2 + 9/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 )*B*b^2 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^3/a + 1/2*(b^2*x^2 + 2*a*b *x + a^2)^(3/2)*B*b^2/a^2 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^3/a^3 - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b/(a*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^2/(a^2*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B/(a^2*x^2 ) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^2) - 1/3*(b^2*x^2 + 2*a *b*x + a^2)^(5/2)*A/(a^2*x^3)
Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.59 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^4} \, dx=B b^{3} x \mathrm {sgn}\left (b x + a\right ) + {\left (3 \, B a b^{2} \mathrm {sgn}\left (b x + a\right ) + A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) - \frac {2 \, A a^{3} \mathrm {sgn}\left (b x + a\right ) + 18 \, {\left (B a^{2} b \mathrm {sgn}\left (b x + a\right ) + A a b^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 3 \, {\left (B a^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{2} b \mathrm {sgn}\left (b x + a\right )\right )} x}{6 \, x^{3}} \]
B*b^3*x*sgn(b*x + a) + (3*B*a*b^2*sgn(b*x + a) + A*b^3*sgn(b*x + a))*log(a bs(x)) - 1/6*(2*A*a^3*sgn(b*x + a) + 18*(B*a^2*b*sgn(b*x + a) + A*a*b^2*sg n(b*x + a))*x^2 + 3*(B*a^3*sgn(b*x + a) + 3*A*a^2*b*sgn(b*x + a))*x)/x^3
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^4} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{x^4} \,d x \]