Integrand size = 29, antiderivative size = 200 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {3 a b (A b+a B) x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b^2 (A b+3 a B) x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {b^3 B x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \]
-a^3*A*((b*x+a)^2)^(1/2)/x/(b*x+a)+3*a*b*(A*b+B*a)*x*((b*x+a)^2)^(1/2)/(b* x+a)+1/2*b^2*(A*b+3*B*a)*x^2*((b*x+a)^2)^(1/2)/(b*x+a)+1/3*b^3*B*x^3*((b*x +a)^2)^(1/2)/(b*x+a)+a^2*(3*A*b+B*a)*ln(x)*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 1.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.44 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (-6 a^3 A+18 a^2 b B x^2+9 a b^2 x^2 (2 A+B x)+b^3 x^3 (3 A+2 B x)+6 a^2 (3 A b+a B) x \log (x)\right )}{6 x (a+b x)} \]
(Sqrt[(a + b*x)^2]*(-6*a^3*A + 18*a^2*b*B*x^2 + 9*a*b^2*x^2*(2*A + B*x) + b^3*x^3*(3*A + 2*B*x) + 6*a^2*(3*A*b + a*B)*x*Log[x]))/(6*x*(a + b*x))
Time = 0.25 (sec) , antiderivative size = 93, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^2} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{x^2}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{x^2}dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^3}{x^2}+\frac {(3 A b+a B) a^2}{x}+3 b (A b+a B) a+b^3 B x^2+b^2 (A b+3 a B) x\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^3 A}{x}+a^2 \log (x) (a B+3 A b)+\frac {1}{2} b^2 x^2 (3 a B+A b)+3 a b x (a B+A b)+\frac {1}{3} b^3 B x^3\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-((a^3*A)/x) + 3*a*b*(A*b + a*B)*x + (b^2* (A*b + 3*a*B)*x^2)/2 + (b^3*B*x^3)/3 + a^2*(3*A*b + a*B)*Log[x]))/(a + b*x )
3.7.72.3.1 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 = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.48
method | result | size |
default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (2 x^{4} B \,b^{3}+3 A \,b^{3} x^{3}+9 B a \,b^{2} x^{3}+18 A \ln \left (x \right ) x \,a^{2} b +18 A a \,b^{2} x^{2}+6 B \,a^{3} \ln \left (x \right ) x +18 B \,a^{2} b \,x^{2}-6 A \,a^{3}\right )}{6 \left (b x +a \right )^{3} x}\) | \(96\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b \left (\frac {1}{3} B \,b^{2} x^{3}+\frac {1}{2} A \,b^{2} x^{2}+\frac {3}{2} B a b \,x^{2}+3 a A b x +3 a^{2} B x \right )}{b x +a}-\frac {a^{3} A \sqrt {\left (b x +a \right )^{2}}}{x \left (b x +a \right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A \,a^{2} b +B \,a^{3}\right ) \ln \left (x \right )}{b x +a}\) | \(117\) |
1/6*((b*x+a)^2)^(3/2)*(2*x^4*B*b^3+3*A*b^3*x^3+9*B*a*b^2*x^3+18*A*ln(x)*x* a^2*b+18*A*a*b^2*x^2+6*B*a^3*ln(x)*x+18*B*a^2*b*x^2-6*A*a^3)/(b*x+a)^3/x
Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.38 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=\frac {2 \, B b^{3} x^{4} - 6 \, A a^{3} + 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x \log \left (x\right )}{6 \, x} \]
1/6*(2*B*b^3*x^4 - 6*A*a^3 + 3*(3*B*a*b^2 + A*b^3)*x^3 + 18*(B*a^2*b + A*a *b^2)*x^2 + 6*(B*a^3 + 3*A*a^2*b)*x*log(x))/x
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{2}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (141) = 282\).
Time = 0.21 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.42 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=\left (-1\right )^{2 \, b^{2} x + 2 \, a b} B a^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) + 3 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A a^{2} b \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B a^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - 3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A a^{2} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a b x + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{2} x + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2} + \frac {9}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a b + \frac {1}{3} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A}{x} \]
(-1)^(2*b^2*x + 2*a*b)*B*a^3*log(2*b^2*x + 2*a*b) + 3*(-1)^(2*b^2*x + 2*a* b)*A*a^2*b*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*B*a^3*log(2*a*b*x /abs(x) + 2*a^2/abs(x)) - 3*(-1)^(2*a*b*x + 2*a^2)*A*a^2*b*log(2*a*b*x/abs (x) + 2*a^2/abs(x)) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a*b*x + 3/2*sqrt (b^2*x^2 + 2*a*b*x + a^2)*A*b^2*x + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^ 2 + 9/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a*b + 1/3*(b^2*x^2 + 2*a*b*x + a^2 )^(3/2)*B - (b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A/x
Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=\frac {1}{3} \, B b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, B a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A b^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 3 \, A a b^{2} x \mathrm {sgn}\left (b x + a\right ) - \frac {A a^{3} \mathrm {sgn}\left (b x + a\right )}{x} + {\left (B a^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{2} b \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) \]
1/3*B*b^3*x^3*sgn(b*x + a) + 3/2*B*a*b^2*x^2*sgn(b*x + a) + 1/2*A*b^3*x^2* sgn(b*x + a) + 3*B*a^2*b*x*sgn(b*x + a) + 3*A*a*b^2*x*sgn(b*x + a) - A*a^3 *sgn(b*x + a)/x + (B*a^3*sgn(b*x + a) + 3*A*a^2*b*sgn(b*x + a))*log(abs(x) )
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{x^2} \,d x \]