Integrand size = 29, antiderivative size = 162 \[ \int \frac {A+B x}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {A (a+b x)}{2 a x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) (a+b x)}{a^2 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (A b-a B) (a+b x) \log (x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (A b-a B) (a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]
-1/2*A*(b*x+a)/a/x^2/((b*x+a)^2)^(1/2)+(A*b-B*a)*(b*x+a)/a^2/x/((b*x+a)^2) ^(1/2)+b*(A*b-B*a)*(b*x+a)*ln(x)/a^3/((b*x+a)^2)^(1/2)-b*(A*b-B*a)*(b*x+a) *ln(b*x+a)/a^3/((b*x+a)^2)^(1/2)
Time = 0.36 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {-\frac {a \sqrt {(a+b x)^2} (-3 A b x+a (A+2 B x))}{x^2}+\frac {a^3 (-2 A b x+a (A+2 B x))}{\sqrt {a^2} x^2}+4 \sqrt {a^2} b (-A b+a B) \log (x)-2 \left (-a+\sqrt {a^2}\right ) b (-A b+a B) \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )-2 \left (a+\sqrt {a^2}\right ) b (-A b+a B) \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )}{4 a^4} \]
(-((a*Sqrt[(a + b*x)^2]*(-3*A*b*x + a*(A + 2*B*x)))/x^2) + (a^3*(-2*A*b*x + a*(A + 2*B*x)))/(Sqrt[a^2]*x^2) + 4*Sqrt[a^2]*b*(-(A*b) + a*B)*Log[x] - 2*(-a + Sqrt[a^2])*b*(-(A*b) + a*B)*Log[Sqrt[a^2] - b*x - Sqrt[(a + b*x)^2 ]] - 2*(a + Sqrt[a^2])*b*(-(A*b) + a*B)*Log[Sqrt[a^2] + b*x - Sqrt[(a + b* x)^2]])/(4*a^4)
Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.54, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1187, 27, 86, 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.
\(\displaystyle \int \frac {A+B x}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b (a+b x) \int \frac {A+B x}{b x^3 (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {A+B x}{x^3 (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {(a+b x) \int \left (\frac {(a B-A b) b^2}{a^3 (a+b x)}-\frac {(a B-A b) b}{a^3 x}+\frac {a B-A b}{a^2 x^2}+\frac {A}{a x^3}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (\frac {b \log (x) (A b-a B)}{a^3}-\frac {b (A b-a B) \log (a+b x)}{a^3}+\frac {A b-a B}{a^2 x}-\frac {A}{2 a x^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/2*A/(a*x^2) + (A*b - a*B)/(a^2*x) + (b*(A*b - a*B)*Log[x])/ a^3 - (b*(A*b - a*B)*Log[a + b*x])/a^3))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.8.11.3.1 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[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.57
method | result | size |
default | \(\frac {\left (b x +a \right ) \left (2 A \ln \left (x \right ) x^{2} b^{2}-2 A \ln \left (b x +a \right ) b^{2} x^{2}-2 B \ln \left (x \right ) x^{2} a b +2 B \ln \left (b x +a \right ) a b \,x^{2}+2 a A b x -2 a^{2} B x -A \,a^{2}\right )}{2 \sqrt {\left (b x +a \right )^{2}}\, a^{3} x^{2}}\) | \(93\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {\left (A b -B a \right ) x}{a^{2}}-\frac {A}{2 a}\right )}{\left (b x +a \right ) x^{2}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A b -B a \right ) b \ln \left (b x +a \right )}{\left (b x +a \right ) a^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A b -B a \right ) b \ln \left (-x \right )}{\left (b x +a \right ) a^{3}}\) | \(111\) |
1/2*(b*x+a)*(2*A*ln(x)*x^2*b^2-2*A*ln(b*x+a)*b^2*x^2-2*B*ln(x)*x^2*a*b+2*B *ln(b*x+a)*a*b*x^2+2*a*A*b*x-2*a^2*B*x-A*a^2)/((b*x+a)^2)^(1/2)/a^3/x^2
Time = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.43 \[ \int \frac {A+B x}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \, {\left (B a b - A b^{2}\right )} x^{2} \log \left (b x + a\right ) - 2 \, {\left (B a b - A b^{2}\right )} x^{2} \log \left (x\right ) - A a^{2} - 2 \, {\left (B a^{2} - A a b\right )} x}{2 \, a^{3} x^{2}} \]
1/2*(2*(B*a*b - A*b^2)*x^2*log(b*x + a) - 2*(B*a*b - A*b^2)*x^2*log(x) - A *a^2 - 2*(B*a^2 - A*a*b)*x)/(a^3*x^2)
\[ \int \frac {A+B x}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A + B x}{x^{3} \sqrt {\left (a + b x\right )^{2}}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{2}} - \frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B}{a^{2} x} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b}{2 \, a^{3} x} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A}{2 \, a^{2} x^{2}} \]
(-1)^(2*a*b*x + 2*a^2)*B*b*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^2 - (-1)^( 2*a*b*x + 2*a^2)*A*b^2*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^3 - sqrt(b^2*x ^2 + 2*a*b*x + a^2)*B/(a^2*x) + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b/(a^3 *x) - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A/(a^2*x^2)
Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {{\left (B a b \mathrm {sgn}\left (b x + a\right ) - A b^{2} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {{\left (B a b^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{a^{3} b} - \frac {A a^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, {\left (B a^{2} \mathrm {sgn}\left (b x + a\right ) - A a b \mathrm {sgn}\left (b x + a\right )\right )} x}{2 \, a^{3} x^{2}} \]
-(B*a*b*sgn(b*x + a) - A*b^2*sgn(b*x + a))*log(abs(x))/a^3 + (B*a*b^2*sgn( b*x + a) - A*b^3*sgn(b*x + a))*log(abs(b*x + a))/(a^3*b) - 1/2*(A*a^2*sgn( b*x + a) + 2*(B*a^2*sgn(b*x + a) - A*a*b*sgn(b*x + a))*x)/(a^3*x^2)
Timed out. \[ \int \frac {A+B x}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A+B\,x}{x^3\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]