Integrand size = 29, antiderivative size = 211 \[ \int \frac {A+B x}{x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {A (a+b x)}{3 a x^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) (a+b x)}{2 a^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (A b-a B) (a+b x)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (A b-a B) (a+b x) \log (x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (A b-a B) (a+b x) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:
-1/3*A*(b*x+a)/a/x^3/((b*x+a)^2)^(1/2)+1/2*(A*b-B*a)*(b*x+a)/a^2/x^2/((b*x +a)^2)^(1/2)-b*(A*b-B*a)*(b*x+a)/a^3/x/((b*x+a)^2)^(1/2)-b^2*(A*b-B*a)*(b* x+a)*ln(x)/a^4/((b*x+a)^2)^(1/2)+b^2*(A*b-B*a)*(b*x+a)*ln(b*x+a)/a^4/((b*x +a)^2)^(1/2)
Time = 0.72 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x}{x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {1}{12} \left (\frac {6 A b^2 x^2-3 a b x (A+2 B x)+a^2 (2 A+3 B x)}{\left (a^2\right )^{3/2} x^3}+\frac {\sqrt {(a+b x)^2} \left (-11 A b^2 x^2-a^2 (2 A+3 B x)+a b x (5 A+9 B x)\right )}{a^4 x^3}-\frac {12 \sqrt {a^2} b^2 (-A b+a B) \log (x)}{a^5}+\frac {6 \left (-a+\sqrt {a^2}\right ) b^2 (-A b+a B) \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )}{a^5}+\frac {6 \left (a+\sqrt {a^2}\right ) b^2 (-A b+a B) \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )}{a^5}\right ) \] Input:
Integrate[(A + B*x)/(x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]
Output:
((6*A*b^2*x^2 - 3*a*b*x*(A + 2*B*x) + a^2*(2*A + 3*B*x))/((a^2)^(3/2)*x^3) + (Sqrt[(a + b*x)^2]*(-11*A*b^2*x^2 - a^2*(2*A + 3*B*x) + a*b*x*(5*A + 9* B*x)))/(a^4*x^3) - (12*Sqrt[a^2]*b^2*(-(A*b) + a*B)*Log[x])/a^5 + (6*(-a + Sqrt[a^2])*b^2*(-(A*b) + a*B)*Log[Sqrt[a^2] - b*x - Sqrt[(a + b*x)^2]])/a ^5 + (6*(a + Sqrt[a^2])*b^2*(-(A*b) + a*B)*Log[Sqrt[a^2] + b*x - Sqrt[(a + b*x)^2]])/a^5)/12
Time = 0.47 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.53, 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.
\(\displaystyle \int \frac {A+B x}{x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b (a+b x) \int \frac {A+B x}{b x^4 (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {A+B x}{x^4 (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {(a+b x) \int \left (-\frac {(a B-A b) b^3}{a^4 (a+b x)}+\frac {(a B-A b) b^2}{a^4 x}-\frac {(a B-A b) b}{a^3 x^2}+\frac {a B-A b}{a^2 x^3}+\frac {A}{a x^4}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (-\frac {b^2 \log (x) (A b-a B)}{a^4}+\frac {b^2 (A b-a B) \log (a+b x)}{a^4}-\frac {b (A b-a B)}{a^3 x}+\frac {A b-a B}{2 a^2 x^2}-\frac {A}{3 a x^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[(A + B*x)/(x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]
Output:
((a + b*x)*(-1/3*A/(a*x^3) + (A*b - a*B)/(2*a^2*x^2) - (b*(A*b - a*B))/(a^ 3*x) - (b^2*(A*b - a*B)*Log[x])/a^4 + (b^2*(A*b - a*B)*Log[a + b*x])/a^4)) /Sqrt[a^2 + 2*a*b*x + b^2*x^2]
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 = 0.87 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.56
method | result | size |
default | \(\frac {\left (b x +a \right ) \left (6 A \ln \left (b x +a \right ) x^{3} b^{3}-6 A \ln \left (x \right ) x^{3} b^{3}-6 B \ln \left (b x +a \right ) a \,b^{2} x^{3}+6 B \ln \left (x \right ) a \,b^{2} x^{3}-6 A a \,b^{2} x^{2}+6 B \,a^{2} b \,x^{2}+3 A \,a^{2} b x -3 B \,a^{3} x -2 a^{3} A \right )}{6 \sqrt {\left (b x +a \right )^{2}}\, a^{4} x^{3}}\) | \(119\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {\left (A b -B a \right ) b \,x^{2}}{a^{3}}+\frac {\left (A b -B a \right ) x}{2 a^{2}}-\frac {A}{3 a}\right )}{\left (b x +a \right ) x^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A b -B a \right ) b^{2} \ln \left (-b x -a \right )}{\left (b x +a \right ) a^{4}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A b -B a \right ) b^{2} \ln \left (x \right )}{\left (b x +a \right ) a^{4}}\) | \(134\) |
Input:
int((B*x+A)/x^4/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/6*(b*x+a)*(6*A*ln(b*x+a)*x^3*b^3-6*A*ln(x)*x^3*b^3-6*B*ln(b*x+a)*a*b^2*x ^3+6*B*ln(x)*a*b^2*x^3-6*A*a*b^2*x^2+6*B*a^2*b*x^2+3*A*a^2*b*x-3*B*a^3*x-2 *a^3*A)/((b*x+a)^2)^(1/2)/a^4/x^3
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.45 \[ \int \frac {A+B x}{x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {6 \, {\left (B a b^{2} - A b^{3}\right )} x^{3} \log \left (b x + a\right ) - 6 \, {\left (B a b^{2} - A b^{3}\right )} x^{3} \log \left (x\right ) + 2 \, A a^{3} - 6 \, {\left (B a^{2} b - A a b^{2}\right )} x^{2} + 3 \, {\left (B a^{3} - A a^{2} b\right )} x}{6 \, a^{4} x^{3}} \] Input:
integrate((B*x+A)/x^4/((b*x+a)^2)^(1/2),x, algorithm="fricas")
Output:
-1/6*(6*(B*a*b^2 - A*b^3)*x^3*log(b*x + a) - 6*(B*a*b^2 - A*b^3)*x^3*log(x ) + 2*A*a^3 - 6*(B*a^2*b - A*a*b^2)*x^2 + 3*(B*a^3 - A*a^2*b)*x)/(a^4*x^3)
\[ \int \frac {A+B x}{x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A + B x}{x^{4} \sqrt {\left (a + b x\right )^{2}}}\, dx \] Input:
integrate((B*x+A)/x**4/((b*x+a)**2)**(1/2),x)
Output:
Integral((A + B*x)/(x**4*sqrt((a + b*x)**2)), x)
Time = 0.04 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x}{x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} + \frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{4}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b}{2 \, a^{3} x} - \frac {11 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{2}}{6 \, a^{4} x} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B}{2 \, a^{2} x^{2}} + \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b}{6 \, a^{3} x^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A}{3 \, a^{2} x^{3}} \] Input:
integrate((B*x+A)/x^4/((b*x+a)^2)^(1/2),x, algorithm="maxima")
Output:
-(-1)^(2*a*b*x + 2*a^2)*B*b^2*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^3 + (-1 )^(2*a*b*x + 2*a^2)*A*b^3*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^4 + 3/2*sqr t(b^2*x^2 + 2*a*b*x + a^2)*B*b/(a^3*x) - 11/6*sqrt(b^2*x^2 + 2*a*b*x + a^2 )*A*b^2/(a^4*x) - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B/(a^2*x^2) + 5/6*sqrt (b^2*x^2 + 2*a*b*x + a^2)*A*b/(a^3*x^2) - 1/3*sqrt(b^2*x^2 + 2*a*b*x + a^2 )*A/(a^2*x^3)
Time = 0.23 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.73 \[ \int \frac {A+B x}{x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {{\left (B a b^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {{\left (B a b^{3} \mathrm {sgn}\left (b x + a\right ) - A b^{4} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{a^{4} b} - \frac {2 \, A a^{3} \mathrm {sgn}\left (b x + a\right ) - 6 \, {\left (B a^{2} b \mathrm {sgn}\left (b x + a\right ) - A a b^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 3 \, {\left (B a^{3} \mathrm {sgn}\left (b x + a\right ) - A a^{2} b \mathrm {sgn}\left (b x + a\right )\right )} x}{6 \, a^{4} x^{3}} \] Input:
integrate((B*x+A)/x^4/((b*x+a)^2)^(1/2),x, algorithm="giac")
Output:
(B*a*b^2*sgn(b*x + a) - A*b^3*sgn(b*x + a))*log(abs(x))/a^4 - (B*a*b^3*sgn (b*x + a) - A*b^4*sgn(b*x + a))*log(abs(b*x + a))/(a^4*b) - 1/6*(2*A*a^3*s gn(b*x + a) - 6*(B*a^2*b*sgn(b*x + a) - A*a*b^2*sgn(b*x + a))*x^2 + 3*(B*a ^3*sgn(b*x + a) - A*a^2*b*sgn(b*x + a))*x)/(a^4*x^3)
Timed out. \[ \int \frac {A+B x}{x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A+B\,x}{x^4\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \] Input:
int((A + B*x)/(x^4*((a + b*x)^2)^(1/2)),x)
Output:
int((A + B*x)/(x^4*((a + b*x)^2)^(1/2)), x)
Time = 0.28 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.02 \[ \int \frac {A+B x}{x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {1}{3 x^{3}} \] Input:
int((B*x+A)/x^4/((b*x+a)^2)^(1/2),x)
Output:
( - 1)/(3*x**3)