\(\int \frac {1}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [188]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 24, antiderivative size = 145 \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {-a-b x}{2 a x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x)}{a^2 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (a+b x) \log (x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

1/2*(-b*x-a)/a/x^2/((b*x+a)^2)^(1/2)+b*(b*x+a)/a^2/x/((b*x+a)^2)^(1/2)+b^2*(b*x+a)*ln(x)/a^3/((b*x+a)^2)^(1/2)
-b^2*(b*x+a)*ln(b*x+a)/a^3/((b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 46} \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {b (a+b x)}{a^2 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 \log (x) (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

Int[1/(x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

-1/2*(a + b*x)/(a*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (b*(a + b*x))/(a^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (
b^2*(a + b*x)*Log[x])/(a^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b^2*(a + b*x)*Log[a + b*x])/(a^3*Sqrt[a^2 + 2*a*b
*x + b^2*x^2])

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
 c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a b+b^2 x\right ) \int \frac {1}{x^3 \left (a b+b^2 x\right )} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {\left (a b+b^2 x\right ) \int \left (\frac {1}{a b x^3}-\frac {1}{a^2 x^2}+\frac {b}{a^3 x}-\frac {b^2}{a^3 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {a+b x}{2 a x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x)}{a^2 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (a+b x) \log (x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\frac {a^3 (a-2 b x)}{\sqrt {a^2} x^2}-\frac {a (a-3 b x) \sqrt {(a+b x)^2}}{x^2}-4 \sqrt {a^2} b^2 \log (x)+2 \left (-a+\sqrt {a^2}\right ) b^2 \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )+2 \left (a+\sqrt {a^2}\right ) b^2 \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )}{4 a^4} \]

[In]

Integrate[1/(x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

((a^3*(a - 2*b*x))/(Sqrt[a^2]*x^2) - (a*(a - 3*b*x)*Sqrt[(a + b*x)^2])/x^2 - 4*Sqrt[a^2]*b^2*Log[x] + 2*(-a +
Sqrt[a^2])*b^2*Log[Sqrt[a^2] - b*x - Sqrt[(a + b*x)^2]] + 2*(a + Sqrt[a^2])*b^2*Log[Sqrt[a^2] + b*x - Sqrt[(a
+ b*x)^2]])/(4*a^4)

Maple [A] (verified)

Time = 1.96 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.40

method result size
default \(\frac {\left (b x +a \right ) \left (2 b^{2} \ln \left (x \right ) x^{2}-2 b^{2} \ln \left (b x +a \right ) x^{2}+2 a b x -a^{2}\right )}{2 \sqrt {\left (b x +a \right )^{2}}\, a^{3} x^{2}}\) \(58\)
risch \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {b x}{a^{2}}-\frac {1}{2 a}\right )}{\left (b x +a \right ) x^{2}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{2} \ln \left (-x \right )}{\left (b x +a \right ) a^{3}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{2} \ln \left (b x +a \right )}{\left (b x +a \right ) a^{3}}\) \(91\)

[In]

int(1/x^3/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*(b*x+a)*(2*b^2*ln(x)*x^2-2*b^2*ln(b*x+a)*x^2+2*a*b*x-a^2)/((b*x+a)^2)^(1/2)/a^3/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.28 \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 \, b^{2} x^{2} \log \left (b x + a\right ) - 2 \, b^{2} x^{2} \log \left (x\right ) - 2 \, a b x + a^{2}}{2 \, a^{3} x^{2}} \]

[In]

integrate(1/x^3/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/2*(2*b^2*x^2*log(b*x + a) - 2*b^2*x^2*log(x) - 2*a*b*x + a^2)/(a^3*x^2)

Sympy [F]

\[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{x^{3} \sqrt {\left (a + b x\right )^{2}}}\, dx \]

[In]

integrate(1/x**3/((b*x+a)**2)**(1/2),x)

[Out]

Integral(1/(x**3*sqrt((a + b*x)**2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b}{2 \, a^{3} x} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}{2 \, a^{2} x^{2}} \]

[In]

integrate(1/x^3/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

-(-1)^(2*a*b*x + 2*a^2)*b^2*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^3 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b/(a^3*
x) - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)/(a^2*x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.37 \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {1}{2} \, {\left (\frac {2 \, b^{2} \log \left ({\left | b x + a \right |}\right )}{a^{3}} - \frac {2 \, b^{2} \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {2 \, a b x - a^{2}}{a^{3} x^{2}}\right )} \mathrm {sgn}\left (b x + a\right ) \]

[In]

integrate(1/x^3/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

-1/2*(2*b^2*log(abs(b*x + a))/a^3 - 2*b^2*log(abs(x))/a^3 - (2*a*b*x - a^2)/(a^3*x^2))*sgn(b*x + a)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{x^3\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]

[In]

int(1/(x^3*((a + b*x)^2)^(1/2)),x)

[Out]

int(1/(x^3*((a + b*x)^2)^(1/2)), x)