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

Optimal. Leaf size=145 \[ \frac {b (a+b x)}{a^2 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {-a-b x}{2 a x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 \log (x) (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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)

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Rubi [A]  time = 0.04, antiderivative size = 142, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {646, 44} \[ \frac {b (a+b x)}{a^2 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 \log (x) (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a + b*x)/(2*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 44

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] && L
tQ[m + n + 2, 0])

Rule 646

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*} \int \frac {1}{x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\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*}

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Mathematica [A]  time = 0.02, size = 59, normalized size = 0.41 \[ -\frac {(a+b x) \left (2 b^2 x^2 \log (a+b x)+a (a-2 b x)-2 b^2 x^2 \log (x)\right )}{2 a^3 x^2 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.99, size = 41, normalized size = 0.28 \[ -\frac {2 \, b^{2} x^{2} \log \left (b x + a\right ) - 2 \, b^{2} x^{2} \log \relax (x) - 2 \, a b x + a^{2}}{2 \, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.15, size = 54, normalized size = 0.37 \[ -\frac {1}{2} \, {\left (\frac {2 \, b^{2} \log \left ({\left | b x + a \right |}\right )}{a^{3}} - \frac {2 \, b^{2} \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {2 \, a b x - a^{2}}{a^{3} x^{2}}\right )} \mathrm {sgn}\left (b x + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.06, size = 56, normalized size = 0.39 \[ -\frac {\left (b x +a \right ) \left (-2 b^{2} x^{2} \ln \relax (x )+2 b^{2} x^{2} \ln \left (b x +a \right )-2 a b x +a^{2}\right )}{2 \sqrt {\left (b x +a \right )^{2}}\, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 95, normalized size = 0.66 \[ -\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b}{2 \, a^{3} x} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}{2 \, a^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^3\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*((a + b*x)^2)^(1/2)),x)

[Out]

int(1/(x^3*((a + b*x)^2)^(1/2)), x)

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sympy [A]  time = 0.24, size = 31, normalized size = 0.21 \[ \frac {- a + 2 b x}{2 a^{2} x^{2}} + \frac {b^{2} \left (\log {\relax (x )} - \log {\left (\frac {a}{b} + x \right )}\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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