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

Optimal. Leaf size=209 \[ \frac {6 b^2 \log (x) (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {6 b^2 (a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b^2}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (a+b x)}{a^4 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

3*b^2/a^4/((b*x+a)^2)^(1/2)+1/2*b^2/a^3/(b*x+a)/((b*x+a)^2)^(1/2)+1/2*(-b*x-a)/a^3/x^2/((b*x+a)^2)^(1/2)+3*b*(
b*x+a)/a^4/x/((b*x+a)^2)^(1/2)+6*b^2*(b*x+a)*ln(x)/a^5/((b*x+a)^2)^(1/2)-6*b^2*(b*x+a)*ln(b*x+a)/a^5/((b*x+a)^
2)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 209, normalized size of antiderivative = 1.00, 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} \[ \frac {b^2}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b^2}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (a+b x)}{a^4 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {6 b^2 \log (x) (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {6 b^2 (a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2)/(a^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + b^2/(2*a^3*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a + b*x)/
(2*a^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*b*(a + b*x))/(a^4*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (6*b^2*(a
+ b*x)*Log[x])/(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (6*b^2*(a + b*x)*Log[a + b*x])/(a^5*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 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{x^3 \left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac {1}{a^3 b^3 x^3}-\frac {3}{a^4 b^2 x^2}+\frac {6}{a^5 b x}-\frac {1}{a^3 (a+b x)^3}-\frac {3}{a^4 (a+b x)^2}-\frac {6}{a^5 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {3 b^2}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (a+b x)}{a^4 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {6 b^2 (a+b x) \log (x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {6 b^2 (a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 99, normalized size = 0.47 \[ \frac {a \left (-a^3+4 a^2 b x+18 a b^2 x^2+12 b^3 x^3\right )+12 b^2 x^2 \log (x) (a+b x)^2-12 b^2 x^2 (a+b x)^2 \log (a+b x)}{2 a^5 x^2 (a+b x) \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-a^3 + 4*a^2*b*x + 18*a*b^2*x^2 + 12*b^3*x^3) + 12*b^2*x^2*(a + b*x)^2*Log[x] - 12*b^2*x^2*(a + b*x)^2*Log
[a + b*x])/(2*a^5*x^2*(a + b*x)*Sqrt[(a + b*x)^2])

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fricas [A]  time = 0.82, size = 130, normalized size = 0.62 \[ \frac {12 \, a b^{3} x^{3} + 18 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x - a^{4} - 12 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \log \left (b x + a\right ) + 12 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \log \relax (x)}{2 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(12*a*b^3*x^3 + 18*a^2*b^2*x^2 + 4*a^3*b*x - a^4 - 12*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*log(b*x + a) +
12*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*log(x))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.06, size = 134, normalized size = 0.64 \[ -\frac {\left (-12 b^{4} x^{4} \ln \relax (x )+12 b^{4} x^{4} \ln \left (b x +a \right )-24 a \,b^{3} x^{3} \ln \relax (x )+24 a \,b^{3} x^{3} \ln \left (b x +a \right )-12 a^{2} b^{2} x^{2} \ln \relax (x )+12 a^{2} b^{2} x^{2} \ln \left (b x +a \right )-12 a \,b^{3} x^{3}-18 a^{2} b^{2} x^{2}-4 a^{3} b x +a^{4}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} a^{5} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(12*ln(b*x+a)*x^4*b^4-12*ln(x)*x^4*b^4+24*ln(b*x+a)*x^3*a*b^3-24*ln(x)*x^3*a*b^3+12*a^2*b^2*x^2*ln(b*x+a)
-12*ln(x)*x^2*a^2*b^2-12*a*b^3*x^3-18*a^2*b^2*x^2-4*a^3*b*x+a^4)*(b*x+a)/x^2/a^5/((b*x+a)^2)^(3/2)

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maxima [A]  time = 1.44, size = 135, normalized size = 0.65 \[ -\frac {6 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{5}} + \frac {6 \, b^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4}} + \frac {5 \, b}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} x} - \frac {1}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} x^{2}} + \frac {1}{2 \, a^{3} {\left (x + \frac {a}{b}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*(-1)^(2*a*b*x + 2*a^2)*b^2*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^5 + 6*b^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^
4) + 5/2*b/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3*x) - 1/2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*x^2) + 1/2/(a^3*(x +
a/b)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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