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

Optimal. Leaf size=165 \[ -\frac{b}{2 a^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 b}{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a+b x}{a^3 x \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 b \log (x) (a+b x)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 b (a+b x) \log (a+b x)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0675738, antiderivative size = 165, normalized size of antiderivative = 1., 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} \[ -\frac{b}{2 a^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 b}{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a+b x}{a^3 x \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 b \log (x) (a+b x)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 b (a+b x) \log (a+b x)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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])

Rubi steps

\begin{align*} \int \frac{1}{x^2 \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^2 \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^2}-\frac{3}{a^4 b^2 x}+\frac{1}{a^2 b (a+b x)^3}+\frac{2}{a^3 b (a+b x)^2}+\frac{3}{a^4 b (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 b}{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b}{2 a^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a+b x}{a^3 x \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 b (a+b x) \log (x)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 b (a+b x) \log (a+b x)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0287591, size = 81, normalized size = 0.49 \[ \frac{-a \left (2 a^2+9 a b x+6 b^2 x^2\right )-6 b x \log (x) (a+b x)^2+6 b x (a+b x)^2 \log (a+b x)}{2 a^4 x (a+b x) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*(2*a^2 + 9*a*b*x + 6*b^2*x^2)) - 6*b*x*(a + b*x)^2*Log[x] + 6*b*x*(a + b*x)^2*Log[a + b*x])/(2*a^4*x*(a +
b*x)*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.242, size = 117, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 6\,{b}^{3}\ln \left ( x \right ){x}^{3}-6\,{b}^{3}\ln \left ( bx+a \right ){x}^{3}+12\,{b}^{2}a\ln \left ( x \right ){x}^{2}-12\,\ln \left ( bx+a \right ){x}^{2}a{b}^{2}+6\,b{a}^{2}\ln \left ( x \right ) x-6\,\ln \left ( bx+a \right ) x{a}^{2}b+6\,a{b}^{2}{x}^{2}+9\,b{a}^{2}x+2\,{a}^{3} \right ) \left ( bx+a \right ) }{2\,{a}^{4}x} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*b^3*ln(x)*x^3-6*b^3*ln(b*x+a)*x^3+12*b^2*a*ln(x)*x^2-12*ln(b*x+a)*x^2*a*b^2+6*b*a^2*ln(x)*x-6*ln(b*x+a
)*x*a^2*b+6*a*b^2*x^2+9*b*a^2*x+2*a^3)*(b*x+a)/a^4/x/((b*x+a)^2)^(3/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72541, size = 232, normalized size = 1.41 \begin{align*} -\frac{6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3} - 6 \,{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (b x + a\right ) + 6 \,{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (x\right )}{2 \,{\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*a*b^2*x^2 + 9*a^2*b*x + 2*a^3 - 6*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*log(b*x + a) + 6*(b^3*x^3 + 2*a*b^
2*x^2 + a^2*b*x)*log(x))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x