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

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

[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])

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Rubi [A]  time = 0.0795661, antiderivative size = 209, 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} \[ \frac{b^2}{2 a^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 b^2}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 b (a+b x)}{a^4 x \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a+b x}{2 a^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{6 b^2 \log (x) (a+b x)}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{6 b^2 (a+b x) \log (a+b x)}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}} \]

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 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^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*}

Mathematica [A]  time = 0.0362402, size = 99, normalized size = 0.47 \[ \frac{a \left (4 a^2 b x-a^3+18 a b^2 x^2+12 b^3 x^3\right )+12 b^2 x^2 \log (x) (a+b x)^2-12 b^2 x^2 (a+b x)^2 \log (a+b x)}{2 a^5 x^2 (a+b x) \sqrt{(a+b x)^2}} \]

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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Maple [A]  time = 0.239, size = 136, normalized size = 0.7 \begin{align*}{\frac{ \left ( 12\,\ln \left ( x \right ){x}^{4}{b}^{4}-12\,\ln \left ( bx+a \right ){x}^{4}{b}^{4}+24\,\ln \left ( x \right ){x}^{3}a{b}^{3}-24\,\ln \left ( bx+a \right ){x}^{3}a{b}^{3}+12\,\ln \left ( x \right ){x}^{2}{a}^{2}{b}^{2}-12\,\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}+12\,a{b}^{3}{x}^{3}+18\,{x}^{2}{a}^{2}{b}^{2}+4\,x{a}^{3}b-{a}^{4} \right ) \left ( bx+a \right ) }{2\,{x}^{2}{a}^{5}} \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^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/2*(12*ln(x)*x^4*b^4-12*ln(b*x+a)*x^4*b^4+24*ln(x)*x^3*a*b^3-24*ln(b*x+a)*x^3*a*b^3+12*ln(x)*x^2*a^2*b^2-12*l
n(b*x+a)*x^2*a^2*b^2+12*a*b^3*x^3+18*x^2*a^2*b^2+4*x*a^3*b-a^4)*(b*x+a)/x^2/a^5/((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^3/(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 = 1.69949, size = 269, normalized size = 1.29 \begin{align*} \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 \left (x\right )}{2 \,{\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \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**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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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^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x