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

Optimal. Leaf size=142 $\frac{b (a+b x)}{a^2 x \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a+b x}{2 a x^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{b^2 \log (x) (a+b x)}{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b^2 (a+b x) \log (a+b x)}{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}$

[Out]

-(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])

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Rubi [A]  time = 0.0447885, antiderivative size = 142, 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} $\frac{b (a+b x)}{a^2 x \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a+b x}{2 a x^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{b^2 \log (x) (a+b x)}{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b^2 (a+b x) \log (a+b x)}{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}$

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 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 \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*}

Mathematica [A]  time = 0.0179001, size = 59, normalized size = 0.42 $-\frac{(a+b x) \left (2 b^2 x^2 \log (a+b x)+a (a-2 b x)-2 b^2 x^2 \log (x)\right )}{2 a^3 x^2 \sqrt{(a+b x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.178, size = 58, normalized size = 0.4 \begin{align*}{\frac{ \left ( bx+a \right ) \left ( 2\,{b}^{2}\ln \left ( x \right ){x}^{2}-2\,{b}^{2}\ln \left ( bx+a \right ){x}^{2}+2\,abx-{a}^{2} \right ) }{2\,{x}^{2}{a}^{3}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

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(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

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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*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.76156, size = 103, normalized size = 0.73 \begin{align*} -\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}} \end{align*}

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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Sympy [A]  time = 0.911211, size = 31, normalized size = 0.22 \begin{align*} \frac{- a + 2 b x}{2 a^{2} x^{2}} + \frac{b^{2} \left (\log{\left (x \right )} - \log{\left (\frac{a}{b} + x \right )}\right )}{a^{3}} \end{align*}

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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Giac [A]  time = 1.3977, size = 73, normalized size = 0.51 \begin{align*} -\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 ) \end{align*}

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)