### 3.144 $$\int \frac{\sqrt{a^2+2 a b x+b^2 x^2}}{x^3} \, dx$$

Optimal. Leaf size=35 $-\frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{2 a x^2}$

[Out]

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

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Rubi [A]  time = 0.0130166, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {646, 37} $-\frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{2 a x^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*a*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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0069257, size = 31, normalized size = 0.89 $-\frac{\sqrt{(a+b x)^2} (a+2 b x)}{2 x^2 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.045, size = 28, normalized size = 0.8 \begin{align*} -{\frac{2\,bx+a}{2\,{x}^{2} \left ( bx+a \right ) }\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.02235, size = 30, normalized size = 0.86 \begin{align*} -\frac{2 \, b x + a}{2 \, x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*x + a)/x^2

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Sympy [A]  time = 0.582048, size = 12, normalized size = 0.34 \begin{align*} - \frac{a + 2 b x}{2 x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a + 2*b*x)/(2*x**2)

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Giac [A]  time = 1.23143, size = 53, normalized size = 1.51 \begin{align*} -\frac{b^{2} \mathrm{sgn}\left (b x + a\right )}{2 \, a} - \frac{2 \, b x \mathrm{sgn}\left (b x + a\right ) + a \mathrm{sgn}\left (b x + a\right )}{2 \, x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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