### 3.193 $$\int \frac{x}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=61 $\frac{a}{2 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}$

[Out]

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

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Rubi [A]  time = 0.0146494, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {640, 607} $\frac{a}{2 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 607

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(2*(a + b*x + c*x^2)^(p + 1))/((2*p + 1)*(b + 2
*c*x)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0096303, size = 33, normalized size = 0.54 $\frac{-a-2 b x}{2 b^2 (a+b x) \sqrt{(a+b x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.176, size = 26, normalized size = 0.4 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( 2\,bx+a \right ) }{2\,{b}^{2}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.83118, size = 59, normalized size = 0.97 \begin{align*} -\frac{1}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac{a}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 1/2*a/((b^2)^(3/2)*b*(x + a/b)^2)

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Fricas [A]  time = 1.58122, size = 68, normalized size = 1.11 \begin{align*} -\frac{2 \, b x + a}{2 \,{\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

sage0*x