### 3.203 $$\int \frac{x}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=63 $\frac{a}{4 b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{1}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0145728, antiderivative size = 63, 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} $\frac{a}{4 b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{1}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[x/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

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 )^{5/2}} \, dx &=-\frac{1}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{a \int \frac{1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx}{b}\\ &=-\frac{1}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{a}{4 b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0100249, size = 33, normalized size = 0.52 $\frac{-a-4 b x}{12 b^2 (a+b x)^3 \sqrt{(a+b x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-a - 4*b*x)/(12*b^2*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.174, size = 26, normalized size = 0.4 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( 4\,bx+a \right ) }{12\,{b}^{2}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

-1/12*(b*x+a)*(4*b*x+a)/b^2/((b*x+a)^2)^(5/2)

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Maxima [A]  time = 1.10861, size = 59, normalized size = 0.94 \begin{align*} -\frac{1}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{2}} + \frac{a}{4 \,{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{4}} \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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51872, size = 112, normalized size = 1.78 \begin{align*} -\frac{4 \, b x + a}{12 \,{\left (b^{6} x^{4} + 4 \, a b^{5} x^{3} + 6 \, a^{2} b^{4} x^{2} + 4 \, a^{3} b^{3} x + a^{4} 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)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(4*b*x + a)/(b^6*x^4 + 4*a*b^5*x^3 + 6*a^2*b^4*x^2 + 4*a^3*b^3*x + a^4*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{5}{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)**(5/2),x)

[Out]

Integral(x/((a + b*x)**2)**(5/2), 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)^(5/2),x, algorithm="giac")

[Out]

sage0*x