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

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

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Rubi [A]  time = 0.01, antiderivative size = 63, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.52 \begin {gather*} \frac {-a-4 b x}{12 b^2 (a+b x)^3 \sqrt {(a+b x)^2}} \end {gather*}

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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IntegrateAlgebraic [B]  time = 0.68, size = 219, normalized size = 3.48 \begin {gather*} \frac {-2 \left (3 a^5 b-a b^5 x^4-4 b^6 x^5\right )-2 \sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (3 a^4-3 a^3 b x+3 a^2 b^2 x^2-3 a b^3 x^3+4 b^4 x^4\right )}{3 x^4 \sqrt {a^2+2 a b x+b^2 x^2} \left (-8 a^3 b^7-24 a^2 b^8 x-24 a b^9 x^2-8 b^{10} x^3\right )+3 \sqrt {b^2} x^4 \left (8 a^4 b^6+32 a^3 b^7 x+48 a^2 b^8 x^2+32 a b^9 x^3+8 b^{10} x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(3*a^4 - 3*a^3*b*x + 3*a^2*b^2*x^2 - 3*a*b^3*x^3 + 4*b^4*x^4) - 2*
(3*a^5*b - a*b^5*x^4 - 4*b^6*x^5))/(3*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-8*a^3*b^7 - 24*a^2*b^8*x - 24*a*b^9*
x^2 - 8*b^10*x^3) + 3*Sqrt[b^2]*x^4*(8*a^4*b^6 + 32*a^3*b^7*x + 48*a^2*b^8*x^2 + 32*a*b^9*x^3 + 8*b^10*x^4))

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fricas [A]  time = 0.42, size = 54, normalized size = 0.86 \begin {gather*} -\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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

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

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maple [A]  time = 0.05, size = 26, normalized size = 0.41 \begin {gather*} -\frac {\left (b x +a \right ) \left (4 b x +a \right )}{12 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{2}} \end {gather*}

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 = 1.24, size = 39, normalized size = 0.62 \begin {gather*} -\frac {1}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {a}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} \end {gather*}

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^6*(x + a/b)^4)

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mupad [B]  time = 0.22, size = 36, normalized size = 0.57 \begin {gather*} -\frac {\left (a+4\,b\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{12\,b^2\,{\left (a+b\,x\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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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