3.16.51 \(\int \frac {A+B x}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=69 \[ -\frac {A b-a B}{2 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {640, 607} \begin {gather*} -\frac {A b-a B}{2 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 39, normalized size = 0.57 \begin {gather*} \frac {-B (a+2 b x)-A b}{2 b^2 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 0.00, size = 178, normalized size = 2.58 \begin {gather*} \frac {-a^3 b B+\sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (a^2 (-B)+a A b+a b B x-A b^2 x-2 b^2 B x^2\right )+a^2 A b^2+a b^3 B x^2+A b^4 x^2+2 b^4 B x^3}{x^2 \left (-2 a b^5-2 b^6 x\right ) \sqrt {a^2+2 a b x+b^2 x^2}+\sqrt {b^2} x^2 \left (2 a^2 b^4+4 a b^5 x+2 b^6 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(a^2*A*b^2 - a^3*b*B + A*b^4*x^2 + a*b^3*B*x^2 + 2*b^4*B*x^3 + Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(a*A*b
- a^2*B - A*b^2*x + a*b*B*x - 2*b^2*B*x^2))/(x^2*(-2*a*b^5 - 2*b^6*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + Sqrt[b^2
]*x^2*(2*a^2*b^4 + 4*a*b^5*x + 2*b^6*x^2))

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fricas [A]  time = 0.42, size = 38, normalized size = 0.55 \begin {gather*} -\frac {2 \, B b x + B a + A b}{2 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*B*b*x + B*a + A*b)/(b^4*x^2 + 2*a*b^3*x + a^2*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((B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.04, size = 32, normalized size = 0.46 \begin {gather*} -\frac {\left (b x +a \right ) \left (2 B b x +A b +B a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 56, normalized size = 0.81 \begin {gather*} -\frac {B}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {B a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {A}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.05, size = 42, normalized size = 0.61 \begin {gather*} -\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (A\,b+B\,a+2\,B\,b\,x\right )}{2\,b^2\,{\left (a+b\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)/((a + b*x)**2)**(3/2), x)

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