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

Optimal. Leaf size=71 \[ -\frac {A b-a B}{4 b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {B}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 71, 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}{4 b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {B}{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[(A + B*x)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

-B/(3*b^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - (A*b - a*B)/(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 {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=-\frac {B}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\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 )^{5/2}} \, dx}{2 b^2}\\ &=-\frac {B}{3 b^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {A b-a B}{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 = 0.02, size = 39, normalized size = 0.55 \begin {gather*} \frac {-B (a+4 b x)-3 A b}{12 b^2 (a+b x)^3 \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)^(5/2),x]

[Out]

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

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IntegrateAlgebraic [B]  time = 0.98, size = 281, normalized size = 3.96 \begin {gather*} \frac {-2 \left (3 a^5 b B-3 a^4 A b^2-a b^5 B x^4-3 A b^6 x^4-4 b^6 B x^5\right )-2 \sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (3 a^4 B-3 a^3 A b-3 a^3 b B x+3 a^2 A b^2 x+3 a^2 b^2 B x^2-3 a A b^3 x^2-3 a b^3 B x^3+3 A b^4 x^3+4 b^4 B 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[(A + B*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^3*A*b + 3*a^4*B + 3*a^2*A*b^2*x - 3*a^3*b*B*x - 3*a*A*b^3*x^
2 + 3*a^2*b^2*B*x^2 + 3*A*b^4*x^3 - 3*a*b^3*B*x^3 + 4*b^4*B*x^4) - 2*(-3*a^4*A*b^2 + 3*a^5*b*B - 3*A*b^6*x^4 -
a*b^5*B*x^4 - 4*b^6*B*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 = 61, normalized size = 0.86 \begin {gather*} -\frac {4 \, B b x + B a + 3 \, A b}{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((B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(4*B*b*x + B*a + 3*A*b)/(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((B*x+A)/(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 = 0.06, size = 33, normalized size = 0.46 \begin {gather*} -\frac {\left (b x +a \right ) \left (4 B b x +3 A b +B 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((B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

-1/12*(b*x+a)/b^2*(4*B*b*x+3*A*b+B*a)/((b*x+a)^2)^(5/2)

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maxima [A]  time = 0.52, size = 56, normalized size = 0.79 \begin {gather*} -\frac {B}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {B a}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {A}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} \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)^(5/2),x, algorithm="maxima")

[Out]

-1/3*B/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 1/4*B*a/(b^6*(x + a/b)^4) - 1/4*A/(b^5*(x + a/b)^4)

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mupad [B]  time = 1.20, size = 43, normalized size = 0.61 \begin {gather*} -\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (3\,A\,b+B\,a+4\,B\,b\,x\right )}{12\,b^2\,{\left (a+b\,x\right )}^5} \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)^(5/2),x)

[Out]

-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(3*A*b + B*a + 4*B*b*x))/(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 {A + B 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((B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((A + B*x)/((a + b*x)**2)**(5/2), x)

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