3.18.80 \(\int \frac {A+B x}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [1780]

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

[Out]

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

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Rubi [A]
time = 0.01, 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 = {654, 621} \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]

-1/3*B/(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 621

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 654

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 = 0.01, size = 39, normalized size = 0.55 \begin {gather*} \frac {-3 A b-B (a+4 b x)}{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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Maple [A]
time = 0.81, size = 33, normalized size = 0.46

method	result	size
gosper	\(-\frac {\left (b x +a \right ) \left (4 b B x +3 A b +B a \right )}{12 b^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(33\)
default	\(-\frac {\left (b x +a \right ) \left (4 b B x +3 A b +B a \right )}{12 b^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(33\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B x}{3 b}-\frac {3 A b +B a}{12 b^{2}}\right )}{\left (b x +a \right )^{5}}\)	\(39\)

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,method=_RETURNVERBOSE)

[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 = 0.27, 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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Fricas [A]
time = 0.70, 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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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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Giac [A]
time = 1.07, size = 33, normalized size = 0.46 \begin {gather*} -\frac {4 \, B b x + B a + 3 \, A b}{12 \, {\left (b x + a\right )}^{4} b^{2} \mathrm {sgn}\left (b x + a\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="giac")

[Out]

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

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Mupad [B]
time = 2.12, 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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