3.16.99 \(\int \frac {d+e x}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [1599]

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

[Out]

-e/b^2/((b*x+a)^2)^(1/2)+1/2*(a*e-b*d)/b^2/(b*x+a)/((b*x+a)^2)^(1/2)

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Rubi [A]
time = 0.03, 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 = {654, 621} \begin {gather*} -\frac {b d-a e}{2 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

-(e/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (b*d - a*e)/(2*b^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^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 {d+e x}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=-\frac {e}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 b^2 d-2 a b e\right ) \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx}{2 b^2}\\ &=-\frac {e}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b d-a e}{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 = 0.01, size = 39, normalized size = 0.57 \begin {gather*} \frac {-a e-b (d+2 e x)}{2 b^2 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(-(a*e) - b*(d + 2*e*x))/(2*b^2*(a + b*x)*Sqrt[(a + b*x)^2])

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Maple [A]
time = 0.60, size = 32, normalized size = 0.46

method	result	size
gosper	\(-\frac {\left (b x +a \right ) \left (2 b e x +a e +b d \right )}{2 b^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(32\)
default	\(-\frac {\left (b x +a \right ) \left (2 b e x +a e +b d \right )}{2 b^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(32\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {e x}{b}-\frac {a e +b d}{2 b^{2}}\right )}{\left (b x +a \right )^{3}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 58, normalized size = 0.84 \begin {gather*} -\frac {e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {d}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {a e}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.57, size = 39, normalized size = 0.57 \begin {gather*} -\frac {b d + {\left (2 \, b x + a\right )} e}{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((e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d + e 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((e*x+d)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral((d + e*x)/((a + b*x)**2)**(3/2), x)

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Giac [A]
time = 1.46, size = 34, normalized size = 0.49 \begin {gather*} -\frac {2 \, b x e + b d + a e}{2 \, {\left (b x + a\right )}^{2} b^{2} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

-1/2*(2*b*x*e + b*d + a*e)/((b*x + a)^2*b^2*sgn(b*x + a))

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Mupad [B]
time = 0.62, size = 42, normalized size = 0.61 \begin {gather*} -\frac {\left (a\,e+b\,d+2\,b\,e\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,b^2\,{\left (a+b\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

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

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