3.20.40 \(\int \frac {a+b x}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [1940]

Optimal. Leaf size=14 \[ -\frac {1}{2 b (a+b x)^2} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {27, 32} \begin {gather*} -\frac {1}{2 b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*1/(b*(a + b*x)^2)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {1}{(a+b x)^3} \, dx\\ &=-\frac {1}{2 b (a+b x)^2}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} -\frac {1}{2 b (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)^2,x]

[Out]

-1/2*1/(b*(a + b*x)^2)

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Maple [A]
time = 0.65, size = 13, normalized size = 0.93

method	result	size
gosper	\(-\frac {1}{2 b \left (b x +a \right )^{2}}\)	\(13\)
default	\(-\frac {1}{2 b \left (b x +a \right )^{2}}\)	\(13\)
risch	\(-\frac {1}{2 b \left (b x +a \right )^{2}}\)	\(13\)
norman	\(\frac {-\frac {a}{2 b}-\frac {x}{2}}{\left (b x +a \right )^{3}}\)	\(19\)

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)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 23, normalized size = 1.64 \begin {gather*} -\frac {1}{2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} b} \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)^2,x, algorithm="maxima")

[Out]

-1/2/((b^2*x^2 + 2*a*b*x + a^2)*b)

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Fricas [A]
time = 2.00, size = 24, normalized size = 1.71 \begin {gather*} -\frac {1}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\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)^2,x, algorithm="fricas")

[Out]

-1/2/(b^3*x^2 + 2*a*b^2*x + a^2*b)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (12) = 24\).
time = 0.08, size = 26, normalized size = 1.86 \begin {gather*} - \frac {1}{2 a^{2} b + 4 a b^{2} x + 2 b^{3} x^{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)**2,x)

[Out]

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

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Giac [A]
time = 1.32, size = 23, normalized size = 1.64 \begin {gather*} -\frac {1}{2 \, {\left (a^{2} + {\left (b x^{2} + 2 \, a x\right )} b\right )} b} \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)^2,x, algorithm="giac")

[Out]

-1/2/((a^2 + (b*x^2 + 2*a*x)*b)*b)

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Mupad [B]
time = 0.03, size = 26, normalized size = 1.86 \begin {gather*} -\frac {1}{2\,a^2\,b+4\,a\,b^2\,x+2\,b^3\,x^2} \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)^2,x)

[Out]

-1/(2*a^2*b + 2*b^3*x^2 + 4*a*b^2*x)

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