3.18.24 \(\int \frac {a+b x}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=14 \[ -\frac {1}{4 b (a+b x)^4} \]

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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}{4 b (a+b x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(4*b*(a + b*x)^4)

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 )^3} \, dx &=\int \frac {1}{(a+b x)^5} \, dx\\ &=-\frac {1}{4 b (a+b x)^4}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} -\frac {1}{4 b (a+b x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*1/(b*(a + b*x)^4)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.38, size = 46, normalized size = 3.29 \begin {gather*} -\frac {1}{4 \, {\left (b^{5} x^{4} + 4 \, a b^{4} x^{3} + 6 \, a^{2} b^{3} x^{2} + 4 \, a^{3} b^{2} x + a^{4} 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)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 23, normalized size = 1.64 \begin {gather*} -\frac {1}{4 \, {\left (a^{2} + {\left (b x^{2} + 2 \, a x\right )} b\right )}^{2} 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)^3,x, algorithm="giac")

[Out]

-1/4/((a^2 + (b*x^2 + 2*a*x)*b)^2*b)

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maple [A]  time = 0.06, size = 13, normalized size = 0.93 \begin {gather*} -\frac {1}{4 \left (b x +a \right )^{4} b} \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,x)

[Out]

-1/4/b/(b*x+a)^4

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maxima [A]  time = 0.56, size = 23, normalized size = 1.64 \begin {gather*} -\frac {1}{4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2} 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)^3,x, algorithm="maxima")

[Out]

-1/4/((b^2*x^2 + 2*a*b*x + a^2)^2*b)

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mupad [B]  time = 1.99, size = 48, normalized size = 3.43 \begin {gather*} -\frac {1}{4\,a^4\,b+16\,a^3\,b^2\,x+24\,a^2\,b^3\,x^2+16\,a\,b^4\,x^3+4\,b^5\,x^4} \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,x)

[Out]

-1/(4*a^4*b + 4*b^5*x^4 + 16*a^3*b^2*x + 16*a*b^4*x^3 + 24*a^2*b^3*x^2)

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sympy [B]  time = 0.32, size = 49, normalized size = 3.50 \begin {gather*} - \frac {1}{4 a^{4} b + 16 a^{3} b^{2} x + 24 a^{2} b^{3} x^{2} + 16 a b^{4} x^{3} + 4 b^{5} x^{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)**3,x)

[Out]

-1/(4*a**4*b + 16*a**3*b**2*x + 24*a**2*b**3*x**2 + 16*a*b**4*x**3 + 4*b**5*x**4)

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