3.7.12 \(\int \frac {(a+b x)^4}{(a^2-b^2 x^2)^3} \, dx\)

Optimal. Leaf size=10 \[ \frac {x}{(a-b x)^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {627, 34} \begin {gather*} \frac {x}{(a-b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^4/(a^2 - b^2*x^2)^3,x]

[Out]

x/(a - b*x)^2

Rule 34

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_)), x_Symbol] :> Simp[(d*x*(a + b*x)^(m + 1))/(b*(m + 2)), x] /
; FreeQ[{a, b, c, d, m}, x] && EqQ[a*d - b*c*(m + 2), 0]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rubi steps

\begin {align*} \int \frac {(a+b x)^4}{\left (a^2-b^2 x^2\right )^3} \, dx &=\int \frac {a+b x}{(a-b x)^3} \, dx\\ &=\frac {x}{(a-b x)^2}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 1.00 \begin {gather*} \frac {x}{(a-b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^4/(a^2 - b^2*x^2)^3,x]

[Out]

x/(a - b*x)^2

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x)^4}{\left (a^2-b^2 x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x)^4/(a^2 - b^2*x^2)^3,x]

[Out]

IntegrateAlgebraic[(a + b*x)^4/(a^2 - b^2*x^2)^3, x]

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fricas [A]  time = 0.38, size = 20, normalized size = 2.00 \begin {gather*} \frac {x}{b^{2} x^{2} - 2 \, a b x + a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^4/(-b^2*x^2+a^2)^3,x, algorithm="fricas")

[Out]

x/(b^2*x^2 - 2*a*b*x + a^2)

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giac [A]  time = 0.20, size = 11, normalized size = 1.10 \begin {gather*} \frac {x}{{\left (b x - a\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^4/(-b^2*x^2+a^2)^3,x, algorithm="giac")

[Out]

x/(b*x - a)^2

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maple [B]  time = 0.04, size = 29, normalized size = 2.90 \begin {gather*} \frac {a}{\left (b x -a \right )^{2} b}+\frac {1}{\left (b x -a \right ) b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^4/(-b^2*x^2+a^2)^3,x)

[Out]

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

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maxima [A]  time = 1.34, size = 20, normalized size = 2.00 \begin {gather*} \frac {x}{b^{2} x^{2} - 2 \, a b x + a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^4/(-b^2*x^2+a^2)^3,x, algorithm="maxima")

[Out]

x/(b^2*x^2 - 2*a*b*x + a^2)

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mupad [B]  time = 0.40, size = 10, normalized size = 1.00 \begin {gather*} \frac {x}{{\left (a-b\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^4/(a^2 - b^2*x^2)^3,x)

[Out]

x/(a - b*x)^2

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sympy [B]  time = 0.25, size = 17, normalized size = 1.70 \begin {gather*} \frac {x}{a^{2} - 2 a b x + b^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(a**2 - 2*a*b*x + b**2*x**2)

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