### 3.771 $$\int \frac{(a+b x)^4}{(a^2-b^2 x^2)^3} \, dx$$

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

[Out]

x/(a - b*x)^2

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Rubi [A]  time = 0.0070928, antiderivative size = 10, normalized size of antiderivative = 1., 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} $\frac{x}{(a-b x)^2}$

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 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]))

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]

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*}

Mathematica [A]  time = 0.0035456, size = 10, normalized size = 1. $\frac{x}{(a-b x)^2}$

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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Maple [B]  time = 0.044, size = 29, normalized size = 2.9 \begin{align*}{\frac{a}{b \left ( bx-a \right ) ^{2}}}+{\frac{1}{b \left ( bx-a \right ) }} \end{align*}

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]

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

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Maxima [A]  time = 1.00771, size = 27, normalized size = 2.7 \begin{align*} \frac{x}{b^{2} x^{2} - 2 \, a b x + a^{2}} \end{align*}

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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Fricas [A]  time = 1.65899, size = 39, normalized size = 3.9 \begin{align*} \frac{x}{b^{2} x^{2} - 2 \, a b x + a^{2}} \end{align*}

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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Sympy [B]  time = 0.357719, size = 17, normalized size = 1.7 \begin{align*} \frac{x}{a^{2} - 2 a b x + b^{2} x^{2}} \end{align*}

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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Giac [A]  time = 1.19041, size = 15, normalized size = 1.5 \begin{align*} \frac{x}{{\left (b x - a\right )}^{2}} \end{align*}

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