### 3.770 $$\int \frac{(a+b x)^5}{(a^2-b^2 x^2)^3} \, dx$$

Optimal. Leaf size=43 $\frac{2 a^2}{b (a-b x)^2}-\frac{4 a}{b (a-b x)}-\frac{\log (a-b x)}{b}$

[Out]

(2*a^2)/(b*(a - b*x)^2) - (4*a)/(b*(a - b*x)) - Log[a - b*x]/b

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Rubi [A]  time = 0.0246914, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {627, 43} $\frac{2 a^2}{b (a-b x)^2}-\frac{4 a}{b (a-b x)}-\frac{\log (a-b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x)^5/(a^2 - b^2*x^2)^3,x]

[Out]

(2*a^2)/(b*(a - b*x)^2) - (4*a)/(b*(a - b*x)) - Log[a - b*x]/b

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(a+b x)^5}{\left (a^2-b^2 x^2\right )^3} \, dx &=\int \frac{(a+b x)^2}{(a-b x)^3} \, dx\\ &=\int \left (\frac{4 a^2}{(a-b x)^3}-\frac{4 a}{(a-b x)^2}+\frac{1}{a-b x}\right ) \, dx\\ &=\frac{2 a^2}{b (a-b x)^2}-\frac{4 a}{b (a-b x)}-\frac{\log (a-b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0201887, size = 30, normalized size = 0.7 $-\frac{\frac{2 a (a-2 b x)}{(a-b x)^2}+\log (a-b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x)^5/(a^2 - b^2*x^2)^3,x]

[Out]

-(((2*a*(a - 2*b*x))/(a - b*x)^2 + Log[a - b*x])/b)

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Maple [A]  time = 0.044, size = 47, normalized size = 1.1 \begin{align*} 2\,{\frac{{a}^{2}}{b \left ( bx-a \right ) ^{2}}}+4\,{\frac{a}{b \left ( bx-a \right ) }}-{\frac{\ln \left ( bx-a \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^5/(-b^2*x^2+a^2)^3,x)

[Out]

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

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Maxima [A]  time = 1.00333, size = 66, normalized size = 1.53 \begin{align*} \frac{2 \,{\left (2 \, a b x - a^{2}\right )}}{b^{3} x^{2} - 2 \, a b^{2} x + a^{2} b} - \frac{\log \left (b x - a\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(-b^2*x^2+a^2)^3,x, algorithm="maxima")

[Out]

2*(2*a*b*x - a^2)/(b^3*x^2 - 2*a*b^2*x + a^2*b) - log(b*x - a)/b

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Fricas [A]  time = 1.7725, size = 122, normalized size = 2.84 \begin{align*} \frac{4 \, a b x - 2 \, a^{2} -{\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \log \left (b x - a\right )}{b^{3} x^{2} - 2 \, a b^{2} x + a^{2} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(-b^2*x^2+a^2)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.420473, size = 39, normalized size = 0.91 \begin{align*} \frac{- 2 a^{2} + 4 a b x}{a^{2} b - 2 a b^{2} x + b^{3} x^{2}} - \frac{\log{\left (- a + b x \right )}}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**5/(-b**2*x**2+a**2)**3,x)

[Out]

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

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Giac [A]  time = 1.1591, size = 54, normalized size = 1.26 \begin{align*} -\frac{\log \left ({\left | b x - a \right |}\right )}{b} + \frac{2 \,{\left (2 \, a b x - a^{2}\right )}}{{\left (b x - a\right )}^{2} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(-b^2*x^2+a^2)^3,x, algorithm="giac")

[Out]

-log(abs(b*x - a))/b + 2*(2*a*b*x - a^2)/((b*x - a)^2*b)