### 3.759 $$\int \frac{(a+b x)^5}{(a^2-b^2 x^2)^2} \, dx$$

Optimal. Leaf size=44 $\frac{8 a^3}{b (a-b x)}+\frac{12 a^2 \log (a-b x)}{b}+5 a x+\frac{b x^2}{2}$

[Out]

5*a*x + (b*x^2)/2 + (8*a^3)/(b*(a - b*x)) + (12*a^2*Log[a - b*x])/b

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Rubi [A]  time = 0.0309178, antiderivative size = 44, 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} $\frac{8 a^3}{b (a-b x)}+\frac{12 a^2 \log (a-b x)}{b}+5 a x+\frac{b x^2}{2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x)^5/(a^2 - b^2*x^2)^2,x]

[Out]

5*a*x + (b*x^2)/2 + (8*a^3)/(b*(a - b*x)) + (12*a^2*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 )^2} \, dx &=\int \frac{(a+b x)^3}{(a-b x)^2} \, dx\\ &=\int \left (5 a+b x+\frac{8 a^3}{(a-b x)^2}-\frac{12 a^2}{a-b x}\right ) \, dx\\ &=5 a x+\frac{b x^2}{2}+\frac{8 a^3}{b (a-b x)}+\frac{12 a^2 \log (a-b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0197676, size = 45, normalized size = 1.02 $-\frac{8 a^3}{b (b x-a)}+\frac{12 a^2 \log (a-b x)}{b}+5 a x+\frac{b x^2}{2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

5*a*x + (b*x^2)/2 - (8*a^3)/(b*(-a + b*x)) + (12*a^2*Log[a - b*x])/b

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Maple [A]  time = 0.046, size = 45, normalized size = 1. \begin{align*}{\frac{b{x}^{2}}{2}}+5\,ax-8\,{\frac{{a}^{3}}{b \left ( bx-a \right ) }}+12\,{\frac{{a}^{2}\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)^2,x)

[Out]

1/2*b*x^2+5*a*x-8/b*a^3/(b*x-a)+12/b*a^2*ln(b*x-a)

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Maxima [A]  time = 1.04831, size = 59, normalized size = 1.34 \begin{align*} \frac{1}{2} \, b x^{2} - \frac{8 \, a^{3}}{b^{2} x - a b} + 5 \, a x + \frac{12 \, a^{2} \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)^2,x, algorithm="maxima")

[Out]

1/2*b*x^2 - 8*a^3/(b^2*x - a*b) + 5*a*x + 12*a^2*log(b*x - a)/b

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Fricas [A]  time = 1.73501, size = 134, normalized size = 3.05 \begin{align*} \frac{b^{3} x^{3} + 9 \, a b^{2} x^{2} - 10 \, a^{2} b x - 16 \, a^{3} + 24 \,{\left (a^{2} b x - a^{3}\right )} \log \left (b x - a\right )}{2 \,{\left (b^{2} x - a b\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^3*x^3 + 9*a*b^2*x^2 - 10*a^2*b*x - 16*a^3 + 24*(a^2*b*x - a^3)*log(b*x - a))/(b^2*x - a*b)

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Sympy [A]  time = 0.383747, size = 37, normalized size = 0.84 \begin{align*} - \frac{8 a^{3}}{- a b + b^{2} x} + \frac{12 a^{2} \log{\left (- a + b x \right )}}{b} + 5 a x + \frac{b x^{2}}{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*a**3/(-a*b + b**2*x) + 12*a**2*log(-a + b*x)/b + 5*a*x + b*x**2/2

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Giac [A]  time = 1.22885, size = 74, normalized size = 1.68 \begin{align*} \frac{12 \, a^{2} \log \left ({\left | b x - a \right |}\right )}{b} - \frac{8 \, a^{3}}{{\left (b x - a\right )} b} + \frac{b^{5} x^{2} + 10 \, a b^{4} x}{2 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*a^2*log(abs(b*x - a))/b - 8*a^3/((b*x - a)*b) + 1/2*(b^5*x^2 + 10*a*b^4*x)/b^4