### 3.749 $$\int \frac{(a+b x)^4}{a^2-b^2 x^2} \, dx$$

Optimal. Leaf size=49 $-\frac{8 a^3 \log (a-b x)}{b}-4 a^2 x-\frac{a (a+b x)^2}{b}-\frac{(a+b x)^3}{3 b}$

[Out]

-4*a^2*x - (a*(a + b*x)^2)/b - (a + b*x)^3/(3*b) - (8*a^3*Log[a - b*x])/b

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Rubi [A]  time = 0.0205057, antiderivative size = 49, 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} $-\frac{8 a^3 \log (a-b x)}{b}-4 a^2 x-\frac{a (a+b x)^2}{b}-\frac{(a+b x)^3}{3 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x)^4/(a^2 - b^2*x^2),x]

[Out]

-4*a^2*x - (a*(a + b*x)^2)/b - (a + b*x)^3/(3*b) - (8*a^3*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)^4}{a^2-b^2 x^2} \, dx &=\int \frac{(a+b x)^3}{a-b x} \, dx\\ &=\int \left (-4 a^2+\frac{8 a^3}{a-b x}-2 a (a+b x)-(a+b x)^2\right ) \, dx\\ &=-4 a^2 x-\frac{a (a+b x)^2}{b}-\frac{(a+b x)^3}{3 b}-\frac{8 a^3 \log (a-b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0054987, size = 39, normalized size = 0.8 $-\frac{8 a^3 \log (a-b x)}{b}-7 a^2 x-2 a b x^2-\frac{b^2 x^3}{3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-7*a^2*x - 2*a*b*x^2 - (b^2*x^3)/3 - (8*a^3*Log[a - b*x])/b

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Maple [A]  time = 0.041, size = 39, normalized size = 0.8 \begin{align*} -{\frac{{b}^{2}{x}^{3}}{3}}-2\,ab{x}^{2}-7\,{a}^{2}x-8\,{\frac{{a}^{3}\ln \left ( bx-a \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b^2*x^3-2*a*b*x^2-7*a^2*x-8/b*a^3*ln(b*x-a)

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Maxima [A]  time = 1.23004, size = 51, normalized size = 1.04 \begin{align*} -\frac{1}{3} \, b^{2} x^{3} - 2 \, a b x^{2} - 7 \, a^{2} x - \frac{8 \, a^{3} \log \left (b x - a\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b^2*x^3 - 2*a*b*x^2 - 7*a^2*x - 8*a^3*log(b*x - a)/b

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Fricas [A]  time = 1.71158, size = 90, normalized size = 1.84 \begin{align*} -\frac{b^{3} x^{3} + 6 \, a b^{2} x^{2} + 21 \, a^{2} b x + 24 \, a^{3} \log \left (b x - a\right )}{3 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^3*x^3 + 6*a*b^2*x^2 + 21*a^2*b*x + 24*a^3*log(b*x - a))/b

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Sympy [A]  time = 0.315249, size = 37, normalized size = 0.76 \begin{align*} - \frac{8 a^{3} \log{\left (- a + b x \right )}}{b} - 7 a^{2} x - 2 a b x^{2} - \frac{b^{2} x^{3}}{3} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*a**3*log(-a + b*x)/b - 7*a**2*x - 2*a*b*x**2 - b**2*x**3/3

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Giac [A]  time = 1.22229, size = 66, normalized size = 1.35 \begin{align*} -\frac{8 \, a^{3} \log \left ({\left | b x - a \right |}\right )}{b} - \frac{b^{5} x^{3} + 6 \, a b^{4} x^{2} + 21 \, a^{2} b^{3} x}{3 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*a^3*log(abs(b*x - a))/b - 1/3*(b^5*x^3 + 6*a*b^4*x^2 + 21*a^2*b^3*x)/b^3