3.8.72 \(\int \frac {(a+b x)^3}{(a^2-b^2 x^2)^3} \, dx\) [772]

Optimal. Leaf size=15 \[ \frac {1}{2 b (a-b x)^2} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 15, 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 = {641, 32} \begin {gather*} \frac {1}{2 b (a-b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3/(a^2 - b^2*x^2)^3,x]

[Out]

1/(2*b*(a - b*x)^2)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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)^3}{\left (a^2-b^2 x^2\right )^3} \, dx &=\int \frac {1}{(a-b x)^3} \, dx\\ &=\frac {1}{2 b (a-b x)^2}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} \frac {1}{2 b (a-b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(2*b*(a - b*x)^2)

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Maple [A]
time = 0.44, size = 14, normalized size = 0.93

method	result	size
gosper	\(\frac {1}{2 b \left (-b x +a \right )^{2}}\)	\(14\)
default	\(\frac {1}{2 b \left (-b x +a \right )^{2}}\)	\(14\)
risch	\(\frac {1}{2 b \left (-b x +a \right )^{2}}\)	\(14\)
norman	\(\frac {a x +\frac {a^{2}}{2 b}+\frac {b \,x^{2}}{2}}{\left (-b^{2} x^{2}+a^{2}\right )^{2}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3/(-b^2*x^2+a^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 24, normalized size = 1.60 \begin {gather*} \frac {1}{2 \, {\left (b^{3} x^{2} - 2 \, a b^{2} x + a^{2} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(b^3*x^2 - 2*a*b^2*x + a^2*b)

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Fricas [A]
time = 2.38, size = 24, normalized size = 1.60 \begin {gather*} \frac {1}{2 \, {\left (b^{3} x^{2} - 2 \, a b^{2} x + a^{2} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(b^3*x^2 - 2*a*b^2*x + a^2*b)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 24 vs. \(2 (10) = 20\).
time = 0.08, size = 24, normalized size = 1.60 \begin {gather*} \frac {1}{2 a^{2} b - 4 a b^{2} x + 2 b^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(2*a**2*b - 4*a*b**2*x + 2*b**3*x**2)

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Giac [A]
time = 1.46, size = 14, normalized size = 0.93 \begin {gather*} \frac {1}{2 \, {\left (b x - a\right )}^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/((b*x - a)^2*b)

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Mupad [B]
time = 0.39, size = 24, normalized size = 1.60 \begin {gather*} \frac {1}{2\,a^2\,b-4\,a\,b^2\,x+2\,b^3\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^3/(a^2 - b^2*x^2)^3,x)

[Out]

1/(2*a^2*b + 2*b^3*x^2 - 4*a*b^2*x)

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