\(\int \frac {(a+b x)^4}{(a^2-b^2 x^2)^3} \, dx\) [771]

Optimal result

Integrand size = 22, antiderivative size = 10 \[ \int \frac {(a+b x)^4}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {x}{(a-b x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {641, 34} \[ \int \frac {(a+b x)^4}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {x}{(a-b x)^2} \]

[In]

Int[(a + b*x)^4/(a^2 - b^2*x^2)^3,x]

[Out]

x/(a - b*x)^2

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]

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*} \text {integral}& = \int \frac {a+b x}{(a-b x)^3} \, dx \\ & = \frac {x}{(a-b x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^4}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {x}{(a-b x)^2} \]

[In]

Integrate[(a + b*x)^4/(a^2 - b^2*x^2)^3,x]

[Out]

x/(a - b*x)^2

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.10

method	result	size
gosper	\(\frac {x}{\left (-b x +a \right )^{2}}\)	\(11\)
risch	\(\frac {x}{\left (-b x +a \right )^{2}}\)	\(11\)
parallelrisch	\(\frac {x}{\left (b x -a \right )^{2}}\)	\(12\)
default	\(\frac {a}{b \left (-b x +a \right )^{2}}-\frac {1}{b \left (-b x +a \right )}\)	\(28\)
norman	\(\frac {b^{2} x^{3}+2 a b \,x^{2}+a^{2} x}{\left (-b^{2} x^{2}+a^{2}\right )^{2}}\)	\(36\)

[In]

int((b*x+a)^4/(-b^2*x^2+a^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 2.00 \[ \int \frac {(a+b x)^4}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {x}{b^{2} x^{2} - 2 \, a b x + a^{2}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (7) = 14\).

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.70 \[ \int \frac {(a+b x)^4}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {x}{a^{2} - 2 a b x + b^{2} x^{2}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 2.00 \[ \int \frac {(a+b x)^4}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {x}{b^{2} x^{2} - 2 \, a b x + a^{2}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b x)^4}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {x}{{\left (b x - a\right )}^{2}} \]

[In]

integrate((b*x+a)^4/(-b^2*x^2+a^2)^3,x, algorithm="giac")

[Out]

x/(b*x - a)^2

Mupad [B] (verification not implemented)

Time = 9.72 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^4}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {x}{{\left (a-b\,x\right )}^2} \]

[In]

int((a + b*x)^4/(a^2 - b^2*x^2)^3,x)

[Out]

x/(a - b*x)^2