\(\int \frac {(a+b x)^2}{(a c-b c x)^4} \, dx\) [1045]

Optimal result

Integrand size = 19, antiderivative size = 28 \[ \int \frac {(a+b x)^2}{(a c-b c x)^4} \, dx=\frac {(a+b x)^3}{6 a b c^4 (a-b x)^3} \]

[Out]

1/6*(b*x+a)^3/a/b/c^4/(-b*x+a)^3

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {37} \[ \int \frac {(a+b x)^2}{(a c-b c x)^4} \, dx=\frac {(a+b x)^3}{6 a b c^4 (a-b x)^3} \]

[In]

Int[(a + b*x)^2/(a*c - b*c*x)^4,x]

[Out]

(a + b*x)^3/(6*a*b*c^4*(a - b*x)^3)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {(a+b x)^3}{6 a b c^4 (a-b x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b x)^2}{(a c-b c x)^4} \, dx=-\frac {a^2+3 b^2 x^2}{3 b c^4 (-a+b x)^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {b \,x^{2}+\frac {a^{2}}{3 b}}{c^{4} \left (-b x +a \right )^{3}}\)	\(27\)
gosper	\(\frac {3 b^{2} x^{2}+a^{2}}{3 \left (-b x +a \right )^{3} c^{4} b}\)	\(29\)
norman	\(\frac {\frac {a^{2}}{3 b c}+\frac {b \,x^{2}}{c}}{c^{3} \left (-b x +a \right )^{3}}\)	\(33\)
parallelrisch	\(\frac {-3 x^{2} b^{4}-a^{2} b^{2}}{3 b^{3} c^{4} \left (b x -a \right )^{3}}\)	\(35\)
default	\(\frac {\frac {4 a^{2}}{3 b \left (-b x +a \right )^{3}}+\frac {1}{b \left (-b x +a \right )}-\frac {2 a}{b \left (-b x +a \right )^{2}}}{c^{4}}\)	\(48\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).

Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.14 \[ \int \frac {(a+b x)^2}{(a c-b c x)^4} \, dx=-\frac {3 \, b^{2} x^{2} + a^{2}}{3 \, {\left (b^{4} c^{4} x^{3} - 3 \, a b^{3} c^{4} x^{2} + 3 \, a^{2} b^{2} c^{4} x - a^{3} b c^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (20) = 40\).

Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.18 \[ \int \frac {(a+b x)^2}{(a c-b c x)^4} \, dx=\frac {- a^{2} - 3 b^{2} x^{2}}{- 3 a^{3} b c^{4} + 9 a^{2} b^{2} c^{4} x - 9 a b^{3} c^{4} x^{2} + 3 b^{4} c^{4} x^{3}} \]

[In]

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

[Out]

(-a**2 - 3*b**2*x**2)/(-3*a**3*b*c**4 + 9*a**2*b**2*c**4*x - 9*a*b**3*c**4*x**2 + 3*b**4*c**4*x**3)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).

Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.14 \[ \int \frac {(a+b x)^2}{(a c-b c x)^4} \, dx=-\frac {3 \, b^{2} x^{2} + a^{2}}{3 \, {\left (b^{4} c^{4} x^{3} - 3 \, a b^{3} c^{4} x^{2} + 3 \, a^{2} b^{2} c^{4} x - a^{3} b c^{4}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b x)^2}{(a c-b c x)^4} \, dx=-\frac {3 \, b^{2} x^{2} + a^{2}}{3 \, {\left (b x - a\right )}^{3} b c^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.07 \[ \int \frac {(a+b x)^2}{(a c-b c x)^4} \, dx=\frac {b\,x^2+\frac {a^2}{3\,b}}{a^3\,c^4-3\,a^2\,b\,c^4\,x+3\,a\,b^2\,c^4\,x^2-b^3\,c^4\,x^3} \]

[In]

int((a + b*x)^2/(a*c - b*c*x)^4,x)

[Out]

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