\(\int \frac {A+B x}{(a^2+2 a b x+b^2 x^2)^3} \, dx\) [649]

Optimal result

Integrand size = 24, antiderivative size = 38 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {-A b+a B}{5 b^2 (a+b x)^5}-\frac {B}{4 b^2 (a+b x)^4} \]

[Out]

1/5*(-A*b+B*a)/b^2/(b*x+a)^5-1/4*B/b^2/(b*x+a)^4

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {27, 45} \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {A b-a B}{5 b^2 (a+b x)^5}-\frac {B}{4 b^2 (a+b x)^4} \]

[In]

Int[(A + B*x)/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

-1/5*(A*b - a*B)/(b^2*(a + b*x)^5) - B/(4*b^2*(a + b*x)^4)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 45

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*} \text {integral}& = \int \frac {A+B x}{(a+b x)^6} \, dx \\ & = \int \left (\frac {A b-a B}{b (a+b x)^6}+\frac {B}{b (a+b x)^5}\right ) \, dx \\ & = -\frac {A b-a B}{5 b^2 (a+b x)^5}-\frac {B}{4 b^2 (a+b x)^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {4 A b+B (a+5 b x)}{20 b^2 (a+b x)^5} \]

[In]

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

[Out]

-1/20*(4*A*b + B*(a + 5*b*x))/(b^2*(a + b*x)^5)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {B}{4 b^{2} \left (b x +a \right )^{4}}-\frac {A b -B a}{5 b^{2} \left (b x +a \right )^{5}}\)	\(35\)
norman	\(\frac {-\frac {B x}{4 b}-\frac {4 A \,b^{4}+B \,b^{3} a}{20 b^{5}}}{\left (b x +a \right )^{5}}\)	\(35\)
gosper	\(-\frac {5 B b x +4 A b +B a}{20 b^{2} \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}\)	\(44\)
risch	\(\frac {-\frac {B x}{4 b}-\frac {4 A b +B a}{20 b^{2}}}{\left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}\)	\(48\)
parallelrisch	\(-\frac {5 b^{4} B x +4 A \,b^{4}+B \,b^{3} a}{20 b^{5} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2} \left (b x +a \right )}\)	\(51\)

[In]

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

[Out]

-1/4*B/b^2/(b*x+a)^4-1/5*(A*b-B*a)/b^2/(b*x+a)^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (34) = 68\).

Time = 0.32 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.89 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {5 \, B b x + B a + 4 \, A b}{20 \, {\left (b^{7} x^{5} + 5 \, a b^{6} x^{4} + 10 \, a^{2} b^{5} x^{3} + 10 \, a^{3} b^{4} x^{2} + 5 \, a^{4} b^{3} x + a^{5} b^{2}\right )}} \]

[In]

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

[Out]

-1/20*(5*B*b*x + B*a + 4*A*b)/(b^7*x^5 + 5*a*b^6*x^4 + 10*a^2*b^5*x^3 + 10*a^3*b^4*x^2 + 5*a^4*b^3*x + a^5*b^2
)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (32) = 64\).

Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.00 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {- 4 A b - B a - 5 B b x}{20 a^{5} b^{2} + 100 a^{4} b^{3} x + 200 a^{3} b^{4} x^{2} + 200 a^{2} b^{5} x^{3} + 100 a b^{6} x^{4} + 20 b^{7} x^{5}} \]

[In]

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

[Out]

(-4*A*b - B*a - 5*B*b*x)/(20*a**5*b**2 + 100*a**4*b**3*x + 200*a**3*b**4*x**2 + 200*a**2*b**5*x**3 + 100*a*b**
6*x**4 + 20*b**7*x**5)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (34) = 68\).

Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.89 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {5 \, B b x + B a + 4 \, A b}{20 \, {\left (b^{7} x^{5} + 5 \, a b^{6} x^{4} + 10 \, a^{2} b^{5} x^{3} + 10 \, a^{3} b^{4} x^{2} + 5 \, a^{4} b^{3} x + a^{5} b^{2}\right )}} \]

[In]

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

[Out]

-1/20*(5*B*b*x + B*a + 4*A*b)/(b^7*x^5 + 5*a*b^6*x^4 + 10*a^2*b^5*x^3 + 10*a^3*b^4*x^2 + 5*a^4*b^3*x + a^5*b^2
)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {5 \, B b x + B a + 4 \, A b}{20 \, {\left (b x + a\right )}^{5} b^{2}} \]

[In]

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

[Out]

-1/20*(5*B*b*x + B*a + 4*A*b)/((b*x + a)^5*b^2)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.95 \[ \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {\frac {4\,A\,b+B\,a}{20\,b^2}+\frac {B\,x}{4\,b}}{a^5+5\,a^4\,b\,x+10\,a^3\,b^2\,x^2+10\,a^2\,b^3\,x^3+5\,a\,b^4\,x^4+b^5\,x^5} \]

[In]

int((A + B*x)/(a^2 + b^2*x^2 + 2*a*b*x)^3,x)

[Out]

-((4*A*b + B*a)/(20*b^2) + (B*x)/(4*b))/(a^5 + b^5*x^5 + 5*a*b^4*x^4 + 10*a^3*b^2*x^2 + 10*a^2*b^3*x^3 + 5*a^4
*b*x)