\(\int \frac {d+e x}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [1520]

Optimal result

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

[Out]

1/3*(a*e-b*d)/b^2/(b*x+a)^3-1/2*e/b^2/(b*x+a)^2

Rubi [A] (verified)

Time = 0.01 (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} \[ \int \frac {d+e x}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {b d-a e}{3 b^2 (a+b x)^3}-\frac {e}{2 b^2 (a+b x)^2} \]

[In]

Int[(d + e*x)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

-1/3*(b*d - a*e)/(b^2*(a + b*x)^3) - e/(2*b^2*(a + b*x)^2)

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {d+e x}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {2 b d+a e+3 b e x}{6 b^2 (a+b x)^3} \]

[In]

Integrate[(d + e*x)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

-1/6*(2*b*d + a*e + 3*b*e*x)/(b^2*(a + b*x)^3)

Maple [A] (verified)

Time = 2.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89

method	result	size
norman	\(\frac {-\frac {e x}{2 b}+\frac {-a e b -2 b^{2} d}{6 b^{3}}}{\left (b x +a \right )^{3}}\)	\(34\)
default	\(-\frac {-a e +b d}{3 b^{2} \left (b x +a \right )^{3}}-\frac {e}{2 b^{2} \left (b x +a \right )^{2}}\)	\(35\)
gosper	\(-\frac {3 b e x +a e +2 b d}{6 b^{2} \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )}\)	\(44\)
risch	\(\frac {-\frac {e x}{2 b}-\frac {a e +2 b d}{6 b^{2}}}{\left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )}\)	\(48\)
parallelrisch	\(\frac {-3 b^{2} e x -a e b -2 b^{2} d}{6 b^{3} \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \left (b x +a \right )}\)	\(50\)

[In]

int((e*x+d)/(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)

[Out]

(-1/2*e*x/b+1/6*(-a*b*e-2*b^2*d)/b^3)/(b*x+a)^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.32 \[ \int \frac {d+e x}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {3 \, b e x + 2 \, b d + a e}{6 \, {\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} \]

[In]

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

[Out]

-1/6*(3*b*e*x + 2*b*d + a*e)/(b^5*x^3 + 3*a*b^4*x^2 + 3*a^2*b^3*x + a^3*b^2)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.39 \[ \int \frac {d+e x}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {- a e - 2 b d - 3 b e x}{6 a^{3} b^{2} + 18 a^{2} b^{3} x + 18 a b^{4} x^{2} + 6 b^{5} x^{3}} \]

[In]

integrate((e*x+d)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

(-a*e - 2*b*d - 3*b*e*x)/(6*a**3*b**2 + 18*a**2*b**3*x + 18*a*b**4*x**2 + 6*b**5*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.32 \[ \int \frac {d+e x}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {3 \, b e x + 2 \, b d + a e}{6 \, {\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} \]

[In]

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

[Out]

-1/6*(3*b*e*x + 2*b*d + a*e)/(b^5*x^3 + 3*a*b^4*x^2 + 3*a^2*b^3*x + a^3*b^2)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int \frac {d+e x}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {3 \, b e x + 2 \, b d + a e}{6 \, {\left (b x + a\right )}^{3} b^{2}} \]

[In]

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

[Out]

-1/6*(3*b*e*x + 2*b*d + a*e)/((b*x + a)^3*b^2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.37 \[ \int \frac {d+e x}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {\frac {a\,e+2\,b\,d}{6\,b^2}+\frac {e\,x}{2\,b}}{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3} \]

[In]

int((d + e*x)/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)

[Out]

-((a*e + 2*b*d)/(6*b^2) + (e*x)/(2*b))/(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x)