\(\int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^4} \, dx\) [1766]

Optimal result

Integrand size = 27, antiderivative size = 28 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^4} \, dx=-\frac {(c+d x)^2}{2 (b c-a d) (a+b x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {24, 37} \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^4} \, dx=-\frac {(c+d x)^2}{2 (a+b x)^2 (b c-a d)} \]

[In]

Int[(a*c + (b*c + a*d)*x + b*d*x^2)/(a + b*x)^4,x]

[Out]

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

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^4} \, dx=-\frac {a d+b (c+2 d x)}{2 b^2 (a+b x)^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.42 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
gosper	\(-\frac {2 b d x +a d +b c}{2 b^{2} \left (b x +a \right )^{2}}\)	\(25\)
parallelrisch	\(\frac {-2 b d x -a d -b c}{2 b^{2} \left (b x +a \right )^{2}}\)	\(27\)
risch	\(\frac {-\frac {d x}{b}-\frac {a d +b c}{2 b^{2}}}{\left (b x +a \right )^{2}}\)	\(29\)
default	\(-\frac {-a d +b c}{2 b^{2} \left (b x +a \right )^{2}}-\frac {d}{b^{2} \left (b x +a \right )}\)	\(35\)
norman	\(\frac {-d \,x^{2}+\frac {a \left (-a b d -b^{2} c \right )}{2 b^{3}}+\frac {\left (-3 a b d -b^{2} c \right ) x}{2 b^{2}}}{\left (b x +a \right )^{3}}\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^4} \, dx=-\frac {2 \, b d x + b c + a d}{2 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^4} \, dx=\frac {- a d - b c - 2 b d x}{2 a^{2} b^{2} + 4 a b^{3} x + 2 b^{4} x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^4} \, dx=-\frac {2 \, b d x + b c + a d}{2 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^4} \, dx=-\frac {2 \, b d x + b c + a d}{2 \, {\left (b x + a\right )}^{2} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.84 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^4} \, dx=-\frac {\frac {a\,d+b\,c}{2\,b^2}+\frac {d\,x}{b}}{a^2+2\,a\,b\,x+b^2\,x^2} \]

[In]

int((a*c + x*(a*d + b*c) + b*d*x^2)/(a + b*x)^4,x)

[Out]

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