\(\int \frac {1}{(1+(c+d x)^2)^2} \, dx\) [90]

Optimal result

Integrand size = 11, antiderivative size = 37 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^2} \, dx=\frac {c+d x}{2 d \left (1+(c+d x)^2\right )}+\frac {\arctan (c+d x)}{2 d} \]

[Out]

1/2*(d*x+c)/d/(1+(d*x+c)^2)+1/2*arctan(d*x+c)/d

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {253, 205, 209} \[ \int \frac {1}{\left (1+(c+d x)^2\right )^2} \, dx=\frac {\arctan (c+d x)}{2 d}+\frac {c+d x}{2 d \left ((c+d x)^2+1\right )} \]

[In]

Int[(1 + (c + d*x)^2)^(-2),x]

[Out]

(c + d*x)/(2*d*(1 + (c + d*x)^2)) + ArcTan[c + d*x]/(2*d)

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 253

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2} \, dx,x,c+d x\right )}{d} \\ & = \frac {c+d x}{2 d \left (1+(c+d x)^2\right )}+\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,c+d x\right )}{2 d} \\ & = \frac {c+d x}{2 d \left (1+(c+d x)^2\right )}+\frac {\tan ^{-1}(c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^2} \, dx=\frac {\frac {c+d x}{1+(c+d x)^2}+\arctan (c+d x)}{2 d} \]

[In]

Integrate[(1 + (c + d*x)^2)^(-2),x]

[Out]

((c + d*x)/(1 + (c + d*x)^2) + ArcTan[c + d*x])/(2*d)

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.16

method	result	size
risch	\(\frac {\frac {x}{2}+\frac {c}{2 d}}{d^{2} x^{2}+2 c d x +c^{2}+1}+\frac {\arctan \left (d x +c \right )}{2 d}\)	\(43\)
default	\(\frac {2 d^{2} x +2 c d}{4 d^{2} \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}+\frac {\arctan \left (\frac {2 d^{2} x +2 c d}{2 d}\right )}{2 d}\)	\(59\)
parallelrisch	\(-\frac {i \ln \left (d x +c -i\right ) x^{2} d^{3}-i \ln \left (d x +c +i\right ) x^{2} d^{3}+2 i \ln \left (d x +c -i\right ) x c \,d^{2}-2 i \ln \left (d x +c +i\right ) x c \,d^{2}+i \ln \left (d x +c -i\right ) c^{2} d -i \ln \left (d x +c +i\right ) c^{2} d +i \ln \left (d x +c -i\right ) d -i \ln \left (d x +c +i\right ) d -2 d^{2} x -2 c d}{4 d^{2} \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}\)	\(156\)

[In]

int(1/(1+(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

(1/2*x+1/2*c/d)/(d^2*x^2+2*c*d*x+c^2+1)+1/2*arctan(d*x+c)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.49 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^2} \, dx=\frac {d x + {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )} \arctan \left (d x + c\right ) + c}{2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + {\left (c^{2} + 1\right )} d\right )}} \]

[In]

integrate(1/(1+(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/2*(d*x + (d^2*x^2 + 2*c*d*x + c^2 + 1)*arctan(d*x + c) + c)/(d^3*x^2 + 2*c*d^2*x + (c^2 + 1)*d)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.51 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^2} \, dx=\frac {c + d x}{2 c^{2} d + 4 c d^{2} x + 2 d^{3} x^{2} + 2 d} + \frac {- \frac {i \log {\left (x + \frac {c - i}{d} \right )}}{4} + \frac {i \log {\left (x + \frac {c + i}{d} \right )}}{4}}{d} \]

[In]

integrate(1/(1+(d*x+c)**2)**2,x)

[Out]

(c + d*x)/(2*c**2*d + 4*c*d**2*x + 2*d**3*x**2 + 2*d) + (-I*log(x + (c - I)/d)/4 + I*log(x + (c + I)/d)/4)/d

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^2} \, dx=\frac {d x + c}{2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + {\left (c^{2} + 1\right )} d\right )}} + \frac {\arctan \left (\frac {d^{2} x + c d}{d}\right )}{2 \, d} \]

[In]

integrate(1/(1+(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

1/2*(d*x + c)/(d^3*x^2 + 2*c*d^2*x + (c^2 + 1)*d) + 1/2*arctan((d^2*x + c*d)/d)/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^2} \, dx=\frac {\arctan \left (d x + c\right )}{2 \, d} + \frac {d x + c}{2 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )} d} \]

[In]

integrate(1/(1+(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/2*arctan(d*x + c)/d + 1/2*(d*x + c)/((d^2*x^2 + 2*c*d*x + c^2 + 1)*d)

Mupad [B] (verification not implemented)

Time = 10.89 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^2} \, dx=\frac {\frac {x}{2}+\frac {c}{2\,d}}{c^2+2\,c\,d\,x+d^2\,x^2+1}+\frac {\mathrm {atan}\left (c+d\,x\right )}{2\,d} \]

[In]

int(1/((c + d*x)^2 + 1)^2,x)

[Out]

(x/2 + c/(2*d))/(c^2 + d^2*x^2 + 2*c*d*x + 1) + atan(c + d*x)/(2*d)