\(\int \frac {1}{(1+(c+d x)^2)^3} \, dx\) [91]

Optimal result

Integrand size = 11, antiderivative size = 60 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^3} \, dx=\frac {c+d x}{4 d \left (1+(c+d x)^2\right )^2}+\frac {3 (c+d x)}{8 d \left (1+(c+d x)^2\right )}+\frac {3 \arctan (c+d x)}{8 d} \]

[Out]

1/4*(d*x+c)/d/(1+(d*x+c)^2)^2+3/8*(d*x+c)/d/(1+(d*x+c)^2)+3/8*arctan(d*x+c)/d

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {253, 205, 209} \[ \int \frac {1}{\left (1+(c+d x)^2\right )^3} \, dx=\frac {3 \arctan (c+d x)}{8 d}+\frac {3 (c+d x)}{8 d \left ((c+d x)^2+1\right )}+\frac {c+d x}{4 d \left ((c+d x)^2+1\right )^2} \]

[In]

Int[(1 + (c + d*x)^2)^(-3),x]

[Out]

(c + d*x)/(4*d*(1 + (c + d*x)^2)^2) + (3*(c + d*x))/(8*d*(1 + (c + d*x)^2)) + (3*ArcTan[c + d*x])/(8*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 )^3} \, dx,x,c+d x\right )}{d} \\ & = \frac {c+d x}{4 d \left (1+(c+d x)^2\right )^2}+\frac {3 \text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2} \, dx,x,c+d x\right )}{4 d} \\ & = \frac {c+d x}{4 d \left (1+(c+d x)^2\right )^2}+\frac {3 (c+d x)}{8 d \left (1+(c+d x)^2\right )}+\frac {3 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,c+d x\right )}{8 d} \\ & = \frac {c+d x}{4 d \left (1+(c+d x)^2\right )^2}+\frac {3 (c+d x)}{8 d \left (1+(c+d x)^2\right )}+\frac {3 \tan ^{-1}(c+d x)}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^3} \, dx=\frac {\frac {2 (c+d x)}{\left (1+(c+d x)^2\right )^2}+\frac {3 (c+d x)}{1+(c+d x)^2}+3 \arctan (c+d x)}{8 d} \]

[In]

Integrate[(1 + (c + d*x)^2)^(-3),x]

[Out]

((2*(c + d*x))/(1 + (c + d*x)^2)^2 + (3*(c + d*x))/(1 + (c + d*x)^2) + 3*ArcTan[c + d*x])/(8*d)

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18

method	result	size
risch	\(\frac {\frac {3 d^{2} x^{3}}{8}+\frac {9 x^{2} c d}{8}+\left (\frac {9 c^{2}}{8}+\frac {5}{8}\right ) x +\frac {c \left (3 c^{2}+5\right )}{8 d}}{\left (d^{2} x^{2}+2 c d x +c^{2}+1\right )^{2}}+\frac {3 \arctan \left (d x +c \right )}{8 d}\)	\(71\)
default	\(\frac {2 d^{2} x +2 c d}{8 d^{2} \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )^{2}}+\frac {\frac {3}{8} d^{2} x +\frac {3}{8} c d}{d^{2} \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}+\frac {3 \arctan \left (\frac {2 d^{2} x +2 c d}{2 d}\right )}{8 d}\)	\(94\)
parallelrisch	\(-\frac {-3 i \ln \left (d x +c +i\right ) c^{4} d^{3}+6 i \ln \left (d x +c -i\right ) c^{2} d^{3}-6 i \ln \left (d x +c +i\right ) c^{2} d^{3}+3 i \ln \left (d x +c -i\right ) x^{4} d^{7}-3 i \ln \left (d x +c +i\right ) x^{4} d^{7}+6 i \ln \left (d x +c -i\right ) x^{2} d^{5}+3 i \ln \left (d x +c -i\right ) c^{4} d^{3}-6 i \ln \left (d x +c +i\right ) x^{2} d^{5}-6 d^{6} x^{3}-18 x^{2} c \,d^{5}-18 x \,c^{2} d^{4}+3 i \ln \left (d x +c -i\right ) d^{3}-3 i \ln \left (d x +c +i\right ) d^{3}+12 i \ln \left (d x +c -i\right ) x c \,d^{4}-12 i \ln \left (d x +c +i\right ) x c \,d^{4}-6 c^{3} d^{3}-10 d^{3} c +12 i \ln \left (d x +c -i\right ) x^{3} c \,d^{6}-12 i \ln \left (d x +c +i\right ) x^{3} c \,d^{6}+18 i \ln \left (d x +c -i\right ) x^{2} c^{2} d^{5}-18 i \ln \left (d x +c +i\right ) x^{2} c^{2} d^{5}+12 i \ln \left (d x +c -i\right ) x \,c^{3} d^{4}-12 i \ln \left (d x +c +i\right ) x \,c^{3} d^{4}-10 x \,d^{4}}{16 d^{4} \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )^{2}}\)	\(380\)

[In]

int(1/(1+(d*x+c)^2)^3,x,method=_RETURNVERBOSE)

[Out]

(3/8*d^2*x^3+9/8*x^2*c*d+(9/8*c^2+5/8)*x+1/8*c/d*(3*c^2+5))/(d^2*x^2+2*c*d*x+c^2+1)^2+3/8*arctan(d*x+c)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (54) = 108\).

Time = 0.25 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.55 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^3} \, dx=\frac {3 \, d^{3} x^{3} + 9 \, c d^{2} x^{2} + 3 \, c^{3} + {\left (9 \, c^{2} + 5\right )} d x + 3 \, {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} x^{2} + c^{4} + 4 \, {\left (c^{3} + c\right )} d x + 2 \, c^{2} + 1\right )} \arctan \left (d x + c\right ) + 5 \, c}{8 \, {\left (d^{5} x^{4} + 4 \, c d^{4} x^{3} + 2 \, {\left (3 \, c^{2} + 1\right )} d^{3} x^{2} + 4 \, {\left (c^{3} + c\right )} d^{2} x + {\left (c^{4} + 2 \, c^{2} + 1\right )} d\right )}} \]

[In]

integrate(1/(1+(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/8*(3*d^3*x^3 + 9*c*d^2*x^2 + 3*c^3 + (9*c^2 + 5)*d*x + 3*(d^4*x^4 + 4*c*d^3*x^3 + 2*(3*c^2 + 1)*d^2*x^2 + c^
4 + 4*(c^3 + c)*d*x + 2*c^2 + 1)*arctan(d*x + c) + 5*c)/(d^5*x^4 + 4*c*d^4*x^3 + 2*(3*c^2 + 1)*d^3*x^2 + 4*(c^
3 + c)*d^2*x + (c^4 + 2*c^2 + 1)*d)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.43 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^3} \, dx=\frac {3 c^{3} + 9 c d^{2} x^{2} + 5 c + 3 d^{3} x^{3} + x \left (9 c^{2} d + 5 d\right )}{8 c^{4} d + 16 c^{2} d + 32 c d^{4} x^{3} + 8 d^{5} x^{4} + 8 d + x^{2} \cdot \left (48 c^{2} d^{3} + 16 d^{3}\right ) + x \left (32 c^{3} d^{2} + 32 c d^{2}\right )} + \frac {- \frac {3 i \log {\left (x + \frac {3 c - 3 i}{3 d} \right )}}{16} + \frac {3 i \log {\left (x + \frac {3 c + 3 i}{3 d} \right )}}{16}}{d} \]

[In]

integrate(1/(1+(d*x+c)**2)**3,x)

[Out]

(3*c**3 + 9*c*d**2*x**2 + 5*c + 3*d**3*x**3 + x*(9*c**2*d + 5*d))/(8*c**4*d + 16*c**2*d + 32*c*d**4*x**3 + 8*d
**5*x**4 + 8*d + x**2*(48*c**2*d**3 + 16*d**3) + x*(32*c**3*d**2 + 32*c*d**2)) + (-3*I*log(x + (3*c - 3*I)/(3*
d))/16 + 3*I*log(x + (3*c + 3*I)/(3*d))/16)/d

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (54) = 108\).

Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.92 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^3} \, dx=\frac {3 \, d^{3} x^{3} + 9 \, c d^{2} x^{2} + 3 \, c^{3} + {\left (9 \, c^{2} + 5\right )} d x + 5 \, c}{8 \, {\left (d^{5} x^{4} + 4 \, c d^{4} x^{3} + 2 \, {\left (3 \, c^{2} + 1\right )} d^{3} x^{2} + 4 \, {\left (c^{3} + c\right )} d^{2} x + {\left (c^{4} + 2 \, c^{2} + 1\right )} d\right )}} + \frac {3 \, \arctan \left (\frac {d^{2} x + c d}{d}\right )}{8 \, d} \]

[In]

integrate(1/(1+(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/8*(3*d^3*x^3 + 9*c*d^2*x^2 + 3*c^3 + (9*c^2 + 5)*d*x + 5*c)/(d^5*x^4 + 4*c*d^4*x^3 + 2*(3*c^2 + 1)*d^3*x^2 +
4*(c^3 + c)*d^2*x + (c^4 + 2*c^2 + 1)*d) + 3/8*arctan((d^2*x + c*d)/d)/d

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^3} \, dx=\frac {3 \, \arctan \left (d x + c\right )}{8 \, d} + \frac {3 \, d^{3} x^{3} + 9 \, c d^{2} x^{2} + 9 \, c^{2} d x + 3 \, c^{3} + 5 \, d x + 5 \, c}{8 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}^{2} d} \]

[In]

integrate(1/(1+(d*x+c)^2)^3,x, algorithm="giac")

[Out]

3/8*arctan(d*x + c)/d + 1/8*(3*d^3*x^3 + 9*c*d^2*x^2 + 9*c^2*d*x + 3*c^3 + 5*d*x + 5*c)/((d^2*x^2 + 2*c*d*x +
c^2 + 1)^2*d)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.85 \[ \int \frac {1}{\left (1+(c+d x)^2\right )^3} \, dx=\frac {3\,\mathrm {atan}\left (c+d\,x\right )}{8\,d}+\frac {x\,\left (\frac {9\,c^2}{8}+\frac {5}{8}\right )+\frac {3\,c^3+5\,c}{8\,d}+\frac {3\,d^2\,x^3}{8}+\frac {9\,c\,d\,x^2}{8}}{x^2\,\left (6\,c^2\,d^2+2\,d^2\right )+2\,c^2+c^4+x\,\left (4\,d\,c^3+4\,d\,c\right )+d^4\,x^4+4\,c\,d^3\,x^3+1} \]

[In]

int(1/((c + d*x)^2 + 1)^3,x)

[Out]

(3*atan(c + d*x))/(8*d) + (x*((9*c^2)/8 + 5/8) + (5*c + 3*c^3)/(8*d) + (3*d^2*x^3)/8 + (9*c*d*x^2)/8)/(x^2*(2*
d^2 + 6*c^2*d^2) + 2*c^2 + c^4 + x*(4*c*d + 4*c^3*d) + d^4*x^4 + 4*c*d^3*x^3 + 1)