\(\int (a+b \cot ^2(c+d x))^2 \, dx\) [3]

Optimal result

Integrand size = 14, antiderivative size = 47 \[ \int \left (a+b \cot ^2(c+d x)\right )^2 \, dx=(a-b)^2 x-\frac {(2 a-b) b \cot (c+d x)}{d}-\frac {b^2 \cot ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3742, 398, 209} \[ \int \left (a+b \cot ^2(c+d x)\right )^2 \, dx=-\frac {b (2 a-b) \cot (c+d x)}{d}+x (a-b)^2-\frac {b^2 \cot ^3(c+d x)}{3 d} \]

[In]

Int[(a + b*Cot[c + d*x]^2)^2,x]

[Out]

(a - b)^2*x - ((2*a - b)*b*Cot[c + d*x])/d - (b^2*Cot[c + d*x]^3)/(3*d)

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 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 3742

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[c*(ff/f), Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^2}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \left ((2 a-b) b+b^2 x^2+\frac {(a-b)^2}{1+x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {(2 a-b) b \cot (c+d x)}{d}-\frac {b^2 \cot ^3(c+d x)}{3 d}-\frac {(a-b)^2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d} \\ & = (a-b)^2 x-\frac {(2 a-b) b \cot (c+d x)}{d}-\frac {b^2 \cot ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.51 \[ \int \left (a+b \cot ^2(c+d x)\right )^2 \, dx=-\frac {\cot (c+d x) \left (b \left (6 a-3 b+b \cot ^2(c+d x)\right )+3 (a-b)^2 \text {arctanh}\left (\sqrt {-\tan ^2(c+d x)}\right ) \sqrt {-\tan ^2(c+d x)}\right )}{3 d} \]

[In]

Integrate[(a + b*Cot[c + d*x]^2)^2,x]

[Out]

-1/3*(Cot[c + d*x]*(b*(6*a - 3*b + b*Cot[c + d*x]^2) + 3*(a - b)^2*ArcTanh[Sqrt[-Tan[c + d*x]^2]]*Sqrt[-Tan[c
+ d*x]^2]))/d

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02

method	result	size
parallelrisch	\(\frac {-b^{2} \cot \left (d x +c \right )^{3}+\left (-6 a b +3 b^{2}\right ) \cot \left (d x +c \right )+3 d x \left (a -b \right )^{2}}{3 d}\)	\(48\)
norman	\(\frac {\left (a^{2}-2 a b +b^{2}\right ) x \tan \left (d x +c \right )^{3}-\frac {b^{2}}{3 d}-\frac {b \left (2 a -b \right ) \tan \left (d x +c \right )^{2}}{d}}{\tan \left (d x +c \right )^{3}}\)	\(61\)
derivativedivides	\(\frac {-\frac {b^{2} \cot \left (d x +c \right )^{3}}{3}-2 \cot \left (d x +c \right ) a b +\cot \left (d x +c \right ) b^{2}+\left (-a^{2}+2 a b -b^{2}\right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\)	\(68\)
default	\(\frac {-\frac {b^{2} \cot \left (d x +c \right )^{3}}{3}-2 \cot \left (d x +c \right ) a b +\cot \left (d x +c \right ) b^{2}+\left (-a^{2}+2 a b -b^{2}\right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\)	\(68\)
parts	\(a^{2} x +\frac {b^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )-\frac {\pi }{2}+\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}+\frac {2 a b \left (-\cot \left (d x +c \right )+\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\)	\(69\)
risch	\(a^{2} x -2 a b x +b^{2} x +\frac {4 i b \left (-3 a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{2 i \left (d x +c \right )} a -3 \,{\mathrm e}^{2 i \left (d x +c \right )} b -3 a +2 b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}\)	\(92\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (45) = 90\).

Time = 0.27 (sec) , antiderivative size = 127, normalized size of antiderivative = 2.70 \[ \int \left (a+b \cot ^2(c+d x)\right )^2 \, dx=\frac {2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) - 2 \, {\left (3 \, a b - 2 \, b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )^{2} + 6 \, a b - 2 \, b^{2} + 3 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} d x \cos \left (2 \, d x + 2 \, c\right ) - {\left (a^{2} - 2 \, a b + b^{2}\right )} d x\right )} \sin \left (2 \, d x + 2 \, c\right )}{3 \, {\left (d \cos \left (2 \, d x + 2 \, c\right ) - d\right )} \sin \left (2 \, d x + 2 \, c\right )} \]

[In]

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

[Out]

1/3*(2*b^2*cos(2*d*x + 2*c) - 2*(3*a*b - 2*b^2)*cos(2*d*x + 2*c)^2 + 6*a*b - 2*b^2 + 3*((a^2 - 2*a*b + b^2)*d*
x*cos(2*d*x + 2*c) - (a^2 - 2*a*b + b^2)*d*x)*sin(2*d*x + 2*c))/((d*cos(2*d*x + 2*c) - d)*sin(2*d*x + 2*c))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.45 \[ \int \left (a+b \cot ^2(c+d x)\right )^2 \, dx=\begin {cases} a^{2} x - 2 a b x - \frac {2 a b \cot {\left (c + d x \right )}}{d} + b^{2} x - \frac {b^{2} \cot ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{2} \cot {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cot ^{2}{\left (c \right )}\right )^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*x - 2*a*b*x - 2*a*b*cot(c + d*x)/d + b**2*x - b**2*cot(c + d*x)**3/(3*d) + b**2*cot(c + d*x)/d
, Ne(d, 0)), (x*(a + b*cot(c)**2)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.34 \[ \int \left (a+b \cot ^2(c+d x)\right )^2 \, dx=a^{2} x - \frac {2 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a b}{d} + \frac {{\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} b^{2}}{3 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (45) = 90\).

Time = 0.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.43 \[ \int \left (a+b \cot ^2(c+d x)\right )^2 \, dx=\frac {b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (d x + c\right )} - \frac {24 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

[In]

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

[Out]

1/24*(b^2*tan(1/2*d*x + 1/2*c)^3 + 24*a*b*tan(1/2*d*x + 1/2*c) - 15*b^2*tan(1/2*d*x + 1/2*c) + 24*(a^2 - 2*a*b
+ b^2)*(d*x + c) - (24*a*b*tan(1/2*d*x + 1/2*c)^2 - 15*b^2*tan(1/2*d*x + 1/2*c)^2 + b^2)/tan(1/2*d*x + 1/2*c)
^3)/d

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96 \[ \int \left (a+b \cot ^2(c+d x)\right )^2 \, dx=x\,{\left (a-b\right )}^2-\frac {b^2\,{\mathrm {cot}\left (c+d\,x\right )}^3}{3\,d}-\frac {b\,\mathrm {cot}\left (c+d\,x\right )\,\left (2\,a-b\right )}{d} \]

[In]

int((a + b*cot(c + d*x)^2)^2,x)

[Out]

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