\(\int (a+b \cot ^2(c+d x))^3 \, dx\) [4]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 14, antiderivative size = 78 \[ \int \left (a+b \cot ^2(c+d x)\right )^3 \, dx=(a-b)^3 x-\frac {b \left (3 a^2-3 a b+b^2\right ) \cot (c+d x)}{d}-\frac {(3 a-b) b^2 \cot ^3(c+d x)}{3 d}-\frac {b^3 \cot ^5(c+d x)}{5 d} \]

[Out]

(a-b)^3*x-b*(3*a^2-3*a*b+b^2)*cot(d*x+c)/d-1/3*(3*a-b)*b^2*cot(d*x+c)^3/d-1/5*b^3*cot(d*x+c)^5/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 78, 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} \[ \int \left (a+b \cot ^2(c+d x)\right )^3 \, dx=-\frac {b \left (3 a^2-3 a b+b^2\right ) \cot (c+d x)}{d}-\frac {b^2 (3 a-b) \cot ^3(c+d x)}{3 d}+x (a-b)^3-\frac {b^3 \cot ^5(c+d x)}{5 d} \]

[In]

Int[(a + b*Cot[c + d*x]^2)^3,x]

[Out]

(a - b)^3*x - (b*(3*a^2 - 3*a*b + b^2)*Cot[c + d*x])/d - ((3*a - b)*b^2*Cot[c + d*x]^3)/(3*d) - (b^3*Cot[c + d
*x]^5)/(5*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 )^3}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \left (b \left (3 a^2-3 a b+b^2\right )+(3 a-b) b^2 x^2+b^3 x^4+\frac {(a-b)^3}{1+x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {b \left (3 a^2-3 a b+b^2\right ) \cot (c+d x)}{d}-\frac {(3 a-b) b^2 \cot ^3(c+d x)}{3 d}-\frac {b^3 \cot ^5(c+d x)}{5 d}-\frac {(a-b)^3 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d} \\ & = (a-b)^3 x-\frac {b \left (3 a^2-3 a b+b^2\right ) \cot (c+d x)}{d}-\frac {(3 a-b) b^2 \cot ^3(c+d x)}{3 d}-\frac {b^3 \cot ^5(c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.95 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.42 \[ \int \left (a+b \cot ^2(c+d x)\right )^3 \, dx=-\frac {\cot ^5(c+d x) \left (\frac {15 (a-b)^3 \text {arctanh}\left (\sqrt {-\tan ^2(c+d x)}\right ) \tan ^8(c+d x)}{\left (-\tan ^2(c+d x)\right )^{3/2}}+b \left (3 b^2+5 (3 a-b) b \tan ^2(c+d x)+15 \left (3 a^2-3 a b+b^2\right ) \tan ^4(c+d x)\right )\right )}{15 d} \]

[In]

Integrate[(a + b*Cot[c + d*x]^2)^3,x]

[Out]

-1/15*(Cot[c + d*x]^5*((15*(a - b)^3*ArcTanh[Sqrt[-Tan[c + d*x]^2]]*Tan[c + d*x]^8)/(-Tan[c + d*x]^2)^(3/2) +
b*(3*b^2 + 5*(3*a - b)*b*Tan[c + d*x]^2 + 15*(3*a^2 - 3*a*b + b^2)*Tan[c + d*x]^4)))/d

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.99

method result size
parallelrisch \(\frac {-3 b^{3} \cot \left (d x +c \right )^{5}+5 \left (-3 a \,b^{2}+b^{3}\right ) \cot \left (d x +c \right )^{3}+15 \left (-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \cot \left (d x +c \right )+15 d x \left (a -b \right )^{3}}{15 d}\) \(77\)
norman \(\frac {\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) x \tan \left (d x +c \right )^{5}-\frac {b^{3}}{5 d}-\frac {b \left (3 a^{2}-3 a b +b^{2}\right ) \tan \left (d x +c \right )^{4}}{d}-\frac {b^{2} \left (3 a -b \right ) \tan \left (d x +c \right )^{2}}{3 d}}{\tan \left (d x +c \right )^{5}}\) \(100\)
derivativedivides \(\frac {-\frac {b^{3} \cot \left (d x +c \right )^{5}}{5}-a \,b^{2} \cot \left (d x +c \right )^{3}+\frac {b^{3} \cot \left (d x +c \right )^{3}}{3}-3 a^{2} b \cot \left (d x +c \right )+3 \cot \left (d x +c \right ) a \,b^{2}-\cot \left (d x +c \right ) b^{3}+\left (-a^{3}+3 a^{2} b -3 a \,b^{2}+b^{3}\right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\) \(116\)
default \(\frac {-\frac {b^{3} \cot \left (d x +c \right )^{5}}{5}-a \,b^{2} \cot \left (d x +c \right )^{3}+\frac {b^{3} \cot \left (d x +c \right )^{3}}{3}-3 a^{2} b \cot \left (d x +c \right )+3 \cot \left (d x +c \right ) a \,b^{2}-\cot \left (d x +c \right ) b^{3}+\left (-a^{3}+3 a^{2} b -3 a \,b^{2}+b^{3}\right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\) \(116\)
parts \(a^{3} x +\frac {b^{3} \left (-\frac {\cot \left (d x +c \right )^{5}}{5}+\frac {\cot \left (d x +c \right )^{3}}{3}-\cot \left (d x +c \right )+\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}+\frac {3 a \,b^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )-\frac {\pi }{2}+\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}+\frac {3 a^{2} b \left (-\cot \left (d x +c \right )+\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\) \(121\)
risch \(a^{3} x -3 a^{2} b x +3 a \,b^{2} x -b^{3} x -\frac {2 i b \left (45 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-90 a b \,{\mathrm e}^{8 i \left (d x +c \right )}+45 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-180 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+270 a b \,{\mathrm e}^{6 i \left (d x +c \right )}-90 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+270 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-330 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+140 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-180 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+210 a b \,{\mathrm e}^{2 i \left (d x +c \right )}-70 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+45 a^{2}-60 a b +23 b^{2}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}\) \(226\)

[In]

int((a+b*cot(d*x+c)^2)^3,x,method=_RETURNVERBOSE)

[Out]

1/15*(-3*b^3*cot(d*x+c)^5+5*(-3*a*b^2+b^3)*cot(d*x+c)^3+15*(-3*a^2*b+3*a*b^2-b^3)*cot(d*x+c)+15*d*x*(a-b)^3)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (74) = 148\).

Time = 0.26 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.24 \[ \int \left (a+b \cot ^2(c+d x)\right )^3 \, dx=-\frac {{\left (45 \, a^{2} b - 60 \, a b^{2} + 23 \, b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )^{3} + 45 \, a^{2} b - 30 \, a b^{2} + 13 \, b^{3} - {\left (45 \, a^{2} b - 30 \, a b^{2} + b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )^{2} - {\left (45 \, a^{2} b - 60 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right ) - 15 \, {\left ({\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d x \cos \left (2 \, d x + 2 \, c\right )^{2} - 2 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d x \cos \left (2 \, d x + 2 \, c\right ) + {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d x\right )} \sin \left (2 \, d x + 2 \, c\right )}{15 \, {\left (d \cos \left (2 \, d x + 2 \, c\right )^{2} - 2 \, d \cos \left (2 \, d x + 2 \, c\right ) + d\right )} \sin \left (2 \, d x + 2 \, c\right )} \]

[In]

integrate((a+b*cot(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-1/15*((45*a^2*b - 60*a*b^2 + 23*b^3)*cos(2*d*x + 2*c)^3 + 45*a^2*b - 30*a*b^2 + 13*b^3 - (45*a^2*b - 30*a*b^2
 + b^3)*cos(2*d*x + 2*c)^2 - (45*a^2*b - 60*a*b^2 + 11*b^3)*cos(2*d*x + 2*c) - 15*((a^3 - 3*a^2*b + 3*a*b^2 -
b^3)*d*x*cos(2*d*x + 2*c)^2 - 2*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*x*cos(2*d*x + 2*c) + (a^3 - 3*a^2*b + 3*a*b^
2 - b^3)*d*x)*sin(2*d*x + 2*c))/((d*cos(2*d*x + 2*c)^2 - 2*d*cos(2*d*x + 2*c) + d)*sin(2*d*x + 2*c))

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.62 \[ \int \left (a+b \cot ^2(c+d x)\right )^3 \, dx=\begin {cases} a^{3} x - 3 a^{2} b x - \frac {3 a^{2} b \cot {\left (c + d x \right )}}{d} + 3 a b^{2} x - \frac {a b^{2} \cot ^{3}{\left (c + d x \right )}}{d} + \frac {3 a b^{2} \cot {\left (c + d x \right )}}{d} - b^{3} x - \frac {b^{3} \cot ^{5}{\left (c + d x \right )}}{5 d} + \frac {b^{3} \cot ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{3} \cot {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cot ^{2}{\left (c \right )}\right )^{3} & \text {otherwise} \end {cases} \]

[In]

integrate((a+b*cot(d*x+c)**2)**3,x)

[Out]

Piecewise((a**3*x - 3*a**2*b*x - 3*a**2*b*cot(c + d*x)/d + 3*a*b**2*x - a*b**2*cot(c + d*x)**3/d + 3*a*b**2*co
t(c + d*x)/d - b**3*x - b**3*cot(c + d*x)**5/(5*d) + b**3*cot(c + d*x)**3/(3*d) - b**3*cot(c + d*x)/d, Ne(d, 0
)), (x*(a + b*cot(c)**2)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.44 \[ \int \left (a+b \cot ^2(c+d x)\right )^3 \, dx=a^{3} x - \frac {3 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{2} b}{d} + \frac {{\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a b^{2}}{d} - \frac {{\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} b^{3}}{15 \, d} \]

[In]

integrate((a+b*cot(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

a^3*x - 3*(d*x + c + 1/tan(d*x + c))*a^2*b/d + (3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a*b^2/d -
 1/15*(15*d*x + 15*c + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/tan(d*x + c)^5)*b^3/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (74) = 148\).

Time = 0.38 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.94 \[ \int \left (a+b \cot ^2(c+d x)\right )^3 \, dx=\frac {3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 60 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 900 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 330 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 480 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (d x + c\right )} - \frac {720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 900 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 330 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 60 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 35 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, b^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]

[In]

integrate((a+b*cot(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/480*(3*b^3*tan(1/2*d*x + 1/2*c)^5 + 60*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 35*b^3*tan(1/2*d*x + 1/2*c)^3 + 720*a^
2*b*tan(1/2*d*x + 1/2*c) - 900*a*b^2*tan(1/2*d*x + 1/2*c) + 330*b^3*tan(1/2*d*x + 1/2*c) + 480*(a^3 - 3*a^2*b
+ 3*a*b^2 - b^3)*(d*x + c) - (720*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 900*a*b^2*tan(1/2*d*x + 1/2*c)^4 + 330*b^3*ta
n(1/2*d*x + 1/2*c)^4 + 60*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 35*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*b^3)/tan(1/2*d*x +
1/2*c)^5)/d

Mupad [B] (verification not implemented)

Time = 12.92 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97 \[ \int \left (a+b \cot ^2(c+d x)\right )^3 \, dx=x\,{\left (a-b\right )}^3-\frac {b^3\,{\mathrm {cot}\left (c+d\,x\right )}^5}{5\,d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (3\,a\,b^2-b^3\right )}{3\,d}-\frac {b\,\mathrm {cot}\left (c+d\,x\right )\,\left (3\,a^2-3\,a\,b+b^2\right )}{d} \]

[In]

int((a + b*cot(c + d*x)^2)^3,x)

[Out]

x*(a - b)^3 - (b^3*cot(c + d*x)^5)/(5*d) - (cot(c + d*x)^3*(3*a*b^2 - b^3))/(3*d) - (b*cot(c + d*x)*(3*a^2 - 3
*a*b + b^2))/d