\(\int \cos ^3(a+b x) \cot ^2(a+b x) \, dx\) [135]

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

Optimal result

Integrand size = 17, antiderivative size = 38 \[ \int \cos ^3(a+b x) \cot ^2(a+b x) \, dx=-\frac {\csc (a+b x)}{b}-\frac {2 \sin (a+b x)}{b}+\frac {\sin ^3(a+b x)}{3 b} \]

[Out]

-csc(b*x+a)/b-2*sin(b*x+a)/b+1/3*sin(b*x+a)^3/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2670, 276} \[ \int \cos ^3(a+b x) \cot ^2(a+b x) \, dx=\frac {\sin ^3(a+b x)}{3 b}-\frac {2 \sin (a+b x)}{b}-\frac {\csc (a+b x)}{b} \]

[In]

Int[Cos[a + b*x]^3*Cot[a + b*x]^2,x]

[Out]

-(Csc[a + b*x]/b) - (2*Sin[a + b*x])/b + Sin[a + b*x]^3/(3*b)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
 x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2670

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \cos ^3(a+b x) \cot ^2(a+b x) \, dx=-\frac {\csc (a+b x)}{b}-\frac {2 \sin (a+b x)}{b}+\frac {\sin ^3(a+b x)}{3 b} \]

[In]

Integrate[Cos[a + b*x]^3*Cot[a + b*x]^2,x]

[Out]

-(Csc[a + b*x]/b) - (2*Sin[a + b*x])/b + Sin[a + b*x]^3/(3*b)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.37

method result size
derivativedivides \(\frac {-\frac {\cos ^{6}\left (b x +a \right )}{\sin \left (b x +a \right )}-\left (\frac {8}{3}+\cos ^{4}\left (b x +a \right )+\frac {4 \left (\cos ^{2}\left (b x +a \right )\right )}{3}\right ) \sin \left (b x +a \right )}{b}\) \(52\)
default \(\frac {-\frac {\cos ^{6}\left (b x +a \right )}{\sin \left (b x +a \right )}-\left (\frac {8}{3}+\cos ^{4}\left (b x +a \right )+\frac {4 \left (\cos ^{2}\left (b x +a \right )\right )}{3}\right ) \sin \left (b x +a \right )}{b}\) \(52\)
risch \(\frac {7 i {\mathrm e}^{i \left (b x +a \right )}}{8 b}-\frac {7 i {\mathrm e}^{-i \left (b x +a \right )}}{8 b}-\frac {2 i {\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {\sin \left (3 b x +3 a \right )}{12 b}\) \(74\)
parallelrisch \(\frac {-3 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-36 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-50 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-3 \cot \left (\frac {b x}{2}+\frac {a}{2}\right )-36 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{6 b \left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{3}}\) \(83\)
norman \(\frac {-\frac {1}{2 b}-\frac {6 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {25 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {6 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )}{2 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{3} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\) \(98\)

[In]

int(cos(b*x+a)^5/sin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b*(-cos(b*x+a)^6/sin(b*x+a)-(8/3+cos(b*x+a)^4+4/3*cos(b*x+a)^2)*sin(b*x+a))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \cos ^3(a+b x) \cot ^2(a+b x) \, dx=\frac {\cos \left (b x + a\right )^{4} + 4 \, \cos \left (b x + a\right )^{2} - 8}{3 \, b \sin \left (b x + a\right )} \]

[In]

integrate(cos(b*x+a)^5/sin(b*x+a)^2,x, algorithm="fricas")

[Out]

1/3*(cos(b*x + a)^4 + 4*cos(b*x + a)^2 - 8)/(b*sin(b*x + a))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (29) = 58\).

Time = 0.53 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.61 \[ \int \cos ^3(a+b x) \cot ^2(a+b x) \, dx=\begin {cases} - \frac {8 \sin ^{3}{\left (a + b x \right )}}{3 b} - \frac {4 \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} - \frac {\cos ^{4}{\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{5}{\left (a \right )}}{\sin ^{2}{\left (a \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)**5/sin(b*x+a)**2,x)

[Out]

Piecewise((-8*sin(a + b*x)**3/(3*b) - 4*sin(a + b*x)*cos(a + b*x)**2/b - cos(a + b*x)**4/(b*sin(a + b*x)), Ne(
b, 0)), (x*cos(a)**5/sin(a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \cos ^3(a+b x) \cot ^2(a+b x) \, dx=\frac {\sin \left (b x + a\right )^{3} - \frac {3}{\sin \left (b x + a\right )} - 6 \, \sin \left (b x + a\right )}{3 \, b} \]

[In]

integrate(cos(b*x+a)^5/sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/3*(sin(b*x + a)^3 - 3/sin(b*x + a) - 6*sin(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \cos ^3(a+b x) \cot ^2(a+b x) \, dx=\frac {\sin \left (b x + a\right )^{3} - \frac {3}{\sin \left (b x + a\right )} - 6 \, \sin \left (b x + a\right )}{3 \, b} \]

[In]

integrate(cos(b*x+a)^5/sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/3*(sin(b*x + a)^3 - 3/sin(b*x + a) - 6*sin(b*x + a))/b

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92 \[ \int \cos ^3(a+b x) \cot ^2(a+b x) \, dx=-\frac {-{\sin \left (a+b\,x\right )}^4+6\,{\sin \left (a+b\,x\right )}^2+3}{3\,b\,\sin \left (a+b\,x\right )} \]

[In]

int(cos(a + b*x)^5/sin(a + b*x)^2,x)

[Out]

-(6*sin(a + b*x)^2 - sin(a + b*x)^4 + 3)/(3*b*sin(a + b*x))