\(\int \cos ^2(a+b x) \cot ^4(a+b x) \, dx\) [160]

Optimal result

Integrand size = 17, antiderivative size = 57 \[ \int \cos ^2(a+b x) \cot ^4(a+b x) \, dx=\frac {5 x}{2}+\frac {5 \cot (a+b x)}{2 b}-\frac {5 \cot ^3(a+b x)}{6 b}+\frac {\cos ^2(a+b x) \cot ^3(a+b x)}{2 b} \]

[Out]

5/2*x+5/2*cot(b*x+a)/b-5/6*cot(b*x+a)^3/b+1/2*cos(b*x+a)^2*cot(b*x+a)^3/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2671, 294, 308, 209} \[ \int \cos ^2(a+b x) \cot ^4(a+b x) \, dx=-\frac {5 \cot ^3(a+b x)}{6 b}+\frac {5 \cot (a+b x)}{2 b}+\frac {\cos ^2(a+b x) \cot ^3(a+b x)}{2 b}+\frac {5 x}{2} \]

[In]

Int[Cos[a + b*x]^2*Cot[a + b*x]^4,x]

[Out]

(5*x)/2 + (5*Cot[a + b*x])/(2*b) - (5*Cot[a + b*x]^3)/(6*b) + (Cos[a + b*x]^2*Cot[a + b*x]^3)/(2*b)

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 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2671

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[b*(ff/f), Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^2} \, dx,x,\cot (a+b x)\right )}{b} \\ & = \frac {\cos ^2(a+b x) \cot ^3(a+b x)}{2 b}-\frac {5 \text {Subst}\left (\int \frac {x^4}{1+x^2} \, dx,x,\cot (a+b x)\right )}{2 b} \\ & = \frac {\cos ^2(a+b x) \cot ^3(a+b x)}{2 b}-\frac {5 \text {Subst}\left (\int \left (-1+x^2+\frac {1}{1+x^2}\right ) \, dx,x,\cot (a+b x)\right )}{2 b} \\ & = \frac {5 \cot (a+b x)}{2 b}-\frac {5 \cot ^3(a+b x)}{6 b}+\frac {\cos ^2(a+b x) \cot ^3(a+b x)}{2 b}-\frac {5 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (a+b x)\right )}{2 b} \\ & = \frac {5 x}{2}+\frac {5 \cot (a+b x)}{2 b}-\frac {5 \cot ^3(a+b x)}{6 b}+\frac {\cos ^2(a+b x) \cot ^3(a+b x)}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.75 \[ \int \cos ^2(a+b x) \cot ^4(a+b x) \, dx=\frac {30 (a+b x)-4 \cot (a+b x) \left (-7+\csc ^2(a+b x)\right )+3 \sin (2 (a+b x))}{12 b} \]

[In]

Integrate[Cos[a + b*x]^2*Cot[a + b*x]^4,x]

[Out]

(30*(a + b*x) - 4*Cot[a + b*x]*(-7 + Csc[a + b*x]^2) + 3*Sin[2*(a + b*x)])/(12*b)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.37

method	result	size
risch	\(\frac {5 x}{2}-\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{8 b}+\frac {i {\mathrm e}^{-2 i \left (b x +a \right )}}{8 b}+\frac {2 i \left (9 \,{\mathrm e}^{4 i \left (b x +a \right )}-12 \,{\mathrm e}^{2 i \left (b x +a \right )}+7\right )}{3 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}}\)	\(78\)
parallelrisch	\(\frac {\left (180 b x \sin \left (b x +a \right )-60 b x \sin \left (3 b x +3 a \right )+30 \cos \left (b x +a \right )-65 \cos \left (3 b x +3 a \right )+3 \cos \left (5 b x +5 a \right )\right ) \left (\sec ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\csc ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{768 b}\)	\(82\)
derivativedivides	\(\frac {-\frac {\cos ^{7}\left (b x +a \right )}{3 \sin \left (b x +a \right )^{3}}+\frac {4 \left (\cos ^{7}\left (b x +a \right )\right )}{3 \sin \left (b x +a \right )}+\frac {4 \left (\cos ^{5}\left (b x +a \right )+\frac {5 \left (\cos ^{3}\left (b x +a \right )\right )}{4}+\frac {15 \cos \left (b x +a \right )}{8}\right ) \sin \left (b x +a \right )}{3}+\frac {5 b x}{2}+\frac {5 a}{2}}{b}\)	\(84\)
default	\(\frac {-\frac {\cos ^{7}\left (b x +a \right )}{3 \sin \left (b x +a \right )^{3}}+\frac {4 \left (\cos ^{7}\left (b x +a \right )\right )}{3 \sin \left (b x +a \right )}+\frac {4 \left (\cos ^{5}\left (b x +a \right )+\frac {5 \left (\cos ^{3}\left (b x +a \right )\right )}{4}+\frac {15 \cos \left (b x +a \right )}{8}\right ) \sin \left (b x +a \right )}{3}+\frac {5 b x}{2}+\frac {5 a}{2}}{b}\)	\(84\)
norman	\(\frac {-\frac {1}{24 b}+\frac {25 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}+\frac {25 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{12 b}-\frac {25 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{12 b}-\frac {25 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}+\frac {\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )}{24 b}+\frac {5 x \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2}+5 x \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\frac {5 x \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{2} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}\)	\(156\)

[In]

int(cos(b*x+a)^6/sin(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

5/2*x-1/8*I/b*exp(2*I*(b*x+a))+1/8*I/b*exp(-2*I*(b*x+a))+2/3*I*(9*exp(4*I*(b*x+a))-12*exp(2*I*(b*x+a))+7)/b/(e
xp(2*I*(b*x+a))-1)^3

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.39 \[ \int \cos ^2(a+b x) \cot ^4(a+b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{5} - 20 \, \cos \left (b x + a\right )^{3} - 15 \, {\left (b x \cos \left (b x + a\right )^{2} - b x\right )} \sin \left (b x + a\right ) + 15 \, \cos \left (b x + a\right )}{6 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \]

[In]

integrate(cos(b*x+a)^6/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/6*(3*cos(b*x + a)^5 - 20*cos(b*x + a)^3 - 15*(b*x*cos(b*x + a)^2 - b*x)*sin(b*x + a) + 15*cos(b*x + a))/((b
*cos(b*x + a)^2 - b)*sin(b*x + a))

Sympy [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.70 \[ \int \cos ^2(a+b x) \cot ^4(a+b x) \, dx=\begin {cases} \frac {5 x \sin ^{2}{\left (a + b x \right )}}{2} + \frac {5 x \cos ^{2}{\left (a + b x \right )}}{2} + \frac {5 \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b} + \frac {5 \cos ^{3}{\left (a + b x \right )}}{3 b \sin {\left (a + b x \right )}} - \frac {\cos ^{5}{\left (a + b x \right )}}{3 b \sin ^{3}{\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{6}{\left (a \right )}}{\sin ^{4}{\left (a \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)**6/sin(b*x+a)**4,x)

[Out]

Piecewise((5*x*sin(a + b*x)**2/2 + 5*x*cos(a + b*x)**2/2 + 5*sin(a + b*x)*cos(a + b*x)/(2*b) + 5*cos(a + b*x)*
*3/(3*b*sin(a + b*x)) - cos(a + b*x)**5/(3*b*sin(a + b*x)**3), Ne(b, 0)), (x*cos(a)**6/sin(a)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \cos ^2(a+b x) \cot ^4(a+b x) \, dx=\frac {15 \, b x + 15 \, a + \frac {15 \, \tan \left (b x + a\right )^{4} + 10 \, \tan \left (b x + a\right )^{2} - 2}{\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}}}{6 \, b} \]

[In]

integrate(cos(b*x+a)^6/sin(b*x+a)^4,x, algorithm="maxima")

[Out]

1/6*(15*b*x + 15*a + (15*tan(b*x + a)^4 + 10*tan(b*x + a)^2 - 2)/(tan(b*x + a)^5 + tan(b*x + a)^3))/b

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \cos ^2(a+b x) \cot ^4(a+b x) \, dx=\frac {15 \, b x + 15 \, a + \frac {3 \, \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{2} + 1} + \frac {2 \, {\left (6 \, \tan \left (b x + a\right )^{2} - 1\right )}}{\tan \left (b x + a\right )^{3}}}{6 \, b} \]

[In]

integrate(cos(b*x+a)^6/sin(b*x+a)^4,x, algorithm="giac")

[Out]

1/6*(15*b*x + 15*a + 3*tan(b*x + a)/(tan(b*x + a)^2 + 1) + 2*(6*tan(b*x + a)^2 - 1)/tan(b*x + a)^3)/b

Mupad [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.81 \[ \int \cos ^2(a+b x) \cot ^4(a+b x) \, dx=\frac {5\,x}{2}+\frac {{\cos \left (a+b\,x\right )}^2\,\left (\frac {5\,{\mathrm {tan}\left (a+b\,x\right )}^4}{2}+\frac {5\,{\mathrm {tan}\left (a+b\,x\right )}^2}{3}-\frac {1}{3}\right )}{b\,{\mathrm {tan}\left (a+b\,x\right )}^3} \]

[In]

int(cos(a + b*x)^6/sin(a + b*x)^4,x)

[Out]

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