Integrand size = 17, antiderivative size = 57 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=-\frac {15 x}{8}-\frac {\cot (a+b x)}{b}-\frac {9 \cos (a+b x) \sin (a+b x)}{8 b}+\frac {\cos (a+b x) \sin ^3(a+b x)}{4 b} \] Output:
-15/8*x-cot(b*x+a)/b-9/8*cos(b*x+a)*sin(b*x+a)/b+1/4*cos(b*x+a)*sin(b*x+a) ^3/b
Time = 0.32 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.72 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=-\frac {60 a+60 b x+32 \cot (a+b x)+16 \sin (2 (a+b x))+\sin (4 (a+b x))}{32 b} \] Input:
Integrate[Cos[a + b*x]^4*Cot[a + b*x]^2,x]
Output:
-1/32*(60*a + 60*b*x + 32*Cot[a + b*x] + 16*Sin[2*(a + b*x)] + Sin[4*(a + b*x)])/b
Time = 0.39 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.39, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3042, 3071, 252, 252, 262, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (a+b x+\frac {\pi }{2}\right )^4 \tan \left (a+b x+\frac {\pi }{2}\right )^2dx\) |
\(\Big \downarrow \) 3071 |
\(\displaystyle -\frac {\int \frac {\cot ^6(a+b x)}{\left (\cot ^2(a+b x)+1\right )^3}d\cot (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {5}{4} \int \frac {\cot ^4(a+b x)}{\left (\cot ^2(a+b x)+1\right )^2}d\cot (a+b x)-\frac {\cot ^5(a+b x)}{4 \left (\cot ^2(a+b x)+1\right )^2}}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {5}{4} \left (\frac {3}{2} \int \frac {\cot ^2(a+b x)}{\cot ^2(a+b x)+1}d\cot (a+b x)-\frac {\cot ^3(a+b x)}{2 \left (\cot ^2(a+b x)+1\right )}\right )-\frac {\cot ^5(a+b x)}{4 \left (\cot ^2(a+b x)+1\right )^2}}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {\frac {5}{4} \left (\frac {3}{2} \left (\cot (a+b x)-\int \frac {1}{\cot ^2(a+b x)+1}d\cot (a+b x)\right )-\frac {\cot ^3(a+b x)}{2 \left (\cot ^2(a+b x)+1\right )}\right )-\frac {\cot ^5(a+b x)}{4 \left (\cot ^2(a+b x)+1\right )^2}}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {5}{4} \left (\frac {3}{2} (\cot (a+b x)-\arctan (\cot (a+b x)))-\frac {\cot ^3(a+b x)}{2 \left (\cot ^2(a+b x)+1\right )}\right )-\frac {\cot ^5(a+b x)}{4 \left (\cot ^2(a+b x)+1\right )^2}}{b}\) |
Input:
Int[Cos[a + b*x]^4*Cot[a + b*x]^2,x]
Output:
-((-1/4*Cot[a + b*x]^5/(1 + Cot[a + b*x]^2)^2 + (5*((3*(-ArcTan[Cot[a + b* x]] + Cot[a + b*x]))/2 - Cot[a + b*x]^3/(2*(1 + Cot[a + b*x]^2))))/4)/b)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[I nt[(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]
Time = 4.62 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.16
method | result | size |
derivativedivides | \(\frac {-\frac {\cos \left (b x +a \right )^{7}}{\sin \left (b x +a \right )}-\left (\cos \left (b x +a \right )^{5}+\frac {5 \cos \left (b x +a \right )^{3}}{4}+\frac {15 \cos \left (b x +a \right )}{8}\right ) \sin \left (b x +a \right )-\frac {15 b x}{8}-\frac {15 a}{8}}{b}\) | \(66\) |
default | \(\frac {-\frac {\cos \left (b x +a \right )^{7}}{\sin \left (b x +a \right )}-\left (\cos \left (b x +a \right )^{5}+\frac {5 \cos \left (b x +a \right )^{3}}{4}+\frac {15 \cos \left (b x +a \right )}{8}\right ) \sin \left (b x +a \right )-\frac {15 b x}{8}-\frac {15 a}{8}}{b}\) | \(66\) |
risch | \(-\frac {15 x}{8}+\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{4 b}-\frac {i {\mathrm e}^{-2 i \left (b x +a \right )}}{4 b}-\frac {2 i}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {\sin \left (4 b x +4 a \right )}{32 b}\) | \(68\) |
Input:
int(cos(b*x+a)^4*cot(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/b*(-1/sin(b*x+a)*cos(b*x+a)^7-(cos(b*x+a)^5+5/4*cos(b*x+a)^3+15/8*cos(b* x+a))*sin(b*x+a)-15/8*b*x-15/8*a)
Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=\frac {2 \, \cos \left (b x + a\right )^{5} + 5 \, \cos \left (b x + a\right )^{3} - 15 \, b x \sin \left (b x + a\right ) - 15 \, \cos \left (b x + a\right )}{8 \, b \sin \left (b x + a\right )} \] Input:
integrate(cos(b*x+a)^4*cot(b*x+a)^2,x, algorithm="fricas")
Output:
1/8*(2*cos(b*x + a)^5 + 5*cos(b*x + a)^3 - 15*b*x*sin(b*x + a) - 15*cos(b* x + a))/(b*sin(b*x + a))
\[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=\int \cos ^{4}{\left (a + b x \right )} \cot ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate(cos(b*x+a)**4*cot(b*x+a)**2,x)
Output:
Integral(cos(a + b*x)**4*cot(a + b*x)**2, x)
Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=-\frac {15 \, b x + 15 \, a + \frac {15 \, \tan \left (b x + a\right )^{4} + 25 \, \tan \left (b x + a\right )^{2} + 8}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}}{8 \, b} \] Input:
integrate(cos(b*x+a)^4*cot(b*x+a)^2,x, algorithm="maxima")
Output:
-1/8*(15*b*x + 15*a + (15*tan(b*x + a)^4 + 25*tan(b*x + a)^2 + 8)/(tan(b*x + a)^5 + 2*tan(b*x + a)^3 + tan(b*x + a)))/b
Time = 0.14 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=-\frac {15 \, b x + 15 \, a + \frac {7 \, \tan \left (b x + a\right )^{3} + 9 \, \tan \left (b x + a\right )}{{\left (\tan \left (b x + a\right )^{2} + 1\right )}^{2}} + \frac {8}{\tan \left (b x + a\right )}}{8 \, b} \] Input:
integrate(cos(b*x+a)^4*cot(b*x+a)^2,x, algorithm="giac")
Output:
-1/8*(15*b*x + 15*a + (7*tan(b*x + a)^3 + 9*tan(b*x + a))/(tan(b*x + a)^2 + 1)^2 + 8/tan(b*x + a))/b
Time = 25.90 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=-\frac {15\,x}{8}-\frac {{\cos \left (a+b\,x\right )}^4\,\left (\frac {15\,{\mathrm {tan}\left (a+b\,x\right )}^4}{8}+\frac {25\,{\mathrm {tan}\left (a+b\,x\right )}^2}{8}+1\right )}{b\,\mathrm {tan}\left (a+b\,x\right )} \] Input:
int(cos(a + b*x)^4*cot(a + b*x)^2,x)
Output:
- (15*x)/8 - (cos(a + b*x)^4*((25*tan(a + b*x)^2)/8 + (15*tan(a + b*x)^4)/ 8 + 1))/(b*tan(a + b*x))
Time = 0.16 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.12 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=\frac {2 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{4}-9 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}-8 \cos \left (b x +a \right )-15 \sin \left (b x +a \right ) b x}{8 \sin \left (b x +a \right ) b} \] Input:
int(cos(b*x+a)^4*cot(b*x+a)^2,x)
Output:
(2*cos(a + b*x)*sin(a + b*x)**4 - 9*cos(a + b*x)*sin(a + b*x)**2 - 8*cos(a + b*x) - 15*sin(a + b*x)*b*x)/(8*sin(a + b*x)*b)