3.120 \(\int \cos ^5(a+b x) \cot (a+b x) \, dx\)

Optimal. Leaf size=53 \[ \frac {\cos ^5(a+b x)}{5 b}+\frac {\cos ^3(a+b x)}{3 b}+\frac {\cos (a+b x)}{b}-\frac {\tanh ^{-1}(\cos (a+b x))}{b} \]

[Out]

-arctanh(cos(b*x+a))/b+cos(b*x+a)/b+1/3*cos(b*x+a)^3/b+1/5*cos(b*x+a)^5/b

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2592, 302, 206} \[ \frac {\cos ^5(a+b x)}{5 b}+\frac {\cos ^3(a+b x)}{3 b}+\frac {\cos (a+b x)}{b}-\frac {\tanh ^{-1}(\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^5*Cot[a + b*x],x]

[Out]

-(ArcTanh[Cos[a + b*x]]/b) + Cos[a + b*x]/b + Cos[a + b*x]^3/(3*b) + Cos[a + b*x]^5/(5*b)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 302

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 2592

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 75, normalized size = 1.42 \[ \frac {11 \cos (a+b x)}{8 b}+\frac {7 \cos (3 (a+b x))}{48 b}+\frac {\cos (5 (a+b x))}{80 b}+\frac {\log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{b}-\frac {\log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^5*Cot[a + b*x],x]

[Out]

(11*Cos[a + b*x])/(8*b) + (7*Cos[3*(a + b*x)])/(48*b) + Cos[5*(a + b*x)]/(80*b) - Log[Cos[(a + b*x)/2]]/b + Lo
g[Sin[(a + b*x)/2]]/b

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 60, normalized size = 1.13 \[ \frac {6 \, \cos \left (b x + a\right )^{5} + 10 \, \cos \left (b x + a\right )^{3} + 30 \, \cos \left (b x + a\right ) - 15 \, \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 15 \, \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right )}{30 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(6*cos(b*x + a)^5 + 10*cos(b*x + a)^3 + 30*cos(b*x + a) - 15*log(1/2*cos(b*x + a) + 1/2) + 15*log(-1/2*co
s(b*x + a) + 1/2))/b

________________________________________________________________________________________

giac [B]  time = 0.27, size = 145, normalized size = 2.74 \[ \frac {\frac {4 \, {\left (\frac {70 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {140 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {90 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - \frac {45 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} - 23\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{5}} + 15 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{30 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(4*(70*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 140*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 90*(cos(b*
x + a) - 1)^3/(cos(b*x + a) + 1)^3 - 45*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 - 23)/((cos(b*x + a) - 1)/(c
os(b*x + a) + 1) - 1)^5 + 15*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b

________________________________________________________________________________________

maple [A]  time = 0.03, size = 58, normalized size = 1.09 \[ \frac {\cos ^{5}\left (b x +a \right )}{5 b}+\frac {\cos ^{3}\left (b x +a \right )}{3 b}+\frac {\cos \left (b x +a \right )}{b}+\frac {\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^6/sin(b*x+a),x)

[Out]

1/5*cos(b*x+a)^5/b+1/3*cos(b*x+a)^3/b+cos(b*x+a)/b+1/b*ln(csc(b*x+a)-cot(b*x+a))

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 56, normalized size = 1.06 \[ \frac {6 \, \cos \left (b x + a\right )^{5} + 10 \, \cos \left (b x + a\right )^{3} + 30 \, \cos \left (b x + a\right ) - 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{30 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(6*cos(b*x + a)^5 + 10*cos(b*x + a)^3 + 30*cos(b*x + a) - 15*log(cos(b*x + a) + 1) + 15*log(cos(b*x + a)
- 1))/b

________________________________________________________________________________________

mupad [B]  time = 5.37, size = 88, normalized size = 1.66 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{b}+\frac {6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8+12\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+\frac {56\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4}{3}+\frac {28\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{3}+\frac {46}{15}}{b\,{\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^6/sin(a + b*x),x)

[Out]

log(tan(a/2 + (b*x)/2))/b + ((28*tan(a/2 + (b*x)/2)^2)/3 + (56*tan(a/2 + (b*x)/2)^4)/3 + 12*tan(a/2 + (b*x)/2)
^6 + 6*tan(a/2 + (b*x)/2)^8 + 46/15)/(b*(tan(a/2 + (b*x)/2)^2 + 1)^5)

________________________________________________________________________________________

sympy [A]  time = 8.16, size = 1085, normalized size = 20.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((15*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**10/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8
+ 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 75*log(tan(a/2 +
b*x/2))*tan(a/2 + b*x/2)**8/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6
+ 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 150*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**
6/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**
4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 150*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(15*b*tan(a/2 + b*x/2)**1
0 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**
2 + 15*b) + 75*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8
+ 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 15*log(tan(a/2 +
b*x/2))/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b
*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 90*tan(a/2 + b*x/2)**8/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2
+ b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 180*
tan(a/2 + b*x/2)**6/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*
tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 280*tan(a/2 + b*x/2)**4/(15*b*tan(a/2 + b*x/2)**10 +
75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 +
15*b) + 140*tan(a/2 + b*x/2)**2/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)
**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 46/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(
a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b), Ne
(b, 0)), (x*cos(a)**6/sin(a), True))

________________________________________________________________________________________