3.173 \(\int \cos ^2(a+b x) \cot ^5(a+b x) \, dx\)

Optimal. Leaf size=58 \[ -\frac {\sin ^2(a+b x)}{2 b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {3 \csc ^2(a+b x)}{2 b}+\frac {3 \log (\sin (a+b x))}{b} \]

[Out]

3/2*csc(b*x+a)^2/b-1/4*csc(b*x+a)^4/b+3*ln(sin(b*x+a))/b-1/2*sin(b*x+a)^2/b

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Rubi [A]  time = 0.04, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2590, 266, 43} \[ -\frac {\sin ^2(a+b x)}{2 b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {3 \csc ^2(a+b x)}{2 b}+\frac {3 \log (\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Cot[a + b*x]^5,x]

[Out]

(3*Csc[a + b*x]^2)/(2*b) - Csc[a + b*x]^4/(4*b) + (3*Log[Sin[a + b*x]])/b - Sin[a + b*x]^2/(2*b)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2590

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

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Mathematica [A]  time = 0.15, size = 47, normalized size = 0.81 \[ \frac {-2 \sin ^2(a+b x)-\csc ^4(a+b x)+6 \csc ^2(a+b x)+12 \log (\sin (a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2*Cot[a + b*x]^5,x]

[Out]

(6*Csc[a + b*x]^2 - Csc[a + b*x]^4 + 12*Log[Sin[a + b*x]] - 2*Sin[a + b*x]^2)/(4*b)

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fricas [A]  time = 0.48, size = 90, normalized size = 1.55 \[ \frac {2 \, \cos \left (b x + a\right )^{6} - 5 \, \cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 12 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (b x + a\right )\right ) + 4}{4 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^7/sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/4*(2*cos(b*x + a)^6 - 5*cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 12*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(1
/2*sin(b*x + a)) + 4)/(b*cos(b*x + a)^4 - 2*b*cos(b*x + a)^2 + b)

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giac [B]  time = 0.29, size = 232, normalized size = 4.00 \[ -\frac {\frac {20 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {\frac {18 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {111 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {36 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - \frac {72 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 1}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}}\right )}^{2}} - 96 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 192 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1 \right |}\right )}{64 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^7/sin(b*x+a)^5,x, algorithm="giac")

[Out]

-1/64*(20*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + (18*(cos(b*x + a
) - 1)/(cos(b*x + a) + 1) - 111*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 36*(cos(b*x + a) - 1)^3/(cos(b*x +
a) + 1)^3 - 72*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 1)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - (cos(b
*x + a) - 1)^2/(cos(b*x + a) + 1)^2)^2 - 96*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 192*log(abs(-(
cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)))/b

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maple [A]  time = 0.02, size = 95, normalized size = 1.64 \[ -\frac {\cos ^{8}\left (b x +a \right )}{4 b \sin \left (b x +a \right )^{4}}+\frac {\cos ^{8}\left (b x +a \right )}{2 b \sin \left (b x +a \right )^{2}}+\frac {\cos ^{6}\left (b x +a \right )}{2 b}+\frac {3 \left (\cos ^{4}\left (b x +a \right )\right )}{4 b}+\frac {3 \left (\cos ^{2}\left (b x +a \right )\right )}{2 b}+\frac {3 \ln \left (\sin \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^7/sin(b*x+a)^5,x)

[Out]

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

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maxima [A]  time = 0.48, size = 49, normalized size = 0.84 \[ -\frac {2 \, \sin \left (b x + a\right )^{2} - \frac {6 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{4}} - 6 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^7/sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/4*(2*sin(b*x + a)^2 - (6*sin(b*x + a)^2 - 1)/sin(b*x + a)^4 - 6*log(sin(b*x + a)^2))/b

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mupad [B]  time = 0.64, size = 74, normalized size = 1.28 \[ \frac {3\,\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b}-\frac {3\,\ln \left ({\mathrm {tan}\left (a+b\,x\right )}^2+1\right )}{2\,b}+\frac {\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^4}{2}+\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^2}{4}-\frac {1}{4}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^6+{\mathrm {tan}\left (a+b\,x\right )}^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^7/sin(a + b*x)^5,x)

[Out]

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

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sympy [A]  time = 11.57, size = 733, normalized size = 12.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**7/sin(b*x+a)**5,x)

[Out]

Piecewise((-192*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**8/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b
*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 384*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**6/(64*b*tan(a/2 + b*
x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 192*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b
*x/2)**4/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 192*log(tan(a/2 +
b*x/2))*tan(a/2 + b*x/2)**8/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4)
+ 384*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*
tan(a/2 + b*x/2)**4) + 192*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2
+ b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - tan(a/2 + b*x/2)**12/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*
x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 18*tan(a/2 + b*x/2)**10/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/
2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 166*tan(a/2 + b*x/2)**6/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)
**6 + 64*b*tan(a/2 + b*x/2)**4) + 18*tan(a/2 + b*x/2)**2/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6
+ 64*b*tan(a/2 + b*x/2)**4) - 1/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)
**4), Ne(b, 0)), (x*cos(a)**7/sin(a)**5, True))

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