3.172 \(\int \cos ^3(a+b x) \cot ^5(a+b x) \, dx\)

Optimal. Leaf size=89 \[ \frac{35 \cos ^3(a+b x)}{24 b}+\frac{35 \cos (a+b x)}{8 b}-\frac{\cos ^3(a+b x) \cot ^4(a+b x)}{4 b}+\frac{7 \cos ^3(a+b x) \cot ^2(a+b x)}{8 b}-\frac{35 \tanh ^{-1}(\cos (a+b x))}{8 b} \]

[Out]

(-35*ArcTanh[Cos[a + b*x]])/(8*b) + (35*Cos[a + b*x])/(8*b) + (35*Cos[a + b*x]^3)/(24*b) + (7*Cos[a + b*x]^3*C
ot[a + b*x]^2)/(8*b) - (Cos[a + b*x]^3*Cot[a + b*x]^4)/(4*b)

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Rubi [A]  time = 0.0501559, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2592, 288, 302, 206} \[ \frac{35 \cos ^3(a+b x)}{24 b}+\frac{35 \cos (a+b x)}{8 b}-\frac{\cos ^3(a+b x) \cot ^4(a+b x)}{4 b}+\frac{7 \cos ^3(a+b x) \cot ^2(a+b x)}{8 b}-\frac{35 \tanh ^{-1}(\cos (a+b x))}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Cot[a + b*x]^5,x]

[Out]

(-35*ArcTanh[Cos[a + b*x]])/(8*b) + (35*Cos[a + b*x])/(8*b) + (35*Cos[a + b*x]^3)/(24*b) + (7*Cos[a + b*x]^3*C
ot[a + b*x]^2)/(8*b) - (Cos[a + b*x]^3*Cot[a + b*x]^4)/(4*b)

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]

Rule 288

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 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 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])

Rubi steps

\begin{align*} \int \cos ^3(a+b x) \cot ^5(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^8}{\left (1-x^2\right )^3} \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\cos ^3(a+b x) \cot ^4(a+b x)}{4 b}+\frac{7 \operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^2} \, dx,x,\cos (a+b x)\right )}{4 b}\\ &=\frac{7 \cos ^3(a+b x) \cot ^2(a+b x)}{8 b}-\frac{\cos ^3(a+b x) \cot ^4(a+b x)}{4 b}-\frac{35 \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\cos (a+b x)\right )}{8 b}\\ &=\frac{7 \cos ^3(a+b x) \cot ^2(a+b x)}{8 b}-\frac{\cos ^3(a+b x) \cot ^4(a+b x)}{4 b}-\frac{35 \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\cos (a+b x)\right )}{8 b}\\ &=\frac{35 \cos (a+b x)}{8 b}+\frac{35 \cos ^3(a+b x)}{24 b}+\frac{7 \cos ^3(a+b x) \cot ^2(a+b x)}{8 b}-\frac{\cos ^3(a+b x) \cot ^4(a+b x)}{4 b}-\frac{35 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (a+b x)\right )}{8 b}\\ &=-\frac{35 \tanh ^{-1}(\cos (a+b x))}{8 b}+\frac{35 \cos (a+b x)}{8 b}+\frac{35 \cos ^3(a+b x)}{24 b}+\frac{7 \cos ^3(a+b x) \cot ^2(a+b x)}{8 b}-\frac{\cos ^3(a+b x) \cot ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0406061, size = 141, normalized size = 1.58 \[ \frac{13 \cos (a+b x)}{4 b}+\frac{\cos (3 (a+b x))}{12 b}-\frac{\csc ^4\left (\frac{1}{2} (a+b x)\right )}{64 b}+\frac{13 \csc ^2\left (\frac{1}{2} (a+b x)\right )}{32 b}+\frac{\sec ^4\left (\frac{1}{2} (a+b x)\right )}{64 b}-\frac{13 \sec ^2\left (\frac{1}{2} (a+b x)\right )}{32 b}+\frac{35 \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{8 b}-\frac{35 \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*Cot[a + b*x]^5,x]

[Out]

(13*Cos[a + b*x])/(4*b) + Cos[3*(a + b*x)]/(12*b) + (13*Csc[(a + b*x)/2]^2)/(32*b) - Csc[(a + b*x)/2]^4/(64*b)
- (35*Log[Cos[(a + b*x)/2]])/(8*b) + (35*Log[Sin[(a + b*x)/2]])/(8*b) - (13*Sec[(a + b*x)/2]^2)/(32*b) + Sec[
(a + b*x)/2]^4/(64*b)

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Maple [A]  time = 0.011, size = 115, normalized size = 1.3 \begin{align*} -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{9}}{4\,b \left ( \sin \left ( bx+a \right ) \right ) ^{4}}}+{\frac{5\, \left ( \cos \left ( bx+a \right ) \right ) ^{9}}{8\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}}}+{\frac{5\, \left ( \cos \left ( bx+a \right ) \right ) ^{7}}{8\,b}}+{\frac{7\, \left ( \cos \left ( bx+a \right ) \right ) ^{5}}{8\,b}}+{\frac{35\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{24\,b}}+{\frac{35\,\cos \left ( bx+a \right ) }{8\,b}}+{\frac{35\,\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{8\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^8/sin(b*x+a)^5,x)

[Out]

-1/4/b*cos(b*x+a)^9/sin(b*x+a)^4+5/8/b/sin(b*x+a)^2*cos(b*x+a)^9+5/8*cos(b*x+a)^7/b+7/8*cos(b*x+a)^5/b+35/24*c
os(b*x+a)^3/b+35/8*cos(b*x+a)/b+35/8/b*ln(csc(b*x+a)-cot(b*x+a))

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Maxima [A]  time = 0.967444, size = 120, normalized size = 1.35 \begin{align*} \frac{16 \, \cos \left (b x + a\right )^{3} - \frac{6 \,{\left (13 \, \cos \left (b x + a\right )^{3} - 11 \, \cos \left (b x + a\right )\right )}}{\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1} + 144 \, \cos \left (b x + a\right ) - 105 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 105 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^8/sin(b*x+a)^5,x, algorithm="maxima")

[Out]

1/48*(16*cos(b*x + a)^3 - 6*(13*cos(b*x + a)^3 - 11*cos(b*x + a))/(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1) + 14
4*cos(b*x + a) - 105*log(cos(b*x + a) + 1) + 105*log(cos(b*x + a) - 1))/b

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Fricas [A]  time = 2.44378, size = 378, normalized size = 4.25 \begin{align*} \frac{16 \, \cos \left (b x + a\right )^{7} + 112 \, \cos \left (b x + a\right )^{5} - 350 \, \cos \left (b x + a\right )^{3} - 105 \,{\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) + 105 \,{\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) + 210 \, \cos \left (b x + a\right )}{48 \,{\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^8/sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/48*(16*cos(b*x + a)^7 + 112*cos(b*x + a)^5 - 350*cos(b*x + a)^3 - 105*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1
)*log(1/2*cos(b*x + a) + 1/2) + 105*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(-1/2*cos(b*x + a) + 1/2) + 210
*cos(b*x + a))/(b*cos(b*x + a)^4 - 2*b*cos(b*x + a)^2 + b)

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Sympy [A]  time = 33.4691, size = 869, normalized size = 9.76 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**8/sin(b*x+a)**5,x)

[Out]

Piecewise((840*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**10/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)
**8 + 576*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) + 2520*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**8/
(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**
4) + 2520*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 +
576*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) + 840*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(192*b*
tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) + 3*
tan(a/2 + b*x/2)**14/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b*x/2)**6 + 192
*b*tan(a/2 + b*x/2)**4) - 63*tan(a/2 + b*x/2)**12/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 57
6*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) + 2016*tan(a/2 + b*x/2)**8/(192*b*tan(a/2 + b*x/2)**10 +
576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) + 3066*tan(a/2 + b*x/2)**6/
(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**
4) + 1694*tan(a/2 + b*x/2)**4/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b*x/2)
**6 + 192*b*tan(a/2 + b*x/2)**4) + 63*tan(a/2 + b*x/2)**2/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)
**8 + 576*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) - 3/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 +
b*x/2)**8 + 576*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4), Ne(b, 0)), (x*cos(a)**8/sin(a)**5, True))

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Giac [B]  time = 1.20934, size = 282, normalized size = 3.17 \begin{align*} -\frac{\frac{3 \,{\left (\frac{24 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{210 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )}{\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} - \frac{72 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{3 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - \frac{256 \,{\left (\frac{9 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{6 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 5\right )}}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{3}} - 420 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{192 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^8/sin(b*x+a)^5,x, algorithm="giac")

[Out]

-1/192*(3*(24*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 210*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 1)*(cos(
b*x + a) + 1)^2/(cos(b*x + a) - 1)^2 - 72*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 3*(cos(b*x + a) - 1)^2/(cos(
b*x + a) + 1)^2 - 256*(9*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 6*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 -
5)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)^3 - 420*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b