3.149 \(\int \cot ^6(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=122 \[ -\frac{a \cot ^5(c+d x)}{5 d}+\frac{a \cot ^3(c+d x)}{3 d}-\frac{a \cot (c+d x)}{d}-a x+\frac{15 b \cos (c+d x)}{8 d}-\frac{b \cos (c+d x) \cot ^4(c+d x)}{4 d}+\frac{5 b \cos (c+d x) \cot ^2(c+d x)}{8 d}-\frac{15 b \tanh ^{-1}(\cos (c+d x))}{8 d} \]

[Out]

-(a*x) - (15*b*ArcTanh[Cos[c + d*x]])/(8*d) + (15*b*Cos[c + d*x])/(8*d) - (a*Cot[c + d*x])/d + (5*b*Cos[c + d*
x]*Cot[c + d*x]^2)/(8*d) + (a*Cot[c + d*x]^3)/(3*d) - (b*Cos[c + d*x]*Cot[c + d*x]^4)/(4*d) - (a*Cot[c + d*x]^
5)/(5*d)

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Rubi [A]  time = 0.0972308, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {2722, 2592, 288, 321, 206, 3473, 8} \[ -\frac{a \cot ^5(c+d x)}{5 d}+\frac{a \cot ^3(c+d x)}{3 d}-\frac{a \cot (c+d x)}{d}-a x+\frac{15 b \cos (c+d x)}{8 d}-\frac{b \cos (c+d x) \cot ^4(c+d x)}{4 d}+\frac{5 b \cos (c+d x) \cot ^2(c+d x)}{8 d}-\frac{15 b \tanh ^{-1}(\cos (c+d x))}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + b*Sin[c + d*x]),x]

[Out]

-(a*x) - (15*b*ArcTanh[Cos[c + d*x]])/(8*d) + (15*b*Cos[c + d*x])/(8*d) - (a*Cot[c + d*x])/d + (5*b*Cos[c + d*
x]*Cot[c + d*x]^2)/(8*d) + (a*Cot[c + d*x]^3)/(3*d) - (b*Cos[c + d*x]*Cot[c + d*x]^4)/(4*d) - (a*Cot[c + d*x]^
5)/(5*d)

Rule 2722

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2
, 0] && IGtQ[m, 0]

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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \cot ^6(c+d x) (a+b \sin (c+d x)) \, dx &=\int \left (b \cos (c+d x) \cot ^5(c+d x)+a \cot ^6(c+d x)\right ) \, dx\\ &=a \int \cot ^6(c+d x) \, dx+b \int \cos (c+d x) \cot ^5(c+d x) \, dx\\ &=-\frac{a \cot ^5(c+d x)}{5 d}-a \int \cot ^4(c+d x) \, dx-\frac{b \operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^3} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{a \cot ^3(c+d x)}{3 d}-\frac{b \cos (c+d x) \cot ^4(c+d x)}{4 d}-\frac{a \cot ^5(c+d x)}{5 d}+a \int \cot ^2(c+d x) \, dx+\frac{(5 b) \operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{4 d}\\ &=-\frac{a \cot (c+d x)}{d}+\frac{5 b \cos (c+d x) \cot ^2(c+d x)}{8 d}+\frac{a \cot ^3(c+d x)}{3 d}-\frac{b \cos (c+d x) \cot ^4(c+d x)}{4 d}-\frac{a \cot ^5(c+d x)}{5 d}-a \int 1 \, dx-\frac{(15 b) \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{8 d}\\ &=-a x+\frac{15 b \cos (c+d x)}{8 d}-\frac{a \cot (c+d x)}{d}+\frac{5 b \cos (c+d x) \cot ^2(c+d x)}{8 d}+\frac{a \cot ^3(c+d x)}{3 d}-\frac{b \cos (c+d x) \cot ^4(c+d x)}{4 d}-\frac{a \cot ^5(c+d x)}{5 d}-\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{8 d}\\ &=-a x-\frac{15 b \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{15 b \cos (c+d x)}{8 d}-\frac{a \cot (c+d x)}{d}+\frac{5 b \cos (c+d x) \cot ^2(c+d x)}{8 d}+\frac{a \cot ^3(c+d x)}{3 d}-\frac{b \cos (c+d x) \cot ^4(c+d x)}{4 d}-\frac{a \cot ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [C]  time = 0.0598431, size = 164, normalized size = 1.34 \[ -\frac{a \cot ^5(c+d x) \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};-\tan ^2(c+d x)\right )}{5 d}+\frac{b \cos (c+d x)}{d}-\frac{b \csc ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}+\frac{9 b \csc ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{b \sec ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}-\frac{9 b \sec ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{15 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}-\frac{15 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + b*Sin[c + d*x]),x]

[Out]

(b*Cos[c + d*x])/d + (9*b*Csc[(c + d*x)/2]^2)/(32*d) - (b*Csc[(c + d*x)/2]^4)/(64*d) - (a*Cot[c + d*x]^5*Hyper
geometric2F1[-5/2, 1, -3/2, -Tan[c + d*x]^2])/(5*d) - (15*b*Log[Cos[(c + d*x)/2]])/(8*d) + (15*b*Log[Sin[(c +
d*x)/2]])/(8*d) - (9*b*Sec[(c + d*x)/2]^2)/(32*d) + (b*Sec[(c + d*x)/2]^4)/(64*d)

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Maple [A]  time = 0.039, size = 159, normalized size = 1.3 \begin{align*} -{\frac{a \left ( \cot \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{a \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{a\cot \left ( dx+c \right ) }{d}}-ax-{\frac{ca}{d}}-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,b \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,b \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{8\,d}}+{\frac{5\,b \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{8\,d}}+{\frac{15\,b\cos \left ( dx+c \right ) }{8\,d}}+{\frac{15\,b\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+b*sin(d*x+c)),x)

[Out]

-1/5*a*cot(d*x+c)^5/d+1/3*a*cot(d*x+c)^3/d-a*cot(d*x+c)/d-a*x-1/d*c*a-1/4/d*b/sin(d*x+c)^4*cos(d*x+c)^7+3/8/d*
b/sin(d*x+c)^2*cos(d*x+c)^7+3/8/d*b*cos(d*x+c)^5+5/8/d*b*cos(d*x+c)^3+15/8*b*cos(d*x+c)/d+15/8/d*b*ln(csc(d*x+
c)-cot(d*x+c))

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Maxima [A]  time = 2.70003, size = 169, normalized size = 1.39 \begin{align*} -\frac{16 \,{\left (15 \, d x + 15 \, c + \frac{15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a + 15 \, b{\left (\frac{2 \,{\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/240*(16*(15*d*x + 15*c + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/tan(d*x + c)^5)*a + 15*b*(2*(9*cos(d*x
+ c)^3 - 7*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - 16*cos(d*x + c) + 15*log(cos(d*x + c) + 1)
- 15*log(cos(d*x + c) - 1)))/d

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Fricas [B]  time = 1.7059, size = 620, normalized size = 5.08 \begin{align*} -\frac{368 \, a \cos \left (d x + c\right )^{5} - 560 \, a \cos \left (d x + c\right )^{3} + 225 \,{\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 225 \,{\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 240 \, a \cos \left (d x + c\right ) + 30 \,{\left (8 \, a d x \cos \left (d x + c\right )^{4} - 8 \, b \cos \left (d x + c\right )^{5} - 16 \, a d x \cos \left (d x + c\right )^{2} + 25 \, b \cos \left (d x + c\right )^{3} + 8 \, a d x - 15 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/240*(368*a*cos(d*x + c)^5 - 560*a*cos(d*x + c)^3 + 225*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(1/2*
cos(d*x + c) + 1/2)*sin(d*x + c) - 225*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(-1/2*cos(d*x + c) + 1/2
)*sin(d*x + c) + 240*a*cos(d*x + c) + 30*(8*a*d*x*cos(d*x + c)^4 - 8*b*cos(d*x + c)^5 - 16*a*d*x*cos(d*x + c)^
2 + 25*b*cos(d*x + c)^3 + 8*a*d*x - 15*b*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 +
d)*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 2.22426, size = 269, normalized size = 2.2 \begin{align*} \frac{6 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 70 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 240 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 960 \,{\left (d x + c\right )} a + 1800 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 660 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{1920 \, b}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} - \frac{4110 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 660 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 240 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 70 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 15 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/960*(6*a*tan(1/2*d*x + 1/2*c)^5 + 15*b*tan(1/2*d*x + 1/2*c)^4 - 70*a*tan(1/2*d*x + 1/2*c)^3 - 240*b*tan(1/2*
d*x + 1/2*c)^2 - 960*(d*x + c)*a + 1800*b*log(abs(tan(1/2*d*x + 1/2*c))) + 660*a*tan(1/2*d*x + 1/2*c) + 1920*b
/(tan(1/2*d*x + 1/2*c)^2 + 1) - (4110*b*tan(1/2*d*x + 1/2*c)^5 + 660*a*tan(1/2*d*x + 1/2*c)^4 - 240*b*tan(1/2*
d*x + 1/2*c)^3 - 70*a*tan(1/2*d*x + 1/2*c)^2 + 15*b*tan(1/2*d*x + 1/2*c) + 6*a)/tan(1/2*d*x + 1/2*c)^5)/d