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

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Rubi [A]  time = 0.10, antiderivative size = 122, normalized size of antiderivative = 1.00, 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 8

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

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*}

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Mathematica [C]  time = 0.05, 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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fricas [B]  time = 0.50, size = 222, normalized size = 1.82 \[ -\frac {368 \, a \cos \left (d x + c\right )^{5} - 560 \, a \cos \left (d x + c\right )^{3} + 225 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 225 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 240 \, a \cos \left (d x + c\right ) + 30 \, {\left (8 \, a d x \cos \left (d x + c\right )^{4} - 8 \, b \cos \left (d x + c\right )^{5} - 16 \, a d x \cos \left (d x + c\right )^{2} + 25 \, b \cos \left (d x + c\right )^{3} + 8 \, a d x - 15 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

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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giac [A]  time = 0.68, size = 199, normalized size = 1.63 \[ \frac {6 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 70 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 240 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 960 \, {\left (d x + c\right )} a + 1800 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 660 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {1920 \, b}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} - \frac {4110 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 660 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 70 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

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

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maple [A]  time = 0.13, size = 159, normalized size = 1.30 \[ -\frac {a \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a \cot \left (d x +c \right )}{d}-a x -\frac {c a}{d}-\frac {b \left (\cos ^{7}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}+\frac {3 b \left (\cos ^{7}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}+\frac {3 b \left (\cos ^{5}\left (d x +c \right )\right )}{8 d}+\frac {5 b \left (\cos ^{3}\left (d x +c \right )\right )}{8 d}+\frac {15 b \cos \left (d x +c \right )}{8 d}+\frac {15 b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d} \]

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*b*cos(d*x+c)^5/d+5/8*b*cos(d*x+c)^3/d+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 = 1.89, size = 125, normalized size = 1.02 \[ -\frac {16 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a + 15 \, b {\left (\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

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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mupad [B]  time = 6.30, size = 288, normalized size = 2.36 \[ \frac {11\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {22\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-72\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {59\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {15\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {32\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {15\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}+\frac {2\,a\,\mathrm {atan}\left (\frac {4\,a^2}{4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+\frac {15\,b\,a}{2}}-\frac {15\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+\frac {15\,b\,a}{2}\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11*a*tan(c/2 + (d*x)/2))/(16*d) - (a/5 + (b*tan(c/2 + (d*x)/2))/2 - (32*a*tan(c/2 + (d*x)/2)^2)/15 + (59*a*ta
n(c/2 + (d*x)/2)^4)/3 + 22*a*tan(c/2 + (d*x)/2)^6 - (15*b*tan(c/2 + (d*x)/2)^3)/2 - 72*b*tan(c/2 + (d*x)/2)^5)
/(d*(32*tan(c/2 + (d*x)/2)^5 + 32*tan(c/2 + (d*x)/2)^7)) - (7*a*tan(c/2 + (d*x)/2)^3)/(96*d) + (a*tan(c/2 + (d
*x)/2)^5)/(160*d) - (b*tan(c/2 + (d*x)/2)^2)/(4*d) + (b*tan(c/2 + (d*x)/2)^4)/(64*d) + (15*b*log(tan(c/2 + (d*
x)/2)))/(8*d) + (2*a*atan((4*a^2)/((15*a*b)/2 + 4*a^2*tan(c/2 + (d*x)/2)) - (15*a*b*tan(c/2 + (d*x)/2))/(2*((1
5*a*b)/2 + 4*a^2*tan(c/2 + (d*x)/2)))))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right ) \cot ^{6}{\left (c + d x \right )}\, dx \]

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]

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

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