3.576 \(\int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=127 \[ \frac {a \cos ^5(c+d x)}{5 d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos (c+d x)}{d}+\frac {a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 a \sin (c+d x) \cos (c+d x)}{16 d}-\frac {a \tanh ^{-1}(\cos (c+d x))}{d}+\frac {5 a x}{16} \]

[Out]

5/16*a*x-a*arctanh(cos(d*x+c))/d+a*cos(d*x+c)/d+1/3*a*cos(d*x+c)^3/d+1/5*a*cos(d*x+c)^5/d+5/16*a*cos(d*x+c)*si
n(d*x+c)/d+5/24*a*cos(d*x+c)^3*sin(d*x+c)/d+1/6*a*cos(d*x+c)^5*sin(d*x+c)/d

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Rubi [A]  time = 0.11, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2838, 2592, 302, 206, 2635, 8} \[ \frac {a \cos ^5(c+d x)}{5 d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos (c+d x)}{d}+\frac {a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 a \sin (c+d x) \cos (c+d x)}{16 d}-\frac {a \tanh ^{-1}(\cos (c+d x))}{d}+\frac {5 a x}{16} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*Cot[c + d*x]*(a + a*Sin[c + d*x]),x]

[Out]

(5*a*x)/16 - (a*ArcTanh[Cos[c + d*x]])/d + (a*Cos[c + d*x])/d + (a*Cos[c + d*x]^3)/(3*d) + (a*Cos[c + d*x]^5)/
(5*d) + (5*a*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*a*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) + (a*Cos[c + d*x]^5*
Sin[c + d*x])/(6*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 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]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rubi steps

\begin {align*} \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx &=a \int \cos ^6(c+d x) \, dx+a \int \cos ^5(c+d x) \cot (c+d x) \, dx\\ &=\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{6} (5 a) \int \cos ^4(c+d x) \, dx-\frac {a \operatorname {Subst}\left (\int \frac {x^6}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{8} (5 a) \int \cos ^2(c+d x) \, dx-\frac {a \operatorname {Subst}\left (\int \left (-1-x^2-x^4+\frac {1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}+\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{16} (5 a) \int 1 \, dx-\frac {a \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {5 a x}{16}-\frac {a \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}+\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 100, normalized size = 0.79 \[ \frac {a \left (225 \sin (2 (c+d x))+45 \sin (4 (c+d x))+5 \sin (6 (c+d x))+1320 \cos (c+d x)+140 \cos (3 (c+d x))+12 \cos (5 (c+d x))+960 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-960 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+300 c+300 d x\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*Cot[c + d*x]*(a + a*Sin[c + d*x]),x]

[Out]

(a*(300*c + 300*d*x + 1320*Cos[c + d*x] + 140*Cos[3*(c + d*x)] + 12*Cos[5*(c + d*x)] - 960*Log[Cos[(c + d*x)/2
]] + 960*Log[Sin[(c + d*x)/2]] + 225*Sin[2*(c + d*x)] + 45*Sin[4*(c + d*x)] + 5*Sin[6*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.80, size = 110, normalized size = 0.87 \[ \frac {48 \, a \cos \left (d x + c\right )^{5} + 80 \, a \cos \left (d x + c\right )^{3} + 75 \, a d x + 240 \, a \cos \left (d x + c\right ) - 120 \, a \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 120 \, a \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 5 \, {\left (8 \, a \cos \left (d x + c\right )^{5} + 10 \, a \cos \left (d x + c\right )^{3} + 15 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(48*a*cos(d*x + c)^5 + 80*a*cos(d*x + c)^3 + 75*a*d*x + 240*a*cos(d*x + c) - 120*a*log(1/2*cos(d*x + c)
+ 1/2) + 120*a*log(-1/2*cos(d*x + c) + 1/2) + 5*(8*a*cos(d*x + c)^5 + 10*a*cos(d*x + c)^3 + 15*a*cos(d*x + c))
*sin(d*x + c))/d

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giac [A]  time = 0.19, size = 201, normalized size = 1.58 \[ \frac {75 \, {\left (d x + c\right )} a + 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (165 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 720 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 25 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2160 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 450 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3680 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 450 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3360 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1488 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 165 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 368 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(75*(d*x + c)*a + 240*a*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(165*a*tan(1/2*d*x + 1/2*c)^11 - 720*a*tan(1/
2*d*x + 1/2*c)^10 - 25*a*tan(1/2*d*x + 1/2*c)^9 - 2160*a*tan(1/2*d*x + 1/2*c)^8 + 450*a*tan(1/2*d*x + 1/2*c)^7
- 3680*a*tan(1/2*d*x + 1/2*c)^6 - 450*a*tan(1/2*d*x + 1/2*c)^5 - 3360*a*tan(1/2*d*x + 1/2*c)^4 + 25*a*tan(1/2
*d*x + 1/2*c)^3 - 1488*a*tan(1/2*d*x + 1/2*c)^2 - 165*a*tan(1/2*d*x + 1/2*c) - 368*a)/(tan(1/2*d*x + 1/2*c)^2
+ 1)^6)/d

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maple [A]  time = 0.38, size = 131, normalized size = 1.03 \[ \frac {a \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{6 d}+\frac {5 a \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{24 d}+\frac {5 a \cos \left (d x +c \right ) \sin \left (d x +c \right )}{16 d}+\frac {5 a x}{16}+\frac {5 c a}{16 d}+\frac {a \left (\cos ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a \cos \left (d x +c \right )}{d}+\frac {a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)*(a+a*sin(d*x+c)),x)

[Out]

1/6*a*cos(d*x+c)^5*sin(d*x+c)/d+5/24*a*cos(d*x+c)^3*sin(d*x+c)/d+5/16*a*cos(d*x+c)*sin(d*x+c)/d+5/16*a*x+5/16/
d*c*a+1/5*a*cos(d*x+c)^5/d+1/3*a*cos(d*x+c)^3/d+a*cos(d*x+c)/d+1/d*a*ln(csc(d*x+c)-cot(d*x+c))

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maxima [A]  time = 0.39, size = 106, normalized size = 0.83 \[ \frac {32 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(32*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 30*cos(d*x + c) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x
+ c) - 1))*a - 5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a)/d

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mupad [B]  time = 10.71, size = 327, normalized size = 2.57 \[ \frac {-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+18\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {92\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+28\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {62\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}+\frac {11\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {46\,a}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {5\,a\,\mathrm {atan}\left (\frac {25\,a^2}{64\,\left (\frac {5\,a^2}{4}-\frac {25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}+\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {5\,a^2}{4}-\frac {25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((46*a)/15 + (11*a*tan(c/2 + (d*x)/2))/8 + (62*a*tan(c/2 + (d*x)/2)^2)/5 - (5*a*tan(c/2 + (d*x)/2)^3)/24 + 28*
a*tan(c/2 + (d*x)/2)^4 + (15*a*tan(c/2 + (d*x)/2)^5)/4 + (92*a*tan(c/2 + (d*x)/2)^6)/3 - (15*a*tan(c/2 + (d*x)
/2)^7)/4 + 18*a*tan(c/2 + (d*x)/2)^8 + (5*a*tan(c/2 + (d*x)/2)^9)/24 + 6*a*tan(c/2 + (d*x)/2)^10 - (11*a*tan(c
/2 + (d*x)/2)^11)/8)/(d*(6*tan(c/2 + (d*x)/2)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c
/2 + (d*x)/2)^8 + 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) + (a*log(tan(c/2 + (d*x)/2)))/d + (5*a
*atan((25*a^2)/(64*((5*a^2)/4 - (25*a^2*tan(c/2 + (d*x)/2))/64)) + (5*a^2*tan(c/2 + (d*x)/2))/(4*((5*a^2)/4 -
(25*a^2*tan(c/2 + (d*x)/2))/64))))/(8*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*csc(d*x+c)*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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