3.77 \(\int \frac{\cot ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=123 \[ -\frac{1}{16 a^2 d (1-\cos (c+d x))}-\frac{23}{16 a^2 d (\cos (c+d x)+1)}+\frac{1}{2 a^2 d (\cos (c+d x)+1)^2}-\frac{1}{12 a^2 d (\cos (c+d x)+1)^3}-\frac{3 \log (1-\cos (c+d x))}{16 a^2 d}-\frac{13 \log (\cos (c+d x)+1)}{16 a^2 d} \]

[Out]

-1/(16*a^2*d*(1 - Cos[c + d*x])) - 1/(12*a^2*d*(1 + Cos[c + d*x])^3) + 1/(2*a^2*d*(1 + Cos[c + d*x])^2) - 23/(
16*a^2*d*(1 + Cos[c + d*x])) - (3*Log[1 - Cos[c + d*x]])/(16*a^2*d) - (13*Log[1 + Cos[c + d*x]])/(16*a^2*d)

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Rubi [A]  time = 0.086958, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3879, 88} \[ -\frac{1}{16 a^2 d (1-\cos (c+d x))}-\frac{23}{16 a^2 d (\cos (c+d x)+1)}+\frac{1}{2 a^2 d (\cos (c+d x)+1)^2}-\frac{1}{12 a^2 d (\cos (c+d x)+1)^3}-\frac{3 \log (1-\cos (c+d x))}{16 a^2 d}-\frac{13 \log (\cos (c+d x)+1)}{16 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3/(a + a*Sec[c + d*x])^2,x]

[Out]

-1/(16*a^2*d*(1 - Cos[c + d*x])) - 1/(12*a^2*d*(1 + Cos[c + d*x])^3) + 1/(2*a^2*d*(1 + Cos[c + d*x])^2) - 23/(
16*a^2*d*(1 + Cos[c + d*x])) - (3*Log[1 - Cos[c + d*x]])/(16*a^2*d) - (13*Log[1 + Cos[c + d*x]])/(16*a^2*d)

Rule 3879

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[((a - b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x^(m + n), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\cot ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=-\frac{a^4 \operatorname{Subst}\left (\int \frac{x^5}{(a-a x)^2 (a+a x)^4} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a^4 \operatorname{Subst}\left (\int \left (\frac{1}{16 a^6 (-1+x)^2}+\frac{3}{16 a^6 (-1+x)}-\frac{1}{4 a^6 (1+x)^4}+\frac{1}{a^6 (1+x)^3}-\frac{23}{16 a^6 (1+x)^2}+\frac{13}{16 a^6 (1+x)}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{1}{16 a^2 d (1-\cos (c+d x))}-\frac{1}{12 a^2 d (1+\cos (c+d x))^3}+\frac{1}{2 a^2 d (1+\cos (c+d x))^2}-\frac{23}{16 a^2 d (1+\cos (c+d x))}-\frac{3 \log (1-\cos (c+d x))}{16 a^2 d}-\frac{13 \log (1+\cos (c+d x))}{16 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.369981, size = 121, normalized size = 0.98 \[ -\frac{\cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (3 \csc ^2\left (\frac{1}{2} (c+d x)\right )+\sec ^6\left (\frac{1}{2} (c+d x)\right )-12 \sec ^4\left (\frac{1}{2} (c+d x)\right )+69 \sec ^2\left (\frac{1}{2} (c+d x)\right )+36 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+156 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{24 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3/(a + a*Sec[c + d*x])^2,x]

[Out]

-(Cos[(c + d*x)/2]^4*(3*Csc[(c + d*x)/2]^2 + 156*Log[Cos[(c + d*x)/2]] + 36*Log[Sin[(c + d*x)/2]] + 69*Sec[(c
+ d*x)/2]^2 - 12*Sec[(c + d*x)/2]^4 + Sec[(c + d*x)/2]^6)*Sec[c + d*x]^2)/(24*a^2*d*(1 + Sec[c + d*x])^2)

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Maple [A]  time = 0.077, size = 108, normalized size = 0.9 \begin{align*} -{\frac{1}{12\,d{a}^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{3}}}+{\frac{1}{2\,d{a}^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}}-{\frac{23}{16\,d{a}^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{13\,\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{16\,d{a}^{2}}}+{\frac{1}{16\,d{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) }}-{\frac{3\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{16\,d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3/(a+a*sec(d*x+c))^2,x)

[Out]

-1/12/d/a^2/(cos(d*x+c)+1)^3+1/2/d/a^2/(cos(d*x+c)+1)^2-23/16/d/a^2/(cos(d*x+c)+1)-13/16*ln(cos(d*x+c)+1)/a^2/
d+1/16/d/a^2/(-1+cos(d*x+c))-3/16/d/a^2*ln(-1+cos(d*x+c))

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Maxima [A]  time = 1.14885, size = 149, normalized size = 1.21 \begin{align*} -\frac{\frac{2 \,{\left (33 \, \cos \left (d x + c\right )^{3} + 18 \, \cos \left (d x + c\right )^{2} - 37 \, \cos \left (d x + c\right ) - 26\right )}}{a^{2} \cos \left (d x + c\right )^{4} + 2 \, a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right ) - a^{2}} + \frac{39 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{9 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/48*(2*(33*cos(d*x + c)^3 + 18*cos(d*x + c)^2 - 37*cos(d*x + c) - 26)/(a^2*cos(d*x + c)^4 + 2*a^2*cos(d*x +
c)^3 - 2*a^2*cos(d*x + c) - a^2) + 39*log(cos(d*x + c) + 1)/a^2 + 9*log(cos(d*x + c) - 1)/a^2)/d

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Fricas [A]  time = 1.21099, size = 444, normalized size = 3.61 \begin{align*} -\frac{66 \, \cos \left (d x + c\right )^{3} + 36 \, \cos \left (d x + c\right )^{2} + 39 \,{\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right ) - 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 9 \,{\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 74 \, \cos \left (d x + c\right ) - 52}{48 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/48*(66*cos(d*x + c)^3 + 36*cos(d*x + c)^2 + 39*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 - 2*cos(d*x + c) - 1)*log
(1/2*cos(d*x + c) + 1/2) + 9*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 - 2*cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) +
1/2) - 74*cos(d*x + c) - 52)/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c) - a^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cot ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(cot(c + d*x)**3/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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Giac [A]  time = 1.39026, size = 251, normalized size = 2.04 \begin{align*} \frac{\frac{3 \,{\left (\frac{6 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}{a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}} - \frac{18 \, \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}} + \frac{96 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}} + \frac{\frac{48 \, a^{4}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{9 \, a^{4}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{4}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{6}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/96*(3*(6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)*(cos(d*x + c) + 1)/(a^2*(cos(d*x + c) - 1)) - 18*log(abs
(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^2 + 96*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^2 +
(48*a^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 9*a^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + a^4*(cos(d*
x + c) - 1)^3/(cos(d*x + c) + 1)^3)/a^6)/d