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

Optimal. Leaf size=103 \[ -\frac{1}{8 a d (1-\cos (c+d x))}-\frac{3}{4 a d (\cos (c+d x)+1)}+\frac{1}{8 a d (\cos (c+d x)+1)^2}-\frac{5 \log (1-\cos (c+d x))}{16 a d}-\frac{11 \log (\cos (c+d x)+1)}{16 a d} \]

[Out]

-1/(8*a*d*(1 - Cos[c + d*x])) + 1/(8*a*d*(1 + Cos[c + d*x])^2) - 3/(4*a*d*(1 + Cos[c + d*x])) - (5*Log[1 - Cos
[c + d*x]])/(16*a*d) - (11*Log[1 + Cos[c + d*x]])/(16*a*d)

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Rubi [A]  time = 0.0762475, antiderivative size = 103, 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} \[ -\frac{1}{8 a d (1-\cos (c+d x))}-\frac{3}{4 a d (\cos (c+d x)+1)}+\frac{1}{8 a d (\cos (c+d x)+1)^2}-\frac{5 \log (1-\cos (c+d x))}{16 a d}-\frac{11 \log (\cos (c+d x)+1)}{16 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(8*a*d*(1 - Cos[c + d*x])) + 1/(8*a*d*(1 + Cos[c + d*x])^2) - 3/(4*a*d*(1 + Cos[c + d*x])) - (5*Log[1 - Cos
[c + d*x]])/(16*a*d) - (11*Log[1 + Cos[c + d*x]])/(16*a*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)} \, dx &=-\frac{a^4 \operatorname{Subst}\left (\int \frac{x^4}{(a-a x)^2 (a+a x)^3} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a^4 \operatorname{Subst}\left (\int \left (\frac{1}{8 a^5 (-1+x)^2}+\frac{5}{16 a^5 (-1+x)}+\frac{1}{4 a^5 (1+x)^3}-\frac{3}{4 a^5 (1+x)^2}+\frac{11}{16 a^5 (1+x)}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{1}{8 a d (1-\cos (c+d x))}+\frac{1}{8 a d (1+\cos (c+d x))^2}-\frac{3}{4 a d (1+\cos (c+d x))}-\frac{5 \log (1-\cos (c+d x))}{16 a d}-\frac{11 \log (1+\cos (c+d x))}{16 a d}\\ \end{align*}

Mathematica [A]  time = 0.634093, size = 107, normalized size = 1.04 \[ -\frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (2 \csc ^2\left (\frac{1}{2} (c+d x)\right )-\sec ^4\left (\frac{1}{2} (c+d x)\right )+12 \sec ^2\left (\frac{1}{2} (c+d x)\right )+20 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+44 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{16 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[(c + d*x)/2]^2*(2*Csc[(c + d*x)/2]^2 + 44*Log[Cos[(c + d*x)/2]] + 20*Log[Sin[(c + d*x)/2]] + 12*Sec[(c +
d*x)/2]^2 - Sec[(c + d*x)/2]^4)*Sec[c + d*x])/(16*a*d*(1 + Sec[c + d*x]))

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Maple [A]  time = 0.067, size = 90, normalized size = 0.9 \begin{align*}{\frac{1}{8\,da \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}}-{\frac{3}{4\,da \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{11\,\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{16\,da}}+{\frac{1}{8\,da \left ( -1+\cos \left ( dx+c \right ) \right ) }}-{\frac{5\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{16\,da}} \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)),x)

[Out]

1/8/a/d/(cos(d*x+c)+1)^2-3/4/a/d/(cos(d*x+c)+1)-11/16*ln(cos(d*x+c)+1)/a/d+1/8/a/d/(-1+cos(d*x+c))-5/16/d/a*ln
(-1+cos(d*x+c))

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Maxima [A]  time = 1.12747, size = 123, normalized size = 1.19 \begin{align*} -\frac{\frac{2 \,{\left (5 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 6\right )}}{a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - a} + \frac{11 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac{5 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{16 \, 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)),x, algorithm="maxima")

[Out]

-1/16*(2*(5*cos(d*x + c)^2 - 3*cos(d*x + c) - 6)/(a*cos(d*x + c)^3 + a*cos(d*x + c)^2 - a*cos(d*x + c) - a) +
11*log(cos(d*x + c) + 1)/a + 5*log(cos(d*x + c) - 1)/a)/d

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Fricas [A]  time = 1.22542, size = 389, normalized size = 3.78 \begin{align*} -\frac{10 \, \cos \left (d x + c\right )^{2} + 11 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 5 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 6 \, \cos \left (d x + c\right ) - 12}{16 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a 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)),x, algorithm="fricas")

[Out]

-1/16*(10*cos(d*x + c)^2 + 11*(cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*log(1/2*cos(d*x + c) + 1/2)
+ 5*(cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) - 6*cos(d*x + c) - 12)/
(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*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{\left (c + d x \right )} + 1}\, dx}{a} \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)),x)

[Out]

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

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Giac [A]  time = 1.53013, size = 212, normalized size = 2.06 \begin{align*} \frac{\frac{2 \,{\left (\frac{5 \,{\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{\left (\cos \left (d x + c\right ) - 1\right )}} - \frac{10 \, \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} + \frac{32 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a} + \frac{\frac{10 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}}}{32 \, 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)),x, algorithm="giac")

[Out]

1/32*(2*(5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)*(cos(d*x + c) + 1)/(a*(cos(d*x + c) - 1)) - 10*log(abs(-
cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a + 32*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a + (10*a
*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/a^2)/d