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

Optimal. Leaf size=143 \[ -\frac{1}{32 a^3 d (1-\cos (c+d x))}-\frac{9}{4 a^3 d (\cos (c+d x)+1)}+\frac{39}{32 a^3 d (\cos (c+d x)+1)^2}-\frac{5}{12 a^3 d (\cos (c+d x)+1)^3}+\frac{1}{16 a^3 d (\cos (c+d x)+1)^4}-\frac{7 \log (1-\cos (c+d x))}{64 a^3 d}-\frac{57 \log (\cos (c+d x)+1)}{64 a^3 d} \]

[Out]

-1/(32*a^3*d*(1 - Cos[c + d*x])) + 1/(16*a^3*d*(1 + Cos[c + d*x])^4) - 5/(12*a^3*d*(1 + Cos[c + d*x])^3) + 39/
(32*a^3*d*(1 + Cos[c + d*x])^2) - 9/(4*a^3*d*(1 + Cos[c + d*x])) - (7*Log[1 - Cos[c + d*x]])/(64*a^3*d) - (57*
Log[1 + Cos[c + d*x]])/(64*a^3*d)

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Rubi [A]  time = 0.0966191, antiderivative size = 143, 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} \[ -\frac{1}{32 a^3 d (1-\cos (c+d x))}-\frac{9}{4 a^3 d (\cos (c+d x)+1)}+\frac{39}{32 a^3 d (\cos (c+d x)+1)^2}-\frac{5}{12 a^3 d (\cos (c+d x)+1)^3}+\frac{1}{16 a^3 d (\cos (c+d x)+1)^4}-\frac{7 \log (1-\cos (c+d x))}{64 a^3 d}-\frac{57 \log (\cos (c+d x)+1)}{64 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(32*a^3*d*(1 - Cos[c + d*x])) + 1/(16*a^3*d*(1 + Cos[c + d*x])^4) - 5/(12*a^3*d*(1 + Cos[c + d*x])^3) + 39/
(32*a^3*d*(1 + Cos[c + d*x])^2) - 9/(4*a^3*d*(1 + Cos[c + d*x])) - (7*Log[1 - Cos[c + d*x]])/(64*a^3*d) - (57*
Log[1 + Cos[c + d*x]])/(64*a^3*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))^3} \, dx &=-\frac{a^4 \operatorname{Subst}\left (\int \frac{x^6}{(a-a x)^2 (a+a x)^5} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a^4 \operatorname{Subst}\left (\int \left (\frac{1}{32 a^7 (-1+x)^2}+\frac{7}{64 a^7 (-1+x)}+\frac{1}{4 a^7 (1+x)^5}-\frac{5}{4 a^7 (1+x)^4}+\frac{39}{16 a^7 (1+x)^3}-\frac{9}{4 a^7 (1+x)^2}+\frac{57}{64 a^7 (1+x)}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{1}{32 a^3 d (1-\cos (c+d x))}+\frac{1}{16 a^3 d (1+\cos (c+d x))^4}-\frac{5}{12 a^3 d (1+\cos (c+d x))^3}+\frac{39}{32 a^3 d (1+\cos (c+d x))^2}-\frac{9}{4 a^3 d (1+\cos (c+d x))}-\frac{7 \log (1-\cos (c+d x))}{64 a^3 d}-\frac{57 \log (1+\cos (c+d x))}{64 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.63356, size = 140, normalized size = 0.98 \[ -\frac{\cos ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (12 \csc ^2\left (\frac{1}{2} (c+d x)\right )-3 \sec ^8\left (\frac{1}{2} (c+d x)\right )+40 \sec ^6\left (\frac{1}{2} (c+d x)\right )-234 \sec ^4\left (\frac{1}{2} (c+d x)\right )+864 \sec ^2\left (\frac{1}{2} (c+d x)\right )+24 \left (7 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+57 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{96 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[(c + d*x)/2]^6*(12*Csc[(c + d*x)/2]^2 + 24*(57*Log[Cos[(c + d*x)/2]] + 7*Log[Sin[(c + d*x)/2]]) + 864*Se
c[(c + d*x)/2]^2 - 234*Sec[(c + d*x)/2]^4 + 40*Sec[(c + d*x)/2]^6 - 3*Sec[(c + d*x)/2]^8)*Sec[c + d*x]^3)/(96*
a^3*d*(1 + Sec[c + d*x])^3)

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Maple [A]  time = 0.088, size = 126, normalized size = 0.9 \begin{align*}{\frac{1}{16\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{4}}}-{\frac{5}{12\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{3}}}+{\frac{39}{32\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}}-{\frac{9}{4\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{57\,\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{64\,d{a}^{3}}}+{\frac{1}{32\,d{a}^{3} \left ( -1+\cos \left ( dx+c \right ) \right ) }}-{\frac{7\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{64\,d{a}^{3}}} \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))^3,x)

[Out]

1/16/d/a^3/(cos(d*x+c)+1)^4-5/12/d/a^3/(cos(d*x+c)+1)^3+39/32/d/a^3/(cos(d*x+c)+1)^2-9/4/d/a^3/(cos(d*x+c)+1)-
57/64*ln(cos(d*x+c)+1)/a^3/d+1/32/d/a^3/(-1+cos(d*x+c))-7/64/d/a^3*ln(-1+cos(d*x+c))

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Maxima [A]  time = 1.18638, size = 197, normalized size = 1.38 \begin{align*} -\frac{\frac{2 \,{\left (213 \, \cos \left (d x + c\right )^{4} + 303 \, \cos \left (d x + c\right )^{3} - 95 \, \cos \left (d x + c\right )^{2} - 333 \, \cos \left (d x + c\right ) - 136\right )}}{a^{3} \cos \left (d x + c\right )^{5} + 3 \, a^{3} \cos \left (d x + c\right )^{4} + 2 \, a^{3} \cos \left (d x + c\right )^{3} - 2 \, a^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{3} \cos \left (d x + c\right ) - a^{3}} + \frac{171 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} + \frac{21 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3}}}{192 \, 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))^3,x, algorithm="maxima")

[Out]

-1/192*(2*(213*cos(d*x + c)^4 + 303*cos(d*x + c)^3 - 95*cos(d*x + c)^2 - 333*cos(d*x + c) - 136)/(a^3*cos(d*x
+ c)^5 + 3*a^3*cos(d*x + c)^4 + 2*a^3*cos(d*x + c)^3 - 2*a^3*cos(d*x + c)^2 - 3*a^3*cos(d*x + c) - a^3) + 171*
log(cos(d*x + c) + 1)/a^3 + 21*log(cos(d*x + c) - 1)/a^3)/d

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Fricas [A]  time = 1.20036, size = 652, normalized size = 4.56 \begin{align*} -\frac{426 \, \cos \left (d x + c\right )^{4} + 606 \, \cos \left (d x + c\right )^{3} - 190 \, \cos \left (d x + c\right )^{2} + 171 \,{\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 21 \,{\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 666 \, \cos \left (d x + c\right ) - 272}{192 \,{\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} - 3 \, a^{3} d \cos \left (d x + c\right ) - a^{3} 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))^3,x, algorithm="fricas")

[Out]

-1/192*(426*cos(d*x + c)^4 + 606*cos(d*x + c)^3 - 190*cos(d*x + c)^2 + 171*(cos(d*x + c)^5 + 3*cos(d*x + c)^4
+ 2*cos(d*x + c)^3 - 2*cos(d*x + c)^2 - 3*cos(d*x + c) - 1)*log(1/2*cos(d*x + c) + 1/2) + 21*(cos(d*x + c)^5 +
3*cos(d*x + c)^4 + 2*cos(d*x + c)^3 - 2*cos(d*x + c)^2 - 3*cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) - 6
66*cos(d*x + c) - 272)/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 2*a^3*d*cos(d*x + c)^3 - 2*a^3*d*cos(d
*x + c)^2 - 3*a^3*d*cos(d*x + c) - a^3*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 ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \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))**3,x)

[Out]

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

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Giac [A]  time = 1.42353, size = 286, normalized size = 2. \begin{align*} \frac{\frac{12 \,{\left (\frac{7 \,{\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^{3}{\left (\cos \left (d x + c\right ) - 1\right )}} - \frac{84 \, \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3}} + \frac{768 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{3}} + \frac{\frac{504 \, a^{9}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{132 \, a^{9}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{28 \, a^{9}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, a^{9}{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{12}}}{768 \, 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))^3,x, algorithm="giac")

[Out]

1/768*(12*(7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)*(cos(d*x + c) + 1)/(a^3*(cos(d*x + c) - 1)) - 84*log(a
bs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^3 + 768*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^
3 + (504*a^9*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 132*a^9*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 28*a^
9*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 3*a^9*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4)/a^12)/d