3.7 \(\int \cot ^3(c+d x) (a+a \sec (c+d x)) \, dx\)

Optimal. Leaf size=57 \[ -\frac{a}{2 d (1-\cos (c+d x))}-\frac{3 a \log (1-\cos (c+d x))}{4 d}-\frac{a \log (\cos (c+d x)+1)}{4 d} \]

[Out]

-a/(2*d*(1 - Cos[c + d*x])) - (3*a*Log[1 - Cos[c + d*x]])/(4*d) - (a*Log[1 + Cos[c + d*x]])/(4*d)

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Rubi [A]  time = 0.0439983, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3879, 88} \[ -\frac{a}{2 d (1-\cos (c+d x))}-\frac{3 a \log (1-\cos (c+d x))}{4 d}-\frac{a \log (\cos (c+d x)+1)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.803195, size = 114, normalized size = 2. \[ -\frac{a \csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{a \sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}-\frac{a \left (\cot ^2(c+d x)+2 \log (\tan (c+d x))+2 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Csc[(c + d*x)/2]^2)/(8*d) + (a*Log[Cos[(c + d*x)/2]])/(2*d) - (a*Log[Sin[(c + d*x)/2]])/(2*d) - (a*(Cot[c
+ d*x]^2 + 2*Log[Cos[c + d*x]] + 2*Log[Tan[c + d*x]]))/(2*d) + (a*Sec[(c + d*x)/2]^2)/(8*d)

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Maple [A]  time = 0.075, size = 60, normalized size = 1.1 \begin{align*} -{\frac{a\ln \left ( 1+\sec \left ( dx+c \right ) \right ) }{4\,d}}-{\frac{a}{2\,d \left ( -1+\sec \left ( dx+c \right ) \right ) }}-{\frac{3\,a\ln \left ( -1+\sec \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{a\ln \left ( \sec \left ( dx+c \right ) \right ) }{d}} \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]

-1/4/d*a*ln(1+sec(d*x+c))-1/2/d*a/(-1+sec(d*x+c))-3/4/d*a*ln(-1+sec(d*x+c))+1/d*a*ln(sec(d*x+c))

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Maxima [A]  time = 1.18442, size = 57, normalized size = 1. \begin{align*} -\frac{a \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, a \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac{2 \, a}{\cos \left (d x + c\right ) - 1}}{4 \, 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]

-1/4*(a*log(cos(d*x + c) + 1) + 3*a*log(cos(d*x + c) - 1) - 2*a/(cos(d*x + c) - 1))/d

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Fricas [A]  time = 0.807956, size = 186, normalized size = 3.26 \begin{align*} -\frac{{\left (a \cos \left (d x + c\right ) - a\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 3 \,{\left (a \cos \left (d x + c\right ) - a\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \, a}{4 \,{\left (d \cos \left (d x + c\right ) - 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]

-1/4*((a*cos(d*x + c) - a)*log(1/2*cos(d*x + c) + 1/2) + 3*(a*cos(d*x + c) - a)*log(-1/2*cos(d*x + c) + 1/2) -
2*a)/(d*cos(d*x + c) - d)

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

[Out]

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

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Giac [B]  time = 1.48934, size = 139, normalized size = 2.44 \begin{align*} -\frac{3 \, a \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 4 \, a \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac{{\left (a + \frac{3 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1}}{4 \, 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]

-1/4*(3*a*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 4*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) +
1) + 1)) - (a + 3*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1))/d