3.25 \(\int \cot ^3(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=40 \[ -\frac{a^2}{d (1-\cos (c+d x))}-\frac{a^2 \log (1-\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.0496429, antiderivative size = 40, 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, 43} \[ -\frac{a^2}{d (1-\cos (c+d x))}-\frac{a^2 \log (1-\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2/(d*(1 - Cos[c + d*x]))) - (a^2*Log[1 - Cos[c + d*x]])/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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0662764, size = 56, normalized size = 1.4 \[ \frac{a^2 \csc ^2\left (\frac{1}{2} (c+d x)\right ) \left (-2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \cos (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-1\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Csc[(c + d*x)/2]^2*(-1 - 2*Log[Sin[(c + d*x)/2]] + 2*Cos[c + d*x]*Log[Sin[(c + d*x)/2]]))/(2*d)

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Maple [A]  time = 0.066, size = 51, normalized size = 1.3 \begin{align*} -{\frac{{a}^{2}}{d \left ( -1+\sec \left ( dx+c \right ) \right ) }}-{\frac{{a}^{2}\ln \left ( -1+\sec \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}\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))^2,x)

[Out]

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

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Maxima [A]  time = 1.12142, size = 46, normalized size = 1.15 \begin{align*} -\frac{a^{2} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac{a^{2}}{\cos \left (d x + c\right ) - 1}}{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]

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

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Fricas [A]  time = 1.19576, size = 113, normalized size = 2.82 \begin{align*} \frac{a^{2} -{\left (a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{d \cos \left (d x + c\right ) - 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="fricas")

[Out]

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

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

[Out]

a**2*(Integral(2*cot(c + d*x)**3*sec(c + d*x), x) + Integral(cot(c + d*x)**3*sec(c + d*x)**2, x) + Integral(co
t(c + d*x)**3, x))

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Giac [B]  time = 1.47374, size = 150, normalized size = 3.75 \begin{align*} -\frac{2 \, a^{2} \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 2 \, a^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac{{\left (a^{2} + \frac{2 \, a^{2}{\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}}{2 \, 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]

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