3.282 \(\int \cot ^2(c+d x) (a+b \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=48 \[ -\frac{a^2 \cot (c+d x)}{d}+a^2 (-x)-\frac{2 a b \csc (c+d x)}{d}-\frac{b^2 \cot (c+d x)}{d} \]

[Out]

-(a^2*x) - (a^2*Cot[c + d*x])/d - (b^2*Cot[c + d*x])/d - (2*a*b*Csc[c + d*x])/d

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Rubi [A]  time = 0.0746202, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3886, 3473, 8, 2606, 3767} \[ -\frac{a^2 \cot (c+d x)}{d}+a^2 (-x)-\frac{2 a b \csc (c+d x)}{d}-\frac{b^2 \cot (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*x) - (a^2*Cot[c + d*x])/d - (b^2*Cot[c + d*x])/d - (2*a*b*Csc[c + d*x])/d

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

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

Mathematica [A]  time = 0.366417, size = 39, normalized size = 0.81 \[ -\frac{\left (a^2+b^2\right ) \cot (c+d x)+a (a (c+d x)+2 b \csc (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 + b^2)*Cot[c + d*x] + a*(a*(c + d*x) + 2*b*Csc[c + d*x]))/d)

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Maple [A]  time = 0.038, size = 49, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -\cot \left ( dx+c \right ) -dx-c \right ) -2\,{\frac{ab}{\sin \left ( dx+c \right ) }}-{b}^{2}\cot \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(-cot(d*x+c)-d*x-c)-2*a*b/sin(d*x+c)-b^2*cot(d*x+c))

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Maxima [A]  time = 1.46353, size = 63, normalized size = 1.31 \begin{align*} -\frac{{\left (d x + c + \frac{1}{\tan \left (d x + c\right )}\right )} a^{2} + \frac{2 \, a b}{\sin \left (d x + c\right )} + \frac{b^{2}}{\tan \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((d*x + c + 1/tan(d*x + c))*a^2 + 2*a*b/sin(d*x + c) + b^2/tan(d*x + c))/d

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Fricas [A]  time = 0.852383, size = 104, normalized size = 2.17 \begin{align*} -\frac{a^{2} d x \sin \left (d x + c\right ) + 2 \, a b +{\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )}{d \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*d*x*sin(d*x + c) + 2*a*b + (a^2 + b^2)*cos(d*x + c))/(d*sin(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(c + d*x))**2*cot(c + d*x)**2, x)

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Giac [A]  time = 1.36982, size = 108, normalized size = 2.25 \begin{align*} -\frac{2 \,{\left (d x + c\right )} a^{2} - a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{a^{2} + 2 \, a b + b^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(d*x + c)*a^2 - a^2*tan(1/2*d*x + 1/2*c) + 2*a*b*tan(1/2*d*x + 1/2*c) - b^2*tan(1/2*d*x + 1/2*c) + (a^
2 + 2*a*b + b^2)/tan(1/2*d*x + 1/2*c))/d