3.283 \(\int \cot ^4(c+d x) (a+b \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=85 \[ -\frac{a^2 \cot ^3(c+d x)}{3 d}+\frac{a^2 \cot (c+d x)}{d}+a^2 x-\frac{2 a b \csc ^3(c+d x)}{3 d}+\frac{2 a b \csc (c+d x)}{d}-\frac{b^2 \cot ^3(c+d x)}{3 d} \]

[Out]

a^2*x + (a^2*Cot[c + d*x])/d - (a^2*Cot[c + d*x]^3)/(3*d) - (b^2*Cot[c + d*x]^3)/(3*d) + (2*a*b*Csc[c + d*x])/
d - (2*a*b*Csc[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.115351, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3886, 3473, 8, 2606, 2607, 30} \[ -\frac{a^2 \cot ^3(c+d x)}{3 d}+\frac{a^2 \cot (c+d x)}{d}+a^2 x-\frac{2 a b \csc ^3(c+d x)}{3 d}+\frac{2 a b \csc (c+d x)}{d}-\frac{b^2 \cot ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + b*Sec[c + d*x])^2,x]

[Out]

a^2*x + (a^2*Cot[c + d*x])/d - (a^2*Cot[c + d*x]^3)/(3*d) - (b^2*Cot[c + d*x]^3)/(3*d) + (2*a*b*Csc[c + d*x])/
d - (2*a*b*Csc[c + d*x]^3)/(3*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 2607

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.460439, size = 122, normalized size = 1.44 \[ -\frac{\csc ^3(c+d x) \left (-9 a^2 c \sin (c+d x)-9 a^2 d x \sin (c+d x)+3 a^2 c \sin (3 (c+d x))+3 a^2 d x \sin (3 (c+d x))+4 a^2 \cos (3 (c+d x))+12 a b \cos (2 (c+d x))-4 a b+3 b^2 \cos (c+d x)+b^2 \cos (3 (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*(a + b*Sec[c + d*x])^2,x]

[Out]

-(Csc[c + d*x]^3*(-4*a*b + 3*b^2*Cos[c + d*x] + 12*a*b*Cos[2*(c + d*x)] + 4*a^2*Cos[3*(c + d*x)] + b^2*Cos[3*(
c + d*x)] - 9*a^2*c*Sin[c + d*x] - 9*a^2*d*x*Sin[c + d*x] + 3*a^2*c*Sin[3*(c + d*x)] + 3*a^2*d*x*Sin[3*(c + d*
x)]))/(12*d)

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Maple [A]  time = 0.047, size = 111, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{ \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3}}+\cot \left ( dx+c \right ) +dx+c \right ) +2\,ab \left ( -1/3\,{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+1/3\,{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{\sin \left ( dx+c \right ) }}+1/3\, \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+b*sec(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/3*cot(d*x+c)^3+cot(d*x+c)+d*x+c)+2*a*b*(-1/3/sin(d*x+c)^3*cos(d*x+c)^4+1/3/sin(d*x+c)*cos(d*x+c)^
4+1/3*(2+cos(d*x+c)^2)*sin(d*x+c))-1/3*b^2/sin(d*x+c)^3*cos(d*x+c)^3)

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Maxima [A]  time = 1.47951, size = 103, normalized size = 1.21 \begin{align*} \frac{{\left (3 \, d x + 3 \, c + \frac{3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a^{2} + \frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{2} - 1\right )} a b}{\sin \left (d x + c\right )^{3}} - \frac{b^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/3*((3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a^2 + 2*(3*sin(d*x + c)^2 - 1)*a*b/sin(d*x + c)^3 -
b^2/tan(d*x + c)^3)/d

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Fricas [A]  time = 0.831908, size = 240, normalized size = 2.82 \begin{align*} \frac{6 \, a b \cos \left (d x + c\right )^{2} +{\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \cos \left (d x + c\right ) - 4 \, a b + 3 \,{\left (a^{2} d x \cos \left (d x + c\right )^{2} - a^{2} d x\right )} \sin \left (d x + c\right )}{3 \,{\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(6*a*b*cos(d*x + c)^2 + (4*a^2 + b^2)*cos(d*x + c)^3 - 3*a^2*cos(d*x + c) - 4*a*b + 3*(a^2*d*x*cos(d*x + c
)^2 - a^2*d*x)*sin(d*x + c))/((d*cos(d*x + c)^2 - 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 ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(a+b*sec(d*x+c))**2,x)

[Out]

Integral((a + b*sec(c + d*x))**2*cot(c + d*x)**4, x)

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Giac [B]  time = 1.39209, size = 238, normalized size = 2.8 \begin{align*} \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \,{\left (d x + c\right )} a^{2} - 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 18 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{2} - 2 \, a b - b^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 - 2*a*b*tan(1/2*d*x + 1/2*c)^3 + b^2*tan(1/2*d*x + 1/2*c)^3 + 24*(d*x + c)*a^
2 - 15*a^2*tan(1/2*d*x + 1/2*c) + 18*a*b*tan(1/2*d*x + 1/2*c) - 3*b^2*tan(1/2*d*x + 1/2*c) + (15*a^2*tan(1/2*d
*x + 1/2*c)^2 + 18*a*b*tan(1/2*d*x + 1/2*c)^2 + 3*b^2*tan(1/2*d*x + 1/2*c)^2 - a^2 - 2*a*b - b^2)/tan(1/2*d*x
+ 1/2*c)^3)/d