3.52 \(\int \cot ^6(c+d x) (a+a \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=107 \[ -\frac{4 a^3 \cot ^5(c+d x)}{5 d}+\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot (c+d x)}{d}-\frac{4 a^3 \csc ^5(c+d x)}{5 d}+\frac{7 a^3 \csc ^3(c+d x)}{3 d}-\frac{3 a^3 \csc (c+d x)}{d}-a^3 x \]

[Out]

-(a^3*x) - (a^3*Cot[c + d*x])/d + (a^3*Cot[c + d*x]^3)/(3*d) - (4*a^3*Cot[c + d*x]^5)/(5*d) - (3*a^3*Csc[c + d
*x])/d + (7*a^3*Csc[c + d*x]^3)/(3*d) - (4*a^3*Csc[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.164053, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3886, 3473, 8, 2606, 194, 2607, 30, 14} \[ -\frac{4 a^3 \cot ^5(c+d x)}{5 d}+\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot (c+d x)}{d}-\frac{4 a^3 \csc ^5(c+d x)}{5 d}+\frac{7 a^3 \csc ^3(c+d x)}{3 d}-\frac{3 a^3 \csc (c+d x)}{d}-a^3 x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + a*Sec[c + d*x])^3,x]

[Out]

-(a^3*x) - (a^3*Cot[c + d*x])/d + (a^3*Cot[c + d*x]^3)/(3*d) - (4*a^3*Cot[c + d*x]^5)/(5*d) - (3*a^3*Csc[c + d
*x])/d + (7*a^3*Csc[c + d*x]^3)/(3*d) - (4*a^3*Csc[c + d*x]^5)/(5*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 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \cot ^6(c+d x) (a+a \sec (c+d x))^3 \, dx &=\int \left (a^3 \cot ^6(c+d x)+3 a^3 \cot ^5(c+d x) \csc (c+d x)+3 a^3 \cot ^4(c+d x) \csc ^2(c+d x)+a^3 \cot ^3(c+d x) \csc ^3(c+d x)\right ) \, dx\\ &=a^3 \int \cot ^6(c+d x) \, dx+a^3 \int \cot ^3(c+d x) \csc ^3(c+d x) \, dx+\left (3 a^3\right ) \int \cot ^5(c+d x) \csc (c+d x) \, dx+\left (3 a^3\right ) \int \cot ^4(c+d x) \csc ^2(c+d x) \, dx\\ &=-\frac{a^3 \cot ^5(c+d x)}{5 d}-a^3 \int \cot ^4(c+d x) \, dx-\frac{a^3 \operatorname{Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\csc (c+d x)\right )}{d}+\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,-\cot (c+d x)\right )}{d}-\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \left (-1+x^2\right )^2 \, dx,x,\csc (c+d x)\right )}{d}\\ &=\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{4 a^3 \cot ^5(c+d x)}{5 d}+a^3 \int \cot ^2(c+d x) \, dx-\frac{a^3 \operatorname{Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\csc (c+d x)\right )}{d}-\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac{a^3 \cot (c+d x)}{d}+\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{4 a^3 \cot ^5(c+d x)}{5 d}-\frac{3 a^3 \csc (c+d x)}{d}+\frac{7 a^3 \csc ^3(c+d x)}{3 d}-\frac{4 a^3 \csc ^5(c+d x)}{5 d}-a^3 \int 1 \, dx\\ &=-a^3 x-\frac{a^3 \cot (c+d x)}{d}+\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{4 a^3 \cot ^5(c+d x)}{5 d}-\frac{3 a^3 \csc (c+d x)}{d}+\frac{7 a^3 \csc ^3(c+d x)}{3 d}-\frac{4 a^3 \csc ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.662737, size = 112, normalized size = 1.05 \[ -\frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (\cot \left (\frac{c}{2}\right ) (13 \cos (c+d x)-10) \csc ^4\left (\frac{1}{2} (c+d x)\right )+\csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) (51 \cos (c+d x)-16 \cos (2 (c+d x))-38) \csc ^5\left (\frac{1}{2} (c+d x)\right )+60 d x\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + a*Sec[c + d*x])^3,x]

[Out]

-(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(60*d*x + (-10 + 13*Cos[c + d*x])*Cot[c/2]*Csc[(c + d*x)/2]^4 +
(-38 + 51*Cos[c + d*x] - 16*Cos[2*(c + d*x)])*Csc[c/2]*Csc[(c + d*x)/2]^5*Sin[(d*x)/2]))/(480*d)

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Maple [B]  time = 0.075, size = 232, normalized size = 2.2 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( -{\frac{ \left ( \cot \left ( dx+c \right ) \right ) ^{5}}{5}}+{\frac{ \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3}}-\cot \left ( dx+c \right ) -dx-c \right ) +3\,{a}^{3} \left ( -1/5\,{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}+1/15\,{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-1/5\,{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{\sin \left ( dx+c \right ) }}-1/5\, \left ( 8/3+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{\frac{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}+{a}^{3} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{15\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{15\,\sin \left ( dx+c \right ) }}+{\frac{ \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{15}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+a*sec(d*x+c))^3,x)

[Out]

1/d*(a^3*(-1/5*cot(d*x+c)^5+1/3*cot(d*x+c)^3-cot(d*x+c)-d*x-c)+3*a^3*(-1/5/sin(d*x+c)^5*cos(d*x+c)^6+1/15/sin(
d*x+c)^3*cos(d*x+c)^6-1/5/sin(d*x+c)*cos(d*x+c)^6-1/5*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-3/5*a^3/
sin(d*x+c)^5*cos(d*x+c)^5+a^3*(-1/5/sin(d*x+c)^5*cos(d*x+c)^4-1/15/sin(d*x+c)^3*cos(d*x+c)^4+1/15/sin(d*x+c)*c
os(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*sin(d*x+c)))

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Maxima [A]  time = 1.76012, size = 165, normalized size = 1.54 \begin{align*} -\frac{{\left (15 \, d x + 15 \, c + \frac{15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a^{3} + \frac{3 \,{\left (15 \, \sin \left (d x + c\right )^{4} - 10 \, \sin \left (d x + c\right )^{2} + 3\right )} a^{3}}{\sin \left (d x + c\right )^{5}} - \frac{{\left (5 \, \sin \left (d x + c\right )^{2} - 3\right )} a^{3}}{\sin \left (d x + c\right )^{5}} + \frac{9 \, a^{3}}{\tan \left (d x + c\right )^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/15*((15*d*x + 15*c + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/tan(d*x + c)^5)*a^3 + 3*(15*sin(d*x + c)^4
- 10*sin(d*x + c)^2 + 3)*a^3/sin(d*x + c)^5 - (5*sin(d*x + c)^2 - 3)*a^3/sin(d*x + c)^5 + 9*a^3/tan(d*x + c)^5
)/d

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Fricas [A]  time = 1.1057, size = 298, normalized size = 2.79 \begin{align*} -\frac{32 \, a^{3} \cos \left (d x + c\right )^{3} - 19 \, a^{3} \cos \left (d x + c\right )^{2} - 29 \, a^{3} \cos \left (d x + c\right ) + 22 \, a^{3} + 15 \,{\left (a^{3} d x \cos \left (d x + c\right )^{2} - 2 \, a^{3} d x \cos \left (d x + c\right ) + a^{3} d x\right )} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) + 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)^6*(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/15*(32*a^3*cos(d*x + c)^3 - 19*a^3*cos(d*x + c)^2 - 29*a^3*cos(d*x + c) + 22*a^3 + 15*(a^3*d*x*cos(d*x + c)
^2 - 2*a^3*d*x*cos(d*x + c) + a^3*d*x)*sin(d*x + c))/((d*cos(d*x + c)^2 - 2*d*cos(d*x + c) + d)*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6*(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.43085, size = 89, normalized size = 0.83 \begin{align*} -\frac{60 \,{\left (d x + c\right )} a^{3} + \frac{105 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 20 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/60*(60*(d*x + c)*a^3 + (105*a^3*tan(1/2*d*x + 1/2*c)^4 - 20*a^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^3)/tan(1/2*d*x
+ 1/2*c)^5)/d