3.35 \(\int \cot ^8(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=139 \[ -\frac {2 a^2 \cot ^7(c+d x)}{7 d}+\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x)}{d}-\frac {2 a^2 \csc ^7(c+d x)}{7 d}+\frac {6 a^2 \csc ^5(c+d x)}{5 d}-\frac {2 a^2 \csc ^3(c+d x)}{d}+\frac {2 a^2 \csc (c+d x)}{d}+a^2 x \]

[Out]

a^2*x+a^2*cot(d*x+c)/d-1/3*a^2*cot(d*x+c)^3/d+1/5*a^2*cot(d*x+c)^5/d-2/7*a^2*cot(d*x+c)^7/d+2*a^2*csc(d*x+c)/d
-2*a^2*csc(d*x+c)^3/d+6/5*a^2*csc(d*x+c)^5/d-2/7*a^2*csc(d*x+c)^7/d

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Rubi [A]  time = 0.14, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3886, 3473, 8, 2606, 194, 2607, 30} \[ -\frac {2 a^2 \cot ^7(c+d x)}{7 d}+\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x)}{d}-\frac {2 a^2 \csc ^7(c+d x)}{7 d}+\frac {6 a^2 \csc ^5(c+d x)}{5 d}-\frac {2 a^2 \csc ^3(c+d x)}{d}+\frac {2 a^2 \csc (c+d x)}{d}+a^2 x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^8*(a + a*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) + (a^2*Cot[c + d*x]^5)/(5*d) - (2*a^2*Cot[c + d*x]^7
)/(7*d) + (2*a^2*Csc[c + d*x])/d - (2*a^2*Csc[c + d*x]^3)/d + (6*a^2*Csc[c + d*x]^5)/(5*d) - (2*a^2*Csc[c + d*
x]^7)/(7*d)

Rule 8

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

Rule 30

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

Rubi steps

\begin {align*} \int \cot ^8(c+d x) (a+a \sec (c+d x))^2 \, dx &=\int \left (a^2 \cot ^8(c+d x)+2 a^2 \cot ^7(c+d x) \csc (c+d x)+a^2 \cot ^6(c+d x) \csc ^2(c+d x)\right ) \, dx\\ &=a^2 \int \cot ^8(c+d x) \, dx+a^2 \int \cot ^6(c+d x) \csc ^2(c+d x) \, dx+\left (2 a^2\right ) \int \cot ^7(c+d x) \csc (c+d x) \, dx\\ &=-\frac {a^2 \cot ^7(c+d x)}{7 d}-a^2 \int \cot ^6(c+d x) \, dx+\frac {a^2 \operatorname {Subst}\left (\int x^6 \, dx,x,-\cot (c+d x)\right )}{d}-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (-1+x^2\right )^3 \, dx,x,\csc (c+d x)\right )}{d}\\ &=\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {2 a^2 \cot ^7(c+d x)}{7 d}+a^2 \int \cot ^4(c+d x) \, dx-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (-1+3 x^2-3 x^4+x^6\right ) \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {2 a^2 \cot ^7(c+d x)}{7 d}+\frac {2 a^2 \csc (c+d x)}{d}-\frac {2 a^2 \csc ^3(c+d x)}{d}+\frac {6 a^2 \csc ^5(c+d x)}{5 d}-\frac {2 a^2 \csc ^7(c+d x)}{7 d}-a^2 \int \cot ^2(c+d x) \, dx\\ &=\frac {a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {2 a^2 \cot ^7(c+d x)}{7 d}+\frac {2 a^2 \csc (c+d x)}{d}-\frac {2 a^2 \csc ^3(c+d x)}{d}+\frac {6 a^2 \csc ^5(c+d x)}{5 d}-\frac {2 a^2 \csc ^7(c+d x)}{7 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 {a^2 \cot ^5(c+d x)}{5 d}-\frac {2 a^2 \cot ^7(c+d x)}{7 d}+\frac {2 a^2 \csc (c+d x)}{d}-\frac {2 a^2 \csc ^3(c+d x)}{d}+\frac {6 a^2 \csc ^5(c+d x)}{5 d}-\frac {2 a^2 \csc ^7(c+d x)}{7 d}\\ \end {align*}

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Mathematica [B]  time = 1.15, size = 312, normalized size = 2.24 \[ \frac {a^2 \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \csc ^7\left (\frac {1}{2} (c+d x)\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) (-16002 \sin (c+d x)+9144 \sin (2 (c+d x))+3429 \sin (3 (c+d x))-4572 \sin (4 (c+d x))+1143 \sin (5 (c+d x))-11760 \sin (2 c+d x)+8864 \sin (c+2 d x)+3360 \sin (3 c+2 d x)+2064 \sin (2 c+3 d x)+2520 \sin (4 c+3 d x)-4432 \sin (3 c+4 d x)-1680 \sin (5 c+4 d x)+1528 \sin (4 c+5 d x)-5880 d x \cos (2 c+d x)-3360 d x \cos (c+2 d x)+3360 d x \cos (3 c+2 d x)-1260 d x \cos (2 c+3 d x)+1260 d x \cos (4 c+3 d x)+1680 d x \cos (3 c+4 d x)-1680 d x \cos (5 c+4 d x)-420 d x \cos (4 c+5 d x)+420 d x \cos (6 c+5 d x)+4032 \sin (c)-9632 \sin (d x)+5880 d x \cos (d x))}{860160 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^8*(a + a*Sec[c + d*x])^2,x]

[Out]

(a^2*Csc[c/2]*Csc[(c + d*x)/2]^7*Sec[c/2]*Sec[(c + d*x)/2]^3*(5880*d*x*Cos[d*x] - 5880*d*x*Cos[2*c + d*x] - 33
60*d*x*Cos[c + 2*d*x] + 3360*d*x*Cos[3*c + 2*d*x] - 1260*d*x*Cos[2*c + 3*d*x] + 1260*d*x*Cos[4*c + 3*d*x] + 16
80*d*x*Cos[3*c + 4*d*x] - 1680*d*x*Cos[5*c + 4*d*x] - 420*d*x*Cos[4*c + 5*d*x] + 420*d*x*Cos[6*c + 5*d*x] + 40
32*Sin[c] - 9632*Sin[d*x] - 16002*Sin[c + d*x] + 9144*Sin[2*(c + d*x)] + 3429*Sin[3*(c + d*x)] - 4572*Sin[4*(c
+ d*x)] + 1143*Sin[5*(c + d*x)] - 11760*Sin[2*c + d*x] + 8864*Sin[c + 2*d*x] + 3360*Sin[3*c + 2*d*x] + 2064*S
in[2*c + 3*d*x] + 2520*Sin[4*c + 3*d*x] - 4432*Sin[3*c + 4*d*x] - 1680*Sin[5*c + 4*d*x] + 1528*Sin[4*c + 5*d*x
]))/(860160*d)

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fricas [A]  time = 0.59, size = 173, normalized size = 1.24 \[ \frac {191 \, a^{2} \cos \left (d x + c\right )^{5} - 172 \, a^{2} \cos \left (d x + c\right )^{4} - 253 \, a^{2} \cos \left (d x + c\right )^{3} + 258 \, a^{2} \cos \left (d x + c\right )^{2} + 87 \, a^{2} \cos \left (d x + c\right ) - 96 \, a^{2} + 105 \, {\left (a^{2} d x \cos \left (d x + c\right )^{4} - 2 \, a^{2} d x \cos \left (d x + c\right )^{3} + 2 \, a^{2} d x \cos \left (d x + c\right ) - a^{2} d x\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^8*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/105*(191*a^2*cos(d*x + c)^5 - 172*a^2*cos(d*x + c)^4 - 253*a^2*cos(d*x + c)^3 + 258*a^2*cos(d*x + c)^2 + 87*
a^2*cos(d*x + c) - 96*a^2 + 105*(a^2*d*x*cos(d*x + c)^4 - 2*a^2*d*x*cos(d*x + c)^3 + 2*a^2*d*x*cos(d*x + c) -
a^2*d*x)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^3 + 2*d*cos(d*x + c) - d)*sin(d*x + c))

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giac [A]  time = 0.39, size = 112, normalized size = 0.81 \[ \frac {35 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3360 \, {\left (d x + c\right )} a^{2} - 735 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {4410 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 770 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 147 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}}}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^8*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/3360*(35*a^2*tan(1/2*d*x + 1/2*c)^3 + 3360*(d*x + c)*a^2 - 735*a^2*tan(1/2*d*x + 1/2*c) + (4410*a^2*tan(1/2*
d*x + 1/2*c)^6 - 770*a^2*tan(1/2*d*x + 1/2*c)^4 + 147*a^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2)/tan(1/2*d*x + 1/2*c
)^7)/d

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maple [A]  time = 1.10, size = 188, normalized size = 1.35 \[ \frac {a^{2} \left (-\frac {\left (\cot ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+2 a^{2} \left (-\frac {\cos ^{8}\left (d x +c \right )}{7 \sin \left (d x +c \right )^{7}}+\frac {\cos ^{8}\left (d x +c \right )}{35 \sin \left (d x +c \right )^{5}}-\frac {\cos ^{8}\left (d x +c \right )}{35 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{8}\left (d x +c \right )}{7 \sin \left (d x +c \right )}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}\right )-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{7 \sin \left (d x +c \right )^{7}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^8*(a+a*sec(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/7*cot(d*x+c)^7+1/5*cot(d*x+c)^5-1/3*cot(d*x+c)^3+cot(d*x+c)+d*x+c)+2*a^2*(-1/7/sin(d*x+c)^7*cos(d
*x+c)^8+1/35/sin(d*x+c)^5*cos(d*x+c)^8-1/35/sin(d*x+c)^3*cos(d*x+c)^8+1/7/sin(d*x+c)*cos(d*x+c)^8+1/7*(16/5+co
s(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))-1/7*a^2/sin(d*x+c)^7*cos(d*x+c)^7)

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maxima [A]  time = 0.83, size = 117, normalized size = 0.84 \[ \frac {{\left (105 \, d x + 105 \, c + \frac {105 \, \tan \left (d x + c\right )^{6} - 35 \, \tan \left (d x + c\right )^{4} + 21 \, \tan \left (d x + c\right )^{2} - 15}{\tan \left (d x + c\right )^{7}}\right )} a^{2} + \frac {6 \, {\left (35 \, \sin \left (d x + c\right )^{6} - 35 \, \sin \left (d x + c\right )^{4} + 21 \, \sin \left (d x + c\right )^{2} - 5\right )} a^{2}}{\sin \left (d x + c\right )^{7}} - \frac {15 \, a^{2}}{\tan \left (d x + c\right )^{7}}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^8*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/105*((105*d*x + 105*c + (105*tan(d*x + c)^6 - 35*tan(d*x + c)^4 + 21*tan(d*x + c)^2 - 15)/tan(d*x + c)^7)*a^
2 + 6*(35*sin(d*x + c)^6 - 35*sin(d*x + c)^4 + 21*sin(d*x + c)^2 - 5)*a^2/sin(d*x + c)^7 - 15*a^2/tan(d*x + c)
^7)/d

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mupad [B]  time = 1.75, size = 182, normalized size = 1.31 \[ \frac {a^2\,\left (35\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-735\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4410\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-770\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+147\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3360\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (c+d\,x\right )\right )}{3360\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^8*(a + a/cos(c + d*x))^2,x)

[Out]

(a^2*(35*sin(c/2 + (d*x)/2)^10 - 15*cos(c/2 + (d*x)/2)^10 - 735*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 + 44
10*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 - 770*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 + 147*cos(c/2 + (
d*x)/2)^8*sin(c/2 + (d*x)/2)^2 + 3360*cos(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^7*(c + d*x)))/(3360*d*cos(c/2 +
(d*x)/2)^3*sin(c/2 + (d*x)/2)^7)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**8*(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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