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

Optimal. Leaf size=153 \[ -\frac {a^2 \cot ^7(c+d x)}{7 d}+\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x)}{d}+a^2 x-\frac {2 a b \csc ^7(c+d x)}{7 d}+\frac {6 a b \csc ^5(c+d x)}{5 d}-\frac {2 a b \csc ^3(c+d x)}{d}+\frac {2 a b \csc (c+d x)}{d}-\frac {b^2 \cot ^7(c+d x)}{7 d} \]

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

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Rubi [A]  time = 0.15, antiderivative size = 153, 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} \[ -\frac {a^2 \cot ^7(c+d x)}{7 d}+\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x)}{d}+a^2 x-\frac {2 a b \csc ^7(c+d x)}{7 d}+\frac {6 a b \csc ^5(c+d x)}{5 d}-\frac {2 a b \csc ^3(c+d x)}{d}+\frac {2 a b \csc (c+d x)}{d}-\frac {b^2 \cot ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

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Mathematica [A]  time = 0.84, size = 257, normalized size = 1.68 \[ -\frac {\csc ^7(c+d x) \left (-3675 a^2 c \sin (c+d x)-3675 a^2 d x \sin (c+d x)+2205 a^2 c \sin (3 (c+d x))+2205 a^2 d x \sin (3 (c+d x))-735 a^2 c \sin (5 (c+d x))-735 a^2 d x \sin (5 (c+d x))+105 a^2 c \sin (7 (c+d x))+105 a^2 d x \sin (7 (c+d x))+1176 a^2 \cos (3 (c+d x))-392 a^2 \cos (5 (c+d x))+176 a^2 \cos (7 (c+d x))+3612 a b \cos (2 (c+d x))-840 a b \cos (4 (c+d x))+420 a b \cos (6 (c+d x))-1272 a b+525 b^2 \cos (c+d x)+315 b^2 \cos (3 (c+d x))+105 b^2 \cos (5 (c+d x))+15 b^2 \cos (7 (c+d x))\right )}{6720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6720*(Csc[c + d*x]^7*(-1272*a*b + 525*b^2*Cos[c + d*x] + 3612*a*b*Cos[2*(c + d*x)] + 1176*a^2*Cos[3*(c + d*
x)] + 315*b^2*Cos[3*(c + d*x)] - 840*a*b*Cos[4*(c + d*x)] - 392*a^2*Cos[5*(c + d*x)] + 105*b^2*Cos[5*(c + d*x)
] + 420*a*b*Cos[6*(c + d*x)] + 176*a^2*Cos[7*(c + d*x)] + 15*b^2*Cos[7*(c + d*x)] - 3675*a^2*c*Sin[c + d*x] -
3675*a^2*d*x*Sin[c + d*x] + 2205*a^2*c*Sin[3*(c + d*x)] + 2205*a^2*d*x*Sin[3*(c + d*x)] - 735*a^2*c*Sin[5*(c +
d*x)] - 735*a^2*d*x*Sin[5*(c + d*x)] + 105*a^2*c*Sin[7*(c + d*x)] + 105*a^2*d*x*Sin[7*(c + d*x)]))/d

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fricas [A]  time = 0.50, size = 206, normalized size = 1.35 \[ \frac {210 \, a b \cos \left (d x + c\right )^{6} + {\left (176 \, a^{2} + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{7} - 406 \, a^{2} \cos \left (d x + c\right )^{5} - 420 \, a b \cos \left (d x + c\right )^{4} + 350 \, a^{2} \cos \left (d x + c\right )^{3} + 336 \, a b \cos \left (d x + c\right )^{2} - 105 \, a^{2} \cos \left (d x + c\right ) - 96 \, a b + 105 \, {\left (a^{2} d x \cos \left (d x + c\right )^{6} - 3 \, a^{2} d x \cos \left (d x + c\right )^{4} + 3 \, a^{2} d x \cos \left (d x + c\right )^{2} - a^{2} d x\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(210*a*b*cos(d*x + c)^6 + (176*a^2 + 15*b^2)*cos(d*x + c)^7 - 406*a^2*cos(d*x + c)^5 - 420*a*b*cos(d*x +
c)^4 + 350*a^2*cos(d*x + c)^3 + 336*a*b*cos(d*x + c)^2 - 105*a^2*cos(d*x + c) - 96*a*b + 105*(a^2*d*x*cos(d*x
+ c)^6 - 3*a^2*d*x*cos(d*x + c)^4 + 3*a^2*d*x*cos(d*x + c)^2 - a^2*d*x)*sin(d*x + c))/((d*cos(d*x + c)^6 - 3*
d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)*sin(d*x + c))

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giac [B]  time = 0.43, size = 366, normalized size = 2.39 \[ \frac {15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 189 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 294 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1295 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1470 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 315 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 13440 \, {\left (d x + c\right )} a^{2} - 9765 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7350 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 525 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {9765 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 7350 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 525 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 1295 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1470 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 315 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 189 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 294 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 105 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} - 30 \, a b - 15 \, b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}}}{13440 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13440*(15*a^2*tan(1/2*d*x + 1/2*c)^7 - 30*a*b*tan(1/2*d*x + 1/2*c)^7 + 15*b^2*tan(1/2*d*x + 1/2*c)^7 - 189*a
^2*tan(1/2*d*x + 1/2*c)^5 + 294*a*b*tan(1/2*d*x + 1/2*c)^5 - 105*b^2*tan(1/2*d*x + 1/2*c)^5 + 1295*a^2*tan(1/2
*d*x + 1/2*c)^3 - 1470*a*b*tan(1/2*d*x + 1/2*c)^3 + 315*b^2*tan(1/2*d*x + 1/2*c)^3 + 13440*(d*x + c)*a^2 - 976
5*a^2*tan(1/2*d*x + 1/2*c) + 7350*a*b*tan(1/2*d*x + 1/2*c) - 525*b^2*tan(1/2*d*x + 1/2*c) + (9765*a^2*tan(1/2*
d*x + 1/2*c)^6 + 7350*a*b*tan(1/2*d*x + 1/2*c)^6 + 525*b^2*tan(1/2*d*x + 1/2*c)^6 - 1295*a^2*tan(1/2*d*x + 1/2
*c)^4 - 1470*a*b*tan(1/2*d*x + 1/2*c)^4 - 315*b^2*tan(1/2*d*x + 1/2*c)^4 + 189*a^2*tan(1/2*d*x + 1/2*c)^2 + 29
4*a*b*tan(1/2*d*x + 1/2*c)^2 + 105*b^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2 - 30*a*b - 15*b^2)/tan(1/2*d*x + 1/2*c)
^7)/d

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maple [A]  time = 0.95, size = 187, normalized size = 1.22 \[ \frac {a^{2} \left (-\frac {\left (\cot ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+2 a b \left (-\frac {\cos ^{8}\left (d x +c \right )}{7 \sin \left (d x +c \right )^{7}}+\frac {\cos ^{8}\left (d x +c \right )}{35 \sin \left (d x +c \right )^{5}}-\frac {\cos ^{8}\left (d x +c \right )}{35 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{8}\left (d x +c \right )}{7 \sin \left (d x +c \right )}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}\right )-\frac {b^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{7 \sin \left (d x +c \right )^{7}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^8*(a+b*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*b*(-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*b^2/sin(d*x+c)^7*cos(d*x+c)^7)

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maxima [A]  time = 0.53, size = 116, normalized size = 0.76 \[ \frac {{\left (105 \, d x + 105 \, c + \frac {105 \, \tan \left (d x + c\right )^{6} - 35 \, \tan \left (d x + c\right )^{4} + 21 \, \tan \left (d x + c\right )^{2} - 15}{\tan \left (d x + c\right )^{7}}\right )} a^{2} + \frac {6 \, {\left (35 \, \sin \left (d x + c\right )^{6} - 35 \, \sin \left (d x + c\right )^{4} + 21 \, \sin \left (d x + c\right )^{2} - 5\right )} a b}{\sin \left (d x + c\right )^{7}} - \frac {15 \, b^{2}}{\tan \left (d x + c\right )^{7}}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^8*(a+b*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*b/sin(d*x + c)^7 - 15*b^2/tan(d*x + c)
^7)/d

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mupad [B]  time = 1.50, size = 258, normalized size = 1.69 \[ a^2\,x+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\left (a-b\right )}^2}{896\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {3\,a^2}{32}-\frac {5\,a\,b}{48}+\frac {b^2}{48}+\frac {{\left (a-b\right )}^2}{384}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {a^2}{80}-\frac {3\,a\,b}{160}+\frac {b^2}{160}+\frac {{\left (a-b\right )}^2}{640}\right )}{d}-\frac {\frac {2\,a\,b}{7}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {37\,a^2}{3}+14\,a\,b+3\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (93\,a^2+70\,a\,b+5\,b^2\right )+\frac {a^2}{7}+\frac {b^2}{7}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {9\,a^2}{5}+\frac {14\,a\,b}{5}+b^2\right )}{128\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {23\,a^2}{32}-\frac {17\,a\,b}{32}+\frac {b^2}{32}+\frac {{\left (a-b\right )}^2}{128}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x + (tan(c/2 + (d*x)/2)^7*(a - b)^2)/(896*d) + (tan(c/2 + (d*x)/2)^3*((3*a^2)/32 - (5*a*b)/48 + b^2/48 + (
a - b)^2/384))/d - (tan(c/2 + (d*x)/2)^5*(a^2/80 - (3*a*b)/160 + b^2/160 + (a - b)^2/640))/d - ((2*a*b)/7 + ta
n(c/2 + (d*x)/2)^4*(14*a*b + (37*a^2)/3 + 3*b^2) - tan(c/2 + (d*x)/2)^6*(70*a*b + 93*a^2 + 5*b^2) + a^2/7 + b^
2/7 - tan(c/2 + (d*x)/2)^2*((14*a*b)/5 + (9*a^2)/5 + b^2))/(128*d*tan(c/2 + (d*x)/2)^7) - (tan(c/2 + (d*x)/2)*
((23*a^2)/32 - (17*a*b)/32 + b^2/32 + (a - b)^2/128))/d

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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+b*sec(d*x+c))**2,x)

[Out]

Timed out

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