3.84 \(\int \frac {\cot ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3888, 3886, 3473, 8, 2606, 194, 2607, 30} \[ -\frac {2 \cot ^7(c+d x)}{7 a^2 d}+\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x)}{a^2 d}+\frac {2 \csc ^7(c+d x)}{7 a^2 d}-\frac {6 \csc ^5(c+d x)}{5 a^2 d}+\frac {2 \csc ^3(c+d x)}{a^2 d}-\frac {2 \csc (c+d x)}{a^2 d}+\frac {x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps

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

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Mathematica [B]  time = 1.07, size = 314, normalized size = 2.26 \[ \frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \csc ^3(c+d x) \sec ^2(c+d x) (16002 \sin (c+d x)+9144 \sin (2 (c+d x))-3429 \sin (3 (c+d x))-4572 \sin (4 (c+d x))-1143 \sin (5 (c+d x))-11760 \sin (2 c+d x)-8864 \sin (c+2 d x)-3360 \sin (3 c+2 d x)+2064 \sin (2 c+3 d x)+2520 \sin (4 c+3 d x)+4432 \sin (3 c+4 d x)+1680 \sin (5 c+4 d x)+1528 \sin (4 c+5 d x)-5880 d x \cos (2 c+d x)+3360 d x \cos (c+2 d x)-3360 d x \cos (3 c+2 d x)-1260 d x \cos (2 c+3 d x)+1260 d x \cos (4 c+3 d x)-1680 d x \cos (3 c+4 d x)+1680 d x \cos (5 c+4 d x)-420 d x \cos (4 c+5 d x)+420 d x \cos (6 c+5 d x)-4032 \sin (c)-9632 \sin (d x)+5880 d x \cos (d x))}{26880 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c/2]*Csc[c + d*x]^3*Sec[c/2]*Sec[c + d*x]^2*(5880*d*x*Cos[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] + 16002*Sin[c + d*x] + 9144*Sin[2*(c + d*x)] - 3429*Sin[3*(c + d*x)] - 4572*Sin[4*(c + d*x)] - 1
143*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]))/(26880*a
^2*d*(1 + Sec[c + d*x])^2)

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fricas [A]  time = 0.77, size = 154, normalized size = 1.11 \[ \frac {191 \, \cos \left (d x + c\right )^{5} + 172 \, \cos \left (d x + c\right )^{4} - 253 \, \cos \left (d x + c\right )^{3} - 258 \, \cos \left (d x + c\right )^{2} + 105 \, {\left (d x \cos \left (d x + c\right )^{4} + 2 \, d x \cos \left (d x + c\right )^{3} - 2 \, d x \cos \left (d x + c\right ) - d x\right )} \sin \left (d x + c\right ) + 87 \, \cos \left (d x + c\right ) + 96}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.31, size = 114, normalized size = 0.82 \[ \frac {\frac {3360 \, {\left (d x + c\right )}}{a^{2}} + \frac {35 \, {\left (21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} + \frac {15 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 147 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 770 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4410 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{14}}}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3360*(3360*(d*x + c)/a^2 + 35*(21*tan(1/2*d*x + 1/2*c)^2 - 1)/(a^2*tan(1/2*d*x + 1/2*c)^3) + (15*a^12*tan(1/
2*d*x + 1/2*c)^7 - 147*a^12*tan(1/2*d*x + 1/2*c)^5 + 770*a^12*tan(1/2*d*x + 1/2*c)^3 - 4410*a^12*tan(1/2*d*x +
1/2*c))/a^14)/d

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maple [A]  time = 0.82, size = 132, normalized size = 0.95 \[ \frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{224 a^{2} d}-\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 a^{2} d}+\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 a^{2} d}-\frac {21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a^{2} d}-\frac {1}{96 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {7}{32 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/224/a^2/d*tan(1/2*d*x+1/2*c)^7-7/160/a^2/d*tan(1/2*d*x+1/2*c)^5+11/48/a^2/d*tan(1/2*d*x+1/2*c)^3-21/16/a^2/d
*tan(1/2*d*x+1/2*c)-1/96/a^2/d/tan(1/2*d*x+1/2*c)^3+7/32/a^2/d/tan(1/2*d*x+1/2*c)+2/a^2/d*arctan(tan(1/2*d*x+1
/2*c))

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maxima [A]  time = 0.60, size = 157, normalized size = 1.13 \[ -\frac {\frac {\frac {4410 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {770 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{2}} - \frac {6720 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {35 \, {\left (\frac {21 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a^{2} \sin \left (d x + c\right )^{3}}}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3360*((4410*sin(d*x + c)/(cos(d*x + c) + 1) - 770*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/
(cos(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^2 - 6720*arctan(sin(d*x + c)/(cos(d*x + c) +
1))/a^2 - 35*(21*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)*(cos(d*x + c) + 1)^3/(a^2*sin(d*x + c)^3))/d

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mupad [B]  time = 1.65, size = 182, normalized size = 1.31 \[ \frac {15\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-35\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-147\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+770\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-4410\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+735\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3360\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (c+d\,x\right )}{3360\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^4/(a + a/cos(c + d*x))^2,x)

[Out]

(15*sin(c/2 + (d*x)/2)^10 - 35*cos(c/2 + (d*x)/2)^10 - 147*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 + 770*cos
(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 - 4410*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 + 735*cos(c/2 + (d*x)/
2)^8*sin(c/2 + (d*x)/2)^2 + 3360*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^3*(c + d*x))/(3360*a^2*d*cos(c/2 + (d
*x)/2)^7*sin(c/2 + (d*x)/2)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cot ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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