3.362 \(\int \frac{\cot ^6(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=207 \[ \frac{b^{7/2} (9 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 f (a+b)^{9/2}}+\frac{\left (2 a^2+6 a b-3 b^2\right ) \cot ^3(e+f x)}{6 a f (a+b)^3}-\frac{\left (8 a^2 b+2 a^3+12 a b^2-b^3\right ) \cot (e+f x)}{2 a f (a+b)^4}-\frac{x}{a^2}-\frac{(2 a-5 b) \cot ^5(e+f x)}{10 a f (a+b)^2}-\frac{b \cot ^5(e+f x)}{2 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

-(x/a^2) + (b^(7/2)*(9*a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*a^2*(a + b)^(9/2)*f) - ((2*a^3
+ 8*a^2*b + 12*a*b^2 - b^3)*Cot[e + f*x])/(2*a*(a + b)^4*f) + ((2*a^2 + 6*a*b - 3*b^2)*Cot[e + f*x]^3)/(6*a*(a
+ b)^3*f) - ((2*a - 5*b)*Cot[e + f*x]^5)/(10*a*(a + b)^2*f) - (b*Cot[e + f*x]^5)/(2*a*(a + b)*f*(a + b + b*Ta
n[e + f*x]^2))

________________________________________________________________________________________

Rubi [A]  time = 0.437015, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4141, 1975, 472, 583, 522, 203, 205} \[ \frac{b^{7/2} (9 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 f (a+b)^{9/2}}+\frac{\left (2 a^2+6 a b-3 b^2\right ) \cot ^3(e+f x)}{6 a f (a+b)^3}-\frac{\left (8 a^2 b+2 a^3+12 a b^2-b^3\right ) \cot (e+f x)}{2 a f (a+b)^4}-\frac{x}{a^2}-\frac{(2 a-5 b) \cot ^5(e+f x)}{10 a f (a+b)^2}-\frac{b \cot ^5(e+f x)}{2 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

-(x/a^2) + (b^(7/2)*(9*a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*a^2*(a + b)^(9/2)*f) - ((2*a^3
+ 8*a^2*b + 12*a*b^2 - b^3)*Cot[e + f*x])/(2*a*(a + b)^4*f) + ((2*a^2 + 6*a*b - 3*b^2)*Cot[e + f*x]^3)/(6*a*(a
+ b)^3*f) - ((2*a - 5*b)*Cot[e + f*x]^5)/(10*a*(a + b)^2*f) - (b*Cot[e + f*x]^5)/(2*a*(a + b)*f*(a + b + b*Ta
n[e + f*x]^2))

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot ^5(e+f x)}{2 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 a-5 b-7 b x^2}{x^6 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a (a+b) f}\\ &=-\frac{(2 a-5 b) \cot ^5(e+f x)}{10 a (a+b)^2 f}-\frac{b \cot ^5(e+f x)}{2 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{5 \left (2 a^2+6 a b-3 b^2\right )+5 (2 a-5 b) b x^2}{x^4 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{10 a (a+b)^2 f}\\ &=\frac{\left (2 a^2+6 a b-3 b^2\right ) \cot ^3(e+f x)}{6 a (a+b)^3 f}-\frac{(2 a-5 b) \cot ^5(e+f x)}{10 a (a+b)^2 f}-\frac{b \cot ^5(e+f x)}{2 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{15 \left (2 a^3+8 a^2 b+12 a b^2-b^3\right )+15 b \left (2 a^2+6 a b-3 b^2\right ) x^2}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{30 a (a+b)^3 f}\\ &=-\frac{\left (2 a^3+8 a^2 b+12 a b^2-b^3\right ) \cot (e+f x)}{2 a (a+b)^4 f}+\frac{\left (2 a^2+6 a b-3 b^2\right ) \cot ^3(e+f x)}{6 a (a+b)^3 f}-\frac{(2 a-5 b) \cot ^5(e+f x)}{10 a (a+b)^2 f}-\frac{b \cot ^5(e+f x)}{2 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{15 \left (2 a^4+10 a^3 b+20 a^2 b^2+20 a b^3+b^4\right )+15 b \left (2 a^3+8 a^2 b+12 a b^2-b^3\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{30 a (a+b)^4 f}\\ &=-\frac{\left (2 a^3+8 a^2 b+12 a b^2-b^3\right ) \cot (e+f x)}{2 a (a+b)^4 f}+\frac{\left (2 a^2+6 a b-3 b^2\right ) \cot ^3(e+f x)}{6 a (a+b)^3 f}-\frac{(2 a-5 b) \cot ^5(e+f x)}{10 a (a+b)^2 f}-\frac{b \cot ^5(e+f x)}{2 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a^2 f}+\frac{\left (b^4 (9 a+2 b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^2 (a+b)^4 f}\\ &=-\frac{x}{a^2}+\frac{b^{7/2} (9 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 (a+b)^{9/2} f}-\frac{\left (2 a^3+8 a^2 b+12 a b^2-b^3\right ) \cot (e+f x)}{2 a (a+b)^4 f}+\frac{\left (2 a^2+6 a b-3 b^2\right ) \cot ^3(e+f x)}{6 a (a+b)^3 f}-\frac{(2 a-5 b) \cot ^5(e+f x)}{10 a (a+b)^2 f}-\frac{b \cot ^5(e+f x)}{2 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 7.32759, size = 3028, normalized size = 14.63 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

((9*a + 2*b)*(a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(-(b^4*ArcTan[Sec[f*x]*(Cos[2*e]/(2*Sqrt[a + b]*S
qrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/2)*Sin[2*e])/(Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]))*(-(a*Sin[f*x
]) - 2*b*Sin[f*x] + a*Sin[2*e + f*x])]*Cos[2*e])/(8*a^2*Sqrt[a + b]*f*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) + ((I/8
)*b^4*ArcTan[Sec[f*x]*(Cos[2*e]/(2*Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/2)*Sin[2*e])/(Sqrt[a + b
]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]))*(-(a*Sin[f*x]) - 2*b*Sin[f*x] + a*Sin[2*e + f*x])]*Sin[2*e])/(a^2*Sqrt[a +
b]*f*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]])))/((a + b)^4*(a + b*Sec[e + f*x]^2)^2) + ((a + 2*b + a*Cos[2*e + 2*f*x]
)*Csc[e]*Csc[e + f*x]^5*Sec[2*e]*Sec[e + f*x]^4*(75*a^5*f*x*Cos[f*x] + 900*a^4*b*f*x*Cos[f*x] + 2850*a^3*b^2*f
*x*Cos[f*x] + 3900*a^2*b^3*f*x*Cos[f*x] + 2475*a*b^4*f*x*Cos[f*x] + 600*b^5*f*x*Cos[f*x] - 15*a^5*f*x*Cos[3*f*
x] + 240*a^4*b*f*x*Cos[3*f*x] + 1110*a^3*b^2*f*x*Cos[3*f*x] + 1740*a^2*b^3*f*x*Cos[3*f*x] + 1185*a*b^4*f*x*Cos
[3*f*x] + 300*b^5*f*x*Cos[3*f*x] - 75*a^5*f*x*Cos[2*e - f*x] - 900*a^4*b*f*x*Cos[2*e - f*x] - 2850*a^3*b^2*f*x
*Cos[2*e - f*x] - 3900*a^2*b^3*f*x*Cos[2*e - f*x] - 2475*a*b^4*f*x*Cos[2*e - f*x] - 600*b^5*f*x*Cos[2*e - f*x]
- 75*a^5*f*x*Cos[2*e + f*x] - 900*a^4*b*f*x*Cos[2*e + f*x] - 2850*a^3*b^2*f*x*Cos[2*e + f*x] - 3900*a^2*b^3*f
*x*Cos[2*e + f*x] - 2475*a*b^4*f*x*Cos[2*e + f*x] - 600*b^5*f*x*Cos[2*e + f*x] + 75*a^5*f*x*Cos[4*e + f*x] + 9
00*a^4*b*f*x*Cos[4*e + f*x] + 2850*a^3*b^2*f*x*Cos[4*e + f*x] + 3900*a^2*b^3*f*x*Cos[4*e + f*x] + 2475*a*b^4*f
*x*Cos[4*e + f*x] + 600*b^5*f*x*Cos[4*e + f*x] + 15*a^5*f*x*Cos[2*e + 3*f*x] - 240*a^4*b*f*x*Cos[2*e + 3*f*x]
- 1110*a^3*b^2*f*x*Cos[2*e + 3*f*x] - 1740*a^2*b^3*f*x*Cos[2*e + 3*f*x] - 1185*a*b^4*f*x*Cos[2*e + 3*f*x] - 30
0*b^5*f*x*Cos[2*e + 3*f*x] - 15*a^5*f*x*Cos[4*e + 3*f*x] + 240*a^4*b*f*x*Cos[4*e + 3*f*x] + 1110*a^3*b^2*f*x*C
os[4*e + 3*f*x] + 1740*a^2*b^3*f*x*Cos[4*e + 3*f*x] + 1185*a*b^4*f*x*Cos[4*e + 3*f*x] + 300*b^5*f*x*Cos[4*e +
3*f*x] + 15*a^5*f*x*Cos[6*e + 3*f*x] - 240*a^4*b*f*x*Cos[6*e + 3*f*x] - 1110*a^3*b^2*f*x*Cos[6*e + 3*f*x] - 17
40*a^2*b^3*f*x*Cos[6*e + 3*f*x] - 1185*a*b^4*f*x*Cos[6*e + 3*f*x] - 300*b^5*f*x*Cos[6*e + 3*f*x] + 45*a^5*f*x*
Cos[2*e + 5*f*x] + 120*a^4*b*f*x*Cos[2*e + 5*f*x] + 30*a^3*b^2*f*x*Cos[2*e + 5*f*x] - 180*a^2*b^3*f*x*Cos[2*e
+ 5*f*x] - 195*a*b^4*f*x*Cos[2*e + 5*f*x] - 60*b^5*f*x*Cos[2*e + 5*f*x] - 45*a^5*f*x*Cos[4*e + 5*f*x] - 120*a^
4*b*f*x*Cos[4*e + 5*f*x] - 30*a^3*b^2*f*x*Cos[4*e + 5*f*x] + 180*a^2*b^3*f*x*Cos[4*e + 5*f*x] + 195*a*b^4*f*x*
Cos[4*e + 5*f*x] + 60*b^5*f*x*Cos[4*e + 5*f*x] + 45*a^5*f*x*Cos[6*e + 5*f*x] + 120*a^4*b*f*x*Cos[6*e + 5*f*x]
+ 30*a^3*b^2*f*x*Cos[6*e + 5*f*x] - 180*a^2*b^3*f*x*Cos[6*e + 5*f*x] - 195*a*b^4*f*x*Cos[6*e + 5*f*x] - 60*b^5
*f*x*Cos[6*e + 5*f*x] - 45*a^5*f*x*Cos[8*e + 5*f*x] - 120*a^4*b*f*x*Cos[8*e + 5*f*x] - 30*a^3*b^2*f*x*Cos[8*e
+ 5*f*x] + 180*a^2*b^3*f*x*Cos[8*e + 5*f*x] + 195*a*b^4*f*x*Cos[8*e + 5*f*x] + 60*b^5*f*x*Cos[8*e + 5*f*x] - 1
5*a^5*f*x*Cos[4*e + 7*f*x] - 60*a^4*b*f*x*Cos[4*e + 7*f*x] - 90*a^3*b^2*f*x*Cos[4*e + 7*f*x] - 60*a^2*b^3*f*x*
Cos[4*e + 7*f*x] - 15*a*b^4*f*x*Cos[4*e + 7*f*x] + 15*a^5*f*x*Cos[6*e + 7*f*x] + 60*a^4*b*f*x*Cos[6*e + 7*f*x]
+ 90*a^3*b^2*f*x*Cos[6*e + 7*f*x] + 60*a^2*b^3*f*x*Cos[6*e + 7*f*x] + 15*a*b^4*f*x*Cos[6*e + 7*f*x] - 15*a^5*
f*x*Cos[8*e + 7*f*x] - 60*a^4*b*f*x*Cos[8*e + 7*f*x] - 90*a^3*b^2*f*x*Cos[8*e + 7*f*x] - 60*a^2*b^3*f*x*Cos[8*
e + 7*f*x] - 15*a*b^4*f*x*Cos[8*e + 7*f*x] + 15*a^5*f*x*Cos[10*e + 7*f*x] + 60*a^4*b*f*x*Cos[10*e + 7*f*x] + 9
0*a^3*b^2*f*x*Cos[10*e + 7*f*x] + 60*a^2*b^3*f*x*Cos[10*e + 7*f*x] + 15*a*b^4*f*x*Cos[10*e + 7*f*x] - 10*a^5*S
in[f*x] + 860*a^4*b*Sin[f*x] + 3120*a^3*b^2*Sin[f*x] + 3600*a^2*b^3*Sin[f*x] - 300*b^5*Sin[f*x] + 46*a^5*Sin[3
*f*x] - 508*a^4*b*Sin[3*f*x] - 2324*a^3*b^2*Sin[3*f*x] - 3120*a^2*b^3*Sin[3*f*x] + 75*a*b^4*Sin[3*f*x] - 150*b
^5*Sin[3*f*x] - 240*a^5*Sin[2*e - f*x] - 1840*a^4*b*Sin[2*e - f*x] - 4840*a^3*b^2*Sin[2*e - f*x] - 5040*a^2*b^
3*Sin[2*e - f*x] - 300*b^5*Sin[2*e - f*x] + 240*a^5*Sin[2*e + f*x] + 1840*a^4*b*Sin[2*e + f*x] + 4840*a^3*b^2*
Sin[2*e + f*x] + 5040*a^2*b^3*Sin[2*e + f*x] - 75*a*b^4*Sin[2*e + f*x] - 300*b^5*Sin[2*e + f*x] - 10*a^5*Sin[4
*e + f*x] + 860*a^4*b*Sin[4*e + f*x] + 3120*a^3*b^2*Sin[4*e + f*x] + 3600*a^2*b^3*Sin[4*e + f*x] + 75*a*b^4*Si
n[4*e + f*x] + 300*b^5*Sin[4*e + f*x] - 240*a^4*b*Sin[2*e + 3*f*x] - 900*a^3*b^2*Sin[2*e + 3*f*x] - 1200*a^2*b
^3*Sin[2*e + 3*f*x] - 75*a*b^4*Sin[2*e + 3*f*x] + 150*b^5*Sin[2*e + 3*f*x] + 46*a^5*Sin[4*e + 3*f*x] - 508*a^4
*b*Sin[4*e + 3*f*x] - 2324*a^3*b^2*Sin[4*e + 3*f*x] - 3120*a^2*b^3*Sin[4*e + 3*f*x] + 60*a*b^4*Sin[4*e + 3*f*x
] + 150*b^5*Sin[4*e + 3*f*x] - 240*a^4*b*Sin[6*e + 3*f*x] - 900*a^3*b^2*Sin[6*e + 3*f*x] - 1200*a^2*b^3*Sin[6*
e + 3*f*x] - 60*a*b^4*Sin[6*e + 3*f*x] - 150*b^5*Sin[6*e + 3*f*x] - 48*a^5*Sin[2*e + 5*f*x] - 32*a^4*b*Sin[2*e
+ 5*f*x] + 340*a^3*b^2*Sin[2*e + 5*f*x] + 864*a^2*b^3*Sin[2*e + 5*f*x] - 60*a*b^4*Sin[2*e + 5*f*x] + 30*b^5*S
in[2*e + 5*f*x] - 90*a^5*Sin[4*e + 5*f*x] - 300*a^4*b*Sin[4*e + 5*f*x] - 300*a^3*b^2*Sin[4*e + 5*f*x] + 60*a*b
^4*Sin[4*e + 5*f*x] - 30*b^5*Sin[4*e + 5*f*x] - 48*a^5*Sin[6*e + 5*f*x] - 32*a^4*b*Sin[6*e + 5*f*x] + 340*a^3*
b^2*Sin[6*e + 5*f*x] + 864*a^2*b^3*Sin[6*e + 5*f*x] - 15*a*b^4*Sin[6*e + 5*f*x] - 30*b^5*Sin[6*e + 5*f*x] - 90
*a^5*Sin[8*e + 5*f*x] - 300*a^4*b*Sin[8*e + 5*f*x] - 300*a^3*b^2*Sin[8*e + 5*f*x] + 15*a*b^4*Sin[8*e + 5*f*x]
+ 30*b^5*Sin[8*e + 5*f*x] + 46*a^5*Sin[4*e + 7*f*x] + 172*a^4*b*Sin[4*e + 7*f*x] + 216*a^3*b^2*Sin[4*e + 7*f*x
] + 15*a*b^4*Sin[4*e + 7*f*x] - 15*a*b^4*Sin[6*e + 7*f*x] + 46*a^5*Sin[8*e + 7*f*x] + 172*a^4*b*Sin[8*e + 7*f*
x] + 216*a^3*b^2*Sin[8*e + 7*f*x]))/(7680*a^2*(a + b)^4*f*(a + b*Sec[e + f*x]^2)^2)

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Maple [A]  time = 0.125, size = 248, normalized size = 1.2 \begin{align*} -{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f{a}^{2}}}-{\frac{1}{5\,f \left ( a+b \right ) ^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{5}}}+{\frac{a}{3\,f \left ( a+b \right ) ^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}+{\frac{b}{f \left ( a+b \right ) ^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{{a}^{2}}{f \left ( a+b \right ) ^{4}\tan \left ( fx+e \right ) }}-4\,{\frac{ab}{f \left ( a+b \right ) ^{4}\tan \left ( fx+e \right ) }}-6\,{\frac{{b}^{2}}{f \left ( a+b \right ) ^{4}\tan \left ( fx+e \right ) }}+{\frac{{b}^{4}\tan \left ( fx+e \right ) }{2\,f \left ( a+b \right ) ^{4}a \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{9\,{b}^{4}}{2\,f \left ( a+b \right ) ^{4}a}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}+{\frac{{b}^{5}}{f \left ( a+b \right ) ^{4}{a}^{2}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x)

[Out]

-1/f/a^2*arctan(tan(f*x+e))-1/5/f/(a+b)^2/tan(f*x+e)^5+1/3/f/(a+b)^3/tan(f*x+e)^3*a+1/f/(a+b)^3/tan(f*x+e)^3*b
-1/f/(a+b)^4/tan(f*x+e)*a^2-4/f/(a+b)^4/tan(f*x+e)*a*b-6/f/(a+b)^4/tan(f*x+e)*b^2+1/2/f*b^4/(a+b)^4/a*tan(f*x+
e)/(a+b+b*tan(f*x+e)^2)+9/2/f*b^4/(a+b)^4/a/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))+1/f*b^5/(a+b)
^4/a^2/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.847581, size = 3366, normalized size = 16.26 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[-1/120*(4*(46*a^5 + 172*a^4*b + 216*a^3*b^2 + 15*a*b^4)*cos(f*x + e)^7 - 4*(70*a^5 + 234*a^4*b + 218*a^3*b^2
- 216*a^2*b^3 + 45*a*b^4)*cos(f*x + e)^5 + 20*(6*a^5 + 10*a^4*b - 20*a^3*b^2 - 78*a^2*b^3 + 9*a*b^4)*cos(f*x +
e)^3 - 15*((9*a^2*b^3 + 2*a*b^4)*cos(f*x + e)^6 + 9*a*b^4 + 2*b^5 - (18*a^2*b^3 - 5*a*b^4 - 2*b^5)*cos(f*x +
e)^4 + (9*a^2*b^3 - 16*a*b^4 - 4*b^5)*cos(f*x + e)^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)
^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 - 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sq
rt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2))*sin(f*x + e) + 60*(2*a^4
*b + 8*a^3*b^2 + 12*a^2*b^3 - a*b^4)*cos(f*x + e) + 120*((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*f*x*c
os(f*x + e)^6 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 + 2*a^2*b^3 - 2*a*b^4 - b^5)*f*x*cos(f*x + e)^4 + (a^5 + 2*a^4*b
- 2*a^3*b^2 - 8*a^2*b^3 - 7*a*b^4 - 2*b^5)*f*x*cos(f*x + e)^2 + (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5
)*f*x)*sin(f*x + e))/(((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^6 - (2*a^7 + 7*a^6*b +
8*a^5*b^2 + 2*a^4*b^3 - 2*a^3*b^4 - a^2*b^5)*f*cos(f*x + e)^4 + (a^7 + 2*a^6*b - 2*a^5*b^2 - 8*a^4*b^3 - 7*a^
3*b^4 - 2*a^2*b^5)*f*cos(f*x + e)^2 + (a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*f)*sin(f*x + e)),
-1/60*(2*(46*a^5 + 172*a^4*b + 216*a^3*b^2 + 15*a*b^4)*cos(f*x + e)^7 - 2*(70*a^5 + 234*a^4*b + 218*a^3*b^2 -
216*a^2*b^3 + 45*a*b^4)*cos(f*x + e)^5 + 10*(6*a^5 + 10*a^4*b - 20*a^3*b^2 - 78*a^2*b^3 + 9*a*b^4)*cos(f*x + e
)^3 + 15*((9*a^2*b^3 + 2*a*b^4)*cos(f*x + e)^6 + 9*a*b^4 + 2*b^5 - (18*a^2*b^3 - 5*a*b^4 - 2*b^5)*cos(f*x + e)
^4 + (9*a^2*b^3 - 16*a*b^4 - 4*b^5)*cos(f*x + e)^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*
sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) + 30*(2*a^4*b + 8*a^3*b^2 + 12*a^2*b^3 - a*b^4)*co
s(f*x + e) + 60*((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*f*x*cos(f*x + e)^6 - (2*a^5 + 7*a^4*b + 8*a^3
*b^2 + 2*a^2*b^3 - 2*a*b^4 - b^5)*f*x*cos(f*x + e)^4 + (a^5 + 2*a^4*b - 2*a^3*b^2 - 8*a^2*b^3 - 7*a*b^4 - 2*b^
5)*f*x*cos(f*x + e)^2 + (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*f*x)*sin(f*x + e))/(((a^7 + 4*a^6*b +
6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^6 - (2*a^7 + 7*a^6*b + 8*a^5*b^2 + 2*a^4*b^3 - 2*a^3*b^4 - a^2
*b^5)*f*cos(f*x + e)^4 + (a^7 + 2*a^6*b - 2*a^5*b^2 - 8*a^4*b^3 - 7*a^3*b^4 - 2*a^2*b^5)*f*cos(f*x + e)^2 + (a
^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*f)*sin(f*x + e))]

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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(f*x+e)**6/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.46434, size = 419, normalized size = 2.02 \begin{align*} \frac{\frac{15 \, b^{4} \tan \left (f x + e\right )}{{\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac{15 \,{\left (9 \, a b^{4} + 2 \, b^{5}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )}}{{\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \sqrt{a b + b^{2}}} - \frac{30 \,{\left (f x + e\right )}}{a^{2}} - \frac{2 \,{\left (15 \, a^{2} \tan \left (f x + e\right )^{4} + 60 \, a b \tan \left (f x + e\right )^{4} + 90 \, b^{2} \tan \left (f x + e\right )^{4} - 5 \, a^{2} \tan \left (f x + e\right )^{2} - 20 \, a b \tan \left (f x + e\right )^{2} - 15 \, b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{5}}}{30 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/30*(15*b^4*tan(f*x + e)/((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*(b*tan(f*x + e)^2 + a + b)) + 15*(9
*a*b^4 + 2*b^5)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^6 + 4*a^5*b
+ 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*sqrt(a*b + b^2)) - 30*(f*x + e)/a^2 - 2*(15*a^2*tan(f*x + e)^4 + 60*a*b*ta
n(f*x + e)^4 + 90*b^2*tan(f*x + e)^4 - 5*a^2*tan(f*x + e)^2 - 20*a*b*tan(f*x + e)^2 - 15*b^2*tan(f*x + e)^2 +
3*a^2 + 6*a*b + 3*b^2)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*tan(f*x + e)^5))/f