Integrand size = 25, antiderivative size = 236 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {b \cot ^3(e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (3 a+b) \cot ^3(e+f x)}{a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(a-b) \left (3 a^2+14 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a^2 (a+b)^4 f}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a^2 (a+b)^3 f} \]
arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(5/2)/f-b*(3*a+b)* cot(f*x+e)^3/a^2/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/3*(a-b)*(3*a^2+14* a*b+3*b^2)*cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/a^2/(a+b)^4/f-1/3*(a^2-10 *a*b-3*b^2)*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/a^2/(a+b)^3/f-1/3*b*co t(f*x+e)^3/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)
Time = 11.19 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.99 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) (a+2 b+a \cos (2 e+2 f x))^{5/2} \sec ^5(e+f x)}{4 \sqrt {2} a^{5/2} f \left (a+b \sec ^2(e+f x)\right )^{5/2}}-\frac {(a+2 b+a \cos (2 (e+f x)))^3 \sec ^5(e+f x) \left (-4 (a+3 b) \csc (e+f x)+(a+b) \csc ^3(e+f x)+\frac {4 b^3 \left (6 a^2+13 a b+3 b^2+2 a (3 a+b) \cos (2 (e+f x))\right ) \sin (e+f x)}{a^2 (a+2 b+a \cos (2 (e+f x)))^2}\right )}{24 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
(ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*(a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^5)/(4*Sqrt[2]*a^(5/2)*f*(a + b*Sec[ e + f*x]^2)^(5/2)) - ((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^5*(-4* (a + 3*b)*Csc[e + f*x] + (a + b)*Csc[e + f*x]^3 + (4*b^3*(6*a^2 + 13*a*b + 3*b^2 + 2*a*(3*a + b)*Cos[2*(e + f*x)])*Sin[e + f*x])/(a^2*(a + 2*b + a*C os[2*(e + f*x)])^2)))/(24*(a + b)^4*f*(a + b*Sec[e + f*x]^2)^(5/2))
Time = 0.56 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4629, 2075, 374, 27, 441, 445, 445, 27, 291, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4629 |
| \(\displaystyle \frac {\int \frac {\cot ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 2075 |
| \(\displaystyle \frac {\int \frac {\cot ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 374 |
| \(\displaystyle \frac {\frac {\int \frac {3 \cot ^4(e+f x) \left (-2 b \tan ^2(e+f x)+a-b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\cot ^4(e+f x) \left (-2 b \tan ^2(e+f x)+a-b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 441 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\cot ^4(e+f x) \left (a^2-10 b a-3 b^2-4 b (3 a+b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {\frac {\frac {-\frac {\int \frac {\cot ^2(e+f x) \left (2 b \left (a^2-10 b a-3 b^2\right ) \tan ^2(e+f x)+(a-b) \left (3 a^2+14 b a+3 b^2\right )\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{3 (a+b)}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {\frac {\frac {-\frac {-\frac {\int \frac {3 (a+b)^4}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a+b}-\frac {(a-b) \left (3 a^2+14 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {-\frac {-3 (a+b)^3 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\frac {(a-b) \left (3 a^2+14 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {-\frac {-3 (a+b)^3 \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\frac {(a-b) \left (3 a^2+14 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {-\frac {-\frac {(a-b) \left (3 a^2+14 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}-\frac {3 (a+b)^3 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a}}}{3 (a+b)}-\frac {\left (a^2-10 a b-3 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{a (a+b)}-\frac {b (3 a+b) \cot ^3(e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{a (a+b)}-\frac {b \cot ^3(e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
(-1/3*(b*Cot[e + f*x]^3)/(a*(a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + (- ((b*(3*a + b)*Cot[e + f*x]^3)/(a*(a + b)*Sqrt[a + b + b*Tan[e + f*x]^2])) + (-1/3*((a^2 - 10*a*b - 3*b^2)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x] ^2])/(a + b) - ((-3*(a + b)^3*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b *Tan[e + f*x]^2]])/Sqrt[a] - ((a - b)*(3*a^2 + 14*a*b + 3*b^2)*Cot[e + f*x ]*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b))/(3*(a + b)))/(a*(a + b)))/(a*(a + b)))/f
3.5.39.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*g*2*(b*c - a*d)*(p + 1))), x] + Si mp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2 )^q*Simp[c*(b*e - a*f)*(m + 1) + e*2*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, q}, x] && LtQ[p, -1]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
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]}, Sim p[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] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Leaf count of result is larger than twice the leaf count of optimal. \(2611\) vs. \(2(216)=432\).
Time = 9.65 (sec) , antiderivative size = 2612, normalized size of antiderivative = 11.07
1/24/f/(a+b)^4/a^2/(-a)^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f* x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^ 2*csc(f*x+e)^2+a+b)*(-18*(-a)^(1/2)*a^5*(1-cos(f*x+e))^2*csc(f*x+e)^2+48*( -a)^(1/2)*b^5*(1-cos(f*x+e))^8*csc(f*x+e)^8+96*(-a)^(1/2)*b^5*(1-cos(f*x+e ))^6*csc(f*x+e)^6+48*(-a)^(1/2)*b^5*(1-cos(f*x+e))^4*csc(f*x+e)^4+(-a)^(1/ 2)*a^5*(1-cos(f*x+e))^12*csc(f*x+e)^12-18*(-a)^(1/2)*a^5*(1-cos(f*x+e))^10 *csc(f*x+e)^10+63*(-a)^(1/2)*a^5*(1-cos(f*x+e))^8*csc(f*x+e)^8-92*a^5*(1-c os(f*x+e))^6*(-a)^(1/2)*csc(f*x+e)^6+63*(-a)^(1/2)*a^5*(1-cos(f*x+e))^4*cs c(f*x+e)^4+(-a)^(1/2)*a^5+240*(-a)^(1/2)*a*b^4*(1-cos(f*x+e))^8*csc(f*x+e) ^8+240*(-a)^(1/2)*a*b^4*(1-cos(f*x+e))^4*csc(f*x+e)^4-42*(-a)^(1/2)*a^2*b^ 3*(1-cos(f*x+e))^2*csc(f*x+e)^2+108*(-a)^(1/2)*a^3*b^2*(1-cos(f*x+e))^6*cs c(f*x+e)^6-652*(-a)^(1/2)*a^2*b^3*(1-cos(f*x+e))^6*csc(f*x+e)^6+224*(-a)^( 1/2)*a*b^4*(1-cos(f*x+e))^6*csc(f*x+e)^6-24*ln(4*((-a)^(1/2)*(a*(1-cos(f*x +e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*c sc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)-2*a*(csc(f*x+e)-c ot(f*x+e)))/((1-cos(f*x+e))^2*csc(f*x+e)^2+1))*(a*(1-cos(f*x+e))^4*csc(f*x +e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2* b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(3/2)*a^4*(1-cos(f*x+e))^3*csc(f*x+e) ^3-24*ln(4*((-a)^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4 *csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*cs...
Leaf count of result is larger than twice the leaf count of optimal. 729 vs. \(2 (216) = 432\).
Time = 7.36 (sec) , antiderivative size = 1579, normalized size of antiderivative = 6.69 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
[-1/24*(3*((a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*cos(f*x + e)^ 6 - a^4*b^2 - 4*a^3*b^3 - 6*a^2*b^4 - 4*a*b^5 - b^6 - (a^6 + 2*a^5*b - 2*a ^4*b^2 - 8*a^3*b^3 - 7*a^2*b^4 - 2*a*b^5)*cos(f*x + e)^4 - (2*a^5*b + 7*a^ 4*b^2 + 8*a^3*b^3 + 2*a^2*b^4 - 2*a*b^5 - b^6)*cos(f*x + e)^2)*sqrt(-a)*lo g(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a* b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16* a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e)) *sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))*sin(f* x + e) - 8*(4*(a^6 + 3*a^5*b + 3*a^3*b^3 + a^2*b^4)*cos(f*x + e)^7 - 3*(a^ 6 + a^5*b - 8*a^4*b^2 + 8*a^3*b^3 - a^2*b^4 - a*b^5)*cos(f*x + e)^5 - 6*(a ^5*b + 3*a^4*b^2 - 4*a^3*b^3 + 3*a^2*b^4 + a*b^5)*cos(f*x + e)^3 - (3*a^4* b^2 + 11*a^3*b^3 - 11*a^2*b^4 - 3*a*b^5)*cos(f*x + e))*sqrt((a*cos(f*x + e )^2 + b)/cos(f*x + e)^2))/(((a^9 + 4*a^8*b + 6*a^7*b^2 + 4*a^6*b^3 + a^5*b ^4)*f*cos(f*x + e)^6 - (a^9 + 2*a^8*b - 2*a^7*b^2 - 8*a^6*b^3 - 7*a^5*b^4 - 2*a^4*b^5)*f*cos(f*x + e)^4 - (2*a^8*b + 7*a^7*b^2 + 8*a^6*b^3 + 2*a^5*b ^4 - 2*a^4*b^5 - a^3*b^6)*f*cos(f*x + e)^2 - (a^7*b^2 + 4*a^6*b^3 + 6*a^5* b^4 + 4*a^4*b^5 + a^3*b^6)*f)*sin(f*x + e)), -1/12*(3*((a^6 + 4*a^5*b + 6* a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*cos(f*x + e)^6 - a^4*b^2 - 4*a^3*b^3 - 6...
\[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\cot ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cot \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Hanged} \]