\(\int \frac {\cot ^6(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [427]

Optimal result

Integrand size = 25, antiderivative size = 241 \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac {b \cot ^5(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (15 a^3+55 a^2 b+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a (a+b)^4 f}+\frac {\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a (a+b)^3 f}-\frac {(a-5 b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f} \]

[Out]

-arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(3/2)/f-b*cot(f*x+e)^5/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)
^(1/2)-1/15*(15*a^3+55*a^2*b+73*a*b^2-15*b^3)*cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/a/(a+b)^4/f+1/15*(5*a^2+14
*a*b-15*b^2)*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/a/(a+b)^3/f-1/5*(a-5*b)*cot(f*x+e)^5*(a+b+b*tan(f*x+e)^2)
^(1/2)/a/(a+b)^2/f

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4226, 2000, 483, 597, 12, 385, 209} \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2} f}+\frac {\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 a f (a+b)^3}-\frac {\left (15 a^3+55 a^2 b+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 a f (a+b)^4}-\frac {(a-5 b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 a f (a+b)^2}-\frac {b \cot ^5(e+f x)}{a f (a+b) \sqrt {a+b \tan ^2(e+f x)+b}} \]

[In]

Int[Cot[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-(ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(a^(3/2)*f)) - (b*Cot[e + f*x]^5)/(a*(a + b)*f
*Sqrt[a + b + b*Tan[e + f*x]^2]) - ((15*a^3 + 55*a^2*b + 73*a*b^2 - 15*b^3)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e
+ f*x]^2])/(15*a*(a + b)^4*f) + ((5*a^2 + 14*a*b - 15*b^2)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(15*
a*(a + b)^3*f) - ((a - 5*b)*Cot[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(5*a*(a + b)^2*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 483

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 597

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 2000

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 4226

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])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x^6 \left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{x^6 \left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {b \cot ^5(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {a-5 b-6 b x^2}{x^6 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a (a+b) f} \\ & = -\frac {b \cot ^5(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {(a-5 b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f}-\frac {\text {Subst}\left (\int \frac {5 a^2+14 a b-15 b^2+4 (a-5 b) b x^2}{x^4 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 a (a+b)^2 f} \\ & = -\frac {b \cot ^5(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a (a+b)^3 f}-\frac {(a-5 b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f}+\frac {\text {Subst}\left (\int \frac {15 a^3+55 a^2 b+73 a b^2-15 b^3+2 b \left (5 a^2+14 a b-15 b^2\right ) x^2}{x^2 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a (a+b)^3 f} \\ & = -\frac {b \cot ^5(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (15 a^3+55 a^2 b+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a (a+b)^4 f}+\frac {\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a (a+b)^3 f}-\frac {(a-5 b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f}-\frac {\text {Subst}\left (\int \frac {15 (a+b)^4}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a (a+b)^4 f} \\ & = -\frac {b \cot ^5(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (15 a^3+55 a^2 b+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a (a+b)^4 f}+\frac {\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a (a+b)^3 f}-\frac {(a-5 b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f}-\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a f} \\ & = -\frac {b \cot ^5(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (15 a^3+55 a^2 b+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a (a+b)^4 f}+\frac {\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a (a+b)^3 f}-\frac {(a-5 b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f}-\frac {\text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a f} \\ & = -\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac {b \cot ^5(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (15 a^3+55 a^2 b+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a (a+b)^4 f}+\frac {\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 a (a+b)^3 f}-\frac {(a-5 b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.09 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.98 \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) (a+2 b+a \cos (2 e+2 f x))^{3/2} \sec ^3(e+f x)}{2 \sqrt {2} a^{3/2} f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {(a+2 b+a \cos (2 (e+f x)))^2 \left (\frac {30 b^4}{a (a+2 b+a \cos (2 (e+f x)))}-\left (23 a^2+80 a b+90 b^2\right ) \csc ^2(e+f x)+(a+b) (11 a+20 b) \csc ^4(e+f x)-3 (a+b)^2 \csc ^6(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{60 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[In]

Integrate[Cot[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-1/2*(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])^(3/2)*Sec[e
+ f*x]^3)/(Sqrt[2]*a^(3/2)*f*(a + b*Sec[e + f*x]^2)^(3/2)) + ((a + 2*b + a*Cos[2*(e + f*x)])^2*((30*b^4)/(a*(
a + 2*b + a*Cos[2*(e + f*x)])) - (23*a^2 + 80*a*b + 90*b^2)*Csc[e + f*x]^2 + (a + b)*(11*a + 20*b)*Csc[e + f*x
]^4 - 3*(a + b)^2*Csc[e + f*x]^6)*Sec[e + f*x]^2*Tan[e + f*x])/(60*(a + b)^4*f*(a + b*Sec[e + f*x]^2)^(3/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2448\) vs. \(2(221)=442\).

Time = 8.82 (sec) , antiderivative size = 2449, normalized size of antiderivative = 10.16

method	result	size
default	\(\text {Expression too large to display}\)	\(2449\)

[In]

int(cot(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/480/f/(a+b)^4/a/(-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)*(3*(-a)^(1/2)*a*b^3+1920*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*c
sc(f*x+e)^2+a+b)^(1/2)-2*a*(csc(f*x+e)-cot(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)^(1/2)*a^3*b*(1-cos(f*x+e))^5*csc(f*x+e)^5+2880*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*csc(f*x+e)^2+a+b)^(1/2)-2*a*(cs
c(f*x+e)-cot(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*c
sc(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)^(1/2)*a^2*b^2*(1-cos(f*x+
e))^5*csc(f*x+e)^5+1920*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*csc(f*x+e)^2+a+b)^(1/2)-2*a*(csc(f*x+e)-cot(f*x+e)))/((1-co
s(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)^(1/2)*a*b^3*(1-cos(f*x+e))^5*csc(f*x+e)^5-1804*a^3*(1
-cos(f*x+e))^6*(-a)^(1/2)*b*csc(f*x+e)^6-796*(-a)^(1/2)*a^2*b^2*(1-cos(f*x+e))^6*csc(f*x+e)^6+2460*(-a)^(1/2)*
a*b^3*(1-cos(f*x+e))^6*csc(f*x+e)^6+480*ln(4*((-a)^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*c
sc(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)^(1/2)-2*a*(csc(f*x+e)-cot
(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)^(1/2)*a^4*(1-cos(f*x+e))^5*csc(f*x+e)
^5+480*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*csc(f*x+e)^2+a+b)^(1/2)-2*a*(csc(f*x+e)-cot(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)^(1/2)*b^4*(1-cos(f*x+e))^5*csc(f*x+e)^5+1511*(-a)^(1/2)*a^3*b*(1-cos(f
*x+e))^4*csc(f*x+e)^4+2311*(-a)^(1/2)*a^2*b^2*(1-cos(f*x+e))^4*csc(f*x+e)^4+1165*(-a)^(1/2)*a*b^3*(1-cos(f*x+e
))^4*csc(f*x+e)^4-138*(-a)^(1/2)*a^3*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+2311*(-a)^(1/2)*a^2*b^2*(1-cos(f*x+e))^8*
csc(f*x+e)^8+1165*(-a)^(1/2)*a*b^3*(1-cos(f*x+e))^8*csc(f*x+e)^8+9*(-a)^(1/2)*a^3*b*(1-cos(f*x+e))^12*csc(f*x+
e)^12+9*(-a)^(1/2)*a^2*b^2*(1-cos(f*x+e))^12*csc(f*x+e)^12+3*(-a)^(1/2)*a*b^3*(1-cos(f*x+e))^12*csc(f*x+e)^12-
138*(-a)^(1/2)*a^3*b*(1-cos(f*x+e))^10*csc(f*x+e)^10-162*(-a)^(1/2)*a^2*b^2*(1-cos(f*x+e))^10*csc(f*x+e)^10-62
*(-a)^(1/2)*a*b^3*(1-cos(f*x+e))^10*csc(f*x+e)^10+1511*(-a)^(1/2)*a^3*b*(1-cos(f*x+e))^8*csc(f*x+e)^8-162*(-a)
^(1/2)*a^2*b^2*(1-cos(f*x+e))^2*csc(f*x+e)^2-62*(-a)^(1/2)*a*b^3*(1-cos(f*x+e))^2*csc(f*x+e)^2+3*(-a)^(1/2)*a^
4+9*(-a)^(1/2)*a^3*b+9*(-a)^(1/2)*a^2*b^2-960*(-a)^(1/2)*b^4*(1-cos(f*x+e))^6*csc(f*x+e)^6+365*(-a)^(1/2)*a^4*
(1-cos(f*x+e))^4*csc(f*x+e)^4-38*(-a)^(1/2)*a^4*(1-cos(f*x+e))^2*csc(f*x+e)^2+3*(-a)^(1/2)*a^4*(1-cos(f*x+e))^
12*csc(f*x+e)^12-38*(-a)^(1/2)*a^4*(1-cos(f*x+e))^10*csc(f*x+e)^10+365*(-a)^(1/2)*a^4*(1-cos(f*x+e))^8*csc(f*x
+e)^8-660*a^4*(1-cos(f*x+e))^6*(-a)^(1/2)*csc(f*x+e)^6)/((a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*c
sc(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)/((1-cos(f*x+e))^2*csc(f*x
+e)^2-1)^2)^(3/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^3/(1-cos(f*x+e))^5*sin(f*x+e)^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 698 vs. \(2 (221) = 442\).

Time = 7.21 (sec) , antiderivative size = 1517, normalized size of antiderivative = 6.29 \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/120*(15*((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*cos(f*x + e)^6 + a^4*b + 4*a^3*b^2 + 6*a^2*b^3 +
4*a*b^4 + b^5 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 + 2*a^2*b^3 - 2*a*b^4 - b^5)*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)*cos(f*x + e)^2)*sqrt(-a)*log(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))*sqr
t(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))*sin(f*x + e) + 8*((23*a^5 + 80*a^4*b + 90*a^3*
b^2 + 15*a*b^4)*cos(f*x + e)^7 - (35*a^5 + 106*a^4*b + 80*a^3*b^2 - 90*a^2*b^3 + 45*a*b^4)*cos(f*x + e)^5 + (1
5*a^5 + 20*a^4*b - 56*a^3*b^2 - 160*a^2*b^3 + 45*a*b^4)*cos(f*x + e)^3 + (15*a^4*b + 55*a^3*b^2 + 73*a^2*b^3 -
15*a*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((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*(15*((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*cos
(f*x + e)^6 + a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 + 2*a^2*b^3 - 2*a*b
^4 - b^5)*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)*cos(f*x + e)^2)*sqrt(a)*a
rctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqr
t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)
^2)*sin(f*x + e)))*sin(f*x + e) - 4*((23*a^5 + 80*a^4*b + 90*a^3*b^2 + 15*a*b^4)*cos(f*x + e)^7 - (35*a^5 + 10
6*a^4*b + 80*a^3*b^2 - 90*a^2*b^3 + 45*a*b^4)*cos(f*x + e)^5 + (15*a^5 + 20*a^4*b - 56*a^3*b^2 - 160*a^2*b^3 +
45*a*b^4)*cos(f*x + e)^3 + (15*a^4*b + 55*a^3*b^2 + 73*a^2*b^3 - 15*a*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)
^2 + b)/cos(f*x + e)^2))/(((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
))]

Sympy [F]

\[ \int \frac {\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cot ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(cot(f*x+e)**6/(a+b*sec(f*x+e)**2)**(3/2),x)

[Out]

Integral(cot(e + f*x)**6/(a + b*sec(e + f*x)**2)**(3/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cot \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Hanged} \]

[In]

int(cot(e + f*x)^6/(a + b/cos(e + f*x)^2)^(3/2),x)

[Out]

\text{Hanged}