3.255 \(\int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=389 \[ \frac {2 \cot ^5(c+d x)}{a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^3(c+d x)}{5 a^2 d (e \cot (c+d x))^{11/2}}-\frac {4 \cot ^4(c+d x) \csc (c+d x)}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}+\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} a^2 d \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}-\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} a^2 d \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}+\frac {2 \sqrt {\sin (2 c+2 d x)} \cot ^5(c+d x) \csc (c+d x) F\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{3 a^2 d (e \cot (c+d x))^{11/2}} \]

[Out]

2/5*cot(d*x+c)^3/a^2/d/(e*cot(d*x+c))^(11/2)+2*cot(d*x+c)^5/a^2/d/(e*cot(d*x+c))^(11/2)-4/3*cot(d*x+c)^4*csc(d
*x+c)/a^2/d/(e*cot(d*x+c))^(11/2)-2/3*cot(d*x+c)^5*csc(d*x+c)*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi+d*x)*El
lipticF(cos(c+1/4*Pi+d*x),2^(1/2))*sin(2*d*x+2*c)^(1/2)/a^2/d/(e*cot(d*x+c))^(11/2)-1/2*arctan(-1+2^(1/2)*tan(
d*x+c)^(1/2))/a^2/d/(e*cot(d*x+c))^(11/2)*2^(1/2)/tan(d*x+c)^(11/2)-1/2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/a^2
/d/(e*cot(d*x+c))^(11/2)*2^(1/2)/tan(d*x+c)^(11/2)+1/4*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a^2/d/(e*cot(
d*x+c))^(11/2)*2^(1/2)/tan(d*x+c)^(11/2)-1/4*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a^2/d/(e*cot(d*x+c))^(1
1/2)*2^(1/2)/tan(d*x+c)^(11/2)

________________________________________________________________________________________

Rubi [A]  time = 0.48, antiderivative size = 389, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 18, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3900, 3888, 3886, 3473, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2611, 2614, 2573, 2641, 2607, 30} \[ \frac {2 \cot ^5(c+d x)}{a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^3(c+d x)}{5 a^2 d (e \cot (c+d x))^{11/2}}-\frac {4 \cot ^4(c+d x) \csc (c+d x)}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}+\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} a^2 d \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}-\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} a^2 d \tan ^{\frac {11}{2}}(c+d x) (e \cot (c+d x))^{11/2}}+\frac {2 \sqrt {\sin (2 c+2 d x)} \cot ^5(c+d x) \csc (c+d x) F\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{3 a^2 d (e \cot (c+d x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cot[c + d*x])^(11/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(2*Cot[c + d*x]^3)/(5*a^2*d*(e*Cot[c + d*x])^(11/2)) + (2*Cot[c + d*x]^5)/(a^2*d*(e*Cot[c + d*x])^(11/2)) - (4
*Cot[c + d*x]^4*Csc[c + d*x])/(3*a^2*d*(e*Cot[c + d*x])^(11/2)) + (2*Cot[c + d*x]^5*Csc[c + d*x]*EllipticF[c -
Pi/4 + d*x, 2]*Sqrt[Sin[2*c + 2*d*x]])/(3*a^2*d*(e*Cot[c + d*x])^(11/2)) + ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*
x]]]/(Sqrt[2]*a^2*d*(e*Cot[c + d*x])^(11/2)*Tan[c + d*x]^(11/2)) - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/(Sqr
t[2]*a^2*d*(e*Cot[c + d*x])^(11/2)*Tan[c + d*x]^(11/2)) + Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(
2*Sqrt[2]*a^2*d*(e*Cot[c + d*x])^(11/2)*Tan[c + d*x]^(11/2)) - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*
x]]/(2*Sqrt[2]*a^2*d*(e*Cot[c + d*x])^(11/2)*Tan[c + d*x]^(11/2))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 204

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

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

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 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

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 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

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]

Rule 3900

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*((a_) + (b_.)*sec[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Dist[(e*Co
t[c + d*x])^m*Tan[c + d*x]^m, Int[(a + b*Sec[c + d*x])^n/Tan[c + d*x]^m, x], x] /; FreeQ[{a, b, c, d, e, m, n}
, x] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx &=\frac {\int \frac {\tan ^{\frac {11}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx}{(e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}\\ &=\frac {\int (-a+a \sec (c+d x))^2 \tan ^{\frac {3}{2}}(c+d x) \, dx}{a^4 (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}\\ &=\frac {\int \left (a^2 \tan ^{\frac {3}{2}}(c+d x)-2 a^2 \sec (c+d x) \tan ^{\frac {3}{2}}(c+d x)+a^2 \sec ^2(c+d x) \tan ^{\frac {3}{2}}(c+d x)\right ) \, dx}{a^4 (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}\\ &=\frac {\int \tan ^{\frac {3}{2}}(c+d x) \, dx}{a^2 (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}+\frac {\int \sec ^2(c+d x) \tan ^{\frac {3}{2}}(c+d x) \, dx}{a^2 (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {2 \int \sec (c+d x) \tan ^{\frac {3}{2}}(c+d x) \, dx}{a^2 (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}\\ &=\frac {2 \cot ^5(c+d x)}{a^2 d (e \cot (c+d x))^{11/2}}-\frac {4 \cot ^4(c+d x) \csc (c+d x)}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \int \frac {\sec (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{3 a^2 (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {\int \frac {1}{\sqrt {\tan (c+d x)}} \, dx}{a^2 (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}+\frac {\operatorname {Subst}\left (\int x^{3/2} \, dx,x,\tan (c+d x)\right )}{a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}\\ &=\frac {2 \cot ^3(c+d x)}{5 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x)}{a^2 d (e \cot (c+d x))^{11/2}}-\frac {4 \cot ^4(c+d x) \csc (c+d x)}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {\left (2 \cos ^{\frac {11}{2}}(c+d x)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}} \, dx}{3 a^2 (e \cot (c+d x))^{11/2} \sin ^{\frac {11}{2}}(c+d x)}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}\\ &=\frac {2 \cot ^3(c+d x)}{5 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x)}{a^2 d (e \cot (c+d x))^{11/2}}-\frac {4 \cot ^4(c+d x) \csc (c+d x)}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {\left (2 \cot ^5(c+d x) \csc (c+d x) \sqrt {\sin (2 c+2 d x)}\right ) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}} \, dx}{3 a^2 (e \cot (c+d x))^{11/2}}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}\\ &=\frac {2 \cot ^3(c+d x)}{5 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x)}{a^2 d (e \cot (c+d x))^{11/2}}-\frac {4 \cot ^4(c+d x) \csc (c+d x)}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x) \csc (c+d x) F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d (e \cot (c+d x))^{11/2}}-\frac {\operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {\operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}\\ &=\frac {2 \cot ^3(c+d x)}{5 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x)}{a^2 d (e \cot (c+d x))^{11/2}}-\frac {4 \cot ^4(c+d x) \csc (c+d x)}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x) \csc (c+d x) F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d (e \cot (c+d x))^{11/2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}\\ &=\frac {2 \cot ^3(c+d x)}{5 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x)}{a^2 d (e \cot (c+d x))^{11/2}}-\frac {4 \cot ^4(c+d x) \csc (c+d x)}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x) \csc (c+d x) F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}\\ &=\frac {2 \cot ^3(c+d x)}{5 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x)}{a^2 d (e \cot (c+d x))^{11/2}}-\frac {4 \cot ^4(c+d x) \csc (c+d x)}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {2 \cot ^5(c+d x) \csc (c+d x) F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d (e \cot (c+d x))^{11/2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {\tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}+\frac {\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}-\frac {\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} a^2 d (e \cot (c+d x))^{11/2} \tan ^{\frac {11}{2}}(c+d x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 13.53, size = 0, normalized size = 0.00 \[ \int \frac {1}{(e \cot (c+d x))^{11/2} (a+a \sec (c+d x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[1/((e*Cot[c + d*x])^(11/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

Integrate[1/((e*Cot[c + d*x])^(11/2)*(a + a*Sec[c + d*x])^2), x]

________________________________________________________________________________________

fricas [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(1/(e*cot(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \cot \left (d x + c\right )\right )^{\frac {11}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*cot(d*x + c))^(11/2)*(a*sec(d*x + c) + a)^2), x)

________________________________________________________________________________________

maple [C]  time = 2.11, size = 721, normalized size = 1.85 \[ -\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (15 i \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )-15 i \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )-15 \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )-15 \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+50 \sin \left (d x +c \right ) \EllipticF \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )-24 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )+44 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-26 \cos \left (d x +c \right ) \sqrt {2}+6 \sqrt {2}\right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {2}}{30 a^{2} d \left (\frac {e \cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )^{\frac {11}{2}} \sin \left (d x +c \right )^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cot(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x)

[Out]

-1/30/a^2/d*(-1+cos(d*x+c))*(15*I*sin(d*x+c)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I
,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+
sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-15*I*sin(d*x+c)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(
1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(
(1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-15*sin(d*x+c)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/s
in(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x
+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-15*sin(d*x+c)*EllipticPi(((1-cos(d*x+c)+s
in(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x
+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+50*sin(d*x+c)*EllipticF(((1-c
os(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*
x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-24*2^(1/2)*cos(d*x+c)^3+44*c
os(d*x+c)^2*2^(1/2)-26*cos(d*x+c)*2^(1/2)+6*2^(1/2))*cos(d*x+c)^3*(1+cos(d*x+c))^2/(e*cos(d*x+c)/sin(d*x+c))^(
11/2)/sin(d*x+c)^9*2^(1/2)

________________________________________________________________________________________

maxima [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(1/(e*cot(d*x+c))^(11/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^2}{a^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{11/2}\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*cot(c + d*x))^(11/2)*(a + a/cos(c + d*x))^2),x)

[Out]

int(cos(c + d*x)^2/(a^2*(e*cot(c + d*x))^(11/2)*(cos(c + d*x) + 1)^2), x)

________________________________________________________________________________________

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(1/(e*cot(d*x+c))**(11/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________