3.252 \(\int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=355 \[ -\frac {4 \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} a^2 d \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}+\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} a^2 d \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^2(c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{a^2 d \sqrt {\sin (2 c+2 d x)} (e \cot (c+d x))^{5/2}} \]

[Out]

-4*cot(d*x+c)^3/a^2/d/(e*cot(d*x+c))^(5/2)+4*cos(d*x+c)*cot(d*x+c)^3/a^2/d/(e*cot(d*x+c))^(5/2)-4*cos(d*x+c)*c
ot(d*x+c)^2*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi+d*x)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2))/a^2/d/(e*cot(d*
x+c))^(5/2)/sin(2*d*x+2*c)^(1/2)-1/2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^2/d/(e*cot(d*x+c))^(5/2)*2^(1/2)/ta
n(d*x+c)^(5/2)-1/2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/a^2/d/(e*cot(d*x+c))^(5/2)*2^(1/2)/tan(d*x+c)^(5/2)-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))^(5/2)*2^(1/2)/tan(d*x+c)^(5/2)+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))^(5/2)*2^(1/2)/tan(d*x+c)^(5/2)

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Rubi [A]  time = 0.42, antiderivative size = 355, 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, 3474, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2608, 2615, 2572, 2639, 2607, 30} \[ -\frac {4 \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} a^2 d \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}+\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} a^2 d \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^2(c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{a^2 d \sqrt {\sin (2 c+2 d x)} (e \cot (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cot[c + d*x])^(5/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(-4*Cot[c + d*x]^3)/(a^2*d*(e*Cot[c + d*x])^(5/2)) + (4*Cos[c + d*x]*Cot[c + d*x]^3)/(a^2*d*(e*Cot[c + d*x])^(
5/2)) + (4*Cos[c + d*x]*Cot[c + d*x]^2*EllipticE[c - Pi/4 + d*x, 2])/(a^2*d*(e*Cot[c + d*x])^(5/2)*Sqrt[Sin[2*
c + 2*d*x]]) + ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/(Sqrt[2]*a^2*d*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2)
) - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/(Sqrt[2]*a^2*d*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2)) - Log[1 -
Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2]*a^2*d*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2)) + Log[
1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2]*a^2*d*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2))

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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{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 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[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 2608

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(n + 1)), x] - Dist[(a^2*(m - 2))/(b^2*(n + 1)), Int[(a*Sec[e
+ f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && (GtQ[m, 1] || (Eq
Q[m, 1] && EqQ[n, -3/2])) && IntegersQ[2*m, 2*n]

Rule 2615

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

Rule 2639

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

Rule 3474

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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))^{5/2} (a+a \sec (c+d x))^2} \, dx &=\frac {\int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx}{(e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}\\ &=\frac {\int \frac {(-a+a \sec (c+d x))^2}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{a^4 (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}\\ &=\frac {\int \left (\frac {a^2}{\tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a^2 \sec (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)}+\frac {a^2 \sec ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x)}\right ) \, dx}{a^4 (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}\\ &=\frac {\int \frac {1}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}+\frac {\int \frac {\sec ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \int \frac {\sec (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}\\ &=-\frac {2 \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}-\frac {\int \sqrt {\tan (c+d x)} \, dx}{a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 \int \cos (c+d x) \sqrt {\tan (c+d x)} \, dx}{a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}+\frac {\operatorname {Subst}\left (\int \frac {1}{x^{3/2}} \, dx,x,\tan (c+d x)\right )}{a^2 d (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}\\ &=-\frac {4 \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {\left (4 \cos ^{\frac {5}{2}}(c+d x)\right ) \int \sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)} \, dx}{a^2 (e \cot (c+d x))^{5/2} \sin ^{\frac {5}{2}}(c+d x)}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{a^2 d (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}\\ &=-\frac {4 \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {\left (4 \cos (c+d x) \cot ^2(c+d x)\right ) \int \sqrt {\sin (2 c+2 d x)} \, dx}{a^2 (e \cot (c+d x))^{5/2} \sqrt {\sin (2 c+2 d x)}}-\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d (e \cot (c+d x))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}\\ &=-\frac {4 \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^2(c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right )}{a^2 d (e \cot (c+d x))^{5/2} \sqrt {\sin (2 c+2 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))^{5/2} \tan ^{\frac {5}{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))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}\\ &=-\frac {4 \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^2(c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right )}{a^2 d (e \cot (c+d x))^{5/2} \sqrt {\sin (2 c+2 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))^{5/2} \tan ^{\frac {5}{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))^{5/2} \tan ^{\frac {5}{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))^{5/2} \tan ^{\frac {5}{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))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}\\ &=-\frac {4 \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^2(c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right )}{a^2 d (e \cot (c+d x))^{5/2} \sqrt {\sin (2 c+2 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))^{5/2} \tan ^{\frac {5}{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))^{5/2} \tan ^{\frac {5}{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))^{5/2} \tan ^{\frac {5}{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))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}\\ &=-\frac {4 \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^3(c+d x)}{a^2 d (e \cot (c+d x))^{5/2}}+\frac {4 \cos (c+d x) \cot ^2(c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right )}{a^2 d (e \cot (c+d x))^{5/2} \sqrt {\sin (2 c+2 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))^{5/2} \tan ^{\frac {5}{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))^{5/2} \tan ^{\frac {5}{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))^{5/2} \tan ^{\frac {5}{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))^{5/2} \tan ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [F]  time = 5.22, size = 0, normalized size = 0.00 \[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[1/((e*Cot[c + d*x])^(5/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

Integrate[1/((e*Cot[c + d*x])^(5/2)*(a + a*Sec[c + d*x])^2), x]

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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))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \cot \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*cot(d*x + c))^(5/2)*(a*sec(d*x + c) + a)^2), x)

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maple [C]  time = 2.14, size = 360, normalized size = 1.01 \[ \frac {\left (1+\cos \left (d x +c \right )\right )^{2} \left (i \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-4 \EllipticF \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )+8 \EllipticE \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}}{2 a^{2} d \left (\frac {e \cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cot(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x)

[Out]

1/2/a^2/d*(1+cos(d*x+c))^2*(I*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))-I
*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))-4*EllipticF(((1-cos(d*x+c)+sin
(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+8*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-
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))-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)*(-1+cos(d*x+c))*cos(d*x+c)^2/(e*cos(d*x+c)/si
n(d*x+c))^(5/2)/sin(d*x+c)^5*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \cot \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate(1/((e*cot(d*x + c))^(5/2)*(a*sec(d*x + c) + a)^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^2}{a^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{5/2}\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*cot(c + d*x))^(5/2)*(a + a/cos(c + d*x))^2),x)

[Out]

int(cos(c + d*x)^2/(a^2*(e*cot(c + d*x))^(5/2)*(cos(c + d*x) + 1)^2), x)

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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))**(5/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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