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[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))

________________________________________________________________________________________

Rubi [A]  time = 0.417798, antiderivative size = 355, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 18, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.72, 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 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]

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 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 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 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 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 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 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 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 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 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 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 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 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 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 30

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

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*}

Mathematica [F]  time = 4.93531, size = 0, normalized size = 0. \[ \int \frac{1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx \]

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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Maple [C]  time = 0.28, size = 367, normalized size = 1. \begin{align*}{\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}} \left ( i{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -i{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) +8\,{\it EllipticE} \left ( \sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},1/2\,\sqrt{2} \right ) -4\,{\it EllipticF} \left ( \sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},1/2\,\sqrt{2} \right ) -{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) \right ) \sqrt{{\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{-1+\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}} \left ({\frac{e\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

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*2^(1/2)*(cos(d*x+c)+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))-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))+8*EllipticE((-(-1+
cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),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))-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+c
os(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)/sin(d*x+c))^(5/2)/sin(d*x+c)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)