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

Optimal. Leaf size=331 \[ -\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{\log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{4 \sqrt{2} a^2 d e^{5/2}}+\frac{\log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{4 \sqrt{2} a^2 d e^{5/2}}-\frac{7 \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 a^2 d e^{5/2}}+\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 \sqrt{2} a^2 d e^{5/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{2 \sqrt{2} a^2 d e^{5/2}}-\frac{1}{2 d e \left (a^2 \cot (c+d x)+a^2\right ) (e \cot (c+d x))^{3/2}}+\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}} \]

[Out]

(-7*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(2*a^2*d*e^(5/2)) + ArcTan[1 - (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e
]]/(2*Sqrt[2]*a^2*d*e^(5/2)) - ArcTan[1 + (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]]/(2*Sqrt[2]*a^2*d*e^(5/2)) +
7/(6*a^2*d*e*(e*Cot[c + d*x])^(3/2)) - 9/(2*a^2*d*e^2*Sqrt[e*Cot[c + d*x]]) - 1/(2*d*e*(e*Cot[c + d*x])^(3/2)*
(a^2 + a^2*Cot[c + d*x])) - Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*Sqrt[e*Cot[c + d*x]]]/(4*Sqrt[2]*a^2*
d*e^(5/2)) + Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] + Sqrt[2]*Sqrt[e*Cot[c + d*x]]]/(4*Sqrt[2]*a^2*d*e^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 1.07539, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 16, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.64, Rules used = {3569, 3649, 3653, 12, 16, 3476, 329, 297, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{\log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{4 \sqrt{2} a^2 d e^{5/2}}+\frac{\log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{4 \sqrt{2} a^2 d e^{5/2}}-\frac{7 \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 a^2 d e^{5/2}}+\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 \sqrt{2} a^2 d e^{5/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{2 \sqrt{2} a^2 d e^{5/2}}-\frac{1}{2 d e \left (a^2 \cot (c+d x)+a^2\right ) (e \cot (c+d x))^{3/2}}+\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(2*a^2*d*e^(5/2)) + ArcTan[1 - (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e
]]/(2*Sqrt[2]*a^2*d*e^(5/2)) - ArcTan[1 + (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]]/(2*Sqrt[2]*a^2*d*e^(5/2)) +
7/(6*a^2*d*e*(e*Cot[c + d*x])^(3/2)) - 9/(2*a^2*d*e^2*Sqrt[e*Cot[c + d*x]]) - 1/(2*d*e*(e*Cot[c + d*x])^(3/2)*
(a^2 + a^2*Cot[c + d*x])) - Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*Sqrt[e*Cot[c + d*x]]]/(4*Sqrt[2]*a^2*
d*e^(5/2)) + Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] + Sqrt[2]*Sqrt[e*Cot[c + d*x]]]/(4*Sqrt[2]*a^2*d*e^(5/2))

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rule 12

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{1}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))^2} \, dx &=-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}-\frac{\int \frac{-\frac{7 a^2 e}{2}+a^2 e \cot (c+d x)-\frac{5}{2} a^2 e \cot ^2(c+d x)}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))} \, dx}{2 a^3 e}\\ &=\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}}-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}-\frac{\int \frac{\frac{27 a^3 e^3}{4}+\frac{3}{2} a^3 e^3 \cot (c+d x)+\frac{21}{4} a^3 e^3 \cot ^2(c+d x)}{(e \cot (c+d x))^{3/2} (a+a \cot (c+d x))} \, dx}{3 a^4 e^4}\\ &=\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}}-\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}-\frac{2 \int \frac{-\frac{21}{8} a^4 e^5-\frac{3}{4} a^4 e^5 \cot (c+d x)-\frac{27}{8} a^4 e^5 \cot ^2(c+d x)}{\sqrt{e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{3 a^5 e^7}\\ &=\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}}-\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}-\frac{\int -\frac{3 a^5 e^5 \cot (c+d x)}{2 \sqrt{e \cot (c+d x)}} \, dx}{3 a^7 e^7}+\frac{7 \int \frac{1+\cot ^2(c+d x)}{\sqrt{e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{4 a e^2}\\ &=\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}}-\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}+\frac{\int \frac{\cot (c+d x)}{\sqrt{e \cot (c+d x)}} \, dx}{2 a^2 e^2}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-e x} (a-a x)} \, dx,x,-\cot (c+d x)\right )}{4 a d e^2}\\ &=\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}}-\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}+\frac{\int \sqrt{e \cot (c+d x)} \, dx}{2 a^2 e^3}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{a+\frac{a x^2}{e}} \, dx,x,\sqrt{e \cot (c+d x)}\right )}{2 a d e^3}\\ &=-\frac{7 \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 a^2 d e^{5/2}}+\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}}-\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{x}}{e^2+x^2} \, dx,x,e \cot (c+d x)\right )}{2 a^2 d e^2}\\ &=-\frac{7 \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 a^2 d e^{5/2}}+\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}}-\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{x^2}{e^2+x^4} \, dx,x,\sqrt{e \cot (c+d x)}\right )}{a^2 d e^2}\\ &=-\frac{7 \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 a^2 d e^{5/2}}+\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}}-\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{e-x^2}{e^2+x^4} \, dx,x,\sqrt{e \cot (c+d x)}\right )}{2 a^2 d e^2}-\frac{\operatorname{Subst}\left (\int \frac{e+x^2}{e^2+x^4} \, dx,x,\sqrt{e \cot (c+d x)}\right )}{2 a^2 d e^2}\\ &=-\frac{7 \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 a^2 d e^{5/2}}+\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}}-\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{e}+2 x}{-e-\sqrt{2} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \cot (c+d x)}\right )}{4 \sqrt{2} a^2 d e^{5/2}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{e}-2 x}{-e+\sqrt{2} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \cot (c+d x)}\right )}{4 \sqrt{2} a^2 d e^{5/2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{e-\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \cot (c+d x)}\right )}{4 a^2 d e^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{e+\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \cot (c+d x)}\right )}{4 a^2 d e^2}\\ &=-\frac{7 \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 a^2 d e^{5/2}}+\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}}-\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}-\frac{\log \left (\sqrt{e}+\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}\right )}{4 \sqrt{2} a^2 d e^{5/2}}+\frac{\log \left (\sqrt{e}+\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}\right )}{4 \sqrt{2} a^2 d e^{5/2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 \sqrt{2} a^2 d e^{5/2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 \sqrt{2} a^2 d e^{5/2}}\\ &=-\frac{7 \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 a^2 d e^{5/2}}+\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 \sqrt{2} a^2 d e^{5/2}}-\frac{\tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 \sqrt{2} a^2 d e^{5/2}}+\frac{7}{6 a^2 d e (e \cot (c+d x))^{3/2}}-\frac{9}{2 a^2 d e^2 \sqrt{e \cot (c+d x)}}-\frac{1}{2 d e (e \cot (c+d x))^{3/2} \left (a^2+a^2 \cot (c+d x)\right )}-\frac{\log \left (\sqrt{e}+\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}\right )}{4 \sqrt{2} a^2 d e^{5/2}}+\frac{\log \left (\sqrt{e}+\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}\right )}{4 \sqrt{2} a^2 d e^{5/2}}\\ \end{align*}

Mathematica [A]  time = 6.32417, size = 467, normalized size = 1.41 \[ \frac{\cot ^3(c+d x) \csc ^2(c+d x) (\sin (c+d x)+\cos (c+d x))^2 \left (-4 \tan (c+d x)+\frac{2}{3} \sec ^2(c+d x)-\frac{\sin (c+d x)}{2 (\sin (c+d x)+\cos (c+d x))}-\frac{2}{3}\right )}{d (a \cot (c+d x)+a)^2 (e \cot (c+d x))^{5/2}}+\frac{\cot ^{\frac{5}{2}}(c+d x) \csc ^2(c+d x) (\sin (c+d x)+\cos (c+d x))^2 \left (-\frac{16 (\cot (c+d x)+1) \csc ^3(c+d x) \sec (c+d x) \tan ^{-1}\left (\sqrt{\cot (c+d x)}\right )}{(\tan (c+d x)+1) \left (\cot ^2(c+d x)+1\right )^2}+\frac{\sin (2 (c+d x)) (\cot (c+d x)+1) \csc ^2(c+d x) \sec ^2(c+d x) \left (2 \tan ^{-1}\left (\sqrt{\cot (c+d x)}\right )-\sqrt{2} \left (\tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )-\tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )\right )\right )}{2 (\tan (c+d x)+1) \left (\cot ^2(c+d x)+1\right )}+\frac{\cos (2 (c+d x)) \csc ^3(c+d x) \sec (c+d x) \left (\log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )-\log \left (-\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}-1\right )\right )}{\sqrt{2} (\tan (c+d x)+1) (\cot (c+d x)-1) \left (\cot ^2(c+d x)+1\right )}\right )}{4 d (a \cot (c+d x)+a)^2 (e \cot (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[c + d*x]^3*Csc[c + d*x]^2*(Cos[c + d*x] + Sin[c + d*x])^2*(-2/3 + (2*Sec[c + d*x]^2)/3 - Sin[c + d*x]/(2*
(Cos[c + d*x] + Sin[c + d*x])) - 4*Tan[c + d*x]))/(d*(e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^2) + (Cot[c +
d*x]^(5/2)*Csc[c + d*x]^2*(Cos[c + d*x] + Sin[c + d*x])^2*((-16*ArcTan[Sqrt[Cot[c + d*x]]]*(1 + Cot[c + d*x])
*Csc[c + d*x]^3*Sec[c + d*x])/((1 + Cot[c + d*x]^2)^2*(1 + Tan[c + d*x])) + (Cos[2*(c + d*x)]*Csc[c + d*x]^3*(
-Log[-1 + Sqrt[2]*Sqrt[Cot[c + d*x]] - Cot[c + d*x]] + Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])*Sec
[c + d*x])/(Sqrt[2]*(-1 + Cot[c + d*x])*(1 + Cot[c + d*x]^2)*(1 + Tan[c + d*x])) + ((-(Sqrt[2]*(-ArcTan[1 - Sq
rt[2]*Sqrt[Cot[c + d*x]]] + ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])) + 2*ArcTan[Sqrt[Cot[c + d*x]]])*(1 + Cot[
c + d*x])*Csc[c + d*x]^2*Sec[c + d*x]^2*Sin[2*(c + d*x)])/(2*(1 + Cot[c + d*x]^2)*(1 + Tan[c + d*x]))))/(4*d*(
e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^2)

________________________________________________________________________________________

Maple [A]  time = 0.039, size = 276, normalized size = 0.8 \begin{align*} -{\frac{\sqrt{2}}{8\,d{a}^{2}{e}^{2}}\ln \left ({ \left ( e\cot \left ( dx+c \right ) -\sqrt [4]{{e}^{2}}\sqrt{e\cot \left ( dx+c \right ) }\sqrt{2}+\sqrt{{e}^{2}} \right ) \left ( e\cot \left ( dx+c \right ) +\sqrt [4]{{e}^{2}}\sqrt{e\cot \left ( dx+c \right ) }\sqrt{2}+\sqrt{{e}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{e}^{2}}}}}-{\frac{\sqrt{2}}{4\,d{a}^{2}{e}^{2}}\arctan \left ({\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{e}^{2}}}}}+{\frac{\sqrt{2}}{4\,d{a}^{2}{e}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{e}^{2}}}}}+{\frac{2}{3\,d{a}^{2}e} \left ( e\cot \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-4\,{\frac{1}{d{a}^{2}{e}^{2}\sqrt{e\cot \left ( dx+c \right ) }}}-{\frac{1}{2\,d{a}^{2}{e}^{2} \left ( e\cot \left ( dx+c \right ) +e \right ) }\sqrt{e\cot \left ( dx+c \right ) }}-{\frac{7}{2\,d{a}^{2}}\arctan \left ({\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt{e}}}} \right ){e}^{-{\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*cot(d*x+c))^2,x)

[Out]

-1/8/d/a^2/e^2/(e^2)^(1/4)*2^(1/2)*ln((e*cot(d*x+c)-(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*c
ot(d*x+c)+(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))-1/4/d/a^2/e^2/(e^2)^(1/4)*2^(1/2)*arctan(2^(1
/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+1/4/d/a^2/e^2/(e^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d
*x+c))^(1/2)+1)+2/3/a^2/d/e/(e*cot(d*x+c))^(3/2)-4/a^2/d/e^2/(e*cot(d*x+c))^(1/2)-1/2/d/a^2/e^2*(e*cot(d*x+c))
^(1/2)/(e*cot(d*x+c)+e)-7/2*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a^2/d/e^(5/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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*cot(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\left (e \cot{\left (c + d x \right )}\right )^{\frac{5}{2}} \cot ^{2}{\left (c + d x \right )} + 2 \left (e \cot{\left (c + d x \right )}\right )^{\frac{5}{2}} \cot{\left (c + d x \right )} + \left (e \cot{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((e*cot(c + d*x))**(5/2)*cot(c + d*x)**2 + 2*(e*cot(c + d*x))**(5/2)*cot(c + d*x) + (e*cot(c + d*x)
)**(5/2)), x)/a**2

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \cot \left (d x + c\right ) + a\right )}^{2} \left (e \cot \left (d x + c\right )\right )^{\frac{5}{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*cot(d*x+c))^2,x, algorithm="giac")

[Out]

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