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

Optimal. Leaf size=351 \[ -\frac{(a+b) \log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2} \left (a^2+b^2\right )}+\frac{(a+b) \log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2} \left (a^2+b^2\right )}-\frac{2 b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cot (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{a^{5/2} d e^{5/2} \left (a^2+b^2\right )}-\frac{(a-b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2} \left (a^2+b^2\right )}+\frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d e^{5/2} \left (a^2+b^2\right )}-\frac{2 b}{a^2 d e^2 \sqrt{e \cot (c+d x)}}+\frac{2}{3 a d e (e \cot (c+d x))^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.962181, antiderivative size = 351, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.52, Rules used = {3569, 3649, 3654, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac{(a+b) \log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2} \left (a^2+b^2\right )}+\frac{(a+b) \log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2} \left (a^2+b^2\right )}-\frac{2 b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cot (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{a^{5/2} d e^{5/2} \left (a^2+b^2\right )}-\frac{(a-b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2} \left (a^2+b^2\right )}+\frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d e^{5/2} \left (a^2+b^2\right )}-\frac{2 b}{a^2 d e^2 \sqrt{e \cot (c+d x)}}+\frac{2}{3 a d e (e \cot (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^(7/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cot[c + d*x]])/(Sqrt[a]*Sqrt[e])])/(a^(5/2)*(a^2 + b^2)*d*e^(5/2)) - ((a -
b)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*(a^2 + b^2)*d*e^(5/2)) + ((a - b)*ArcTan[1 + (
Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*(a^2 + b^2)*d*e^(5/2)) + 2/(3*a*d*e*(e*Cot[c + d*x])^(3/2)) -
(2*b)/(a^2*d*e^2*Sqrt[e*Cot[c + d*x]]) - ((a + b)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*Sqrt[e*Cot[c +
d*x]]])/(2*Sqrt[2]*(a^2 + b^2)*d*e^(5/2)) + ((a + b)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] + Sqrt[2]*Sqrt[e*Cot[
c + d*x]]])/(2*Sqrt[2]*(a^2 + b^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 3654

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*ta
n[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*x])^n*Simp[a*(A - C) - (A*b - b*
C)*Tan[e + f*x], x], x], x] + Dist[(A*b^2 + 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, 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 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

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

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

Mathematica [C]  time = 0.260025, size = 109, normalized size = 0.31 \[ \frac{2 \left (b^2 \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-\frac{b \cot (c+d x)}{a}\right )+a \left (a \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\cot ^2(c+d x)\right )-3 b \cot (c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{4},1,\frac{3}{4},-\cot ^2(c+d x)\right )\right )\right )}{3 a d e \left (a^2+b^2\right ) (e \cot (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b^2*Hypergeometric2F1[-3/2, 1, -1/2, -((b*Cot[c + d*x])/a)] + a*(a*Hypergeometric2F1[-3/4, 1, 1/4, -Cot[c
+ d*x]^2] - 3*b*Cot[c + d*x]*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2])))/(3*a*(a^2 + b^2)*d*e*(e*Cot[c
+ d*x])^(3/2))

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Maple [A]  time = 0.041, size = 481, normalized size = 1.4 \begin{align*}{\frac{a\sqrt{2}}{4\,d{e}^{3} \left ({a}^{2}+{b}^{2} \right ) }\sqrt [4]{{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{a\sqrt{2}}{2\,d{e}^{3} \left ({a}^{2}+{b}^{2} \right ) }\sqrt [4]{{e}^{2}}\arctan \left ({\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ) }-{\frac{a\sqrt{2}}{2\,d{e}^{3} \left ({a}^{2}+{b}^{2} \right ) }\sqrt [4]{{e}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ) }-{\frac{b\sqrt{2}}{4\,d{e}^{2} \left ({a}^{2}+{b}^{2} \right ) }\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{b\sqrt{2}}{2\,d{e}^{2} \left ({a}^{2}+{b}^{2} \right ) }\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{b\sqrt{2}}{2\,d{e}^{2} \left ({a}^{2}+{b}^{2} \right ) }\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}}}}}-2\,{\frac{{b}^{4}}{d{e}^{2}{a}^{2} \left ({a}^{2}+{b}^{2} \right ) \sqrt{aeb}}\arctan \left ({\frac{\sqrt{e\cot \left ( dx+c \right ) }b}{\sqrt{aeb}}} \right ) }+{\frac{2}{3\,ade} \left ( e\cot \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{b}{d{e}^{2}{a}^{2}\sqrt{e\cot \left ( dx+c \right ) }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d/e^3/(a^2+b^2)*a*(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*cot(d*x+c)-(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+1/2/d/e^3/(a^2+b^2)*a*(e^2)^(1/4)*2^(1
/2)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-1/2/d/e^3/(a^2+b^2)*a*(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/e^2/(a^2+b^2)*b/(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*cot(d*x+c)+(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))
-1/2/d/e^2/(a^2+b^2)*b/(e^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+1/2/d/e^2/(a^2+b
^2)*b/(e^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-2/d/e^2/a^2*b^4/(a^2+b^2)/(a*e*b
)^(1/2)*arctan((e*cot(d*x+c))^(1/2)*b/(a*e*b)^(1/2))+2/3/a/d/e/(e*cot(d*x+c))^(3/2)-2*b/a^2/d/e^2/(e*cot(d*x+c
))^(1/2)

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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+b*cot(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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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+b*cot(d*x+c)),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+b*cot(d*x+c)),x)

[Out]

Timed out

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

[Out]

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