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

Optimal. Leaf size=325 \[ \frac{e^{5/2} (a-b) \log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{e^{5/2} (a-b) \log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}+\frac{2 a^{5/2} e^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cot (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{b^{3/2} d \left (a^2+b^2\right )}-\frac{e^{5/2} (a+b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{e^{5/2} (a+b) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )}-\frac{2 e^2 \sqrt{e \cot (c+d x)}}{b d} \]

[Out]

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

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Rubi [A]  time = 0.660319, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {3566, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac{e^{5/2} (a-b) \log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{e^{5/2} (a-b) \log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}+\frac{2 a^{5/2} e^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cot (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{b^{3/2} d \left (a^2+b^2\right )}-\frac{e^{5/2} (a+b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{e^{5/2} (a+b) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )}-\frac{2 e^2 \sqrt{e \cot (c+d x)}}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3566

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 - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[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 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{(e \cot (c+d x))^{5/2}}{a+b \cot (c+d x)} \, dx &=-\frac{2 e^2 \sqrt{e \cot (c+d x)}}{b d}-\frac{2 \int \frac{\frac{a e^3}{2}+\frac{1}{2} b e^3 \cot (c+d x)+\frac{1}{2} a e^3 \cot ^2(c+d x)}{\sqrt{e \cot (c+d x)} (a+b \cot (c+d x))} \, dx}{b}\\ &=-\frac{2 e^2 \sqrt{e \cot (c+d x)}}{b d}-\frac{2 \int \frac{\frac{b^2 e^3}{2}+\frac{1}{2} a b e^3 \cot (c+d x)}{\sqrt{e \cot (c+d x)}} \, dx}{b \left (a^2+b^2\right )}-\frac{\left (a^3 e^3\right ) \int \frac{1+\cot ^2(c+d x)}{\sqrt{e \cot (c+d x)} (a+b \cot (c+d x))} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac{2 e^2 \sqrt{e \cot (c+d x)}}{b d}-\frac{4 \operatorname{Subst}\left (\int \frac{-\frac{1}{2} b^2 e^4-\frac{1}{2} a b e^3 x^2}{e^2+x^4} \, dx,x,\sqrt{e \cot (c+d x)}\right )}{b \left (a^2+b^2\right ) d}-\frac{\left (a^3 e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-e x} (a-b x)} \, dx,x,-\cot (c+d x)\right )}{b \left (a^2+b^2\right ) d}\\ &=-\frac{2 e^2 \sqrt{e \cot (c+d x)}}{b d}+\frac{\left (2 a^3 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+\frac{b x^2}{e}} \, dx,x,\sqrt{e \cot (c+d x)}\right )}{b \left (a^2+b^2\right ) d}-\frac{\left ((a-b) e^3\right ) \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}+\frac{\left ((a+b) e^3\right ) \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}\\ &=\frac{2 a^{5/2} e^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cot (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{b^{3/2} \left (a^2+b^2\right ) d}-\frac{2 e^2 \sqrt{e \cot (c+d x)}}{b d}+\frac{\left ((a-b) e^{5/2}\right ) \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}+\frac{\left ((a-b) e^{5/2}\right ) \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}+\frac{\left ((a+b) e^3\right ) \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}+\frac{\left ((a+b) e^3\right ) \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}\\ &=\frac{2 a^{5/2} e^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cot (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{b^{3/2} \left (a^2+b^2\right ) d}-\frac{2 e^2 \sqrt{e \cot (c+d x)}}{b d}+\frac{(a-b) e^{5/2} \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}-\frac{(a-b) e^{5/2} \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}+\frac{\left ((a+b) e^{5/2}\right ) \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}-\frac{\left ((a+b) e^{5/2}\right ) \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}\\ &=\frac{2 a^{5/2} e^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cot (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{b^{3/2} \left (a^2+b^2\right ) d}-\frac{(a+b) e^{5/2} \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}+\frac{(a+b) e^{5/2} \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}-\frac{2 e^2 \sqrt{e \cot (c+d x)}}{b d}+\frac{(a-b) e^{5/2} \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}-\frac{(a-b) e^{5/2} \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}\\ \end{align*}

Mathematica [C]  time = 0.83518, size = 286, normalized size = 0.88 \[ \frac{(e \cot (c+d x))^{5/2} \left (8 a b^{3/2} \cot ^{\frac{3}{2}}(c+d x) \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},-\cot ^2(c+d x)\right )-3 \left (8 a^2 \sqrt{b} \sqrt{\cot (c+d x)}-8 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\cot (c+d x)}}{\sqrt{a}}\right )+8 b^{5/2} \sqrt{\cot (c+d x)}+\sqrt{2} b^{5/2} \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )-\sqrt{2} b^{5/2} \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )+2 \sqrt{2} b^{5/2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )-2 \sqrt{2} b^{5/2} \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )\right )\right )}{12 b^{3/2} d \left (a^2+b^2\right ) \cot ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Cot[c + d*x])^(5/2)*(8*a*b^(3/2)*Cot[c + d*x]^(3/2)*Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2] - 3*(2
*Sqrt[2]*b^(5/2)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - 2*Sqrt[2]*b^(5/2)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*
x]]] - 8*a^(5/2)*ArcTan[(Sqrt[b]*Sqrt[Cot[c + d*x]])/Sqrt[a]] + 8*a^2*Sqrt[b]*Sqrt[Cot[c + d*x]] + 8*b^(5/2)*S
qrt[Cot[c + d*x]] + Sqrt[2]*b^(5/2)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - Sqrt[2]*b^(5/2)*Log[1
+ Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])))/(12*b^(3/2)*(a^2 + b^2)*d*Cot[c + d*x]^(5/2))

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Maple [A]  time = 0.046, size = 459, normalized size = 1.4 \begin{align*} -2\,{\frac{{e}^{2}\sqrt{e\cot \left ( dx+c \right ) }}{bd}}+{\frac{b{e}^{2}\sqrt{2}}{4\,d \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{b{e}^{2}\sqrt{2}}{2\,d \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{e}^{2}\sqrt{2}}{2\,d \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{e}^{3}\sqrt{2}}{4\,d \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{a{e}^{3}\sqrt{2}}{2\,d \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{a{e}^{3}\sqrt{2}}{2\,d \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{{e}^{3}{a}^{3}}{bd \left ({a}^{2}+{b}^{2} \right ) \sqrt{aeb}}\arctan \left ({\frac{\sqrt{e\cot \left ( dx+c \right ) }b}{\sqrt{aeb}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*e^2*(e*cot(d*x+c))^(1/2)/b/d+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)+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)+
2/d*e^3/b*a^3/(a^2+b^2)/(a*e*b)^(1/2)*arctan((e*cot(d*x+c))^(1/2)*b/(a*e*b)^(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((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((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((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{\left (e \cot \left (d x + c\right )\right )^{\frac{5}{2}}}{b \cot \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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