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

Optimal. Leaf size=281 \[ \frac{e^2 \sqrt{e \cot (c+d x)}}{2 d \left (a^2 \cot (c+d x)+a^2\right )}-\frac{e^{5/2} \log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{4 \sqrt{2} a^2 d}+\frac{e^{5/2} \log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{4 \sqrt{2} a^2 d}-\frac{3 e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 a^2 d}-\frac{e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 \sqrt{2} a^2 d}+\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{2 \sqrt{2} a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.544004, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.56, Rules used = {3565, 3653, 12, 3476, 329, 211, 1165, 628, 1162, 617, 204, 3634, 63, 205} \[ \frac{e^2 \sqrt{e \cot (c+d x)}}{2 d \left (a^2 \cot (c+d x)+a^2\right )}-\frac{e^{5/2} \log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{4 \sqrt{2} a^2 d}+\frac{e^{5/2} \log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{4 \sqrt{2} a^2 d}-\frac{3 e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 a^2 d}-\frac{e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{2 \sqrt{2} a^2 d}+\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{2 \sqrt{2} a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

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 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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

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

Mathematica [A]  time = 1.97762, size = 224, normalized size = 0.8 \[ \frac{(e \cot (c+d x))^{5/2} (\sin (c+d x)+\cos (c+d x)) \left (2 \cot ^{\frac{3}{2}}(c+d x) \sec (c+d x)-\frac{1}{2} (\cot (c+d x)+1) \csc (c+d x) \left (\sqrt{2} \log \left (-\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}-1\right )-\sqrt{2} \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )+2 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )-2 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )+12 \tan ^{-1}\left (\sqrt{\cot (c+d x)}\right )\right )\right )}{4 a^2 d \cot ^{\frac{5}{2}}(c+d x) (\cot (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Cot[c + d*x])^(5/2)*(-((1 + Cot[c + d*x])*Csc[c + d*x]*(2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] -
2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]] + 12*ArcTan[Sqrt[Cot[c + d*x]]] + Sqrt[2]*Log[-1 + Sqrt[2]*S
qrt[Cot[c + d*x]] - Cot[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]))/2 + 2*Cot[c +
d*x]^(3/2)*Sec[c + d*x])*(Cos[c + d*x] + Sin[c + d*x]))/(4*a^2*d*Cot[c + d*x]^(5/2)*(1 + Cot[c + d*x])^2)

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Maple [A]  time = 0.038, size = 234, normalized size = 0.8 \begin{align*}{\frac{{e}^{2}\sqrt{2}}{8\,d{a}^{2}}\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{{e}^{2}\sqrt{2}}{4\,d{a}^{2}}\sqrt [4]{{e}^{2}}\arctan \left ({\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ) }-{\frac{{e}^{2}\sqrt{2}}{4\,d{a}^{2}}\sqrt [4]{{e}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ) }+{\frac{{e}^{3}}{2\,d{a}^{2} \left ( e\cot \left ( dx+c \right ) +e \right ) }\sqrt{e\cot \left ( dx+c \right ) }}-{\frac{3}{2\,d{a}^{2}}{e}^{{\frac{5}{2}}}\arctan \left ({\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt{e}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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*co
t(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)+1/2/d/a^2*e^3*(e*cot(d*x+c))^(1/2)/(e*cot(d*x+c)+e)-3/2*e^(5/2)*arctan((e*cot(d*x+c))^(1/2)/e^(
1/2))/a^2/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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