3.35 \(\int \frac{(e \cot (c+d x))^{5/2}}{(a+a \cot (c+d x))^3} \, dx\)

Optimal. Leaf size=164 \[ -\frac{5 e^2 \sqrt{e \cot (c+d x)}}{8 a^3 d (\cot (c+d x)+1)}-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{8 a^3 d}+\frac{e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} \cot (c+d x)+\sqrt{e}}{\sqrt{2} \sqrt{e \cot (c+d x)}}\right )}{2 \sqrt{2} a^3 d}+\frac{e^2 \sqrt{e \cot (c+d x)}}{4 a d (a \cot (c+d x)+a)^2} \]

[Out]

-(e^(5/2)*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(8*a^3*d) + (e^(5/2)*ArcTanh[(Sqrt[e] + Sqrt[e]*Cot[c + d*x])/
(Sqrt[2]*Sqrt[e*Cot[c + d*x]])])/(2*Sqrt[2]*a^3*d) - (5*e^2*Sqrt[e*Cot[c + d*x]])/(8*a^3*d*(1 + Cot[c + d*x]))
+ (e^2*Sqrt[e*Cot[c + d*x]])/(4*a*d*(a + a*Cot[c + d*x])^2)

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Rubi [A]  time = 0.617429, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3565, 3649, 3654, 3532, 208, 3634, 63, 205} \[ -\frac{5 e^2 \sqrt{e \cot (c+d x)}}{8 a^3 d (\cot (c+d x)+1)}-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{8 a^3 d}+\frac{e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} \cot (c+d x)+\sqrt{e}}{\sqrt{2} \sqrt{e \cot (c+d x)}}\right )}{2 \sqrt{2} a^3 d}+\frac{e^2 \sqrt{e \cot (c+d x)}}{4 a d (a \cot (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cot[c + d*x])^(5/2)/(a + a*Cot[c + d*x])^3,x]

[Out]

-(e^(5/2)*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(8*a^3*d) + (e^(5/2)*ArcTanh[(Sqrt[e] + Sqrt[e]*Cot[c + d*x])/
(Sqrt[2]*Sqrt[e*Cot[c + d*x]])])/(2*Sqrt[2]*a^3*d) - (5*e^2*Sqrt[e*Cot[c + d*x]])/(8*a^3*d*(1 + Cot[c + d*x]))
+ (e^2*Sqrt[e*Cot[c + d*x]])/(4*a*d*(a + a*Cot[c + d*x])^2)

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 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 3532

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*d^2)/f,
Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x
] && EqQ[c^2 - d^2, 0]

Rule 208

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

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))^3} \, dx &=\frac{e^2 \sqrt{e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}-\frac{\int \frac{-\frac{1}{2} a^2 e^3+2 a^2 e^3 \cot (c+d x)-\frac{5}{2} a^2 e^3 \cot ^2(c+d x)}{\sqrt{e \cot (c+d x)} (a+a \cot (c+d x))^2} \, dx}{4 a^3}\\ &=-\frac{5 e^2 \sqrt{e \cot (c+d x)}}{8 a^3 d (1+\cot (c+d x))}+\frac{e^2 \sqrt{e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}+\frac{\int \frac{-\frac{3}{2} a^4 e^4+\frac{5}{2} a^4 e^4 \cot ^2(c+d x)}{\sqrt{e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{8 a^6 e}\\ &=-\frac{5 e^2 \sqrt{e \cot (c+d x)}}{8 a^3 d (1+\cot (c+d x))}+\frac{e^2 \sqrt{e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}+\frac{\int \frac{-4 a^5 e^4+4 a^5 e^4 \cot (c+d x)}{\sqrt{e \cot (c+d x)}} \, dx}{16 a^8 e}+\frac{e^3 \int \frac{1+\cot ^2(c+d x)}{\sqrt{e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{16 a^2}\\ &=-\frac{5 e^2 \sqrt{e \cot (c+d x)}}{8 a^3 d (1+\cot (c+d x))}+\frac{e^2 \sqrt{e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}+\frac{e^3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-e x} (a-a x)} \, dx,x,-\cot (c+d x)\right )}{16 a^2 d}-\frac{\left (2 a^2 e^7\right ) \operatorname{Subst}\left (\int \frac{1}{32 a^{10} e^8-e x^2} \, dx,x,\frac{-4 a^5 e^4-4 a^5 e^4 \cot (c+d x)}{\sqrt{e \cot (c+d x)}}\right )}{d}\\ &=\frac{e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e}+\sqrt{e} \cot (c+d x)}{\sqrt{2} \sqrt{e \cot (c+d x)}}\right )}{2 \sqrt{2} a^3 d}-\frac{5 e^2 \sqrt{e \cot (c+d x)}}{8 a^3 d (1+\cot (c+d x))}+\frac{e^2 \sqrt{e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}-\frac{e^2 \operatorname{Subst}\left (\int \frac{1}{a+\frac{a x^2}{e}} \, dx,x,\sqrt{e \cot (c+d x)}\right )}{8 a^2 d}\\ &=-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{8 a^3 d}+\frac{e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e}+\sqrt{e} \cot (c+d x)}{\sqrt{2} \sqrt{e \cot (c+d x)}}\right )}{2 \sqrt{2} a^3 d}-\frac{5 e^2 \sqrt{e \cot (c+d x)}}{8 a^3 d (1+\cot (c+d x))}+\frac{e^2 \sqrt{e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 2.07205, size = 192, normalized size = 1.17 \[ \frac{\csc (c+d x) (e \cot (c+d x))^{5/2} (\sin (c+d x)+\cos (c+d x))^3 \left (\frac{\sec ^4(c+d x) (-5 \sin (2 (c+d x))+3 \cos (2 (c+d x))-3)}{(\tan (c+d x)+1)^2}-\frac{2 \csc (c+d x) \sec (c+d x) \left (\sqrt{2} \left (\log \left (-\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}-1\right )-\log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )\right )+\tan ^{-1}\left (\sqrt{\cot (c+d x)}\right )\right )}{\cot ^{\frac{3}{2}}(c+d x)}\right )}{16 a^3 d (\cot (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cot[c + d*x])^(5/2)/(a + a*Cot[c + d*x])^3,x]

[Out]

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

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Maple [B]  time = 0.042, size = 440, normalized size = 2.7 \begin{align*}{\frac{{e}^{2}\sqrt{2}}{16\,d{a}^{3}}\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}}{8\,d{a}^{3}}\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}}{8\,d{a}^{3}}\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}\sqrt{2}}{16\,d{a}^{3}}\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{{e}^{3}\sqrt{2}}{8\,d{a}^{3}}\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{{e}^{3}\sqrt{2}}{8\,d{a}^{3}}\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{5\,{e}^{3}}{8\,d{a}^{3} \left ( e\cot \left ( dx+c \right ) +e \right ) ^{2}} \left ( e\cot \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{e}^{4}}{8\,d{a}^{3} \left ( e\cot \left ( dx+c \right ) +e \right ) ^{2}}\sqrt{e\cot \left ( dx+c \right ) }}-{\frac{1}{8\,d{a}^{3}}{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))^3,x)

[Out]

1/16/d/a^3*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)))+1/8/d/a^3*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)-1/8/d/a^3*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)-1/16/d/a^3*e^3/(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)))-1/8/d/a^3*e^3/(e^2)^(1/4)*2^(
1/2)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+1/8/d/a^3*e^3/(e^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(e^2
)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-5/8/d/a^3*e^3/(e*cot(d*x+c)+e)^2*(e*cot(d*x+c))^(3/2)-3/8/d/a^3*e^4/(e*cot(d*x
+c)+e)^2*(e*cot(d*x+c))^(1/2)-1/8*e^(5/2)*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a^3/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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.10143, size = 1407, normalized size = 8.58 \begin{align*} \left [-\frac{4 \,{\left (\sqrt{2} e^{2} \sin \left (2 \, d x + 2 \, c\right ) + \sqrt{2} e^{2}\right )} \sqrt{-e} \arctan \left (\frac{{\left (\sqrt{2} \cos \left (2 \, d x + 2 \, c\right ) + \sqrt{2} \sin \left (2 \, d x + 2 \, c\right ) + \sqrt{2}\right )} \sqrt{-e} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{2 \,{\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) -{\left (e^{2} \sin \left (2 \, d x + 2 \, c\right ) + e^{2}\right )} \sqrt{-e} \log \left (\frac{e \cos \left (2 \, d x + 2 \, c\right ) - e \sin \left (2 \, d x + 2 \, c\right ) - 2 \, \sqrt{-e} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + e}{\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1}\right ) -{\left (3 \, e^{2} \cos \left (2 \, d x + 2 \, c\right ) - 5 \, e^{2} \sin \left (2 \, d x + 2 \, c\right ) - 3 \, e^{2}\right )} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{16 \,{\left (a^{3} d \sin \left (2 \, d x + 2 \, c\right ) + a^{3} d\right )}}, -\frac{2 \,{\left (e^{2} \sin \left (2 \, d x + 2 \, c\right ) + e^{2}\right )} \sqrt{e} \arctan \left (\frac{\sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{\sqrt{e}}\right ) - 2 \,{\left (\sqrt{2} e^{2} \sin \left (2 \, d x + 2 \, c\right ) + \sqrt{2} e^{2}\right )} \sqrt{e} \log \left (-{\left (\sqrt{2} \cos \left (2 \, d x + 2 \, c\right ) - \sqrt{2} \sin \left (2 \, d x + 2 \, c\right ) - \sqrt{2}\right )} \sqrt{e} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} + 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) -{\left (3 \, e^{2} \cos \left (2 \, d x + 2 \, c\right ) - 5 \, e^{2} \sin \left (2 \, d x + 2 \, c\right ) - 3 \, e^{2}\right )} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{16 \,{\left (a^{3} d \sin \left (2 \, d x + 2 \, c\right ) + a^{3} d\right )}}\right ] \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))^3,x, algorithm="fricas")

[Out]

[-1/16*(4*(sqrt(2)*e^2*sin(2*d*x + 2*c) + sqrt(2)*e^2)*sqrt(-e)*arctan(1/2*(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2)
*sin(2*d*x + 2*c) + sqrt(2))*sqrt(-e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))/(e*cos(2*d*x + 2*c) + e)
) - (e^2*sin(2*d*x + 2*c) + e^2)*sqrt(-e)*log((e*cos(2*d*x + 2*c) - e*sin(2*d*x + 2*c) - 2*sqrt(-e)*sqrt((e*co
s(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) + e)/(cos(2*d*x + 2*c) + sin(2*d*x + 2*c) + 1)) - (3*e^
2*cos(2*d*x + 2*c) - 5*e^2*sin(2*d*x + 2*c) - 3*e^2)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(a^3*d*s
in(2*d*x + 2*c) + a^3*d), -1/16*(2*(e^2*sin(2*d*x + 2*c) + e^2)*sqrt(e)*arctan(sqrt((e*cos(2*d*x + 2*c) + e)/s
in(2*d*x + 2*c))/sqrt(e)) - 2*(sqrt(2)*e^2*sin(2*d*x + 2*c) + sqrt(2)*e^2)*sqrt(e)*log(-(sqrt(2)*cos(2*d*x + 2
*c) - sqrt(2)*sin(2*d*x + 2*c) - sqrt(2))*sqrt(e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)) + 2*e*sin(2*
d*x + 2*c) + e) - (3*e^2*cos(2*d*x + 2*c) - 5*e^2*sin(2*d*x + 2*c) - 3*e^2)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(
2*d*x + 2*c)))/(a^3*d*sin(2*d*x + 2*c) + a^3*d)]

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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))**3,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 )}^{3}}\,{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))^3,x, algorithm="giac")

[Out]

integrate((e*cot(d*x + c))^(5/2)/(a*cot(d*x + c) + a)^3, x)