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

Optimal. Leaf size=135 \[ -\frac{2}{a d e^2 \sqrt{e \cot (c+d x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{a d e^{5/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{e} \cot (c+d x)+\sqrt{e}}{\sqrt{2} \sqrt{e \cot (c+d x)}}\right )}{\sqrt{2} a d e^{5/2}}+\frac{2}{3 a d e (e \cot (c+d x))^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.538333, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3569, 3649, 12, 16, 3573, 3532, 208, 3634, 63, 205} \[ -\frac{2}{a d e^2 \sqrt{e \cot (c+d x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{a d e^{5/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{e} \cot (c+d x)+\sqrt{e}}{\sqrt{2} \sqrt{e \cot (c+d x)}}\right )}{\sqrt{2} a d e^{5/2}}+\frac{2}{3 a d e (e \cot (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 3573

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1
/(c^2 + d^2), Int[Simp[a^2*c - b^2*c + 2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e +
f*x]], x], x] + Dist[(b*c - a*d)^2/(c^2 + d^2), Int[(1 + Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan
[e + f*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]

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

Mathematica [A]  time = 1.27724, size = 131, normalized size = 0.97 \[ \frac{8 (\tan (c+d x)-3)-3 \sqrt{2} \sqrt{\cot (c+d x)} \left (\log \left (-\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}-1\right )-\log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )\right )-12 \sqrt{\cot (c+d x)} \tan ^{-1}\left (\sqrt{\cot (c+d x)}\right )}{12 a d e^2 \sqrt{e \cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*ArcTan[Sqrt[Cot[c + d*x]]]*Sqrt[Cot[c + d*x]] - 3*Sqrt[2]*Sqrt[Cot[c + d*x]]*(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]]) + 8*(-3 + Tan[c + d*x]))/(12*a*
d*e^2*Sqrt[e*Cot[c + d*x]])

________________________________________________________________________________________

Maple [B]  time = 0.034, size = 416, normalized size = 3.1 \begin{align*}{\frac{\sqrt{2}}{8\,da{e}^{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{\sqrt{2}}{4\,da{e}^{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{\sqrt{2}}{4\,da{e}^{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{\sqrt{2}}{8\,da{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\,da{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\,da{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\,dae} \left ( e\cot \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{1}{da{e}^{2}\sqrt{e\cot \left ( dx+c \right ) }}}-{\frac{1}{da}\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)),x)

[Out]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.44228, size = 1295, normalized size = 9.59 \begin{align*} \left [-\frac{3 \, \sqrt{2} \sqrt{-e}{\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \arctan \left (\frac{\sqrt{2} \sqrt{-e} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}{\left (\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}}{2 \,{\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) + 3 \, \sqrt{-e}{\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \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 ) + 4 \, \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}{\left (\cos \left (2 \, d x + 2 \, c\right ) + 3 \, \sin \left (2 \, d x + 2 \, c\right ) - 1\right )}}{6 \,{\left (a d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + a d e^{3}\right )}}, \frac{3 \, \sqrt{2} \sqrt{e}{\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \log \left (-\sqrt{2} \sqrt{e} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}{\left (\cos \left (2 \, d x + 2 \, c\right ) - \sin \left (2 \, d x + 2 \, c\right ) - 1\right )} + 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) - 12 \, \sqrt{e}{\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \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 ) - 8 \, \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}{\left (\cos \left (2 \, d x + 2 \, c\right ) + 3 \, \sin \left (2 \, d x + 2 \, c\right ) - 1\right )}}{12 \,{\left (a d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + a d e^{3}\right )}}\right ] \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)),x, algorithm="fricas")

[Out]

[-1/6*(3*sqrt(2)*sqrt(-e)*(cos(2*d*x + 2*c) + 1)*arctan(1/2*sqrt(2)*sqrt(-e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin
(2*d*x + 2*c))*(cos(2*d*x + 2*c) + sin(2*d*x + 2*c) + 1)/(e*cos(2*d*x + 2*c) + e)) + 3*sqrt(-e)*(cos(2*d*x + 2
*c) + 1)*log((e*cos(2*d*x + 2*c) - e*sin(2*d*x + 2*c) + 2*sqrt(-e)*sqrt((e*cos(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)) + 4*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*
d*x + 2*c))*(cos(2*d*x + 2*c) + 3*sin(2*d*x + 2*c) - 1))/(a*d*e^3*cos(2*d*x + 2*c) + a*d*e^3), 1/12*(3*sqrt(2)
*sqrt(e)*(cos(2*d*x + 2*c) + 1)*log(-sqrt(2)*sqrt(e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*(cos(2*d*
x + 2*c) - sin(2*d*x + 2*c) - 1) + 2*e*sin(2*d*x + 2*c) + e) - 12*sqrt(e)*(cos(2*d*x + 2*c) + 1)*arctan(sqrt((
e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))/sqrt(e)) - 8*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*(cos(2*
d*x + 2*c) + 3*sin(2*d*x + 2*c) - 1))/(a*d*e^3*cos(2*d*x + 2*c) + a*d*e^3)]

________________________________________________________________________________________

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{\left (c + d x \right )} + \left (e \cot{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx}{a} \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)),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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