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

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

[Out]

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

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Rubi [A]  time = 0.54, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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 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]

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 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 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 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 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 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])))

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*}

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Mathematica [A]  time = 1.33, 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]])

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fricas [A]  time = 0.94, size = 500, normalized size = 3.70 \[ \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 ] \]

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)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \cot \left (d x + c\right ) + a\right )} \left (e \cot \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

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)

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maple [B]  time = 0.76, size = 416, normalized size = 3.08 \[ -\frac {\arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{a d \,e^{\frac {5}{2}}}+\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{8 d a \,e^{3}}+\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a \,e^{3}}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a \,e^{3}}-\frac {\sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{8 d a \,e^{2} \left (e^{2}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a \,e^{2} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a \,e^{2} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {2}{3 a d e \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {2}{a d \,e^{2} \sqrt {e \cot \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a/d/e^(5/2)+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*2^(1/2)/(e^2)^(1/4)*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*2^(1/2)/(e^2)^(1/4)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+1/4/d/a/e^2*2^(1
/2)/(e^2)^(1/4)*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)

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maxima [A]  time = 0.44, size = 154, normalized size = 1.14 \[ \frac {e {\left (\frac {3 \, {\left (\frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}}\right )}}{a e^{3}} - \frac {12 \, \arctan \left (\frac {\sqrt {\frac {e}{\tan \left (d x + c\right )}}}{\sqrt {e}}\right )}{a e^{\frac {7}{2}}} + \frac {8 \, {\left (e - \frac {3 \, e}{\tan \left (d x + c\right )}\right )}}{a e^{3} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {3}{2}}}\right )}}{12 \, d} \]

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]

1/12*e*(3*(sqrt(2)*log(sqrt(2)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e) - sqrt(2)*log(-sqrt(
2)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e))/(a*e^3) - 12*arctan(sqrt(e/tan(d*x + c))/sqrt(e
))/(a*e^(7/2)) + 8*(e - 3*e/tan(d*x + c))/(a*e^3*(e/tan(d*x + c))^(3/2)))/d

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mupad [B]  time = 0.93, size = 132, normalized size = 0.98 \[ \frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {12\,\sqrt {2}\,a^3\,d^3\,e^{21/2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{12\,a^3\,d^3\,e^{11}+12\,a^3\,d^3\,e^{11}\,\mathrm {cot}\left (c+d\,x\right )}\right )}{2\,a\,d\,e^{5/2}}-\frac {\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{a\,d\,e^{5/2}}-\frac {\frac {2\,\mathrm {cot}\left (c+d\,x\right )}{e}-\frac {2}{3\,e}}{a\,d\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*atanh((12*2^(1/2)*a^3*d^3*e^(21/2)*(e*cot(c + d*x))^(1/2))/(12*a^3*d^3*e^11 + 12*a^3*d^3*e^11*cot(c +
d*x))))/(2*a*d*e^(5/2)) - atan((e*cot(c + d*x))^(1/2)/e^(1/2))/(a*d*e^(5/2)) - ((2*cot(c + d*x))/e - 2/(3*e))
/(a*d*(e*cot(c + d*x))^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}} \cot {\left (c + d x \right )} + \left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx}{a} \]

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

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