3.37 \(\int \frac {\sqrt {e \cot (c+d x)}}{(a+a \cot (c+d x))^3} \, dx\)

Optimal. Leaf size=161 \[ \frac {3 \sqrt {e \cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+a^3\right )}-\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{8 a^3 d}-\frac {\sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} \cot (c+d x)+\sqrt {e}}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d}+\frac {\sqrt {e \cot (c+d x)}}{4 a d (a \cot (c+d x)+a)^2} \]

[Out]

-1/8*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))*e^(1/2)/a^3/d-1/4*arctanh(1/2*(e^(1/2)+cot(d*x+c)*e^(1/2))*2^(1/2)/(
e*cot(d*x+c))^(1/2))*e^(1/2)/a^3/d*2^(1/2)+1/4*(e*cot(d*x+c))^(1/2)/a/d/(a+a*cot(d*x+c))^2+3/8*(e*cot(d*x+c))^
(1/2)/d/(a^3+a^3*cot(d*x+c))

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Rubi [A]  time = 0.59, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3568, 3649, 3654, 3532, 208, 3634, 63, 205} \[ \frac {3 \sqrt {e \cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+a^3\right )}-\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{8 a^3 d}-\frac {\sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} \cot (c+d x)+\sqrt {e}}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d}+\frac {\sqrt {e \cot (c+d x)}}{4 a d (a \cot (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3568

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n)/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(a^2
+ b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*(m + 1) - b*d*n - (b*c - a*d)*
(m + 1)*Tan[e + f*x] - b*d*(m + 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] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[2*m]

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])))

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]

Rubi steps

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

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Mathematica [A]  time = 0.82, size = 181, normalized size = 1.12 \[ -\frac {\sqrt {e \cot (c+d x)} \left (\sqrt {\cot (c+d x)} (-3 \sin (2 (c+d x))+5 \cos (2 (c+d x))-5)+2 (\sin (2 (c+d x))+1) \tan ^{-1}\left (\sqrt {\cot (c+d x)}\right )-2 \sqrt {2} (\sin (c+d x)+\cos (c+d x))^2 \left (\log \left (-\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}-1\right )-\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )\right )}{16 a^3 d \sqrt {\cot (c+d x)} (\sin (c+d x)+\cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 518, normalized size = 3.22 \[ \left [\frac {4 \, {\left (\sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) + \sqrt {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 ) + \sqrt {-e} {\left (\sin \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 ) - \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (5 \, \cos \left (2 \, d x + 2 \, c\right ) - 3 \, \sin \left (2 \, d x + 2 \, c\right ) - 5\right )}}{16 \, {\left (a^{3} d \sin \left (2 \, d x + 2 \, c\right ) + a^{3} d\right )}}, -\frac {2 \, \sqrt {e} {\left (\sin \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 ) - 2 \, {\left (\sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) + \sqrt {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 ) + \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (5 \, \cos \left (2 \, d x + 2 \, c\right ) - 3 \, \sin \left (2 \, d x + 2 \, c\right ) - 5\right )}}{16 \, {\left (a^{3} d \sin \left (2 \, d x + 2 \, c\right ) + a^{3} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(1/2)/(a+a*cot(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \cot \left (d x + c\right )}}{{\left (a \cot \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(1/2)/(a+a*cot(d*x+c))^3,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.86, size = 423, normalized size = 2.63 \[ \frac {3 e \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}{8 d \,a^{3} \left (e \cot \left (d x +c \right )+e \right )^{2}}+\frac {5 e^{2} \sqrt {e \cot \left (d x +c \right )}}{8 d \,a^{3} \left (e \cot \left (d x +c \right )+e \right )^{2}}-\frac {\arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e}}{8 a^{3} d}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{16 d \,a^{3}}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3}}+\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3}}+\frac {e \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{16 d \,a^{3} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {e \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3} \left (e^{2}\right )^{\frac {1}{4}}}-\frac {e \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3} \left (e^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cot(d*x+c))^(1/2)/(a+cot(d*x+c)*a)^3,x)

[Out]

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

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maxima [A]  time = 0.60, size = 190, normalized size = 1.18 \[ \frac {e {\left (\frac {5 \, e \sqrt {\frac {e}{\tan \left (d x + c\right )}} + 3 \, \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {3}{2}}}{a^{3} e^{2} + \frac {2 \, a^{3} e^{2}}{\tan \left (d x + c\right )} + \frac {a^{3} e^{2}}{\tan \left (d x + c\right )^{2}}} - \frac {\frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}}}{a^{3}} - \frac {\arctan \left (\frac {\sqrt {\frac {e}{\tan \left (d x + c\right )}}}{\sqrt {e}}\right )}{a^{3} \sqrt {e}}\right )}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*e*((5*e*sqrt(e/tan(d*x + c)) + 3*(e/tan(d*x + c))^(3/2))/(a^3*e^2 + 2*a^3*e^2/tan(d*x + c) + a^3*e^2/tan(d
*x + c)^2) - (sqrt(2)*log(sqrt(2)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e) - sqrt(2)*log(-sq
rt(2)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e))/a^3 - arctan(sqrt(e/tan(d*x + c))/sqrt(e))/(
a^3*sqrt(e)))/d

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mupad [B]  time = 0.90, size = 151, normalized size = 0.94 \[ \frac {\frac {3\,e\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{8}+\frac {5\,e^2\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{8}}{d\,a^3\,e^2\,{\mathrm {cot}\left (c+d\,x\right )}^2+2\,d\,a^3\,e^2\,\mathrm {cot}\left (c+d\,x\right )+d\,a^3\,e^2}-\frac {\sqrt {e}\,\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{8\,a^3\,d}-\frac {\sqrt {2}\,\sqrt {e}\,\mathrm {atanh}\left (\frac {9\,\sqrt {2}\,e^{17/2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{32\,\left (\frac {9\,e^9\,\mathrm {cot}\left (c+d\,x\right )}{32}+\frac {9\,e^9}{32}\right )}\right )}{4\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cot(c + d*x))^(1/2)/(a + a*cot(c + d*x))^3,x)

[Out]

((3*e*(e*cot(c + d*x))^(3/2))/8 + (5*e^2*(e*cot(c + d*x))^(1/2))/8)/(a^3*d*e^2 + a^3*d*e^2*cot(c + d*x)^2 + 2*
a^3*d*e^2*cot(c + d*x)) - (e^(1/2)*atan((e*cot(c + d*x))^(1/2)/e^(1/2)))/(8*a^3*d) - (2^(1/2)*e^(1/2)*atanh((9
*2^(1/2)*e^(17/2)*(e*cot(c + d*x))^(1/2))/(32*((9*e^9*cot(c + d*x))/32 + (9*e^9)/32))))/(4*a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {e \cot {\left (c + d x \right )}}}{\cot ^{3}{\left (c + d x \right )} + 3 \cot ^{2}{\left (c + d x \right )} + 3 \cot {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))**(1/2)/(a+a*cot(d*x+c))**3,x)

[Out]

Integral(sqrt(e*cot(c + d*x))/(cot(c + d*x)**3 + 3*cot(c + d*x)**2 + 3*cot(c + d*x) + 1), x)/a**3

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