3.31 \(\int \frac {\sqrt {e \cot (c+d x)}}{(a+a \cot (c+d x))^2} \, dx\)

Optimal. Leaf size=278 \[ \frac {\sqrt {e \cot (c+d x)}}{2 d \left (a^2 \cot (c+d x)+a^2\right )}+\frac {\sqrt {e} \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d}-\frac {\sqrt {e} \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d}+\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d}+\frac {\sqrt {e} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 \sqrt {2} a^2 d}-\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{2 \sqrt {2} a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.53, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3568, 3653, 12, 3476, 329, 211, 1165, 628, 1162, 617, 204, 3634, 63, 205} \[ \frac {\sqrt {e \cot (c+d x)}}{2 d \left (a^2 \cot (c+d x)+a^2\right )}+\frac {\sqrt {e} \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d}-\frac {\sqrt {e} \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{4 \sqrt {2} a^2 d}+\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d}+\frac {\sqrt {e} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 \sqrt {2} a^2 d}-\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{2 \sqrt {2} a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e]*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(2*a^2*d) + (Sqrt[e]*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/
Sqrt[e]])/(2*Sqrt[2]*a^2*d) - (Sqrt[e]*ArcTan[1 + (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(2*Sqrt[2]*a^2*d) +
Sqrt[e*Cot[c + d*x]]/(2*d*(a^2 + a^2*Cot[c + d*x])) + (Sqrt[e]*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*S
qrt[e*Cot[c + d*x]]])/(4*Sqrt[2]*a^2*d) - (Sqrt[e]*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] + Sqrt[2]*Sqrt[e*Cot[c +
d*x]]])/(4*Sqrt[2]*a^2*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 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 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 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 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 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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 1.30, size = 207, normalized size = 0.74 \[ \frac {\csc (c+d x) \sqrt {e \cot (c+d x)} (\sin (c+d x)+\cos (c+d x)) \left (\frac {1}{2} (\tan (c+d x)+1) \sqrt {\cot (c+d x)} \left (\sqrt {2} \log \left (-\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}-1\right )-\sqrt {2} \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+4 \tan ^{-1}\left (\sqrt {\cot (c+d x)}\right )\right )+2\right )}{4 a^2 d (\cot (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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))^2,x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   catdef: division by zero

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

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))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.79, size = 223, normalized size = 0.80 \[ -\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{8 d \,a^{2}}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d \,a^{2}}+\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d \,a^{2}}+\frac {e \sqrt {e \cot \left (d x +c \right )}}{2 d \,a^{2} \left (e \cot \left (d x +c \right )+e \right )}+\frac {\arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e}}{2 a^{2} d} \]

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)^2,x)

[Out]

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

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maxima [A]  time = 0.58, size = 231, normalized size = 0.83 \[ \frac {e {\left (\frac {4 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}}{a^{2} e + \frac {a^{2} e}{\tan \left (d x + c\right )}} - \frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}}}{a^{2}} + \frac {4 \, \arctan \left (\frac {\sqrt {\frac {e}{\tan \left (d x + c\right )}}}{\sqrt {e}}\right )}{a^{2} \sqrt {e}}\right )}}{8 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

1/8*e*(4*sqrt(e/tan(d*x + c))/(a^2*e + a^2*e/tan(d*x + c)) - (2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) +
2*sqrt(e/tan(d*x + c)))/sqrt(e))/sqrt(e) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(e/tan(d*x +
c)))/sqrt(e))/sqrt(e) + 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^2 + 4*arctan(sqrt(e/tan(d*x + c
))/sqrt(e))/(a^2*sqrt(e)))/d

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mupad [B]  time = 0.72, size = 366, normalized size = 1.32 \[ \frac {\mathrm {atan}\left (\frac {4\,e^{12}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {e^2}{a^8\,d^4}\right )}^{1/4}}{\frac {4\,e^{13}}{a^2\,d}-4\,a^2\,d\,e^{12}\,\sqrt {-\frac {e^2}{a^8\,d^4}}}+\frac {4\,e^{11}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {e^2}{a^8\,d^4}\right )}^{3/4}}{\frac {4\,e^{13}}{a^6\,d^3}-\frac {4\,e^{12}\,\sqrt {-\frac {e^2}{a^8\,d^4}}}{a^2\,d}}\right )\,{\left (-\frac {e^2}{a^8\,d^4}\right )}^{1/4}}{2}+\mathrm {atan}\left (\frac {e^{12}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {e^2}{256\,a^8\,d^4}\right )}^{1/4}\,16{}\mathrm {i}}{\frac {4\,e^{13}}{a^2\,d}+64\,a^2\,d\,e^{12}\,\sqrt {-\frac {e^2}{256\,a^8\,d^4}}}-\frac {e^{11}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {e^2}{256\,a^8\,d^4}\right )}^{3/4}\,256{}\mathrm {i}}{\frac {4\,e^{13}}{a^6\,d^3}+\frac {64\,e^{12}\,\sqrt {-\frac {e^2}{256\,a^8\,d^4}}}{a^2\,d}}\right )\,{\left (-\frac {e^2}{256\,a^8\,d^4}\right )}^{1/4}\,2{}\mathrm {i}+\frac {e\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\left (a^2\,d\,e+a^2\,d\,e\,\mathrm {cot}\left (c+d\,x\right )\right )}-\frac {\sqrt {-e}\,\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,1{}\mathrm {i}}{\sqrt {-e}}\right )\,1{}\mathrm {i}}{2\,a^2\,d} \]

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))^2,x)

[Out]

(atan((4*e^12*(e*cot(c + d*x))^(1/2)*(-e^2/(a^8*d^4))^(1/4))/((4*e^13)/(a^2*d) - 4*a^2*d*e^12*(-e^2/(a^8*d^4))
^(1/2)) + (4*e^11*(e*cot(c + d*x))^(1/2)*(-e^2/(a^8*d^4))^(3/4))/((4*e^13)/(a^6*d^3) - (4*e^12*(-e^2/(a^8*d^4)
)^(1/2))/(a^2*d)))*(-e^2/(a^8*d^4))^(1/4))/2 + atan((e^12*(e*cot(c + d*x))^(1/2)*(-e^2/(256*a^8*d^4))^(1/4)*16
i)/((4*e^13)/(a^2*d) + 64*a^2*d*e^12*(-e^2/(256*a^8*d^4))^(1/2)) - (e^11*(e*cot(c + d*x))^(1/2)*(-e^2/(256*a^8
*d^4))^(3/4)*256i)/((4*e^13)/(a^6*d^3) + (64*e^12*(-e^2/(256*a^8*d^4))^(1/2))/(a^2*d)))*(-e^2/(256*a^8*d^4))^(
1/4)*2i + (e*(e*cot(c + d*x))^(1/2))/(2*(a^2*d*e + a^2*d*e*cot(c + d*x))) - ((-e)^(1/2)*atan(((e*cot(c + d*x))
^(1/2)*1i)/(-e)^(1/2))*1i)/(2*a^2*d)

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

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))**2,x)

[Out]

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

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