3.1.32 \(\int \frac {1}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))^2} \, dx\) [32]

Optimal. Leaf size=281 \[ -\frac {3 \text {ArcTan}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d \sqrt {e}}-\frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 \sqrt {2} a^2 d \sqrt {e}}+\frac {\text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 \sqrt {2} a^2 d \sqrt {e}}-\frac {\sqrt {e \cot (c+d x)}}{2 d e \left (a^2+a^2 \cot (c+d x)\right )}+\frac {\log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{4 \sqrt {2} a^2 d \sqrt {e}}-\frac {\log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{4 \sqrt {2} a^2 d \sqrt {e}} \]

[Out]

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

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Rubi [A]
time = 0.37, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3650, 3734, 12, 16, 3557, 335, 303, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} -\frac {3 \text {ArcTan}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 a^2 d \sqrt {e}}-\frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{2 \sqrt {2} a^2 d \sqrt {e}}+\frac {\text {ArcTan}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{2 \sqrt {2} a^2 d \sqrt {e}}-\frac {\sqrt {e \cot (c+d x)}}{2 d e \left (a^2 \cot (c+d x)+a^2\right )}+\frac {\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 \sqrt {e}}-\frac {\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 \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*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]) -
Sqrt[e*Cot[c + d*x]]/(2*d*e*(a^2 + a^2*Cot[c + d*x])) + Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*Sqrt[e*Co
t[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*Sqrt[e])

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 65

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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

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

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 335

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 631

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 642

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 1176

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 1179

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 3557

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 3650

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 3715

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 3734

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

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Mathematica [A]
time = 0.88, size = 337, normalized size = 1.20 \begin {gather*} -\frac {\sqrt {\cot (c+d x)} \left (12 \text {ArcTan}\left (\sqrt {\cot (c+d x)}\right ) \cos (c+d x)-\sqrt {2} \cos (c+d x) \log \left (-1+\sqrt {2} \sqrt {\cot (c+d x)}-\cot (c+d x)\right )+\sqrt {2} \cos (c+d x) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )+12 \text {ArcTan}\left (\sqrt {\cot (c+d x)}\right ) \sin (c+d x)+4 \sqrt {\cot (c+d x)} \sin (c+d x)-\sqrt {2} \log \left (-1+\sqrt {2} \sqrt {\cot (c+d x)}-\cot (c+d x)\right ) \sin (c+d x)+\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right ) \sin (c+d x)+2 \sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right ) (\cos (c+d x)+\sin (c+d x))-2 \sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right ) (\cos (c+d x)+\sin (c+d x))\right )}{8 a^2 d \sqrt {e \cot (c+d x)} (\cos (c+d x)+\sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(Sqrt[Cot[c + d*x]]*(12*ArcTan[Sqrt[Cot[c + d*x]]]*Cos[c + d*x] - Sqrt[2]*Cos[c + d*x]*Log[-1 + Sqrt[2]*S
qrt[Cot[c + d*x]] - Cot[c + d*x]] + Sqrt[2]*Cos[c + d*x]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] +
12*ArcTan[Sqrt[Cot[c + d*x]]]*Sin[c + d*x] + 4*Sqrt[Cot[c + d*x]]*Sin[c + d*x] - Sqrt[2]*Log[-1 + Sqrt[2]*Sqrt
[Cot[c + d*x]] - Cot[c + d*x]]*Sin[c + d*x] + Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]*Sin[c
+ d*x] + 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]]*(Cos[c + d*x] + Sin[c + d*x]) - 2*Sqrt[2]*ArcTan[1
+ Sqrt[2]*Sqrt[Cot[c + d*x]]]*(Cos[c + d*x] + Sin[c + d*x])))/(a^2*d*Sqrt[e*Cot[c + d*x]]*(Cos[c + d*x] + Sin[
c + d*x]))

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Maple [A]
time = 0.54, size = 197, normalized size = 0.70

method	result	size
derivativedivides	\(-\frac {2 e^{3} \left (\frac {\frac {\sqrt {e \cot \left (d x +c \right )}}{2 e \cot \left (d x +c \right )+2 e}+\frac {3 \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2 \sqrt {e}}}{2 e^{3}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 e^{3} \left (e^{2}\right )^{\frac {1}{4}}}\right )}{d \,a^{2}}\)	\(197\)
default	\(-\frac {2 e^{3} \left (\frac {\frac {\sqrt {e \cot \left (d x +c \right )}}{2 e \cot \left (d x +c \right )+2 e}+\frac {3 \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2 \sqrt {e}}}{2 e^{3}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 e^{3} \left (e^{2}\right )^{\frac {1}{4}}}\right )}{d \,a^{2}}\)	\(197\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.54, size = 161, normalized size = 0.57 \begin {gather*} \frac {{\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{2}} - \frac {12 \, \arctan \left (\frac {1}{\sqrt {\tan \left (d x + c\right )}}\right )}{a^{2}} - \frac {4}{{\left (a^{2} + \frac {a^{2}}{\tan \left (d x + c\right )}\right )} \sqrt {\tan \left (d x + c\right )}}\right )} e^{\left (-\frac {1}{2}\right )}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)
- 2/sqrt(tan(d*x + c)))) - sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(2)*log(-sqrt(2)
/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))/a^2 - 12*arctan(1/sqrt(tan(d*x + c)))/a^2 - 4/((a^2 + a^2/tan(d*x +
c))*sqrt(tan(d*x + c))))*e^(-1/2)/d

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{\sqrt {e \cot {\left (c + d x \right )}} \cot ^{2}{\left (c + d x \right )} + 2 \sqrt {e \cot {\left (c + d x \right )}} \cot {\left (c + d x \right )} + \sqrt {e \cot {\left (c + d x \right )}}}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.81, size = 366, normalized size = 1.30 \begin {gather*} \frac {\mathrm {atan}\left (\frac {4\,e^8\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {1}{a^8\,d^4\,e^2}\right )}^{1/4}}{\frac {4\,e^8}{a^2\,d}+36\,a^2\,d\,e^9\,\sqrt {-\frac {1}{a^8\,d^4\,e^2}}}+\frac {36\,e^9\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {1}{a^8\,d^4\,e^2}\right )}^{3/4}}{\frac {4\,e^8}{a^6\,d^3}+\frac {36\,e^9\,\sqrt {-\frac {1}{a^8\,d^4\,e^2}}}{a^2\,d}}\right )\,{\left (-\frac {1}{a^8\,d^4\,e^2}\right )}^{1/4}}{2}+\mathrm {atan}\left (\frac {e^8\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^4\,e^2}\right )}^{1/4}\,16{}\mathrm {i}}{\frac {4\,e^8}{a^2\,d}-576\,a^2\,d\,e^9\,\sqrt {-\frac {1}{256\,a^8\,d^4\,e^2}}}-\frac {e^9\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^4\,e^2}\right )}^{3/4}\,2304{}\mathrm {i}}{\frac {4\,e^8}{a^6\,d^3}-\frac {576\,e^9\,\sqrt {-\frac {1}{256\,a^8\,d^4\,e^2}}}{a^2\,d}}\right )\,{\left (-\frac {1}{256\,a^8\,d^4\,e^2}\right )}^{1/4}\,2{}\mathrm {i}-\frac {\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 {\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,1{}\mathrm {i}}{\sqrt {-e}}\right )\,3{}\mathrm {i}}{2\,a^2\,d\,\sqrt {-e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((4*e^8*(e*cot(c + d*x))^(1/2)*(-1/(a^8*d^4*e^2))^(1/4))/((4*e^8)/(a^2*d) + 36*a^2*d*e^9*(-1/(a^8*d^4*e^2
))^(1/2)) + (36*e^9*(e*cot(c + d*x))^(1/2)*(-1/(a^8*d^4*e^2))^(3/4))/((4*e^8)/(a^6*d^3) + (36*e^9*(-1/(a^8*d^4
*e^2))^(1/2))/(a^2*d)))*(-1/(a^8*d^4*e^2))^(1/4))/2 + atan((e^8*(e*cot(c + d*x))^(1/2)*(-1/(256*a^8*d^4*e^2))^
(1/4)*16i)/((4*e^8)/(a^2*d) - 576*a^2*d*e^9*(-1/(256*a^8*d^4*e^2))^(1/2)) - (e^9*(e*cot(c + d*x))^(1/2)*(-1/(2
56*a^8*d^4*e^2))^(3/4)*2304i)/((4*e^8)/(a^6*d^3) - (576*e^9*(-1/(256*a^8*d^4*e^2))^(1/2))/(a^2*d)))*(-1/(256*a
^8*d^4*e^2))^(1/4)*2i - (e*cot(c + d*x))^(1/2)/(2*(a^2*d*e + a^2*d*e*cot(c + d*x))) - (atan(((e*cot(c + d*x))^
(1/2)*1i)/(-e)^(1/2))*3i)/(2*a^2*d*(-e)^(1/2))

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