3.5.3 \(\int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\) [403]

Optimal. Leaf size=161 \[ \frac {\sqrt {-1+\sqrt {2}} \text {ArcTan}\left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}-\frac {2 \tanh ^{-1}\left (\sqrt {1+\tan (e+f x)}\right )}{f}+\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f} \]

[Out]

-2*arctanh((1+tan(f*x+e))^(1/2))/f+1/2*arctan((3-2*2^(1/2)+(1-2^(1/2))*tan(f*x+e))/(-14+10*2^(1/2))^(1/2)/(1+t
an(f*x+e))^(1/2))*(2^(1/2)-1)^(1/2)/f+1/2*arctanh((3+2*2^(1/2)+(1+2^(1/2))*tan(f*x+e))/(14+10*2^(1/2))^(1/2)/(
1+tan(f*x+e))^(1/2))*(1+2^(1/2))^(1/2)/f

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Rubi [A]
time = 0.15, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3655, 3617, 3616, 209, 213, 3715, 65} \begin {gather*} \frac {\sqrt {\sqrt {2}-1} \text {ArcTan}\left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f}-\frac {2 \tanh ^{-1}\left (\sqrt {\tan (e+f x)+1}\right )}{f}+\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]/Sqrt[1 + Tan[e + f*x]],x]

[Out]

(Sqrt[-1 + Sqrt[2]]*ArcTan[(3 - 2*Sqrt[2] + (1 - Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(-7 + 5*Sqrt[2])]*Sqrt[1 + Tan
[e + f*x]])])/(2*f) - (2*ArcTanh[Sqrt[1 + Tan[e + f*x]]])/f + (Sqrt[1 + Sqrt[2]]*ArcTanh[(3 + 2*Sqrt[2] + (1 +
Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(7 + 5*Sqrt[2])]*Sqrt[1 + Tan[e + f*x]])])/(2*f)

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 209

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

Rule 213

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

Rule 3616

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2*(
d^2/f), Subst[Int[1/(2*b*c*d - 4*a*d^2 + x^2), x], x, (b*c - 2*a*d - b*d*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*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] && EqQ[2
*a*c*d - b*(c^2 - d^2), 0]

Rule 3617

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> With[{q =
Rt[a^2 + b^2, 2]}, Dist[1/(2*q), Int[(a*c + b*d + c*q + (b*c - a*d + d*q)*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*
x]], x], x] - Dist[1/(2*q), Int[(a*c + b*d - c*q + (b*c - a*d - d*q)*Tan[e + f*x])/Sqrt[a + b*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] && NeQ[2
*a*c*d - b*(c^2 - d^2), 0] && (PerfectSquareQ[a^2 + b^2] || RationalQ[a, b, c, d])

Rule 3655

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/
(c^2 + d^2), Int[(a + b*Tan[e + f*x])^m*(c - d*Tan[e + f*x]), x], x] + Dist[d^2/(c^2 + d^2), Int[(a + b*Tan[e
+ f*x])^m*((1 + Tan[e + f*x]^2)/(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

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]

Rubi steps

\begin {align*} \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx &=-\int \frac {\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx+\int \frac {\cot (e+f x) \left (1+\tan ^2(e+f x)\right )}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=\frac {\int \frac {1+\left (-1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{2 \sqrt {2}}-\frac {\int \frac {1+\left (-1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{2 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}-\frac {\left (4-3 \sqrt {2}\right ) \text {Subst}\left (\int \frac {1}{2 \left (-1+\sqrt {2}\right )-4 \left (-1+\sqrt {2}\right )^2+x^2} \, dx,x,\frac {1-2 \left (-1+\sqrt {2}\right )-\left (-1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}}\right )}{2 f}-\frac {\left (4+3 \sqrt {2}\right ) \text {Subst}\left (\int \frac {1}{2 \left (-1-\sqrt {2}\right )-4 \left (-1-\sqrt {2}\right )^2+x^2} \, dx,x,\frac {1-2 \left (-1-\sqrt {2}\right )-\left (-1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}}\right )}{2 f}\\ &=\frac {\sqrt {-1+\sqrt {2}} \tan ^{-1}\left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}-\frac {2 \tanh ^{-1}\left (\sqrt {1+\tan (e+f x)}\right )}{f}+\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.10, size = 83, normalized size = 0.52 \begin {gather*} -\frac {2 \tanh ^{-1}\left (\sqrt {1+\tan (e+f x)}\right )}{f}+\frac {\tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i} f}+\frac {\tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]/Sqrt[1 + Tan[e + f*x]],x]

[Out]

(-2*ArcTanh[Sqrt[1 + Tan[e + f*x]]])/f + ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]]/(Sqrt[1 - I]*f) + ArcTanh
[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]]/(Sqrt[1 + I]*f)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 1.02, size = 2043, normalized size = 12.69

method	result	size
default	\(\text {Expression too large to display}\)	\(2043\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)/(1+tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/16/f*((cos(f*x+e)+sin(f*x+e))/cos(f*x+e))^(1/2)*(cos(f*x+e)+1)^2*(cos(f*x+e)-1)^2*(1+sin(f*x+e))*(((sin(f*x
+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*2^(1/2)*((2^(1/2)*cos(f*x+e)-2^(1/2)*sin(f*x+e)+2^(1/2)+2*sin(f*x
+e)-2)/cos(f*x+e)*2^(1/2))^(1/2)*((2^(1/2)*cos(f*x+e)-2^(1/2)*sin(f*x+e)+2^(1/2)-2*sin(f*x+e)+2)/cos(f*x+e)*2^
(1/2))^(1/2)*EllipticF(1/2*((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*2^(1/2),I*((2-2^(1/2))/(2+2^(
1/2)))^(1/2))-((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*2^(1/2)*((2^(1/2)*cos(f*x+e)-2^(1/2)*sin(f
*x+e)+2^(1/2)+2*sin(f*x+e)-2)/cos(f*x+e)*2^(1/2))^(1/2)*((2^(1/2)*cos(f*x+e)-2^(1/2)*sin(f*x+e)+2^(1/2)-2*sin(
f*x+e)+2)/cos(f*x+e)*2^(1/2))^(1/2)*EllipticE(1/2*((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*2^(1/2
),I*((2-2^(1/2))/(2+2^(1/2)))^(1/2))+2*2^(1/2)*EllipticE(1/2*((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(
1/2)*2^(1/2),I*((2-2^(1/2))/(2+2^(1/2)))^(1/2))*((sin(f*x+e)-1)*(1+2^(1/2))/cos(f*x+e))^(1/2)*((2^(1/2)*sin(f*
x+e)-2^(1/2)+cos(f*x+e)-sin(f*x+e)+1)/cos(f*x+e))^(1/2)*(-(2^(1/2)*sin(f*x+e)-2^(1/2)-cos(f*x+e)+sin(f*x+e)-1)
/cos(f*x+e))^(1/2)-28*2^(1/2)*EllipticPi(1/2*((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*2^(1/2),2^(
1/2)/(2+2^(1/2)),I*((2-2^(1/2))/(2+2^(1/2)))^(1/2))*((sin(f*x+e)-1)*(1+2^(1/2))/cos(f*x+e))^(1/2)*((2^(1/2)*si
n(f*x+e)-2^(1/2)+cos(f*x+e)-sin(f*x+e)+1)/cos(f*x+e))^(1/2)*(-(2^(1/2)*sin(f*x+e)-2^(1/2)-cos(f*x+e)+sin(f*x+e
)-1)/cos(f*x+e))^(1/2)+16*2^(1/2)*((sin(f*x+e)-1)*(1+2^(1/2))/cos(f*x+e))^(1/2)*((2^(1/2)*sin(f*x+e)-2^(1/2)+c
os(f*x+e)-sin(f*x+e)+1)/cos(f*x+e))^(1/2)*(-(2^(1/2)*sin(f*x+e)-2^(1/2)-cos(f*x+e)+sin(f*x+e)-1)/cos(f*x+e))^(
1/2)*EllipticPi(1/2*((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*2^(1/2),-I*2^(1/2)/(2+2^(1/2)),I*((2
-2^(1/2))/(2+2^(1/2)))^(1/2))+16*2^(1/2)*((sin(f*x+e)-1)*(1+2^(1/2))/cos(f*x+e))^(1/2)*((2^(1/2)*sin(f*x+e)-2^
(1/2)+cos(f*x+e)-sin(f*x+e)+1)/cos(f*x+e))^(1/2)*(-(2^(1/2)*sin(f*x+e)-2^(1/2)-cos(f*x+e)+sin(f*x+e)-1)/cos(f*
x+e))^(1/2)*EllipticPi(1/2*((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*2^(1/2),I*2^(1/2)/(2+2^(1/2))
,I*((2-2^(1/2))/(2+2^(1/2)))^(1/2))-((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*((2^(1/2)*cos(f*x+e)
-2^(1/2)*sin(f*x+e)+2^(1/2)+2*sin(f*x+e)-2)/cos(f*x+e)*2^(1/2))^(1/2)*((2^(1/2)*cos(f*x+e)-2^(1/2)*sin(f*x+e)+
2^(1/2)-2*sin(f*x+e)+2)/cos(f*x+e)*2^(1/2))^(1/2)*EllipticE(1/2*((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2)
)^(1/2)*2^(1/2),I*((2-2^(1/2))/(2+2^(1/2)))^(1/2))+6*((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*((2
^(1/2)*cos(f*x+e)-2^(1/2)*sin(f*x+e)+2^(1/2)+2*sin(f*x+e)-2)/cos(f*x+e)*2^(1/2))^(1/2)*((2^(1/2)*cos(f*x+e)-2^
(1/2)*sin(f*x+e)+2^(1/2)-2*sin(f*x+e)+2)/cos(f*x+e)*2^(1/2))^(1/2)*EllipticPi(1/2*((sin(f*x+e)-1)/cos(f*x+e)*(
2+2^(1/2))*2^(1/2))^(1/2)*2^(1/2),2^(1/2)/(2+2^(1/2)),I*((2-2^(1/2))/(2+2^(1/2)))^(1/2))-8*((sin(f*x+e)-1)/cos
(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*((2^(1/2)*cos(f*x+e)-2^(1/2)*sin(f*x+e)+2^(1/2)+2*sin(f*x+e)-2)/cos(f*x+e)*
2^(1/2))^(1/2)*((2^(1/2)*cos(f*x+e)-2^(1/2)*sin(f*x+e)+2^(1/2)-2*sin(f*x+e)+2)/cos(f*x+e)*2^(1/2))^(1/2)*Ellip
ticPi(1/2*((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*2^(1/2),-2^(1/2)/(2+2^(1/2)),I*((2-2^(1/2))/(2
+2^(1/2)))^(1/2))-4*((sin(f*x+e)-1)*(1+2^(1/2))/cos(f*x+e))^(1/2)*((2^(1/2)*sin(f*x+e)-2^(1/2)+cos(f*x+e)-sin(
f*x+e)+1)/cos(f*x+e))^(1/2)*(-(2^(1/2)*sin(f*x+e)-2^(1/2)-cos(f*x+e)+sin(f*x+e)-1)/cos(f*x+e))^(1/2)*EllipticF
(1/2*((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*2^(1/2),I*((2-2^(1/2))/(2+2^(1/2)))^(1/2))+4*Ellipt
icE(1/2*((sin(f*x+e)-1)/cos(f*x+e)*(2+2^(1/2))*2^(1/2))^(1/2)*2^(1/2),I*((2-2^(1/2))/(2+2^(1/2)))^(1/2))*((sin
(f*x+e)-1)*(1+2^(1/2))/cos(f*x+e))^(1/2)*((2^(1/2)*sin(f*x+e)-2^(1/2)+cos(f*x+e)-sin(f*x+e)+1)/cos(f*x+e))^(1/
2)*(-(2^(1/2)*sin(f*x+e)-2^(1/2)-cos(f*x+e)+sin(f*x+e)-1)/cos(f*x+e))^(1/2))*4^(1/2)/sin(f*x+e)^4/(cos(f*x+e)+
sin(f*x+e))/(2+2^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(1+tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(cot(f*x + e)/sqrt(tan(f*x + e) + 1), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 990 vs. \(2 (127) = 254\).
time = 1.42, size = 990, normalized size = 6.15 \begin {gather*} \frac {4 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {\frac {1}{2}} f^{3}\right )} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (2 \, \sqrt {\frac {1}{2}} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + f \cos \left (f x + e\right )\right )} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + \cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - 2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {\frac {1}{2}} f^{3}\right )} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - f^{2} \sqrt {\frac {1}{f^{4}}} - 2 \, \sqrt {\frac {1}{2}}\right ) + 4 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {\frac {1}{2}} f^{3}\right )} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (2 \, \sqrt {\frac {1}{2}} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + f \cos \left (f x + e\right )\right )} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + \cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - 2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {\frac {1}{2}} f^{3}\right )} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} + f^{2} \sqrt {\frac {1}{f^{4}}} + 2 \, \sqrt {\frac {1}{2}}\right ) + \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (\sqrt {\frac {1}{2}} f^{3} \sqrt {\frac {1}{f^{4}}} + f\right )} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (2 \, \sqrt {\frac {1}{2}} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + f \cos \left (f x + e\right )\right )} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + \cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) - \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (\sqrt {\frac {1}{2}} f^{3} \sqrt {\frac {1}{f^{4}}} + f\right )} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (2 \, \sqrt {\frac {1}{2}} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + f \cos \left (f x + e\right )\right )} \sqrt {-4 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + \cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) - 4 \, \log \left (\sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} + 1\right ) + 4 \, \log \left (\sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} - 1\right )}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(1+tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/4*(4*(1/2)^(3/4)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(2*(1/2)^(3/4)*(f^5*sqrt(f^(
-4)) + sqrt(1/2)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e)
+ (1/2)^(1/4)*(2*sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4))
+ 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + cos(f*x + e) + sin(f*x + e))/cos(f*x +
e))*(f^(-4))^(3/4) - 2*(1/2)^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(1/2)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*
sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) - f^2*sqrt(f^(-4)) - 2*sqrt(1/2)) + 4*(1/2)^(3
/4)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(2*(1/2)^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(1/2
)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) - (1/2)^(1/4)*(
2*sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(
f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3
/4) - 2*(1/2)^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(1/2)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x +
e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) + f^2*sqrt(f^(-4)) + 2*sqrt(1/2)) + (1/2)^(1/4)*(sqrt(1/2)*f^
3*sqrt(f^(-4)) + f)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*log((2*sqrt(1/2)*f^2*sqrt(f^(-4))*c
os(f*x + e) + (1/2)^(1/4)*(2*sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*s
qrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + cos(f*x + e) + sin(f*x + e)
)/cos(f*x + e)) - (1/2)^(1/4)*(sqrt(1/2)*f^3*sqrt(f^(-4)) + f)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4)
)^(1/4)*log((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) - (1/2)^(1/4)*(2*sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e
) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(
f^(-4))^(1/4) + cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) - 4*log(sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x
+ e)) + 1) + 4*log(sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) - 1))/f

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot {\left (e + f x \right )}}{\sqrt {\tan {\left (e + f x \right )} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(1+tan(f*x+e))**(1/2),x)

[Out]

Integral(cot(e + f*x)/sqrt(tan(e + f*x) + 1), x)

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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(cot(f*x+e)/(1+tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(cot(f*x + e)/sqrt(tan(f*x + e) + 1), x)

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Mupad [B]
time = 0.16, size = 85, normalized size = 0.53 \begin {gather*} -\frac {2\,\mathrm {atanh}\left (\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )}{f}+2\,\mathrm {atanh}\left (2\,f\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}+2\,\mathrm {atanh}\left (2\,f\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)/(tan(e + f*x) + 1)^(1/2),x)

[Out]

2*atanh(2*f*((1/8 - 1i/8)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2))*((1/8 - 1i/8)/f^2)^(1/2) - (2*atanh((tan(e + f*
x) + 1)^(1/2)))/f + 2*atanh(2*f*((1/8 + 1i/8)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2))*((1/8 + 1i/8)/f^2)^(1/2)

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