3.5.9 \(\int \frac {\cot ^2(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\) [409]

Optimal. Leaf size=280 \[ \frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 f}-\frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 f}+\frac {\tanh ^{-1}\left (\sqrt {1+\tan (e+f x)}\right )}{f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}-\frac {\cot (e+f x) \sqrt {1+\tan (e+f x)}}{f} \]

[Out]

arctanh((1+tan(f*x+e))^(1/2))/f+1/4*ln(1+2^(1/2)-(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(1+2^(
1/2))^(1/2)-1/4*ln(1+2^(1/2)+(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(1+2^(1/2))^(1/2)+1/2*arct
an(((2+2*2^(1/2))^(1/2)-2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))*(1+2^(1/2))^(1/2)/f-1/2*arctan(((2+2*2^(
1/2))^(1/2)+2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))*(1+2^(1/2))^(1/2)/f-cot(f*x+e)*(1+tan(f*x+e))^(1/2)/
f

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Rubi [A]
time = 0.27, antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 15, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3650, 3713, 21, 3654, 12, 3566, 722, 1108, 648, 632, 210, 642, 3715, 65, 213} \begin {gather*} \frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{2 f}-\frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{2 f}+\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}} f}-\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\tanh ^{-1}\left (\sqrt {\tan (e+f x)+1}\right )}{f}-\frac {\sqrt {\tan (e+f x)+1} \cot (e+f x)}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2/Sqrt[1 + Tan[e + f*x]],x]

[Out]

(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/(2*f) -
(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/(2*f) +
ArcTanh[Sqrt[1 + Tan[e + f*x]]]/f + Log[1 + Sqrt[2] + Tan[e + f*x] - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*
x]]]/(4*Sqrt[1 + Sqrt[2]]*f) - Log[1 + Sqrt[2] + Tan[e + f*x] + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/
(4*Sqrt[1 + Sqrt[2]]*f) - (Cot[e + f*x]*Sqrt[1 + Tan[e + f*x]])/f

Rule 12

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 722

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2*e, Subst[Int[1/(c*d^2 + a*e^2 - 2*c
*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1108

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}
, Dist[1/(2*c*q*r), Int[(r - x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(r + x)/(q + r*x + x^2), x], x
]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]

Rule 3566

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[(a + x)^n/(b^2 + x^2), x], x
, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 + b^2, 0]

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 3654

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 3713

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Tan[e + f*x])
^(m + 1)*(c + d*Tan[e + f*x])^n*(b*B - a*C + b*C*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.26, size = 101, normalized size = 0.36 \begin {gather*} -\frac {-2 \tanh ^{-1}\left (\sqrt {1+\tan (e+f x)}\right )+(1-i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )+(1+i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )+2 \cot (e+f x) \sqrt {1+\tan (e+f x)}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^2/Sqrt[1 + Tan[e + f*x]],x]

[Out]

-1/2*(-2*ArcTanh[Sqrt[1 + Tan[e + f*x]]] + (1 - I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + (1 + I)
^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]] + 2*Cot[e + f*x]*Sqrt[1 + Tan[e + f*x]])/f

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.65, size = 7175, normalized size = 25.62

method	result	size
default	\(\text {Expression too large to display}\)	\(7175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

integrate(cot(f*x + e)^2/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. 1016 vs. \(2 (225) = 450\).
time = 1.00, size = 1016, normalized size = 3.63 \begin {gather*} -\frac {\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} {\left (f \cos \left (f x + e\right )^{2} - \sqrt {\frac {1}{2}} {\left (f^{3} \cos \left (f x + e\right )^{2} - f^{3}\right )} \sqrt {\frac {1}{f^{4}}} - f\right )} \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 ) + 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \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 ) + \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}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} {\left (f \cos \left (f x + e\right )^{2} - \sqrt {\frac {1}{2}} {\left (f^{3} \cos \left (f x + e\right )^{2} - f^{3}\right )} \sqrt {\frac {1}{f^{4}}} - f\right )} \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 ) - 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \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 ) + \cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) - 2 \, \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (\sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} + 1\right ) + {\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (\sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} - 1\right ) - \frac {4 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (f^{5} \cos \left (f x + e\right )^{2} - f^{5}\right )} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \sqrt {\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \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 ) + \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 {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \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 )}{f^{4}} - \frac {4 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (f^{5} \cos \left (f x + e\right )^{2} - f^{5}\right )} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \sqrt {\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \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 ) + \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 {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \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 )}{f^{4}}}{2 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{2}{\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)**2/(1+tan(f*x+e))**(1/2),x)

[Out]

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

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Giac [A]
time = 0.86, size = 346, normalized size = 1.24 \begin {gather*} \frac {\log \left (\sqrt {\tan \left (f x + e\right ) + 1} + 1\right )}{2 \, f} - \frac {\log \left ({\left | \sqrt {\tan \left (f x + e\right ) + 1} - 1 \right |}\right )}{2 \, f} - \frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} - 2} + f \sqrt {2 \, \sqrt {2} + 2} {\left | f \right |}\right )} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{4 \, f^{3}} - \frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} - 2} + f \sqrt {2 \, \sqrt {2} + 2} {\left | f \right |}\right )} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{4 \, f^{3}} - \frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} + 2} - f \sqrt {2 \, \sqrt {2} - 2} {\left | f \right |}\right )} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{8 \, f^{3}} + \frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} + 2} - f \sqrt {2 \, \sqrt {2} - 2} {\left | f \right |}\right )} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{8 \, f^{3}} - \frac {\sqrt {\tan \left (f x + e\right ) + 1}}{f \tan \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(sqrt(tan(f*x + e) + 1) + 1)/f - 1/2*log(abs(sqrt(tan(f*x + e) + 1) - 1))/f - 1/4*(f^2*sqrt(2*sqrt(2) -
2) + f*sqrt(2*sqrt(2) + 2)*abs(f))*arctan(1/2*2^(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) + 2*sqrt(tan(f*x + e) + 1))/
sqrt(-sqrt(2) + 2))/f^3 - 1/4*(f^2*sqrt(2*sqrt(2) - 2) + f*sqrt(2*sqrt(2) + 2)*abs(f))*arctan(-1/2*2^(3/4)*(2^
(1/4)*sqrt(sqrt(2) + 2) - 2*sqrt(tan(f*x + e) + 1))/sqrt(-sqrt(2) + 2))/f^3 - 1/8*(f^2*sqrt(2*sqrt(2) + 2) - f
*sqrt(2*sqrt(2) - 2)*abs(f))*log(2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqrt(2) + tan(f*x + e) + 1
)/f^3 + 1/8*(f^2*sqrt(2*sqrt(2) + 2) - f*sqrt(2*sqrt(2) - 2)*abs(f))*log(-2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f
*x + e) + 1) + sqrt(2) + tan(f*x + e) + 1)/f^3 - sqrt(tan(f*x + e) + 1)/(f*tan(f*x + e))

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Mupad [B]
time = 0.17, size = 117, normalized size = 0.42 \begin {gather*} \frac {\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f-f\,\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}-\frac {\mathrm {atan}\left (\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{f}-\mathrm {atan}\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 {i}+\mathrm {atan}\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 {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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