3.409 \(\int \frac{\cot ^2(e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx\)
Optimal. Leaf size=280 \[ \frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{\tan (e+f x)+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{2 f}-\frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sqrt{\tan (e+f x)+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{2 f}+\frac{\log \left (\tan (e+f x)-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{4 \sqrt{1+\sqrt{2}} f}-\frac{\log \left (\tan (e+f x)+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{4 \sqrt{1+\sqrt{2}} f}+\frac{\tanh ^{-1}\left (\sqrt{\tan (e+f x)+1}\right )}{f}-\frac{\sqrt{\tan (e+f x)+1} \cot (e+f x)}{f} \]
[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
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Rubi [A] time = 0.3989, antiderivative size = 280, normalized size of antiderivative = 1.,
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 = {3569, 3632, 21, 3573, 12, 3485, 708, 1094, 634, 618, 204, 628, 3634, 63, 207}
\[ \frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{\tan (e+f x)+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{2 f}-\frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sqrt{\tan (e+f x)+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{2 f}+\frac{\log \left (\tan (e+f x)-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{4 \sqrt{1+\sqrt{2}} f}-\frac{\log \left (\tan (e+f x)+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{4 \sqrt{1+\sqrt{2}} f}+\frac{\tanh ^{-1}\left (\sqrt{\tan (e+f x)+1}\right )}{f}-\frac{\sqrt{\tan (e+f x)+1} \cot (e+f x)}{f} \]
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 3569
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 3632
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 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 3573
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 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 3485
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 708
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 1094
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 634
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 618
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 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 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 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 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 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
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{\operatorname{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{\operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sqrt{1+\tan (e+f x)}\right )}{f}-\frac{\operatorname{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 \operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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*}
Mathematica [C] time = 0.238339, size = 101, normalized size = 0.36 \[ -\frac{-2 \tanh ^{-1}\left (\sqrt{\tan (e+f x)+1}\right )+(1-i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1-i}}\right )+(1+i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1+i}}\right )+2 \sqrt{\tan (e+f x)+1} \cot (e+f x)}{2 f} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[e + f*x]^2/Sqrt[1 + Tan[e + f*x]],x]
[Out]
-(-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]])/(2*f)
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Maple [C] time = 0.504, size = 7217, normalized size = 25.8 \begin{align*} \text{output too large to display} \end{align*}
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)
[Out]
result too large to display
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{2}}{\sqrt{\tan \left (f x + e\right ) + 1}}\,{d x} \end{align*}
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] time = 2.1368, size = 2851, normalized size = 10.18 \begin{align*} \text{result too large to display} \end{align*}
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., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{2}{\left (e + f x \right )}}{\sqrt{\tan{\left (e + f x \right )} + 1}}\, dx \end{align*}
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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{2}}{\sqrt{\tan \left (f x + e\right ) + 1}}\,{d x} \end{align*}
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]
integrate(cot(f*x + e)^2/sqrt(tan(f*x + e) + 1), x)