3.387 \(\int \cot ^2(e+f x) \sqrt {1+\tan (e+f x)} \, dx\)

Optimal. Leaf size=288 \[ \frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}-\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}+\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\tanh ^{-1}\left (\sqrt {\tan (e+f x)+1}\right )}{f}-\frac {\sqrt {\tan (e+f x)+1} \cot (e+f x)}{f} \]

[Out]

-arctanh((1+tan(f*x+e))^(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))
*(2+2*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))*(2+2*2^(1
/2))^(1/2)/f-1/2*ln(1+2^(1/2)-(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(2+2*2^(1/2))^(1/2)+1/2*l
n(1+2^(1/2)+(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(2+2*2^(1/2))^(1/2)-cot(f*x+e)*(1+tan(f*x+e
))^(1/2)/f

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Rubi [A]  time = 0.35, antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3568, 3653, 3485, 700, 1127, 1161, 618, 204, 1164, 628, 3634, 63, 207} \[ \frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}-\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}+\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\tanh ^{-1}\left (\sqrt {\tan (e+f x)+1}\right )}{f}-\frac {\sqrt {\tan (e+f x)+1} \cot (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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])

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 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 700

Int[Sqrt[(d_) + (e_.)*(x_)]/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(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 1127

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

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[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, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 1164

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e - b/c, 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, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

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 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 \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}+\frac {1}{2} \int \frac {\cot (e+f x) \left (1+\tan ^2(e+f x)\right )}{\sqrt {1+\tan (e+f x)}} \, dx-\int \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}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {1+x}}{1+x^2} \, dx,x,\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}{-1+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}-\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{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}-x^2}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+x^2}{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 {1}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{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 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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} 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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} 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 )}{2 \sqrt {2 \left (1+\sqrt {2}\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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} 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 )}{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 )}{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 )}{\sqrt {2 \left (-1+\sqrt {2}\right )} 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 )}{\sqrt {2 \left (-1+\sqrt {2}\right )} 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 )}{2 \sqrt {2 \left (1+\sqrt {2}\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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\cot (e+f x) \sqrt {1+\tan (e+f x)}}{f}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 102, normalized size = 0.35 \[ -\frac {\tanh ^{-1}\left (\sqrt {\tan (e+f x)+1}\right )-i \sqrt {1-i} \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1-i}}\right )+i \sqrt {1+i} \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1+i}}\right )+\sqrt {\tan (e+f x)+1} \cot (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.56, size = 1000, normalized size = 3.47 \[ \text {result too large to display} \]

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]

1/8*(2^(1/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(2*f*cos(f*x + e)^2 - sqrt(2)*(f^3*cos(f*x + e)^2 - f^3)*sqr
t(f^(-4)) - 2*f)*(f^(-4))^(1/4)*log(1/2*(2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((cos(f*x + e) +
sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4)*cos(f*x + e) + 2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2*cos(f*x
+ e) + 2*sin(f*x + e))/cos(f*x + e)) - 2^(1/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(2*f*cos(f*x + e)^2 - sqr
t(2)*(f^3*cos(f*x + e)^2 - f^3)*sqrt(f^(-4)) - 2*f)*(f^(-4))^(1/4)*log(-1/2*(2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f
^(-4)) + 4)*f^3*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4)*cos(f*x + e) - 2*sqrt(2)*f^2*s
qrt(f^(-4))*cos(f*x + e) - 2*cos(f*x + e) - 2*sin(f*x + e))/cos(f*x + e)) + 8*sqrt((cos(f*x + e) + sin(f*x + e
))/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - 4*(cos(f*x + e)^2 - 1)*log(sqrt((cos(f*x + e) + sin(f*x + e))/cos
(f*x + e)) + 1) + 4*(cos(f*x + e)^2 - 1)*log(sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) - 1) + 4*2^(3/4)
*(f^5*cos(f*x + e)^2 - f^5)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*arctan(1/2*2^(3/4)*sqrt(1/2)*s
qrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^5*sqrt((2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((cos(f*x +
e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4)*cos(f*x + e) + 2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2*co
s(f*x + e) + 2*sin(f*x + e))/cos(f*x + e))*(f^(-4))^(5/4) - 1/2*2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f
^5*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(5/4) - f^2*sqrt(f^(-4)) - sqrt(2))/f^4 + 4*2^(3/
4)*(f^5*cos(f*x + e)^2 - f^5)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*arctan(1/2*2^(3/4)*sqrt(1/2)
*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^5*sqrt(-(2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((cos(f*
x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4)*cos(f*x + e) - 2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2
*cos(f*x + e) - 2*sin(f*x + e))/cos(f*x + e))*(f^(-4))^(5/4) - 1/2*2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4
)*f^5*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(5/4) + f^2*sqrt(f^(-4)) + sqrt(2))/f^4)/(f*co
s(f*x + e)^2 - f)

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

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)(3616*sqrt(sqrt(2)-1)*f^2-3616*sqrt(sqrt(2)+1)*f*abs(f))/14464
/f^3*ln(tan(f*x+exp(1))+1-(2*f/f)^(1/4)*sqrt(2+sqrt(2))*sqrt(tan(f*x+exp(1))+1)+(2*f/f)^(1/4)*(2*f/f)^(1/4))+(
-3616*sqrt(sqrt(2)+1)*f^2-3616*sqrt(sqrt(2)-1)*f*abs(f))/7232/f^3*atan((sqrt(tan(f*x+exp(1))+1)-sqrt(2+sqrt(2)
)/2*(2*f/f)^(1/4))/sqrt(2-sqrt(2))*2/(2*f/f)^(1/4))+(-3616*sqrt(sqrt(2)-1)*f^2+3616*sqrt(sqrt(2)+1)*f*abs(f))/
14464/f^3*ln(tan(f*x+exp(1))+1+(2*f/f)^(1/4)*sqrt(2+sqrt(2))*sqrt(tan(f*x+exp(1))+1)+(2*f/f)^(1/4)*(2*f/f)^(1/
4))-(3616*sqrt(sqrt(2)+1)*f^2+3616*sqrt(sqrt(2)-1)*f*abs(f))/7232/f^3*atan((sqrt(tan(f*x+exp(1))+1)+sqrt(2+sqr
t(2))/2*(2*f/f)^(1/4))/sqrt(2-sqrt(2))*2/(2*f/f)^(1/4))+1/2/f*ln(abs(sqrt(tan(f*x+exp(1))+1)-1))-1/2/f*ln(sqrt
(tan(f*x+exp(1))+1)+1)-sqrt(tan(f*x+exp(1))+1)/f/tan(f*x+exp(1))

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maple [C]  time = 1.30, size = 8262, normalized size = 28.69 \[ \text {output too large to display} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\tan \left (f x + e\right ) + 1} \cot \left (f x + e\right )^{2}\,{d x} \]

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

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mupad [B]  time = 3.89, size = 121, normalized size = 0.42 \[ \frac {\mathrm {atan}\left (\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{f}+\frac {\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f-f\,\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}+\mathrm {atan}\left (f\,\sqrt {\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1-\mathrm {i}\right )\right )\,\sqrt {\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (f\,\sqrt {\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1+1{}\mathrm {i}\right )\right )\,\sqrt {\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \]

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]

(atan((tan(e + f*x) + 1)^(1/2)*1i)*1i)/f + (tan(e + f*x) + 1)^(1/2)/(f - f*(tan(e + f*x) + 1)) + atan(f*((- 1/
4 - 1i/4)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*(1 - 1i))*((- 1/4 - 1i/4)/f^2)^(1/2)*2i - atan(f*((- 1/4 + 1i/4)
/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*(1 + 1i))*((- 1/4 + 1i/4)/f^2)^(1/2)*2i

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

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

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