3.382 \(\int \cot ^3(e+f x) \sqrt{1+\tan (e+f x)} \, dx\)

Optimal. Leaf size=221 \[ -\frac{\sqrt{\frac{1}{2} \left (\sqrt{2}-1\right )} \tan ^{-1}\left (\frac{\left (2-\sqrt{2}\right ) \tan (e+f x)-3 \sqrt{2}+4}{2 \sqrt{5 \sqrt{2}-7} \sqrt{\tan (e+f x)+1}}\right )}{f}+\frac{9 \tanh ^{-1}\left (\sqrt{\tan (e+f x)+1}\right )}{4 f}-\frac{\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tanh ^{-1}\left (\frac{\left (2+\sqrt{2}\right ) \tan (e+f x)+3 \sqrt{2}+4}{2 \sqrt{7+5 \sqrt{2}} \sqrt{\tan (e+f x)+1}}\right )}{f}-\frac{\sqrt{\tan (e+f x)+1} \cot ^2(e+f x)}{2 f}-\frac{\sqrt{\tan (e+f x)+1} \cot (e+f x)}{4 f} \]

[Out]

-((Sqrt[(-1 + Sqrt[2])/2]*ArcTan[(4 - 3*Sqrt[2] + (2 - Sqrt[2])*Tan[e + f*x])/(2*Sqrt[-7 + 5*Sqrt[2]]*Sqrt[1 +
Tan[e + f*x]])])/f) + (9*ArcTanh[Sqrt[1 + Tan[e + f*x]]])/(4*f) - (Sqrt[(1 + Sqrt[2])/2]*ArcTanh[(4 + 3*Sqrt[
2] + (2 + Sqrt[2])*Tan[e + f*x])/(2*Sqrt[7 + 5*Sqrt[2]]*Sqrt[1 + Tan[e + f*x]])])/f - (Cot[e + f*x]*Sqrt[1 + T
an[e + f*x]])/(4*f) - (Cot[e + f*x]^2*Sqrt[1 + Tan[e + f*x]])/(2*f)

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Rubi [A]  time = 0.500701, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3568, 3649, 3653, 3536, 3535, 203, 207, 3634, 63} \[ -\frac{\sqrt{\frac{1}{2} \left (\sqrt{2}-1\right )} \tan ^{-1}\left (\frac{\left (2-\sqrt{2}\right ) \tan (e+f x)-3 \sqrt{2}+4}{2 \sqrt{5 \sqrt{2}-7} \sqrt{\tan (e+f x)+1}}\right )}{f}+\frac{9 \tanh ^{-1}\left (\sqrt{\tan (e+f x)+1}\right )}{4 f}-\frac{\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tanh ^{-1}\left (\frac{\left (2+\sqrt{2}\right ) \tan (e+f x)+3 \sqrt{2}+4}{2 \sqrt{7+5 \sqrt{2}} \sqrt{\tan (e+f x)+1}}\right )}{f}-\frac{\sqrt{\tan (e+f x)+1} \cot ^2(e+f x)}{2 f}-\frac{\sqrt{\tan (e+f x)+1} \cot (e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^3*Sqrt[1 + Tan[e + f*x]],x]

[Out]

-((Sqrt[(-1 + Sqrt[2])/2]*ArcTan[(4 - 3*Sqrt[2] + (2 - Sqrt[2])*Tan[e + f*x])/(2*Sqrt[-7 + 5*Sqrt[2]]*Sqrt[1 +
Tan[e + f*x]])])/f) + (9*ArcTanh[Sqrt[1 + Tan[e + f*x]]])/(4*f) - (Sqrt[(1 + Sqrt[2])/2]*ArcTanh[(4 + 3*Sqrt[
2] + (2 + Sqrt[2])*Tan[e + f*x])/(2*Sqrt[7 + 5*Sqrt[2]]*Sqrt[1 + Tan[e + f*x]])])/f - (Cot[e + f*x]*Sqrt[1 + T
an[e + f*x]])/(4*f) - (Cot[e + f*x]^2*Sqrt[1 + Tan[e + f*x]])/(2*f)

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 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, 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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

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]

Rule 3536

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 3535

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 203

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

Rubi steps

\begin{align*} \int \cot ^3(e+f x) \sqrt{1+\tan (e+f x)} \, dx &=-\frac{\cot ^2(e+f x) \sqrt{1+\tan (e+f x)}}{2 f}-\frac{1}{2} \int \frac{\cot ^2(e+f x) \left (-\frac{1}{2}+2 \tan (e+f x)+\frac{3}{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)}}{4 f}-\frac{\cot ^2(e+f x) \sqrt{1+\tan (e+f x)}}{2 f}+\frac{1}{2} \int \frac{\cot (e+f x) \left (-\frac{9}{4}-2 \tan (e+f x)-\frac{1}{4} \tan ^2(e+f x)\right )}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=-\frac{\cot (e+f x) \sqrt{1+\tan (e+f x)}}{4 f}-\frac{\cot ^2(e+f x) \sqrt{1+\tan (e+f x)}}{2 f}+\frac{1}{2} \int \frac{-2+2 \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx-\frac{9}{8} \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)}}{4 f}-\frac{\cot ^2(e+f x) \sqrt{1+\tan (e+f x)}}{2 f}-\frac{\int \frac{2 \sqrt{2}+\left (-4-2 \sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx}{4 \sqrt{2}}+\frac{\int \frac{-2 \sqrt{2}+\left (-4+2 \sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx}{4 \sqrt{2}}-\frac{9 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+x}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac{\cot (e+f x) \sqrt{1+\tan (e+f x)}}{4 f}-\frac{\cot ^2(e+f x) \sqrt{1+\tan (e+f x)}}{2 f}-\frac{9 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sqrt{1+\tan (e+f x)}\right )}{4 f}+\frac{\left (2 \left (4-3 \sqrt{2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \sqrt{2} \left (-4+2 \sqrt{2}\right )-4 \left (-4+2 \sqrt{2}\right )^2+x^2} \, dx,x,\frac{-2 \sqrt{2}-2 \left (-4+2 \sqrt{2}\right )-\left (-4+2 \sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}}\right )}{f}+\frac{\left (2 \left (4+3 \sqrt{2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 \sqrt{2} \left (-4-2 \sqrt{2}\right )-4 \left (-4-2 \sqrt{2}\right )^2+x^2} \, dx,x,\frac{2 \sqrt{2}-2 \left (-4-2 \sqrt{2}\right )-\left (-4-2 \sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}}\right )}{f}\\ &=-\frac{\sqrt{\frac{1}{2} \left (-1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{4-3 \sqrt{2}+\left (2-\sqrt{2}\right ) \tan (e+f x)}{2 \sqrt{-7+5 \sqrt{2}} \sqrt{1+\tan (e+f x)}}\right )}{f}+\frac{9 \tanh ^{-1}\left (\sqrt{1+\tan (e+f x)}\right )}{4 f}-\frac{\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tanh ^{-1}\left (\frac{4+3 \sqrt{2}+\left (2+\sqrt{2}\right ) \tan (e+f x)}{2 \sqrt{7+5 \sqrt{2}} \sqrt{1+\tan (e+f x)}}\right )}{f}-\frac{\cot (e+f x) \sqrt{1+\tan (e+f x)}}{4 f}-\frac{\cot ^2(e+f x) \sqrt{1+\tan (e+f x)}}{2 f}\\ \end{align*}

Mathematica [C]  time = 0.459726, size = 124, normalized size = 0.56 \[ -\frac{-9 \tanh ^{-1}\left (\sqrt{\tan (e+f x)+1}\right )+4 \sqrt{1-i} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1-i}}\right )+4 \sqrt{1+i} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1+i}}\right )+2 \sqrt{\tan (e+f x)+1} \cot ^2(e+f x)+\sqrt{\tan (e+f x)+1} \cot (e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^3*Sqrt[1 + Tan[e + f*x]],x]

[Out]

-(-9*ArcTanh[Sqrt[1 + Tan[e + f*x]]] + 4*Sqrt[1 - I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + 4*Sqrt[1 +
I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]] + Cot[e + f*x]*Sqrt[1 + Tan[e + f*x]] + 2*Cot[e + f*x]^2*Sqrt[1
+ Tan[e + f*x]])/(4*f)

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Maple [C]  time = 0.64, size = 11217, normalized size = 50.8 \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)^3*(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 \sqrt{\tan \left (f x + e\right ) + 1} \cot \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(1+tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(tan(f*x + e) + 1)*cot(f*x + e)^3, x)

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Fricas [B]  time = 2.15884, size = 3274, normalized size = 14.81 \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)^3*(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)*s
qrt(f^(-4)) - 2*f)*(f^(-4))^(1/4)*log(1/2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2^(1/4)*(sqrt(2)*f^3*sqrt
(f^(-4))*cos(f*x + e) + 2*f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x +
e))/cos(f*x + e))*(f^(-4))^(1/4) + 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 + sqrt(2)*(f^3*cos(f*x + e)^2 - f^3)*sqrt(f^(-4)) - 2*f)*(f^(-4))^(1/
4)*log(1/2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 2*f*cos
(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/
4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e)) - 9*(cos(f*x + e)^2 - 1)*log(sqrt((cos(f*x + e) + sin(f*x
+ e))/cos(f*x + e)) + 1) + 9*(cos(f*x + e)^2 - 1)*log(sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) - 1) -
2*(2*cos(f*x + e)^2 + cos(f*x + e)*sin(f*x + e))*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) - 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)*(
f^5*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x
+ e) + 2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 2*f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) +
4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x
+ e))*(f^(-4))^(3/4) - 1/2*2^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(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)) - 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)*(f^5*
sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e
) - 2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 2*f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*s
qrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e)
)*(f^(-4))^(3/4) - 1/2*2^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((co
s(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) + f^2*sqrt(f^(-4)) + sqrt(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 \sqrt{\tan{\left (e + f x \right )} + 1} \cot ^{3}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**3*(1+tan(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(tan(e + f*x) + 1)*cot(e + f*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\tan \left (f x + e\right ) + 1} \cot \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(1+tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(tan(f*x + e) + 1)*cot(f*x + e)^3, x)