3.393 \(\int \cot ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=307 \[ \frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{\tan (e+f x)+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{f}-\frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sqrt{\tan (e+f x)+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{f}+\frac{\log \left (\tan (e+f x)-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{2 \sqrt{1+\sqrt{2}} f}-\frac{\log \left (\tan (e+f x)+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{2 \sqrt{1+\sqrt{2}} f}+\frac{5 \tanh ^{-1}\left (\sqrt{\tan (e+f x)+1}\right )}{4 f}-\frac{\sqrt{\tan (e+f x)+1} \cot ^2(e+f x)}{2 f}-\frac{5 \sqrt{\tan (e+f x)+1} \cot (e+f x)}{4 f} \]

[Out]

(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f - (Sqr
t[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f + (5*ArcTa
nh[Sqrt[1 + Tan[e + f*x]]])/(4*f) + Log[1 + Sqrt[2] + Tan[e + f*x] - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*
x]]]/(2*Sqrt[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[1 + Sqrt[2]]*f) - (5*Cot[e + f*x]*Sqrt[1 + Tan[e + f*x]])/(4*f) - (Cot[e + f*x]^2*Sqrt[1 + Tan[e + f*x
]])/(2*f)

________________________________________________________________________________________

Rubi [A]  time = 0.469267, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 14, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {3567, 3650, 3653, 12, 3485, 708, 1094, 634, 618, 204, 628, 3634, 63, 207} \[ \frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{\tan (e+f x)+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{f}-\frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sqrt{\tan (e+f x)+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{f}+\frac{\log \left (\tan (e+f x)-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{2 \sqrt{1+\sqrt{2}} f}-\frac{\log \left (\tan (e+f x)+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{2 \sqrt{1+\sqrt{2}} f}+\frac{5 \tanh ^{-1}\left (\sqrt{\tan (e+f x)+1}\right )}{4 f}-\frac{\sqrt{\tan (e+f x)+1} \cot ^2(e+f x)}{2 f}-\frac{5 \sqrt{\tan (e+f x)+1} \cot (e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^3*(1 + Tan[e + f*x])^(3/2),x]

[Out]

(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f - (Sqr
t[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f + (5*ArcTa
nh[Sqrt[1 + Tan[e + f*x]]])/(4*f) + Log[1 + Sqrt[2] + Tan[e + f*x] - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*
x]]]/(2*Sqrt[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[1 + Sqrt[2]]*f) - (5*Cot[e + f*x]*Sqrt[1 + Tan[e + f*x]])/(4*f) - (Cot[e + f*x]^2*Sqrt[1 + Tan[e + f*x
]])/(2*f)

Rule 3567

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1))/(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 - 2)*Simp[a*c^2*(m + 1) + a*
d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*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] && LtQ[1, n, 2] && IntegerQ[2*m]

Rule 3650

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*t
an[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 + a^2*C)*(a + b*Tan[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)) - a*C*(b*c*(m + 1)
+ a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - b*C)*Tan[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 2)*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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] && ( !IntegerQ[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 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 \cot ^3(e+f x) (1+\tan (e+f x))^{3/2} \, 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{5}{2}+\frac{3}{2} \tan ^2(e+f x)\right )}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=-\frac{5 \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{5}{4}-4 \tan (e+f x)-\frac{5}{4} \tan ^2(e+f x)\right )}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=-\frac{5 \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{4}{\sqrt{1+\tan (e+f x)}} \, dx-\frac{5}{8} \int \frac{\cot (e+f x) \left (1+\tan ^2(e+f x)\right )}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=-\frac{5 \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}-2 \int \frac{1}{\sqrt{1+\tan (e+f x)}} \, dx-\frac{5 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+x}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac{5 \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{5 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sqrt{1+\tan (e+f x)}\right )}{4 f}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{5 \tanh ^{-1}\left (\sqrt{1+\tan (e+f x)}\right )}{4 f}-\frac{5 \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{4 \operatorname{Subst}\left (\int \frac{1}{2-2 x^2+x^4} \, dx,x,\sqrt{1+\tan (e+f x)}\right )}{f}\\ &=\frac{5 \tanh ^{-1}\left (\sqrt{1+\tan (e+f x)}\right )}{4 f}-\frac{5 \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{\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 )}{\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 )}{\sqrt{1+\sqrt{2}} f}\\ &=\frac{5 \tanh ^{-1}\left (\sqrt{1+\tan (e+f x)}\right )}{4 f}-\frac{5 \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{\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 )}{\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 )}{\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 )}{2 \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 )}{2 \sqrt{1+\sqrt{2}} f}\\ &=\frac{5 \tanh ^{-1}\left (\sqrt{1+\tan (e+f x)}\right )}{4 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{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 )}{2 \sqrt{1+\sqrt{2}} f}-\frac{5 \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{\sqrt{2} \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{\sqrt{2} \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{-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 )}{\sqrt{-1+\sqrt{2}} f}+\frac{5 \tanh ^{-1}\left (\sqrt{1+\tan (e+f x)}\right )}{4 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{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 )}{2 \sqrt{1+\sqrt{2}} f}-\frac{5 \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 = 0.637587, size = 125, normalized size = 0.41 \[ -\frac{-5 \tanh ^{-1}\left (\sqrt{\tan (e+f x)+1}\right )+4 (1-i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1-i}}\right )+4 (1+i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1+i}}\right )+2 \sqrt{\tan (e+f x)+1} \cot ^2(e+f x)+5 \sqrt{\tan (e+f x)+1} \cot (e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^3*(1 + Tan[e + f*x])^(3/2),x]

[Out]

-(-5*ArcTanh[Sqrt[1 + Tan[e + f*x]]] + 4*(1 - I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + 4*(1 + I)
^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]] + 5*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 = 0.571, size = 9665, normalized size = 31.5 \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))^(3/2),x)

[Out]

result too large to display

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [B]  time = 2.08872, size = 2880, normalized size = 9.38 \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))^(3/2),x, algorithm="fricas")

[Out]

-1/8*(8^(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)*sq
rt(f^(-4)) - 2*f)*(f^(-4))^(1/4)*log(2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 8^(1/4)*sqrt(2*sqrt(2)*f^2*s
qrt(f^(-4)) + 4)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) + 2*cos(f*x +
e) + 2*sin(f*x + e))/cos(f*x + e)) - 8^(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(2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e)
- 8^(1/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1
/4)*cos(f*x + e) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e)) - 5*(cos(f*x + e)^2 - 1)*log(sqrt((cos(f*x +
e) + sin(f*x + e))/cos(f*x + e)) + 1) + 5*(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 + 5*cos(f*x + e)*sin(f*x + e))*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x +
e)) - 4*8^(1/4)*sqrt(2)*(f^5*cos(f*x + e)^2 - f^5)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*arctan(
-1/8*8^(3/4)*sqrt(2)*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) + 1/8*8^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*
x + e) + 8^(1/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-
4))^(1/4)*cos(f*x + e) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) - f^2*sqrt(f^(-4)) - sq
rt(2))/f^4 - 4*8^(1/4)*sqrt(2)*(f^5*cos(f*x + e)^2 - f^5)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*
arctan(-1/8*8^(3/4)*sqrt(2)*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) + 1/8*8^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))
*cos(f*x + e) - 8^(1/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)
)*(f^(-4))^(1/4)*cos(f*x + e) + 2*cos(f*x + e) + 2*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\tan \left (f x + e\right ) + 1\right )}^{\frac{3}{2}} \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))^(3/2),x, algorithm="giac")

[Out]

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