3.4.93 \(\int \cot ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx\) [393]

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

[Out]

5/4*arctanh((1+tan(f*x+e))^(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/(1
+2^(1/2))^(1/2)-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/(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-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-5/4*cot(f*x+e)*(1+tan(f*x+e))^(1/2)/
f-1/2*cot(f*x+e)^2*(1+tan(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.31, antiderivative size = 307, normalized size of antiderivative = 1.00, 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 = {3648, 3731, 3734, 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 )}{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 )}{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} \end {gather*}

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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 3648

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 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]

Rule 3731

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 3734

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

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 27.71, size = 4055, normalized size = 13.21 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[e + f*x]*(1/2 - (5*Cot[e + f*x])/4 - Csc[e + f*x]^2/2)*(1 + Tan[e + f*x])^(3/2))/(f*(Cos[e + f*x] + Sin[e
+ f*x])) + (Cos[e + f*x]*(-5*EllipticF[ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/
Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] + 5*EllipticPi[-1 - Sqrt[2], ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(
-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] + (16*I)*EllipticPi[(-I)*(1 + Sqrt[2]), ArcSin[(2
^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] - (16*I)*Elli
pticPi[I*(1 + Sqrt[2]), ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]
]], -3 - 2*Sqrt[2]] + 5*EllipticPi[1 + Sqrt[2], ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x
)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]])*((-5*Csc[e + f*x]*Sqrt[Sec[e + f*x]])/(8*Sqrt[Cos[e + f*x] + Sin[
e + f*x]]) - (Csc[e + f*x]*Sqrt[Sec[e + f*x]]*Sin[2*(e + f*x)])/Sqrt[Cos[e + f*x] + Sin[e + f*x]])*Sqrt[-((1 +
Tan[(e + f*x)/2])/((-2 + Sqrt[2])*(-1 + Tan[(e + f*x)/2])))]*(1 + Tan[e + f*x])^(3/2))/(2*2^(1/4)*f*(Cos[e +
f*x] + Sin[e + f*x])*Sqrt[(Cos[e + f*x] + Sin[e + f*x])/(-1 + Sin[e + f*x])]*(((-5*EllipticF[ArcSin[(2^(1/4)*S
qrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] + 5*EllipticPi[-1 - S
qrt[2], ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[
2]] + (16*I)*EllipticPi[(-I)*(1 + Sqrt[2]), ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2]
)])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] - (16*I)*EllipticPi[I*(1 + Sqrt[2]), ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e
+ f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] + 5*EllipticPi[1 + Sqrt[2], ArcSin[(2
^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]])*Sqrt[Sec[e +
f*x]]*(Cos[e + f*x] - Sin[e + f*x])*Sqrt[-((1 + Tan[(e + f*x)/2])/((-2 + Sqrt[2])*(-1 + Tan[(e + f*x)/2])))])
/(4*2^(1/4)*Sqrt[Cos[e + f*x] + Sin[e + f*x]]*Sqrt[(Cos[e + f*x] + Sin[e + f*x])/(-1 + Sin[e + f*x])]) + ((-5*
EllipticF[ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqr
t[2]] + 5*EllipticPi[-1 - Sqrt[2], ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[
2 + Sqrt[2]]], -3 - 2*Sqrt[2]] + (16*I)*EllipticPi[(-I)*(1 + Sqrt[2]), ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)
/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] - (16*I)*EllipticPi[I*(1 + Sqrt[2]), ArcSin
[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] + 5*Ellipt
icPi[1 + Sqrt[2], ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3
- 2*Sqrt[2]])*Sec[e + f*x]^(3/2)*Sin[e + f*x]*Sqrt[Cos[e + f*x] + Sin[e + f*x]]*Sqrt[-((1 + Tan[(e + f*x)/2])
/((-2 + Sqrt[2])*(-1 + Tan[(e + f*x)/2])))])/(4*2^(1/4)*Sqrt[(Cos[e + f*x] + Sin[e + f*x])/(-1 + Sin[e + f*x])
]) - ((-5*EllipticF[ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]],
-3 - 2*Sqrt[2]] + 5*EllipticPi[-1 - Sqrt[2], ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2
])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] + (16*I)*EllipticPi[(-I)*(1 + Sqrt[2]), ArcSin[(2^(1/4)*Sqrt[(1 + Tan
[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] - (16*I)*EllipticPi[I*(1 + Sqrt[2
]), ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]]
+ 5*EllipticPi[1 + Sqrt[2], ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqr
t[2]]], -3 - 2*Sqrt[2]])*Sqrt[Sec[e + f*x]]*Sqrt[Cos[e + f*x] + Sin[e + f*x]]*((Cos[e + f*x] - Sin[e + f*x])/(
-1 + Sin[e + f*x]) - (Cos[e + f*x]*(Cos[e + f*x] + Sin[e + f*x]))/(-1 + Sin[e + f*x])^2)*Sqrt[-((1 + Tan[(e +
f*x)/2])/((-2 + Sqrt[2])*(-1 + Tan[(e + f*x)/2])))])/(4*2^(1/4)*((Cos[e + f*x] + Sin[e + f*x])/(-1 + Sin[e + f
*x]))^(3/2)) + ((-5*EllipticF[ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + S
qrt[2]]], -3 - 2*Sqrt[2]] + 5*EllipticPi[-1 - Sqrt[2], ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(
e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] + (16*I)*EllipticPi[(-I)*(1 + Sqrt[2]), ArcSin[(2^(1/4)*Sqr
t[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]] - (16*I)*EllipticPi[I*(
1 + Sqrt[2]), ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sqrt[2 + Sqrt[2]]], -3 - 2
*Sqrt[2]] + 5*EllipticPi[1 + Sqrt[2], ArcSin[(2^(1/4)*Sqrt[(1 + Tan[(e + f*x)/2])/(-1 + Tan[(e + f*x)/2])])/Sq
rt[2 + Sqrt[2]]], -3 - 2*Sqrt[2]])*Sqrt[Sec[e + f*x]]*Sqrt[Cos[e + f*x] + Sin[e + f*x]]*(-1/2*Sec[(e + f*x)/2]
^2/((-2 + Sqrt[2])*(-1 + Tan[(e + f*x)/2])) + (Sec[(e + f*x)/2]^2*(1 + Tan[(e + f*x)/2]))/(2*(-2 + Sqrt[2])*(-
1 + Tan[(e + f*x)/2])^2)))/(4*2^(1/4)*Sqrt[(Cos[e + f*x] + Sin[e + f*x])/(-1 + Sin[e + f*x])]*Sqrt[-((1 + Tan[
(e + f*x)/2])/((-2 + Sqrt[2])*(-1 + Tan[(e + f*...

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.72, size = 9665, normalized size = 31.48

method	result	size
default	\(\text {Expression too large to display}\)	\(9665\)

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1071 vs. \(2 (250) = 500\).
time = 1.13, size = 1071, normalized size = 3.49 \begin {gather*} -\frac {8^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} {\left (2 \, f \cos \left (f x + e\right )^{2} - \sqrt {2} {\left (f^{3} \cos \left (f x + e\right )^{2} - f^{3}\right )} \sqrt {\frac {1}{f^{4}}} - 2 \, f\right )} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, {\left (2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 8^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} 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 ) + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )\right )}}{\cos \left (f x + e\right )}\right ) - 8^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} {\left (2 \, f \cos \left (f x + e\right )^{2} - \sqrt {2} {\left (f^{3} \cos \left (f x + e\right )^{2} - f^{3}\right )} \sqrt {\frac {1}{f^{4}}} - 2 \, f\right )} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, {\left (2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 8^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} 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 ) + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )\right )}}{\cos \left (f x + e\right )}\right ) - 5 \, {\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 ) + 5 \, {\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 ) - 2 \, {\left (2 \, \cos \left (f x + e\right )^{2} + 5 \, \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} - \frac {4 \cdot 8^{\frac {1}{4}} \sqrt {2} {\left (f^{5} \cos \left (f x + e\right )^{2} - f^{5}\right )} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (-\frac {1}{8} \cdot 8^{\frac {3}{4}} \sqrt {2} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} 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}} + \frac {1}{8} \cdot 8^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f^{3} \sqrt {\frac {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 8^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} 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 ) + 2 \, \cos \left (f x + e\right ) + 2 \, \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}}} - \sqrt {2}\right )}{f^{4}} - \frac {4 \cdot 8^{\frac {1}{4}} \sqrt {2} {\left (f^{5} \cos \left (f x + e\right )^{2} - f^{5}\right )} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (-\frac {1}{8} \cdot 8^{\frac {3}{4}} \sqrt {2} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} 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}} + \frac {1}{8} \cdot 8^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f^{3} \sqrt {\frac {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 8^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} 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 ) + 2 \, \cos \left (f x + e\right ) + 2 \, \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}}} + \sqrt {2}\right )}{f^{4}}}{8 \, {\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)^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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (\tan {\left (e + f x \right )} + 1\right )^{\frac {3}{2}} \cot ^{3}{\left (e + f x \right )}\, dx \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 3.97, size = 142, normalized size = 0.46 \begin {gather*} \frac {\frac {3\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{4}-\frac {5\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{3/2}}{4}}{f-2\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+1\right )+f\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^2}-\frac {\mathrm {atan}\left (\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{4\,f}-\mathrm {atan}\left (f\,\sqrt {\frac {-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (f\,\sqrt {\frac {-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{2}+\frac {1}{2}{}\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)^3*(tan(e + f*x) + 1)^(3/2),x)

[Out]

((3*(tan(e + f*x) + 1)^(1/2))/4 - (5*(tan(e + f*x) + 1)^(3/2))/4)/(f - 2*f*(tan(e + f*x) + 1) + f*(tan(e + f*x
) + 1)^2) - (atan((tan(e + f*x) + 1)^(1/2)*1i)*5i)/(4*f) - atan(f*((- 1/2 - 1i/2)/f^2)^(1/2)*(tan(e + f*x) + 1
)^(1/2))*((- 1/2 - 1i/2)/f^2)^(1/2)*2i + atan(f*((- 1/2 + 1i/2)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2))*((- 1/2 +
1i/2)/f^2)^(1/2)*2i

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