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

Optimal. Leaf size=215 \[ -\frac {\sqrt {\sqrt {2}-1} \tan ^{-1}\left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f}+\frac {5 \tanh ^{-1}\left (\sqrt {\tan (e+f x)+1}\right )}{4 f}-\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f}-\frac {\sqrt {\tan (e+f x)+1} \cot ^2(e+f x)}{2 f}+\frac {3 \sqrt {\tan (e+f x)+1} \cot (e+f x)}{4 f} \]

[Out]

5/4*arctanh((1+tan(f*x+e))^(1/2))/f-1/2*arctan((3-2*2^(1/2)+(1-2^(1/2))*tan(f*x+e))/(-14+10*2^(1/2))^(1/2)/(1+
tan(f*x+e))^(1/2))*(2^(1/2)-1)^(1/2)/f-1/2*arctanh((3+2*2^(1/2)+(1+2^(1/2))*tan(f*x+e))/(14+10*2^(1/2))^(1/2)/
(1+tan(f*x+e))^(1/2))*(1+2^(1/2))^(1/2)/f+3/4*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 = 0.38, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3569, 3649, 3654, 12, 3536, 3535, 203, 207, 3634, 63} \[ -\frac {\sqrt {\sqrt {2}-1} \tan ^{-1}\left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f}+\frac {5 \tanh ^{-1}\left (\sqrt {\tan (e+f x)+1}\right )}{4 f}-\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f}-\frac {\sqrt {\tan (e+f x)+1} \cot ^2(e+f x)}{2 f}+\frac {3 \sqrt {\tan (e+f x)+1} \cot (e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[-1 + Sqrt[2]]*ArcTan[(3 - 2*Sqrt[2] + (1 - Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(-7 + 5*Sqrt[2])]*Sqrt[1 + Ta
n[e + f*x]])])/(2*f) + (5*ArcTanh[Sqrt[1 + Tan[e + f*x]]])/(4*f) - (Sqrt[1 + Sqrt[2]]*ArcTanh[(3 + 2*Sqrt[2] +
 (1 + Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(7 + 5*Sqrt[2])]*Sqrt[1 + Tan[e + f*x]])])/(2*f) + (3*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 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 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 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 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 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 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 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 3654

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*ta
n[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*x])^n*Simp[a*(A - C) - (A*b - b*
C)*Tan[e + f*x], x], x], x] + Dist[(A*b^2 + 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, 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 \frac {\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 {3}{2}+2 \tan (e+f x)+\frac {3}{2} \tan ^2(e+f x)\right )}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=\frac {3 \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}+\frac {3}{4} \tan ^2(e+f x)\right )}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=\frac {3 \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 \tan (e+f x)}{\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 {3 \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}{x \sqrt {1+x}} \, dx,x,\tan (e+f x)\right )}{8 f}+\int \frac {\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=\frac {3 \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 {1+\left (-1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{2 \sqrt {2}}+\frac {\int \frac {1+\left (-1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{2 \sqrt {2}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{4 f}\\ &=\frac {5 \tanh ^{-1}\left (\sqrt {1+\tan (e+f x)}\right )}{4 f}+\frac {3 \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 {\left (4-3 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{2 \left (-1+\sqrt {2}\right )-4 \left (-1+\sqrt {2}\right )^2+x^2} \, dx,x,\frac {1-2 \left (-1+\sqrt {2}\right )-\left (-1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}}\right )}{2 f}+\frac {\left (4+3 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{2 \left (-1-\sqrt {2}\right )-4 \left (-1-\sqrt {2}\right )^2+x^2} \, dx,x,\frac {1-2 \left (-1-\sqrt {2}\right )-\left (-1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}}\right )}{2 f}\\ &=-\frac {\sqrt {-1+\sqrt {2}} \tan ^{-1}\left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}+\frac {5 \tanh ^{-1}\left (\sqrt {1+\tan (e+f x)}\right )}{4 f}-\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}+\frac {3 \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]  time = 0.31, size = 125, normalized size = 0.58 \[ \frac {5 \tanh ^{-1}\left (\sqrt {\tan (e+f x)+1}\right )-\frac {4 \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}-\frac {4 \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}-2 \sqrt {\tan (e+f x)+1} \cot ^2(e+f x)+3 \sqrt {\tan (e+f x)+1} \cot (e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[Sqrt[1 + Tan[e + f*x]]] - (4*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]])/Sqrt[1 - I] - (4*ArcTanh[
Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]])/Sqrt[1 + I] + 3*Cot[e + f*x]*Sqrt[1 + Tan[e + f*x]] - 2*Cot[e + f*x]^2*Sq
rt[1 + Tan[e + f*x]])/(4*f)

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fricas [B]  time = 0.87, size = 1121, normalized size = 5.21 \[ \text {result too large to display} \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (f x + e\right )^{3}}{\sqrt {\tan \left (f x + e\right ) + 1}}\,{d x} \]

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

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maple [C]  time = 1.41, size = 8963, normalized size = 41.69 \[ \text {output too large to display} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (f x + e\right )^{3}}{\sqrt {\tan \left (f x + e\right ) + 1}}\,{d x} \]

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

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mupad [B]  time = 4.17, size = 147, normalized size = 0.68 \[ -\frac {\mathrm {atan}\left (\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{4\,f}-\frac {\frac {5\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{4}-\frac {3\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{3/2}}{4}}{f-2\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+1\right )+f\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^2}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,2{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,2{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(f*((1/8 - 1i/8)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*2i)*((1/8 - 1i/8)/f^2)^(1/2)*2i - ((5*(tan(e + f*x) +
 1)^(1/2))/4 - (3*(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*((1/8 + 1i/8)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*2i)*((1/8 + 1
i/8)/f^2)^(1/2)*2i

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{3}{\left (e + f x \right )}}{\sqrt {\tan {\left (e + f x \right )} + 1}}\, dx \]

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

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