3.398 \(\int \cot ^2(e+f x) (1+\tan (e+f x))^{3/2} \, dx\)
Optimal. Leaf size=178 \[ \frac {\sqrt {\sqrt {2}-1} \tan ^{-1}\left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{f}-\frac {3 \tanh ^{-1}\left (\sqrt {\tan (e+f x)+1}\right )}{f}+\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{f}-\frac {\sqrt {\tan (e+f x)+1} \cot (e+f x)}{f} \]
[Out]
-3*arctanh((1+tan(f*x+e))^(1/2))/f+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+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-cot(f*x+e)*(1+tan(f*x+e))^(1/2)/f
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Rubi [A] time = 0.30, antiderivative size = 178, normalized size of antiderivative = 1.00,
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
= {3567, 3654, 12, 3536, 3535, 203, 207, 3634, 63} \[ \frac {\sqrt {\sqrt {2}-1} \tan ^{-1}\left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{f}-\frac {3 \tanh ^{-1}\left (\sqrt {\tan (e+f x)+1}\right )}{f}+\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{f}-\frac {\sqrt {\tan (e+f x)+1} \cot (e+f x)}{f} \]
Antiderivative was successfully verified.
[In]
Int[Cot[e + f*x]^2*(1 + Tan[e + f*x])^(3/2),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 + Tan
[e + f*x]])])/f - (3*ArcTanh[Sqrt[1 + Tan[e + f*x]]])/f + (Sqrt[1 + Sqrt[2]]*ArcTanh[(3 + 2*Sqrt[2] + (1 + Sqr
t[2])*Tan[e + f*x])/(Sqrt[2*(7 + 5*Sqrt[2])]*Sqrt[1 + Tan[e + f*x]])])/f - (Cot[e + f*x]*Sqrt[1 + Tan[e + f*x]
])/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 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 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 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 \cot ^2(e+f x) (1+\tan (e+f x))^{3/2} \, dx &=-\frac {\cot (e+f x) \sqrt {1+\tan (e+f x)}}{f}-\int \frac {\cot (e+f x) \left (-\frac {3}{2}+\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 {3}{2} \int \frac {\cot (e+f x) \left (1+\tan ^2(e+f x)\right )}{\sqrt {1+\tan (e+f x)}} \, dx-\int \frac {2 \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=-\frac {\cot (e+f x) \sqrt {1+\tan (e+f x)}}{f}-2 \int \frac {\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx+\frac {3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {\cot (e+f x) \sqrt {1+\tan (e+f x)}}{f}+\frac {\int \frac {1+\left (-1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{\sqrt {2}}-\frac {\int \frac {1+\left (-1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{\sqrt {2}}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}\\ &=-\frac {3 \tanh ^{-1}\left (\sqrt {1+\tan (e+f x)}\right )}{f}-\frac {\cot (e+f x) \sqrt {1+\tan (e+f x)}}{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 )}{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 )}{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 )}{f}-\frac {3 \tanh ^{-1}\left (\sqrt {1+\tan (e+f x)}\right )}{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 )}{f}-\frac {\cot (e+f x) \sqrt {1+\tan (e+f x)}}{f}\\ \end {align*}
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Mathematica [C] time = 0.20, size = 100, normalized size = 0.56 \[ \frac {-3 \tanh ^{-1}\left (\sqrt {\tan (e+f x)+1}\right )+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}-\sqrt {\tan (e+f x)+1} \cot (e+f x)}{f} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[e + f*x]^2*(1 + Tan[e + f*x])^(3/2),x]
[Out]
(-3*ArcTanh[Sqrt[1 + Tan[e + f*x]]] + (2*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]])/Sqrt[1 - I] + (2*ArcTanh
[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]])/Sqrt[1 + I] - Cot[e + f*x]*Sqrt[1 + Tan[e + f*x]])/f
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fricas [B] time = 0.72, size = 1137, normalized size = 6.39 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^2*(1+tan(f*x+e))^(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)*f^3*sqrt(f^
(-4))*cos(f*x + e) + 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)) - 8^(1/4)*sqrt(-2*sqrt(2)*f^2*sq
rt(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)*lo
g(2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 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*co
s(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)*s
in(f*x + e) - 12*(cos(f*x + e)^2 - 1)*log(sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) + 1) + 12*(cos(f*x
+ e)^2 - 1)*log(sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) - 1) + 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(1/16*8^(3/4)*sqrt(2)*(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) + 8^(1/4)*(sq
rt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 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)
- 1/8*8^(3/4)*(2*f^5*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + si
n(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) - f^2*sqrt(f^(-4)) - sqrt(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(1/16*8^(3/4)*sqrt(2)*(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) - 8^(1/4
)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 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) - 1/8*8^(3/4)*(2*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)/(f*cos(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))^(3/2),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)(-1808*sqrt(2*(sqrt(2)-1))*f^2-1808*sqrt(2*(sqrt(2)+1))*f*abs(
f))/7232/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))+(-1808*sqrt(2*(sqrt(2)+1))*f^2+1808*sqrt(2*(sqrt(2)-1))*f*abs(f))/3616/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))+(1808*sqrt(2*(sqrt(2)-1))*f^2+1808*sqrt(2*
(sqrt(2)+1))*f*abs(f))/7232/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))-(1808*sqrt(2*(sqrt(2)+1))*f^2-1808*sqrt(2*(sqrt(2)-1))*f*abs(f))/3616/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))+3/2/f*ln(abs(sqrt(tan(f*
x+exp(1))+1)-1))-3/2/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 = 1.18, size = 6692, normalized size = 37.60 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(f*x+e)^2*(1+tan(f*x+e))^(3/2),x)
[Out]
result too large to display
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^2*(1+tan(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate((tan(f*x + e) + 1)^(3/2)*cot(f*x + e)^2, x)
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mupad [B] time = 0.18, size = 119, normalized size = 0.67 \[ \frac {\mathrm {atan}\left (\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{f}+\frac {\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f-f\,\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(e + f*x)^2*(tan(e + f*x) + 1)^(3/2),x)
[Out]
(atan((tan(e + f*x) + 1)^(1/2)*1i)*3i)/f + (tan(e + f*x) + 1)^(1/2)/(f - f*(tan(e + f*x) + 1)) - atan(f*((1/2
- 1i/2)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*1i)*((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)*1i)*((1/2 + 1i/2)/f^2)^(1/2)*2i
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\tan {\left (e + f x \right )} + 1\right )^{\frac {3}{2}} \cot ^{2}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)**2*(1+tan(f*x+e))**(3/2),x)
[Out]
Integral((tan(e + f*x) + 1)**(3/2)*cot(e + f*x)**2, x)
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