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

Optimal. Leaf size=173 \[ \frac {\sqrt {\sqrt {2}-1} \tan ^{-1}\left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{f}+\frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\frac {2 \sqrt {\tan (e+f x)+1}}{f}+\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{f} \]

[Out]

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+a
rctanh((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-2*
(1+tan(f*x+e))^(1/2)/f+2/5*(1+tan(f*x+e))^(5/2)/f

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Rubi [A]  time = 0.18, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3543, 3482, 12, 3536, 3535, 203, 207} \[ \frac {\sqrt {\sqrt {2}-1} \tan ^{-1}\left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{f}+\frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\frac {2 \sqrt {\tan (e+f x)+1}}{f}+\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[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 + (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]])])/f - (2*Sqrt[1 + Tan[e + f*x]])/f + (2*(1 + Tan[e + f*x])^(5/2))/(5*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 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 3482

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]

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 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps

\begin {align*} \int \tan ^2(e+f x) (1+\tan (e+f x))^{3/2} \, dx &=\frac {2 (1+\tan (e+f x))^{5/2}}{5 f}-\int (1+\tan (e+f x))^{3/2} \, dx\\ &=-\frac {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {2 (1+\tan (e+f x))^{5/2}}{5 f}-\int \frac {2 \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=-\frac {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {2 (1+\tan (e+f x))^{5/2}}{5 f}-2 \int \frac {\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=-\frac {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {2 (1+\tan (e+f x))^{5/2}}{5 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 {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {2 (1+\tan (e+f x))^{5/2}}{5 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 {\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 {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {2 (1+\tan (e+f x))^{5/2}}{5 f}\\ \end {align*}

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Mathematica [C]  time = 0.38, size = 100, normalized size = 0.58 \[ \frac {2 \sqrt {\tan (e+f x)+1} \left (\tan ^2(e+f x)+2 \tan (e+f x)-4\right )+\frac {10 \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}+\frac {10 \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((10*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]])/Sqrt[1 - I] + (10*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]
])/Sqrt[1 + I] + 2*Sqrt[1 + Tan[e + f*x]]*(-4 + 2*Tan[e + f*x] + Tan[e + f*x]^2))/(5*f)

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fricas [B]  time = 0.95, size = 1025, normalized size = 5.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(20*8^(1/4)*sqrt(2)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(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)*sqr
t((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) - f^2*sqrt(f^(-4)) - sqrt(2))*cos(f*x + e)^2 + 20
*8^(1/4)*sqrt(2)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(1/16*8^(3/4)*sqrt(2)*(2*f^5*sqr
t(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))*cos(f*x + e)^2 + 5*8^(1/4)*
(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e)^2 + 2*f*cos(f*x + e)^2)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(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)) - 5*8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e)^2 + 2*f*
cos(f*x + e)^2)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(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)) - 16*(5*cos(f*x + e)^2 - 2*cos(f*x + e)*sin(f*x + e) - 1)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))
)/(f*cos(f*x + e)^2)

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giac [A]  time = 1.30, size = 237, normalized size = 1.37 \[ -\frac {\sqrt {\sqrt {2} - 1} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{f} - \frac {\sqrt {\sqrt {2} - 1} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{f} + \frac {\sqrt {\sqrt {2} + 1} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{2 \, f} - \frac {\sqrt {\sqrt {2} + 1} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{2 \, f} + \frac {2 \, {\left (f^{4} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {5}{2}} - 5 \, f^{4} \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{5 \, f^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(sqrt(2) - 1)*arctan(1/2*2^(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) + 2*sqrt(tan(f*x + e) + 1))/sqrt(-sqrt(2) + 2
))/f - sqrt(sqrt(2) - 1)*arctan(-1/2*2^(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) - 2*sqrt(tan(f*x + e) + 1))/sqrt(-sqrt
(2) + 2))/f + 1/2*sqrt(sqrt(2) + 1)*log(2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqrt(2) + tan(f*x +
e) + 1)/f - 1/2*sqrt(sqrt(2) + 1)*log(-2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqrt(2) + tan(f*x +
e) + 1)/f + 2/5*(f^4*(tan(f*x + e) + 1)^(5/2) - 5*f^4*sqrt(tan(f*x + e) + 1))/f^5

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maple [B]  time = 0.22, size = 327, normalized size = 1.89 \[ \frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}-\frac {2 \sqrt {1+\tan \left (f x +e \right )}}{f}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}}-\frac {2 \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}}-\frac {2 \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(1+tan(f*x+e))^(5/2)/f-2*(1+tan(f*x+e))^(1/2)/f-1/4/f*(2*2^(1/2)+2)^(1/2)*2^(1/2)*ln(1+2^(1/2)-(2*2^(1/2)+
2)^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))+1/f/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+tan(f*x+e))^(1/2)-(2*2^(1/2)+2
)^(1/2))/(-2+2*2^(1/2))^(1/2))*2^(1/2)-2/f/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+tan(f*x+e))^(1/2)-(2*2^(1/2)+2)^(
1/2))/(-2+2*2^(1/2))^(1/2))+1/4/f*(2*2^(1/2)+2)^(1/2)*2^(1/2)*ln(1+2^(1/2)+(2*2^(1/2)+2)^(1/2)*(1+tan(f*x+e))^
(1/2)+tan(f*x+e))+1/f/(-2+2*2^(1/2))^(1/2)*arctan(((2*2^(1/2)+2)^(1/2)+2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/2))^
(1/2))*2^(1/2)-2/f/(-2+2*2^(1/2))^(1/2)*arctan(((2*2^(1/2)+2)^(1/2)+2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/
2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.88, size = 103, normalized size = 0.60 \[ \frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}-\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(tan(e + f*x) + 1)^(5/2))/(5*f) - (2*(tan(e + f*x) + 1)^(1/2))/f - 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 \[ \int \left (\tan {\left (e + f x \right )} + 1\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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