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]

(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)

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Rubi [A]  time = 0.178385, antiderivative size = 173, normalized size of antiderivative = 1., 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 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])

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 12

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

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 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 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])

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*}

Mathematica [C]  time = 0.363994, 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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Maple [B]  time = 0.029, size = 327, normalized size = 1.9 \begin{align*}{\frac{2}{5\,f} \left ( 1+\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-2\,{\frac{\sqrt{1+\tan \left ( fx+e \right ) }}{f}}-{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{2}}{4\,f}\ln \left ( 1+\sqrt{2}-\sqrt{2\,\sqrt{2}+2}\sqrt{1+\tan \left ( fx+e \right ) }+\tan \left ( fx+e \right ) \right ) }+{\frac{\sqrt{2}}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+\tan \left ( fx+e \right ) }-\sqrt{2\,\sqrt{2}+2} \right ) } \right ) }-2\,{\frac{1}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,\sqrt{1+\tan \left ( fx+e \right ) }-\sqrt{2\,\sqrt{2}+2}}{\sqrt{-2+2\,\sqrt{2}}}} \right ) }+{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{2}}{4\,f}\ln \left ( 1+\sqrt{2}+\sqrt{2\,\sqrt{2}+2}\sqrt{1+\tan \left ( fx+e \right ) }+\tan \left ( fx+e \right ) \right ) }+{\frac{\sqrt{2}}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( \sqrt{2\,\sqrt{2}+2}+2\,\sqrt{1+\tan \left ( fx+e \right ) } \right ) } \right ) }-2\,{\frac{1}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{\sqrt{2\,\sqrt{2}+2}+2\,\sqrt{1+\tan \left ( fx+e \right ) }}{\sqrt{-2+2\,\sqrt{2}}}} \right ) } \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int{\left (\tan \left (f x + e\right ) + 1\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{2}\,{d x} \end{align*}

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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Fricas [B]  time = 2.05657, size = 3000, normalized size = 17.34 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\tan{\left (e + f x \right )} + 1\right )^{\frac{3}{2}} \tan ^{2}{\left (e + f x \right )}\, dx \end{align*}

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

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]

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