3.402 \(\int \frac{\tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx\)

Optimal. Leaf size=143 \[ -\frac{\sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{\left (1-\sqrt{2}\right ) \tan (e+f x)-2 \sqrt{2}+3}{\sqrt{2 \left (5 \sqrt{2}-7\right )} \sqrt{\tan (e+f x)+1}}\right )}{2 f}-\frac{\sqrt{1+\sqrt{2}} \tanh ^{-1}\left (\frac{\left (1+\sqrt{2}\right ) \tan (e+f x)+2 \sqrt{2}+3}{\sqrt{2 \left (7+5 \sqrt{2}\right )} \sqrt{\tan (e+f x)+1}}\right )}{2 f} \]

[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) - (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)

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Rubi [A]  time = 0.115935, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3536, 3535, 203, 207} \[ -\frac{\sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{\left (1-\sqrt{2}\right ) \tan (e+f x)-2 \sqrt{2}+3}{\sqrt{2 \left (5 \sqrt{2}-7\right )} \sqrt{\tan (e+f x)+1}}\right )}{2 f}-\frac{\sqrt{1+\sqrt{2}} \tanh ^{-1}\left (\frac{\left (1+\sqrt{2}\right ) \tan (e+f x)+2 \sqrt{2}+3}{\sqrt{2 \left (7+5 \sqrt{2}\right )} \sqrt{\tan (e+f x)+1}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]/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) - (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)

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 \frac{\tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx &=-\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{\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{\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}\\ \end{align*}

Mathematica [C]  time = 0.0337409, size = 67, normalized size = 0.47 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1-i}}\right )}{\sqrt{1-i} f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1+i}}\right )}{\sqrt{1+i} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]]/(Sqrt[1 - I]*f)) - ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]]/(
Sqrt[1 + I]*f)

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Maple [B]  time = 0.041, size = 297, normalized size = 2.1 \begin{align*}{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{2}}{8\,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}}{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 ) }+{\frac{1}{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 ) }-{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{2}}{8\,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}}{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 ) }+{\frac{1}{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/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))-1/2/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))*2^(1/2)+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))-1/8/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))-1/2/f/(-2+2*2^(1/2))^(1/2)*arcta
n(((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)+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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.03271, size = 2749, normalized size = 19.22 \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)/(1+tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/4*(1/2)^(1/4)*(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*(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)) + 1/4*(1/2)^(1/4)*(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*sqrt(-4*sqrt(
1/2)*f^2*sqrt(f^(-4)) + 4)*(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)) - (1/2)^(3/4)*sqr
t(-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)*sq
rt(-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)) - (1/2)^(3/4)*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*sqr
t(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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