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

Optimal. Leaf size=166 \[ -\frac{\sqrt{\frac{1}{2} \left (\sqrt{2}-1\right )} \tan ^{-1}\left (\frac{\left (2-\sqrt{2}\right ) \tan (e+f x)-3 \sqrt{2}+4}{2 \sqrt{5 \sqrt{2}-7} \sqrt{\tan (e+f x)+1}}\right )}{f}+\frac{2 \sqrt{\tan (e+f x)+1}}{f}-\frac{\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tanh ^{-1}\left (\frac{\left (2+\sqrt{2}\right ) \tan (e+f x)+3 \sqrt{2}+4}{2 \sqrt{7+5 \sqrt{2}} \sqrt{\tan (e+f x)+1}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.16214, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3528, 3536, 3535, 203, 207} \[ -\frac{\sqrt{\frac{1}{2} \left (\sqrt{2}-1\right )} \tan ^{-1}\left (\frac{\left (2-\sqrt{2}\right ) \tan (e+f x)-3 \sqrt{2}+4}{2 \sqrt{5 \sqrt{2}-7} \sqrt{\tan (e+f x)+1}}\right )}{f}+\frac{2 \sqrt{\tan (e+f x)+1}}{f}-\frac{\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tanh ^{-1}\left (\frac{\left (2+\sqrt{2}\right ) \tan (e+f x)+3 \sqrt{2}+4}{2 \sqrt{7+5 \sqrt{2}} \sqrt{\tan (e+f x)+1}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]*Sqrt[1 + Tan[e + f*x]],x]

[Out]

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 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 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 (e+f x) \sqrt{1+\tan (e+f x)} \, dx &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}+\int \frac{-1+\tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}-\frac{\int \frac{\sqrt{2}+\left (-2-\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx}{2 \sqrt{2}}+\frac{\int \frac{-\sqrt{2}+\left (-2+\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx}{2 \sqrt{2}}\\ &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}+\frac{\left (4-3 \sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{-2 \sqrt{2} \left (-2+\sqrt{2}\right )-4 \left (-2+\sqrt{2}\right )^2+x^2} \, dx,x,\frac{-\sqrt{2}-2 \left (-2+\sqrt{2}\right )-\left (-2+\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 \sqrt{2} \left (-2-\sqrt{2}\right )-4 \left (-2-\sqrt{2}\right )^2+x^2} \, dx,x,\frac{\sqrt{2}-2 \left (-2-\sqrt{2}\right )-\left (-2-\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}}\right )}{f}\\ &=-\frac{\sqrt{\frac{1}{2} \left (-1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{4-3 \sqrt{2}+\left (2-\sqrt{2}\right ) \tan (e+f x)}{2 \sqrt{-7+5 \sqrt{2}} \sqrt{1+\tan (e+f x)}}\right )}{f}-\frac{\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tanh ^{-1}\left (\frac{4+3 \sqrt{2}+\left (2+\sqrt{2}\right ) \tan (e+f x)}{2 \sqrt{7+5 \sqrt{2}} \sqrt{1+\tan (e+f x)}}\right )}{f}+\frac{2 \sqrt{1+\tan (e+f x)}}{f}\\ \end{align*}

Mathematica [C]  time = 0.0676101, size = 78, normalized size = 0.47 \[ -\frac{-2 \sqrt{\tan (e+f x)+1}+\sqrt{1-i} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1-i}}\right )+\sqrt{1+i} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1+i}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.036, size = 306, normalized size = 1.8 \begin{align*} 2\,{\frac{\sqrt{1+\tan \left ( fx+e \right ) }}{f}}+{\frac{\sqrt{2\,\sqrt{2}+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 ) }+{\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}}{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{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 ) }-{\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.06091, size = 2761, normalized size = 16.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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