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]

-1/2*arctan(1/2*(4-3*2^(1/2)+(2-2^(1/2))*tan(f*x+e))/(-7+5*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))*(-2+2*2^(1/2))
^(1/2)/f-1/2*arctanh(1/2*(4+3*2^(1/2)+(2+2^(1/2))*tan(f*x+e))/(7+5*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))*(2+2*2
^(1/2))^(1/2)/f+2*(1+tan(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.16, antiderivative size = 166, normalized size of antiderivative = 1.00, 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 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 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 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])

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*}

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Mathematica [C]  time = 0.07, 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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fricas [B]  time = 0.51, size = 936, normalized size = 5.64 \[ \text {result too large to display} \]

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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giac [A]  time = 1.08, size = 224, normalized size = 1.35 \[ -\frac {\sqrt {2 \, \sqrt {2} - 2} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f} - \frac {\sqrt {2 \, \sqrt {2} - 2} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f} - \frac {\sqrt {2 \, \sqrt {2} + 2} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f} + \frac {\sqrt {2 \, \sqrt {2} + 2} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f} + \frac {2 \, \sqrt {\tan \left (f x + e\right ) + 1}}{f} \]

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]

-1/2*sqrt(2*sqrt(2) - 2)*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(2*sqrt(2) - 2)*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 - 1/4*sqrt(2*sqrt(2) + 2)*log(2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqrt(
2) + tan(f*x + e) + 1)/f + 1/4*sqrt(2*sqrt(2) + 2)*log(-2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqr
t(2) + tan(f*x + e) + 1)/f + 2*sqrt(tan(f*x + e) + 1)/f

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maple [B]  time = 0.17, size = 306, normalized size = 1.84 \[ \frac {2 \sqrt {1+\tan \left (f x +e \right )}}{f}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\tan \left (f x + e\right ) + 1} \tan \left (f x + e\right )\,{d x} \]

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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mupad [B]  time = 4.10, size = 90, normalized size = 0.54 \[ \frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1-\mathrm {i}\right )\right )\,\sqrt {\frac {\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1+1{}\mathrm {i}\right )\right )\,\sqrt {\frac {\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\tan {\left (e + f x \right )} + 1} \tan {\left (e + f x \right )}\, dx \]

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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