3.5.8 \(\int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx\) [408]

Optimal. Leaf size=240 \[ -\frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 f}+\frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3566, 722, 1108, 648, 632, 210, 642} \begin {gather*} -\frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{2 f}+\frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{2 f}-\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + Tan[e + f*x]],x]

[Out]

-1/2*(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f +
(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/(2*f) -
Log[1 + Sqrt[2] + Tan[e + f*x] - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(4*Sqrt[1 + Sqrt[2]]*f) + Log[
1 + Sqrt[2] + Tan[e + f*x] + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(4*Sqrt[1 + Sqrt[2]]*f)

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 722

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2*e, Subst[Int[1/(c*d^2 + a*e^2 - 2*c
*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1108

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}
, Dist[1/(2*c*q*r), Int[(r - x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(r + x)/(q + r*x + x^2), x], x
]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]

Rule 3566

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

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 \text {Subst}\left (\int \frac {1}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}\\ &=\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-x}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}+x}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}\\ &=\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2} f}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2} f}-\frac {\text {Subst}\left (\int \frac {-\sqrt {2 \left (1+\sqrt {2}\right )}+2 x}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 x}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}\\ &=-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}-\frac {\text {Subst}\left (\int \frac {1}{2 \left (1-\sqrt {2}\right )-x^2} \, dx,x,-\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}\right )}{\sqrt {2} f}-\frac {\text {Subst}\left (\int \frac {1}{2 \left (1-\sqrt {2}\right )-x^2} \, dx,x,\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}\right )}{\sqrt {2} f}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 \sqrt {-1+\sqrt {2}} f}+\frac {\tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 \sqrt {-1+\sqrt {2}} f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.03, size = 66, normalized size = 0.28 \begin {gather*} \frac {(1-i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )+(1+i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[1 + Tan[e + f*x]],x]

[Out]

((1 - I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + (1 + I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt
[1 + I]])/(2*f)

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Maple [A]
time = 0.14, size = 303, normalized size = 1.26

method	result	size
derivativedivides	\(\frac {-\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8}-\frac {\left (-2 \sqrt {2}+\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \sqrt {2 \sqrt {2}+2}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}+\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8}+\frac {\left (2 \sqrt {2}-\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \sqrt {2 \sqrt {2}+2}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}}{f}\)	\(303\)
default	\(\frac {-\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8}-\frac {\left (-2 \sqrt {2}+\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \sqrt {2 \sqrt {2}+2}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}+\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8}+\frac {\left (2 \sqrt {2}-\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \sqrt {2 \sqrt {2}+2}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}}{f}\)	\(303\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-1/8*(-(2*2^(1/2)+2)^(1/2)*2^(1/2)+2*(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*(-2*2^(1/2)+1/2*(-(2*2^(1/2)+2)^(1/2)*2^(1/2)+2*(2*2^(1/2)+2)^(1/2))*(2*2^(1/2)+2)^(1/2
))/(-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*(-(2*2^(
1/2)+2)^(1/2)*2^(1/2)+2*(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*(2*2^(1/2)-1/2*(-(2*2^(1/2)+2)^(1/2)*2^(1/2)+2*(2*2^(1/2)+2)^(1/2))*(2*2^(1/2)+2)^(1/2))/(-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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  1which is not
of the expected type LIST

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 766 vs. \(2 (186) = 372\).
time = 0.92, size = 766, normalized size = 3.19 \begin {gather*} -\frac {1}{2} \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} {\left (\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} - 1\right )} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} \cos \left (f x + e\right ) + \cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) + \frac {1}{2} \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} {\left (\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} - 1\right )} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} \cos \left (f x + e\right ) + \cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) - 2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \sqrt {\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} \cos \left (f x + e\right ) + \cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - f^{2} \sqrt {\frac {1}{f^{4}}} - 2 \, \sqrt {\frac {1}{2}}\right ) - 2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \sqrt {\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} \cos \left (f x + e\right ) + \cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} + f^{2} \sqrt {\frac {1}{f^{4}}} + 2 \, \sqrt {\frac {1}{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*(sqrt(1/2)*f^2*sqrt(f^(-4)) - 1)*(f^(-4))^(1/4)*log((2*s
qrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*f*sqrt((cos(f*x +
e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) + cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) + 1/
2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*(sqrt(1/2)*f^2*sqrt(f^(-4)) - 1)*(f^(-4))^(1/4)*log((2*sqrt
(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*f*sqrt((cos(f*x + e)
+ sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) + cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) - 2*(1/
2)^(3/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*(f^(-4))^(1/4)*arctan(2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-
4)) + 1)*f^3*sqrt((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) +
1)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) + cos(f*x + e) + sin(f*x +
e))/cos(f*x + e))*(f^(-4))^(3/4) - 2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*f^3*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)) - 2*(1/2)^(3/4)*sqrt(sqrt(1/2)*f
^2*sqrt(f^(-4)) + 1)*(f^(-4))^(1/4)*arctan(2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*f^3*sqrt((2*sqrt
(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*f*sqrt((cos(f*x + e)
+ sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) + cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4)
)^(3/4) - 2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*f^3*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {\tan {\left (e + f x \right )} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(tan(e + f*x) + 1), x)

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Giac [A]
time = 0.75, size = 282, normalized size = 1.18 \begin {gather*} \frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} - 2} + f \sqrt {2 \, \sqrt {2} + 2} {\left | f \right |}\right )} \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 )}{4 \, f^{3}} + \frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} - 2} + f \sqrt {2 \, \sqrt {2} + 2} {\left | f \right |}\right )} \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 )}{4 \, f^{3}} + \frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} + 2} - f \sqrt {2 \, \sqrt {2} - 2} {\left | f \right |}\right )} \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 )}{8 \, f^{3}} - \frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} + 2} - f \sqrt {2 \, \sqrt {2} - 2} {\left | f \right |}\right )} \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 )}{8 \, f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(f^2*sqrt(2*sqrt(2) - 2) + f*sqrt(2*sqrt(2) + 2)*abs(f))*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^3 + 1/4*(f^2*sqrt(2*sqrt(2) - 2) + f*sqrt(2*sqrt(2) + 2)*abs(f)
)*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^3 + 1/8*(f^
2*sqrt(2*sqrt(2) + 2) - f*sqrt(2*sqrt(2) - 2)*abs(f))*log(2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + s
qrt(2) + tan(f*x + e) + 1)/f^3 - 1/8*(f^2*sqrt(2*sqrt(2) + 2) - f*sqrt(2*sqrt(2) - 2)*abs(f))*log(-2^(1/4)*sqr
t(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqrt(2) + tan(f*x + e) + 1)/f^3

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Mupad [B]
time = 4.21, size = 71, normalized size = 0.30 \begin {gather*} \mathrm {atan}\left (2\,f\,\sqrt {\frac {-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (2\,f\,\sqrt {\frac {-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tan(e + f*x) + 1)^(1/2),x)

[Out]

atan(2*f*((- 1/8 - 1i/8)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2))*((- 1/8 - 1i/8)/f^2)^(1/2)*2i - atan(2*f*((- 1/8
+ 1i/8)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2))*((- 1/8 + 1i/8)/f^2)^(1/2)*2i

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