3.4.86 \(\int \sqrt {1+\tan (e+f x)} \, dx\) [386]

Optimal. Leaf size=247 \[ -\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f} \]

[Out]

-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))*(2+2*2^(1/2))^(1/2)/f+1/2*arcta
n(((2+2*2^(1/2))^(1/2)+2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))*(2+2*2^(1/2))^(1/2)/f+1/2*ln(1+2^(1/2)-(2
+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(2+2*2^(1/2))^(1/2)-1/2*ln(1+2^(1/2)+(2+2*2^(1/2))^(1/2)*
(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(2+2*2^(1/2))^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3566, 714, 1141, 1175, 632, 210, 1178, 642} \begin {gather*} -\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \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 )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \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 )}{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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} 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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + Tan[e + f*x]],x]

[Out]

-((Sqrt[(1 + Sqrt[2])/2]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f)
+ (Sqrt[(1 + Sqrt[2])/2]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f
+ Log[1 + Sqrt[2] + Tan[e + f*x] - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(2*Sqrt[2*(1 + Sqrt[2])]*f)
- Log[1 + Sqrt[2] + Tan[e + f*x] + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(2*Sqrt[2*(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 714

Int[Sqrt[(d_) + (e_.)*(x_)]/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(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 1141

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

Rule 1175

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e) - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !Lt
Q[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 1178

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

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.03, size = 67, normalized size = 0.27 \begin {gather*} -\frac {i \left (\sqrt {1-i} \tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )-\sqrt {1+i} \tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + Tan[e + f*x]],x]

[Out]

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

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Maple [A]
time = 0.23, size = 217, normalized size = 0.88

method	result	size
derivativedivides	\(\frac {-\frac {\sqrt {2 \sqrt {2}+2}\, \left (\sqrt {2}-1\right ) \left (-\frac {\ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}-\frac {\sqrt {2 \sqrt {2}+2}\, \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}-\frac {\sqrt {2 \sqrt {2}+2}\, \left (\sqrt {2}-1\right ) \left (\frac {\ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}-\frac {\sqrt {2 \sqrt {2}+2}\, \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}}{f}\)	\(217\)
default	\(\frac {-\frac {\sqrt {2 \sqrt {2}+2}\, \left (\sqrt {2}-1\right ) \left (-\frac {\ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}-\frac {\sqrt {2 \sqrt {2}+2}\, \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}-\frac {\sqrt {2 \sqrt {2}+2}\, \left (\sqrt {2}-1\right ) \left (\frac {\ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}-\frac {\sqrt {2 \sqrt {2}+2}\, \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}}{f}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-1/2*(2*2^(1/2)+2)^(1/2)*(2^(1/2)-1)*(-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))-(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)))-1/2*(2*2^(1/2)+2)^(1/2)*(2^(1/2)-1)*(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))-(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+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. 805 vs. \(2 (194) = 388\).
time = 1.52, size = 805, normalized size = 3.26 \begin {gather*} \frac {1}{8} \cdot 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} {\left (\sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} - 2\right )} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} 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}} \cos \left (f x + e\right ) + 2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )}{2 \, \cos \left (f x + e\right )}\right ) - \frac {1}{8} \cdot 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} {\left (\sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} - 2\right )} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (-\frac {2^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} 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}} \cos \left (f x + e\right ) - 2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 2 \, \sin \left (f x + e\right )}{2 \, \cos \left (f x + e\right )}\right ) - \frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {\frac {1}{2}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f^{5} \sqrt {\frac {2^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} 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}} \cos \left (f x + e\right ) + 2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {5}{4}} - \frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f^{5} \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 {5}{4}} - f^{2} \sqrt {\frac {1}{f^{4}}} - \sqrt {2}\right ) - \frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {\frac {1}{2}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f^{5} \sqrt {-\frac {2^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} 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}} \cos \left (f x + e\right ) - 2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 2 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {5}{4}} - \frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f^{5} \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 {5}{4}} + f^{2} \sqrt {\frac {1}{f^{4}}} + \sqrt {2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*2^(1/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(sqrt(2)*f^2*sqrt(f^(-4)) - 2)*(f^(-4))^(1/4)*log(1/2*(2^(3/4
)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4)*cos
(f*x + e) + 2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e)) - 1/8*2^(
1/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(sqrt(2)*f^2*sqrt(f^(-4)) - 2)*(f^(-4))^(1/4)*log(-1/2*(2^(3/4)*sqrt
(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4)*cos(f*x +
e) - 2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2*cos(f*x + e) - 2*sin(f*x + e))/cos(f*x + e)) - 1/2*2^(3/4)*s
qrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*arctan(1/2*2^(3/4)*sqrt(1/2)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)
) + 4)*f^5*sqrt((2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x +
e))*(f^(-4))^(3/4)*cos(f*x + e) + 2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2*cos(f*x + e) + 2*sin(f*x + e))/
cos(f*x + e))*(f^(-4))^(5/4) - 1/2*2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^5*sqrt((cos(f*x + e) + sin(f
*x + e))/cos(f*x + e))*(f^(-4))^(5/4) - f^2*sqrt(f^(-4)) - sqrt(2)) - 1/2*2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-
4)) + 4)*(f^(-4))^(1/4)*arctan(1/2*2^(3/4)*sqrt(1/2)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^5*sqrt(-(2^(3/4)*s
qrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4)*cos(f*
x + e) - 2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2*cos(f*x + e) - 2*sin(f*x + e))/cos(f*x + e))*(f^(-4))^(5/
4) - 1/2*2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^5*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^
(-4))^(5/4) + f^2*sqrt(f^(-4)) + sqrt(2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(tan(e + f*x) + 1), x)

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Giac [A]
time = 0.79, size = 266, normalized size = 1.08 \begin {gather*} \frac {{\left (f^{2} \sqrt {\sqrt {2} + 1} + f \sqrt {\sqrt {2} - 1} {\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 )}{2 \, f^{3}} + \frac {{\left (f^{2} \sqrt {\sqrt {2} + 1} + f \sqrt {\sqrt {2} - 1} {\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 )}{2 \, f^{3}} + \frac {{\left (f^{2} \sqrt {\sqrt {2} - 1} - f \sqrt {\sqrt {2} + 1} {\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 )}{4 \, f^{3}} - \frac {{\left (f^{2} \sqrt {\sqrt {2} - 1} - f \sqrt {\sqrt {2} + 1} {\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 )}{4 \, f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.18, size = 73, normalized size = 0.30 \begin {gather*} 2\,\mathrm {atanh}\left (4\,f^3\,{\left (\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}+2\,\mathrm {atanh}\left (4\,f^3\,{\left (\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(e + f*x) + 1)^(1/2),x)

[Out]

2*atanh(4*f^3*((- 1/4 - 1i/4)/f^2)^(3/2)*(tan(e + f*x) + 1)^(1/2))*((- 1/4 - 1i/4)/f^2)^(1/2) + 2*atanh(4*f^3*
((- 1/4 + 1i/4)/f^2)^(3/2)*(tan(e + f*x) + 1)^(1/2))*((- 1/4 + 1i/4)/f^2)^(1/2)

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