\(\int \frac {1}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [102]

Optimal result

Integrand size = 16, antiderivative size = 39 \[ \int \frac {1}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{\sqrt {a} f} \]

[Out]

arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f/a^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {4213, 385, 209} \[ \int \frac {1}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a} f} \]

[In]

Int[1/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(Sqrt[a]*f)

Rule 209

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 4213

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f} \\ & = \frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{\sqrt {a} f} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(87\) vs. \(2(39)=78\).

Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.23 \[ \int \frac {1}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) \sqrt {a+2 b+a \cos (2 e+2 f x)} \sec (e+f x)}{\sqrt {2} \sqrt {a} f \sqrt {a+b \sec ^2(e+f x)}} \]

[In]

Integrate[1/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

(ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sec[e + f*x]
)/(Sqrt[2]*Sqrt[a]*f*Sqrt[a + b*Sec[e + f*x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(137\) vs. \(2(33)=66\).

Time = 2.73 (sec) , antiderivative size = 138, normalized size of antiderivative = 3.54

method	result	size
default	\(\frac {\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (\sec \left (f x +e \right )+1\right )}{f \sqrt {-a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}\)	\(138\)

[In]

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

[Out]

1/f/(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^
2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)/(a+b*sec(f*x+e)^2
)^(1/2)*(sec(f*x+e)+1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (33) = 66\).

Time = 0.39 (sec) , antiderivative size = 408, normalized size of antiderivative = 10.46 \[ \int \frac {1}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [-\frac {\sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right )}{8 \, a f}, -\frac {\arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right )}{4 \, \sqrt {a} f}\right ] \]

[In]

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

[Out]

[-1/8*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^
2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(
f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f
*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)
*sin(f*x + e))/(a*f), -1/4*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^
2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 -
(a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e)))/(sqrt(a)*f)]

Sympy [F]

\[ \int \frac {1}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {1}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]

[In]

integrate(1/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Integral(1/sqrt(a + b*sec(e + f*x)**2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 992 vs. \(2 (33) = 66\).

Time = 0.43 (sec) , antiderivative size = 992, normalized size of antiderivative = 25.44 \[ \int \frac {1}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/2*(arctan2(2*a*sin(2*f*x + 2*e) + 2*(a^2*cos(4*f*x + 4*e)^2 + a^2*sin(4*f*x + 4*e)^2 + 4*(a^2 + 4*a*b + 4*b^
2)*cos(2*f*x + 2*e)^2 + 4*(a^2 + 2*a*b)*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*(a^2 + 4*a*b + 4*b^2)*sin(2*f*x
+ 2*e)^2 + a^2 + 2*(a^2 + 2*(a^2 + 2*a*b)*cos(2*f*x + 2*e))*cos(4*f*x + 4*e) + 4*(a^2 + 2*a*b)*cos(2*f*x + 2*e
))^(1/4)*sqrt(a)*sin(1/2*arctan2(a*sin(4*f*x + 4*e) + 2*(a + 2*b)*sin(2*f*x + 2*e), a*cos(4*f*x + 4*e) + 2*(a
+ 2*b)*cos(2*f*x + 2*e) + a)), 2*a*cos(2*f*x + 2*e) + 2*(a^2*cos(4*f*x + 4*e)^2 + a^2*sin(4*f*x + 4*e)^2 + 4*(
a^2 + 4*a*b + 4*b^2)*cos(2*f*x + 2*e)^2 + 4*(a^2 + 2*a*b)*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*(a^2 + 4*a*b +
4*b^2)*sin(2*f*x + 2*e)^2 + a^2 + 2*(a^2 + 2*(a^2 + 2*a*b)*cos(2*f*x + 2*e))*cos(4*f*x + 4*e) + 4*(a^2 + 2*a*
b)*cos(2*f*x + 2*e))^(1/4)*sqrt(a)*cos(1/2*arctan2(a*sin(4*f*x + 4*e) + 2*(a + 2*b)*sin(2*f*x + 2*e), a*cos(4*
f*x + 4*e) + 2*(a + 2*b)*cos(2*f*x + 2*e) + a)) + 2*a + 4*b) - arctan2(2*(a^2*cos(4*f*x + 4*e)^2 + a^2*sin(4*f
*x + 4*e)^2 + 4*(a^2 + 4*a*b + 4*b^2)*cos(2*f*x + 2*e)^2 + 4*(a^2 + 2*a*b)*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) +
4*(a^2 + 4*a*b + 4*b^2)*sin(2*f*x + 2*e)^2 + a^2 + 2*(a^2 + 2*(a^2 + 2*a*b)*cos(2*f*x + 2*e))*cos(4*f*x + 4*e
) + 4*(a^2 + 2*a*b)*cos(2*f*x + 2*e))^(1/4)*sqrt(a)*sin(1/2*arctan2(a*sin(4*f*x + 4*e) + 2*(a + 2*b)*sin(2*f*x
+ 2*e), a*cos(4*f*x + 4*e) + 2*(a + 2*b)*cos(2*f*x + 2*e) + a)), 2*(a^2*cos(4*f*x + 4*e)^2 + a^2*sin(4*f*x +
4*e)^2 + 4*(a^2 + 4*a*b + 4*b^2)*cos(2*f*x + 2*e)^2 + 4*(a^2 + 2*a*b)*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*(a
^2 + 4*a*b + 4*b^2)*sin(2*f*x + 2*e)^2 + a^2 + 2*(a^2 + 2*(a^2 + 2*a*b)*cos(2*f*x + 2*e))*cos(4*f*x + 4*e) + 4
*(a^2 + 2*a*b)*cos(2*f*x + 2*e))^(1/4)*sqrt(a)*cos(1/2*arctan2(a*sin(4*f*x + 4*e) + 2*(a + 2*b)*sin(2*f*x + 2*
e), a*cos(4*f*x + 4*e) + 2*(a + 2*b)*cos(2*f*x + 2*e) + a)) + 4*a + 4*b))/(sqrt(a)*f)

Giac [F]

\[ \int \frac {1}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {1}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(b*sec(f*x + e)^2 + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {1}{\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]

[In]

int(1/(a + b/cos(e + f*x)^2)^(1/2),x)

[Out]

int(1/(a + b/cos(e + f*x)^2)^(1/2), x)