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

Optimal. Leaf size=39 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{\sqrt{a} f} \]

[Out]

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

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Rubi [A]  time = 0.0299862, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {4128, 377, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[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 4128

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]

Rule 377

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 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])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+b \sec ^2(e+f x)}} \, dx &=\frac{\operatorname{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{\operatorname{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{\tan ^{-1}\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]  time = 0.0827838, size = 87, normalized size = 2.23 \[ \frac{\sec (e+f x) \sqrt{a \cos (2 e+2 f x)+a+2 b} \tan ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{-a \sin ^2(e+f x)+a+b}}\right )}{\sqrt{2} \sqrt{a} f \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[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])

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Maple [C]  time = 0.407, size = 380, normalized size = 9.7 \begin{align*} -{\frac{\sqrt{2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{f\cos \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) }\sqrt{{\frac{1}{ \left ( a+b \right ) \left ( 1+\cos \left ( fx+e \right ) \right ) } \left ( i\cos \left ( fx+e \right ) \sqrt{a}\sqrt{b}-i\sqrt{a}\sqrt{b}+a\cos \left ( fx+e \right ) +b \right ) }}\sqrt{-2\,{\frac{i\cos \left ( fx+e \right ) \sqrt{a}\sqrt{b}-i\sqrt{a}\sqrt{b}-a\cos \left ( fx+e \right ) -b}{ \left ( a+b \right ) \left ( 1+\cos \left ( fx+e \right ) \right ) }}} \left ({\it EllipticF} \left ({\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }\sqrt{{\frac{1}{a+b} \left ( 2\,i\sqrt{a}\sqrt{b}+a-b \right ) }}},\sqrt{-{\frac{1}{ \left ( a+b \right ) ^{2}} \left ( 4\,i{a}^{{\frac{3}{2}}}\sqrt{b}-4\,i\sqrt{a}{b}^{{\frac{3}{2}}}-{a}^{2}+6\,ab-{b}^{2} \right ) }} \right ) -2\,{\it EllipticPi} \left ({\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }\sqrt{{\frac{2\,i\sqrt{a}\sqrt{b}+a-b}{a+b}}}},-{\frac{a+b}{2\,i\sqrt{a}\sqrt{b}+a-b}},{\sqrt{-{\frac{2\,i\sqrt{a}\sqrt{b}-a+b}{a+b}}}{\frac{1}{\sqrt{{\frac{2\,i\sqrt{a}\sqrt{b}+a-b}{a+b}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{1}{a+b} \left ( 2\,i\sqrt{a}\sqrt{b}+a-b \right ) }}}}{\frac{1}{\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+
a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b
)/(1+cos(f*x+e)))^(1/2)*(EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a
^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))-2*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b
^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2
)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)))*sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)/cos(f*x+e)/(-
1+cos(f*x+e))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.7916, size = 976, normalized size = 25.03 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)