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

Optimal. Leaf size=42 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{f \sqrt{a+b}} \]

[Out]

ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]]/(Sqrt[a + b]*f)

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Rubi [A]  time = 0.0720426, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3190, 377, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{f \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]]/(Sqrt[a + b]*f)

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

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 206

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

Rubi steps

\begin{align*} \int \frac{\sec (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{1-(a+b) x^2} \, dx,x,\frac{\sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{f}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{\sqrt{a+b} f}\\ \end{align*}

Mathematica [A]  time = 0.0315527, size = 42, normalized size = 1. \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{f \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]]/(Sqrt[a + b]*f)

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Maple [B]  time = 2.483, size = 113, normalized size = 2.7 \begin{align*}{\frac{1}{2\,f}\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ){\frac{1}{\sqrt{a+b}}}}-{\frac{1}{2\,f}\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ){\frac{1}{\sqrt{a+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.7494, size = 595, normalized size = 14.17 \begin{align*} \left [\frac{\log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \,{\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right )}{4 \, \sqrt{a + b} f}, -\frac{\sqrt{-a - b} \arctan \left (\frac{{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a - b}}{2 \,{\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right )}{2 \,{\left (a + b\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 8*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^2 - 4*((a + 2*b)*cos(f*x
+ e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b)*sin(f*x + e) + 8*a^2 + 16*a*b + 8*b^2)/cos(f*
x + e)^4)/(sqrt(a + b)*f), -1/2*sqrt(-a - b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x
+ e)^2 + a + b)*sqrt(-a - b)/(((a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*sin(f*x + e)))/((a + b)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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