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

Optimal. Leaf size=83 \[ \frac{(2 a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{3/2} f (a+b)^{3/2}}-\frac{b \sin (e+f x)}{2 a f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )} \]

[Out]

((2*a + b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(2*a^(3/2)*(a + b)^(3/2)*f) - (b*Sin[e + f*x])/(2*a*(a
+ b)*f*(a + b - a*Sin[e + f*x]^2))

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Rubi [A]  time = 0.0683364, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4147, 385, 208} \[ \frac{(2 a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{3/2} f (a+b)^{3/2}}-\frac{b \sin (e+f x)}{2 a f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

((2*a + b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(2*a^(3/2)*(a + b)^(3/2)*f) - (b*Sin[e + f*x])/(2*a*(a
+ b)*f*(a + b - a*Sin[e + f*x]^2))

Rule 4147

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

Rule 385

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

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sec (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac{b \sin (e+f x)}{2 a (a+b) f \left (a+b-a \sin ^2(e+f x)\right )}+\frac{(2 a+b) \operatorname{Subst}\left (\int \frac{1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{2 a (a+b) f}\\ &=\frac{(2 a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{3/2} (a+b)^{3/2} f}-\frac{b \sin (e+f x)}{2 a (a+b) f \left (a+b-a \sin ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.413117, size = 82, normalized size = 0.99 \[ \frac{\frac{(2 a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{(a+b)^{3/2}}-\frac{2 \sqrt{a} b \sin (e+f x)}{(a+b) (a \cos (2 (e+f x))+a+2 b)}}{2 a^{3/2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

(((2*a + b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2) - (2*Sqrt[a]*b*Sin[e + f*x])/((a + b)*(
a + 2*b + a*Cos[2*(e + f*x)])))/(2*a^(3/2)*f)

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Maple [A]  time = 0.083, size = 80, normalized size = 1. \begin{align*}{\frac{1}{f} \left ({\frac{\sin \left ( fx+e \right ) b}{ \left ( 2\,a+2\,b \right ) a \left ( -a-b+a \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{b+2\,a}{ \left ( 2\,a+2\,b \right ) a}{\it Artanh} \left ({\sin \left ( fx+e \right ) a{\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.558604, size = 676, normalized size = 8.14 \begin{align*} \left [\frac{{\left ({\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + 2 \, a b + b^{2}\right )} \sqrt{a^{2} + a b} \log \left (-\frac{a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) - 2 \,{\left (a^{2} b + a b^{2}\right )} \sin \left (f x + e\right )}{4 \,{\left ({\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}}, -\frac{{\left ({\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + 2 \, a b + b^{2}\right )} \sqrt{-a^{2} - a b} \arctan \left (\frac{\sqrt{-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) +{\left (a^{2} b + a b^{2}\right )} \sin \left (f x + e\right )}{2 \,{\left ({\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(((2*a^2 + a*b)*cos(f*x + e)^2 + 2*a*b + b^2)*sqrt(a^2 + a*b)*log(-(a*cos(f*x + e)^2 - 2*sqrt(a^2 + a*b)*
sin(f*x + e) - 2*a - b)/(a*cos(f*x + e)^2 + b)) - 2*(a^2*b + a*b^2)*sin(f*x + e))/((a^5 + 2*a^4*b + a^3*b^2)*f
*cos(f*x + e)^2 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*f), -1/2*(((2*a^2 + a*b)*cos(f*x + e)^2 + 2*a*b + b^2)*sqrt(-a
^2 - a*b)*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a + b)) + (a^2*b + a*b^2)*sin(f*x + e))/((a^5 + 2*a^4*b + a^3*
b^2)*f*cos(f*x + e)^2 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.32121, size = 127, normalized size = 1.53 \begin{align*} -\frac{\frac{{\left (2 \, a + b\right )} \arctan \left (\frac{a \sin \left (f x + e\right )}{\sqrt{-a^{2} - a b}}\right )}{{\left (a^{2} + a b\right )} \sqrt{-a^{2} - a b}} - \frac{b \sin \left (f x + e\right )}{{\left (a \sin \left (f x + e\right )^{2} - a - b\right )}{\left (a^{2} + a b\right )}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((2*a + b)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a^2 + a*b)*sqrt(-a^2 - a*b)) - b*sin(f*x + e)/((a*si
n(f*x + e)^2 - a - b)*(a^2 + a*b)))/f