3.524 \(\int \frac{a+b \sec (d+e x)}{\sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}} \, dx\)

Optimal. Leaf size=142 \[ \frac{x \left (a^2 \sec (d+e x)+a b\right )}{b \sqrt{a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2}}-\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right ) (a \sec (d+e x)+b)}{b e \sqrt{a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2}} \]

[Out]

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

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Rubi [A]  time = 0.213063, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {4174, 3919, 3831, 2659, 205} \[ \frac{x \left (a^2 \sec (d+e x)+a b\right )}{b \sqrt{a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2}}-\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right ) (a \sec (d+e x)+b)}{b e \sqrt{a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4174

Int[((A_) + (B_.)*sec[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*sec[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[(a + b*Sec[d + e*x] + c*Sec[d + e*x]^2)^n/(b + 2*c*Sec[d + e*x])^(2*n), Int[(A +
B*Sec[d + e*x])*(b + 2*c*Sec[d + e*x])^(2*n), x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0
] &&  !IntegerQ[n]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{a+b \sec (d+e x)}{\sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}} \, dx &=\frac{\left (2 a b+2 a^2 \sec (d+e x)\right ) \int \frac{a+b \sec (d+e x)}{2 a b+2 a^2 \sec (d+e x)} \, dx}{\sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}\\ &=\frac{x \left (a b+a^2 \sec (d+e x)\right )}{b \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}-\frac{\left (\left (2 a^3-2 a b^2\right ) \left (2 a b+2 a^2 \sec (d+e x)\right )\right ) \int \frac{\sec (d+e x)}{2 a b+2 a^2 \sec (d+e x)} \, dx}{2 a b \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}\\ &=\frac{x \left (a b+a^2 \sec (d+e x)\right )}{b \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}-\frac{\left (\left (2 a^3-2 a b^2\right ) \left (2 a b+2 a^2 \sec (d+e x)\right )\right ) \int \frac{1}{1+\frac{b \cos (d+e x)}{a}} \, dx}{4 a^3 b \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}\\ &=\frac{x \left (a b+a^2 \sec (d+e x)\right )}{b \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}-\frac{\left (\left (2 a^3-2 a b^2\right ) \left (2 a b+2 a^2 \sec (d+e x)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{b}{a}+\left (1-\frac{b}{a}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (d+e x)\right )\right )}{2 a^3 b e \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}\\ &=-\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right ) (b+a \sec (d+e x))}{b e \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}+\frac{x \left (a b+a^2 \sec (d+e x)\right )}{b \sqrt{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)}}\\ \end{align*}

Mathematica [A]  time = 0.392007, size = 92, normalized size = 0.65 \[ \frac{\sec (d+e x) (a+b \cos (d+e x)) \left (2 \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{b^2-a^2}}\right )+a (d+e x)\right )}{b e \sqrt{(a \sec (d+e x)+b)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(d + e*x) + 2*Sqrt[-a^2 + b^2]*ArcTanh[((-a + b)*Tan[(d + e*x)/2])/Sqrt[-a^2 + b^2]])*(a + b*Cos[d + e*x])
*Sec[d + e*x])/(b*e*Sqrt[(b + a*Sec[d + e*x])^2])

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Maple [A]  time = 0.238, size = 157, normalized size = 1.1 \begin{align*}{\frac{b\cos \left ( ex+d \right ) +a}{be\cos \left ( ex+d \right ) } \left ( a \left ( ex+d \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }+2\,\arctan \left ({\frac{ \left ( \cos \left ( ex+d \right ) -1 \right ) \left ( a-b \right ) }{\sin \left ( ex+d \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ){a}^{2}-2\,\arctan \left ({\frac{ \left ( \cos \left ( ex+d \right ) -1 \right ) \left ( a-b \right ) }{\sin \left ( ex+d \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ){b}^{2} \right ){\frac{1}{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}}{\frac{1}{\sqrt{{\frac{ \left ( b\cos \left ( ex+d \right ) +a \right ) ^{2}}{ \left ( \cos \left ( ex+d \right ) \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e/b/((a-b)*(a+b))^(1/2)*(b*cos(e*x+d)+a)*(a*(e*x+d)*((a-b)*(a+b))^(1/2)+2*arctan((cos(e*x+d)-1)*(a-b)/sin(e*
x+d)/((a-b)*(a+b))^(1/2))*a^2-2*arctan((cos(e*x+d)-1)*(a-b)/sin(e*x+d)/((a-b)*(a+b))^(1/2))*b^2)/cos(e*x+d)/((
b*cos(e*x+d)+a)^2/cos(e*x+d)^2)^(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((a+b*sec(e*x+d))/(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.17424, size = 421, normalized size = 2.96 \begin{align*} \left [\frac{2 \, a e x + \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (e x + d\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (e x + d\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (e x + d\right ) + b\right )} \sin \left (e x + d\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}}\right )}{2 \, b e}, \frac{a e x - \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (e x + d\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (e x + d\right )}\right )}{b e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.55967, size = 265, normalized size = 1.87 \begin{align*} -{\left (\frac{{\left (x e - 2 \, \pi \left \lfloor \frac{x e + d}{2 \, \pi } + \frac{1}{2} \right \rfloor + d\right )} a}{b \mathrm{sgn}\left (a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{4} - b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} - a - b\right )} - \frac{2 \, \sqrt{a^{2} - b^{2}} \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{\sqrt{a^{2} - b^{2}}}\right )}{b \mathrm{sgn}\left (a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{4} - b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} - a - b\right )}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((x*e - 2*pi*floor(1/2*(x*e + d)/pi + 1/2) + d)*a/(b*sgn(a*tan(1/2*x*e + 1/2*d)^4 - b*tan(1/2*x*e + 1/2*d)^4
+ 2*b*tan(1/2*x*e + 1/2*d)^2 - a - b)) - 2*sqrt(a^2 - b^2)*arctan((a*tan(1/2*x*e + 1/2*d) - b*tan(1/2*x*e + 1/
2*d))/sqrt(a^2 - b^2))/(b*sgn(a*tan(1/2*x*e + 1/2*d)^4 - b*tan(1/2*x*e + 1/2*d)^4 + 2*b*tan(1/2*x*e + 1/2*d)^2
- a - b)))*e^(-1)