3.238 \(\int \frac{\sec (e+f x) \sqrt{a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx\)

Optimal. Leaf size=61 \[ \frac{2 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{d} f \sqrt{c+d}} \]

[Out]

(2*Sqrt[a]*ArcTan[(Sqrt[a]*Sqrt[d]*Tan[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[d]*Sqrt[c + d]
*f)

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Rubi [A]  time = 0.136476, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {3967, 205} \[ \frac{2 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{d} f \sqrt{c+d}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*Sqrt[a + a*Sec[e + f*x]])/(c + d*Sec[e + f*x]),x]

[Out]

(2*Sqrt[a]*ArcTan[(Sqrt[a]*Sqrt[d]*Tan[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[d]*Sqrt[c + d]
*f)

Rule 3967

Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)
), x_Symbol] :> Dist[(-2*b)/f, Subst[Int[1/(b*c + a*d + d*x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x
]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[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{\sec (e+f x) \sqrt{a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx &=-\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{a c+a d+d x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{f}\\ &=\frac{2 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{d} \sqrt{c+d} f}\\ \end{align*}

Mathematica [A]  time = 0.226029, size = 94, normalized size = 1.54 \[ \frac{\sqrt{2} \sqrt{\cos (e+f x)} \sec \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d} \sqrt{\cos (e+f x)}}\right )}{\sqrt{d} f \sqrt{c+d}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*Sqrt[a + a*Sec[e + f*x]])/(c + d*Sec[e + f*x]),x]

[Out]

(Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(e + f*x)/2])/(Sqrt[c + d]*Sqrt[Cos[e + f*x]])]*Sqrt[Cos[e + f*x]]*Sec[(e
+ f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])])/(Sqrt[d]*Sqrt[c + d]*f)

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Maple [B]  time = 0.255, size = 431, normalized size = 7.1 \begin{align*}{\frac{\sqrt{2}}{2\,f} \left ( \ln \left ( -2\,{\frac{1}{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }\sin \left ( fx+e \right ) +c\cos \left ( fx+e \right ) -d\cos \left ( fx+e \right ) -c+d} \left ( \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sqrt{{\frac{d}{c-d}}}\sqrt{2}c\sin \left ( fx+e \right ) -\sqrt{2}\sqrt{{\frac{d}{c-d}}}\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}d\sin \left ( fx+e \right ) -\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }\cos \left ( fx+e \right ) -c\sin \left ( fx+e \right ) +d\sin \left ( fx+e \right ) +\sqrt{ \left ( c+d \right ) \left ( c-d \right ) } \right ) } \right ) -\ln \left ( 2\,{\frac{1}{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }\sin \left ( fx+e \right ) -c\cos \left ( fx+e \right ) +d\cos \left ( fx+e \right ) +c-d} \left ( \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sqrt{{\frac{d}{c-d}}}\sqrt{2}c\sin \left ( fx+e \right ) -\sqrt{2}\sqrt{{\frac{d}{c-d}}}\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}d\sin \left ( fx+e \right ) +\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }\cos \left ( fx+e \right ) -c\sin \left ( fx+e \right ) +d\sin \left ( fx+e \right ) -\sqrt{ \left ( c+d \right ) \left ( c-d \right ) } \right ) } \right ) \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}{\frac{1}{\sqrt{{\frac{d}{c-d}}}}}{\frac{1}{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*2^(1/2)/(d/(c-d))^(1/2)/((c+d)*(c-d))^(1/2)*(ln(-2*((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(d/(c-d))^(1/2)
*2^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*d*sin(f*x+e)-((c+d)*(c-d))^
(1/2)*cos(f*x+e)-c*sin(f*x+e)+d*sin(f*x+e)+((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)+c*cos(f*x+e)-d
*cos(f*x+e)-c+d))-ln(2*((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(d/(c-d))^(1/2)*2^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(
c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*d*sin(f*x+e)+((c+d)*(c-d))^(1/2)*cos(f*x+e)-c*sin(f*x+e)+d*si
n(f*x+e)-((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c-d)))*(-2*cos(f*x+e)/
(1+cos(f*x+e)))^(1/2)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*sec(f*x + e) + a)*sec(f*x + e)/(d*sec(f*x + e) + c), x)

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Fricas [B]  time = 1.04114, size = 801, normalized size = 13.13 \begin{align*} \left [\frac{\sqrt{-\frac{a}{c d + d^{2}}} \log \left (-\frac{{\left (a c^{2} + 8 \, a c d + 8 \, a d^{2}\right )} \cos \left (f x + e\right )^{3} + a d^{2} +{\left (a c^{2} + 2 \, a c d\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (c^{2} d + 3 \, c d^{2} + 2 \, d^{3}\right )} \cos \left (f x + e\right )^{2} -{\left (c d^{2} + d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt{-\frac{a}{c d + d^{2}}} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) -{\left (6 \, a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )}{c^{2} \cos \left (f x + e\right )^{3} +{\left (c^{2} + 2 \, c d\right )} \cos \left (f x + e\right )^{2} + d^{2} +{\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )}\right )}{2 \, f}, \frac{\sqrt{\frac{a}{c d + d^{2}}} \arctan \left (\frac{2 \,{\left (c d + d^{2}\right )} \sqrt{\frac{a}{c d + d^{2}}} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{{\left (a c + 2 \, a d\right )} \cos \left (f x + e\right )^{2} - a d +{\left (a c + a d\right )} \cos \left (f x + e\right )}\right )}{f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(-a/(c*d + d^2))*log(-((a*c^2 + 8*a*c*d + 8*a*d^2)*cos(f*x + e)^3 + a*d^2 + (a*c^2 + 2*a*c*d)*cos(f*x
+ e)^2 - 4*((c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e)^2 - (c*d^2 + d^3)*cos(f*x + e))*sqrt(-a/(c*d + d^2))*sqrt(
(a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - (6*a*c*d + 7*a*d^2)*cos(f*x + e))/(c^2*cos(f*x + e)^3 + (c^2
+ 2*c*d)*cos(f*x + e)^2 + d^2 + (2*c*d + d^2)*cos(f*x + e)))/f, sqrt(a/(c*d + d^2))*arctan(2*(c*d + d^2)*sqrt
(a/(c*d + d^2))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/((a*c + 2*a*d)*cos(f*x + e)^
2 - a*d + (a*c + a*d)*cos(f*x + e)))/f]

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

[Out]

Integral(sqrt(a*(sec(e + f*x) + 1))*sec(e + f*x)/(c + d*sec(e + f*x)), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out