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

Optimal. Leaf size=65 \[ \frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c-d} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{d} f \sqrt{c-d}} \]

[Out]

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

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Rubi [A]  time = 0.159167, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {3967, 208} \[ \frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c-d} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{d} f \sqrt{c-d}} \]

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]*ArcTanh[(Sqrt[a]*Sqrt[d]*Tan[e + f*x])/(Sqrt[c - d]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[c - d]*Sqrt[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 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) \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} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c-d} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{c-d} \sqrt{d} f}\\ \end{align*}

Mathematica [A]  time = 0.260163, size = 98, normalized size = 1.51 \[ \frac{\sqrt{2} \sqrt{\cos (e+f x)} \sec \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c-d} \sqrt{\cos (e+f x)}}\right )}{\sqrt{d} f \sqrt{c-d}} \]

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]*ArcTanh[(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[c - d]*Sqrt[d]*f)

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Maple [B]  time = 0.272, size = 411, normalized size = 6.3 \begin{align*} -{\frac{1}{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{-2\,{\frac{d}{c+d}}}c\sin \left ( fx+e \right ) +\sqrt{-2\,{\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{-2\,{\frac{d}{c+d}}}c\sin \left ( fx+e \right ) +\sqrt{-2\,{\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{-2\,{\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/f/(-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)*(-2*d/(c+d))^(1/2)*c*s
in(f*x+e)+(-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)*(-2*d/(c+d))^(1/2)*c*sin(f*x+e)+(-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)))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/c
os(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 = 1.04891, size = 802, normalized size = 12.34 \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