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

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

[Out]

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

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Rubi [A]  time = 0.154608, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {3980, 206} \[ \frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{d} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3980

Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) +
(c_)], x_Symbol] :> Dist[(-2*b)/f, Subst[Int[1/(1 - b*d*x^2), x], x, Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sq
rt[c + d*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] && NeQ[
c^2 - d^2, 0]

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

Mathematica [A]  time = 0.226638, size = 102, normalized size = 1.67 \[ \frac{\sqrt{2} \sec \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \sqrt{c \cos (e+f x)+d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c \cos (e+f x)+d}}\right )}{\sqrt{d} f \sqrt{c+d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.364, size = 302, normalized size = 5. \begin{align*} -{\frac{\sqrt{2}\cos \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) }{f \left ( \sin \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}\sqrt{{\frac{d+c\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}} \left ( \ln \left ( 2\,{\frac{1}{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) } \left ( \sqrt{2}\sqrt{-d}\sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) -c\sin \left ( fx+e \right ) -d\sin \left ( fx+e \right ) +c\cos \left ( fx+e \right ) -d\cos \left ( fx+e \right ) -c+d \right ) } \right ) -\ln \left ( -2\,{\frac{1}{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) } \left ( \sqrt{2}\sqrt{-d}\sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) -c\sin \left ( fx+e \right ) -d\sin \left ( fx+e \right ) -c\cos \left ( fx+e \right ) +d\cos \left ( fx+e \right ) +c-d \right ) } \right ) \right ){\frac{1}{\sqrt{-d}}}{\frac{1}{\sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \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))^(1/2),x)

[Out]

-1/f*2^(1/2)/(-d)^(1/2)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*((d+c*cos(f*x+e))/cos(f*x+e))^(1/2)*cos(f*x+e)*(
ln(2*(2^(1/2)*(-d)^(1/2)*(-2*(d+c*cos(f*x+e))/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)-c*sin(f*x+e)-d*sin(f*x+e)+c*cos
(f*x+e)-d*cos(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))-ln(-2*(2^(1/2)*(-d)^(1/2)*(-2*(d+c*cos(f*x+e))/(1+cos(f*x
+e)))^(1/2)*sin(f*x+e)-c*sin(f*x+e)-d*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c-d)/(-1+cos(f*x+e)+sin(f*x+e))))*(
-1+cos(f*x+e))/sin(f*x+e)^2/(-2*(d+c*cos(f*x+e))/(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 )}{\sqrt{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 0.966249, size = 738, normalized size = 12.1 \begin{align*} \left [\frac{\sqrt{\frac{a}{d}} \log \left (-\frac{8 \, a c d \cos \left (f x + e\right ) +{\left (a c^{2} - 6 \, a c d + a d^{2}\right )} \cos \left (f x + e\right )^{3} + 4 \,{\left (2 \, d^{2} \cos \left (f x + e\right ) +{\left (c d - d^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{a}{d}} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + 8 \, a d^{2} +{\left (a c^{2} + 2 \, a c d - 7 \, a d^{2}\right )} \cos \left (f x + e\right )^{2}}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2}}\right )}{2 \, f}, \frac{\sqrt{-\frac{a}{d}} \arctan \left (-\frac{2 \, d \sqrt{-\frac{a}{d}} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{{\left (a c - a d\right )} \cos \left (f x + e\right )^{2} + 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))^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

Integral(sqrt(a*(sec(e + f*x) + 1))*sec(e + f*x)/sqrt(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))^(1/2),x, algorithm="giac")

[Out]

Timed out