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

Optimal. Leaf size=141 \[ \frac{2 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{a} f}-\frac{\sqrt{2} \sqrt{c-d} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{a} f} \]

[Out]

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

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Rubi [A]  time = 0.368627, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {3935, 3934, 203, 3983} \[ \frac{2 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{a} f}-\frac{\sqrt{2} \sqrt{c-d} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3935

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)], x_Symbol] :> Dist[a
/c, Int[Sqrt[c + d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[(b*c - a*d)/c, Int[Csc[e + f*x]/(Sqrt
[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &&
NeQ[a^2 - b^2, 0] && EqQ[c^2 - d^2, 0]

Rule 3934

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)], x_Symbol] :> Dist[(
-2*a)/f, Subst[Int[1/(1 + a*c*x^2), x], x, Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3983

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

Rubi steps

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

Mathematica [A]  time = 14.5044, size = 184, normalized size = 1.3 \[ \frac{2 \cos \left (\frac{1}{2} (e+f x)\right ) \sqrt{c+d \sec (e+f x)} \left (\frac{\sqrt{2} \sqrt{c} \sqrt{c+d} \sin ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sin \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right ) \sqrt{\frac{c \cos (e+f x)+d}{c+d}}}{\sqrt{c \cos (e+f x)+d}}+\sqrt{d-c} \tanh ^{-1}\left (\frac{\sqrt{d-c} \sin \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c \cos (e+f x)+d}}\right )\right )}{f \sqrt{a (\sec (e+f x)+1)} \sqrt{c \cos (e+f x)+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.33, size = 494, normalized size = 3.5 \begin{align*} -2\,{\frac{\cos \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) }{af\sqrt{c-d} \left ({c}^{2}-2\,cd+{d}^{2} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{d+c\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( \sqrt{2}\sqrt{- \left ( c-d \right ) ^{4}c}\arctan \left ({\frac{c \left ( c-d \right ) ^{2}\sqrt{2} \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sqrt{- \left ( c-d \right ) ^{4}c}\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}}} \right ) \sqrt{c-d}+\ln \left ( -{\frac{1}{\sin \left ( fx+e \right ) } \left ( \sqrt{c-d}\cos \left ( fx+e \right ) -\sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) -\sqrt{c-d} \right ) } \right ){c}^{3}-3\,\ln \left ( -{\frac{1}{\sin \left ( fx+e \right ) } \left ( \sqrt{c-d}\cos \left ( fx+e \right ) -\sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) -\sqrt{c-d} \right ) } \right ){c}^{2}d+3\,\ln \left ( -{\frac{1}{\sin \left ( fx+e \right ) } \left ( \sqrt{c-d}\cos \left ( fx+e \right ) -\sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) -\sqrt{c-d} \right ) } \right ) c{d}^{2}-\ln \left ( -{\frac{1}{\sin \left ( fx+e \right ) } \left ( \sqrt{c-d}\cos \left ( fx+e \right ) -\sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) -\sqrt{c-d} \right ) } \right ){d}^{3} \right ){\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((c+d*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2),x)

[Out]

-2/f/a/(c-d)^(1/2)/(c^2-2*c*d+d^2)*((d+c*cos(f*x+e))/cos(f*x+e))^(1/2)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*c
os(f*x+e)*(2^(1/2)*(-(c-d)^4*c)^(1/2)*arctan((c-d)^2*c*2^(1/2)/(-(c-d)^4*c)^(1/2)*(-1+cos(f*x+e))/(-2*(d+c*cos
(f*x+e))/(1+cos(f*x+e)))^(1/2)/sin(f*x+e))*(c-d)^(1/2)+ln(-((c-d)^(1/2)*cos(f*x+e)-(-2*(d+c*cos(f*x+e))/(1+cos
(f*x+e)))^(1/2)*sin(f*x+e)-(c-d)^(1/2))/sin(f*x+e))*c^3-3*ln(-((c-d)^(1/2)*cos(f*x+e)-(-2*(d+c*cos(f*x+e))/(1+
cos(f*x+e)))^(1/2)*sin(f*x+e)-(c-d)^(1/2))/sin(f*x+e))*c^2*d+3*ln(-((c-d)^(1/2)*cos(f*x+e)-(-2*(d+c*cos(f*x+e)
)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)-(c-d)^(1/2))/sin(f*x+e))*c*d^2-ln(-((c-d)^(1/2)*cos(f*x+e)-(-2*(d+c*cos(f*x
+e))/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)-(c-d)^(1/2))/sin(f*x+e))*d^3)*(-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(-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((c+d*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.89197, size = 2190, normalized size = 15.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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