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

Optimal. Leaf size=140 \[ \frac{\sqrt{2} \sqrt{c-d} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{a} f}+\frac{2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{a} f} \]

[Out]

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

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Rubi [A]  time = 0.462287, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3981, 3983, 203, 3980, 206} \[ \frac{\sqrt{2} \sqrt{c-d} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{a} f}+\frac{2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a} \sqrt{c+d \sec (e+f x)}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3981

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

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 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{c+d \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}} \, dx &=\frac{d \int \frac{\sec (e+f x) \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-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 d) \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{\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}+\frac{2 \sqrt{d} \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{a} f}\\ \end{align*}

Mathematica [A]  time = 16.9484, size = 187, normalized size = 1.34 \[ \frac{\sqrt{c} \sin (e+f x) \sqrt{c+d \sec (e+f x)} \left (2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{c \cos (e+f x)+d}}{\sqrt{d} \sqrt{c-c \cos (e+f x)}}\right )-\sqrt{2} \sqrt{c-d} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{c \cos (e+f x)+d}}{\sqrt{c-d} \sqrt{c-c \cos (e+f x)}}\right )\right )}{f \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \cos (e+f x)} \sqrt{c \cos (e+f x)+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.366, size = 504, normalized size = 3.6 \begin{align*} -{\frac{\sqrt{2}\cos \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) }{af \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 ( -\ln \left ({\frac{1}{\sin \left ( fx+e \right ) } \left ( \sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sqrt{c-d}\sin \left ( fx+e \right ) -c\cos \left ( fx+e \right ) +d\cos \left ( fx+e \right ) +c-d \right ){\frac{1}{\sqrt{c-d}}}} \right ) \sqrt{2}\sqrt{-d}c+\ln \left ({\frac{1}{\sin \left ( fx+e \right ) } \left ( \sqrt{-2\,{\frac{d+c\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sqrt{c-d}\sin \left ( fx+e \right ) -c\cos \left ( fx+e \right ) +d\cos \left ( fx+e \right ) +c-d \right ){\frac{1}{\sqrt{c-d}}}} \right ) \sqrt{2}\sqrt{-d}d+d\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 ) \sqrt{c-d}-d\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 ) \sqrt{c-d} \right ){\frac{1}{\sqrt{c-d}}}{\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)*(c+d*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2),x)

[Out]

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

[Out]

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

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Fricas [A]  time = 1.14313, size = 2603, normalized size = 18.59 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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