3.169 \(\int \frac{1}{\sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx\)

Optimal. Leaf size=166 \[ \frac{2 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} c f (c-d) \sqrt{c+d}}-\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f (c-d)}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} c f} \]

[Out]

(2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(Sqrt[a]*c*f) - (Sqrt[2]*ArcTan[(Sqrt[a]*Tan[e + f
*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*(c - d)*f) + (2*d^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Tan[e + f*x
])/(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*c*(c - d)*Sqrt[c + d]*f)

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Rubi [A]  time = 0.372448, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {3929, 3920, 3774, 203, 3795, 3967, 205} \[ \frac{2 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} c f (c-d) \sqrt{c+d}}-\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f (c-d)}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} c f} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]

[Out]

(2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(Sqrt[a]*c*f) - (Sqrt[2]*ArcTan[(Sqrt[a]*Tan[e + f
*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*(c - d)*f) + (2*d^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Tan[e + f*x
])/(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*c*(c - d)*Sqrt[c + d]*f)

Rule 3929

Int[1/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))), x_Symbol] :> Dist[1
/(c*(b*c - a*d)), Int[(b*c - a*d - b*d*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d^2/(c*(b*c - a*d
)), Int[(Csc[e + f*x]*Sqrt[a + b*Csc[e + f*x]])/(c + d*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &
& NeQ[b*c - a*d, 0] && (EqQ[a^2 - b^2, 0] || EqQ[c^2 - d^2, 0])

Rule 3920

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

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^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 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

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{1}{\sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx &=\frac{\int \frac{a c-a d-a d \sec (e+f x)}{\sqrt{a+a \sec (e+f x)}} \, dx}{a c (c-d)}+\frac{d^2 \int \frac{\sec (e+f x) \sqrt{a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx}{a c (c-d)}\\ &=\frac{\int \sqrt{a+a \sec (e+f x)} \, dx}{a c}-\frac{\int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)}} \, dx}{c-d}-\frac{\left (2 d^2\right ) \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 )}{c (c-d) f}\\ &=\frac{2 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} c (c-d) \sqrt{c+d} f}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c f}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{(c-d) f}\\ &=\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} c f}-\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} (c-d) f}+\frac{2 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} c (c-d) \sqrt{c+d} f}\\ \end{align*}

Mathematica [C]  time = 38.7076, size = 431238, normalized size = 2597.82 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]

[Out]

Result too large to show

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Maple [B]  time = 0.249, size = 662, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f/(d/(c-d))^(1/2)/(c-d)/c/((c+d)*(c-d))^(1/2)/a*(2*((c+d)*(c-d))^(1/2)*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*co
s(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*(d/(c-d))^(1/2)*c-2*((c+d)*(c-d))^(1/2)*2^(1/2)*arctanh(
1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*(d/(c-d))^(1/2)*d+2*((c+d)*(c-d))^(1/2
)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*(d/(c-d))^(1/2)*c-2^(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))*d^2+2^(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))*d^2)*(-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{1}{\sqrt{a \sec \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right ) + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 100.191, size = 2695, normalized size = 16.23 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(2)*a*c*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1/a)*cos(f*x + e)*
sin(f*x + e) - 3*cos(f*x + e)^2 - 2*cos(f*x + e) + 1)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 2*a*d*sqrt(-d/(
a*c + a*d))*log((2*(c + d)*sqrt(-d/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x +
e) + (c + 2*d)*cos(f*x + e)^2 + (c + d)*cos(f*x + e) - d)/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) + 2*
sqrt(-a)*(c - d)*log((2*a*cos(f*x + e)^2 + 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin
(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)))/((a*c^2 - a*c*d)*f), -1/2*(sqrt(2)*a*c*sqrt(-1/a)*log(-(2
*sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1/a)*cos(f*x + e)*sin(f*x + e) - 3*cos(f*x + e)^2 - 2*c
os(f*x + e) + 1)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 4*a*d*sqrt(d/(a*c + a*d))*arctan((c + d)*sqrt(d/(a*c
+ a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(d*sin(f*x + e))) + 2*sqrt(-a)*(c - d)*log((2*a*
cos(f*x + e)^2 + 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e)
- a)/(cos(f*x + e) + 1)))/((a*c^2 - a*c*d)*f), -(a*d*sqrt(-d/(a*c + a*d))*log((2*(c + d)*sqrt(-d/(a*c + a*d))
*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (c + 2*d)*cos(f*x + e)^2 + (c + d)*cos(f*
x + e) - d)/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) - sqrt(2)*sqrt(a)*c*arctan(sqrt(2)*sqrt((a*cos(f*x
+ e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + 2*sqrt(a)*(c - d)*arctan(sqrt((a*cos(f*x + e) +
a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))))/((a*c^2 - a*c*d)*f), -(2*a*d*sqrt(d/(a*c + a*d))*arcta
n((c + d)*sqrt(d/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(d*sin(f*x + e))) - sqrt(2)
*sqrt(a)*c*arctan(sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + 2*sqr
t(a)*(c - d)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))))/((a*c^2 - a*
c*d)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out