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

Optimal. Leaf size=122 \[ \frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f (c-d)}-\frac{2 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f (c-d) \sqrt{c+d}} \]

[Out]

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

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Rubi [A]  time = 0.301194, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {3972, 3795, 203, 3967, 205} \[ \frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f (c-d)}-\frac{2 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f (c-d) \sqrt{c+d}} \]

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]

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

Rule 3972

Int[csc[(e_.) + (f_.)*(x_)]/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
), x_Symbol] :> Dist[b/(b*c - a*d), Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[d/(b*c - a*d), In
t[(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 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 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 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{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx &=\frac{\int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)}} \, dx}{c-d}-\frac{d \int \frac{\sec (e+f x) \sqrt{a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx}{a (c-d)}\\ &=-\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 d) \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-d) 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 \sqrt{d} \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-d) \sqrt{c+d} f}\\ \end{align*}

Mathematica [C]  time = 33.619, size = 228531, normalized size = 1873.2 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[e + f*x]/(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 = 0.238, size = 518, normalized size = 4.3 \begin{align*}{\frac{1}{2\,f \left ( c-d \right ) a} \left ( 2\,\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }\ln \left ( -{\frac{1}{\sin \left ( fx+e \right ) } \left ( -\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) +\cos \left ( fx+e \right ) -1 \right ) } \right ) \sqrt{{\frac{d}{c-d}}}-d\sqrt{2}\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{{\frac{d}{c-d}}}\sqrt{2}c\sin \left ( fx+e \right ) -\sqrt{2}\sqrt{{\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 ) +d\sqrt{2}\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{{\frac{d}{c-d}}}\sqrt{2}c\sin \left ( fx+e \right ) -\sqrt{2}\sqrt{{\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{{\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)/(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+d)*(c-d))^(1/2)/a*(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)-d*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^(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)))*(-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{\sec \left (f x + e\right )}{\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(sec(f*x+e)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.21029, size = 2403, normalized size = 19.7 \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))/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec{\left (e + f x \right )}}{\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(sec(f*x+e)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))**(1/2),x)

[Out]

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

[Out]

Timed out