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

Optimal. Leaf size=105 \[ \frac{2 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{c f}-\frac{2 \sqrt{a} \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}\right )}{c f \sqrt{c+d}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.22961, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {3925, 3774, 203, 3967, 205} \[ \frac{2 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{c f}-\frac{2 \sqrt{a} \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}\right )}{c f \sqrt{c+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3925

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[1/c,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[d/c, 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 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 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{\sqrt{a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx &=\frac{\int \sqrt{a+a \sec (e+f x)} \, dx}{c}-\frac{d \int \frac{\sec (e+f x) \sqrt{a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx}{c}\\ &=-\frac{(2 a) \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 a 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 f}\\ &=\frac{2 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c f}-\frac{2 \sqrt{a} \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 )}{c \sqrt{c+d} f}\\ \end{align*}

Mathematica [C]  time = 24.9909, size = 2686, normalized size = 25.58 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*Sqrt[2]*Cos[(e + f*x)/4]^2*Sqrt[(-1 + Sqrt[2] - (-2 + Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*(
d + c*Cos[e + f*x])*(c*EllipticF[ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + 2*(c + d)*El
lipticPi[-3 + 2*Sqrt[2], -ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - d*(EllipticPi[-(((-
3 + 2*Sqrt[2])*(c + d))/(3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] - d)), -ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]],
17 - 12*Sqrt[2]] + EllipticPi[((-3 + 2*Sqrt[2])*(c + d))/(-3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] + d), -ArcSin[Tan[
(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]]))*Sec[(e + f*x)/2]*((Cos[(e + f*x)/2]*Sqrt[Sec[e + f*x]])/
(2*(d + c*Cos[e + f*x])) + (Cos[(3*(e + f*x))/2]*Sqrt[Sec[e + f*x]])/(2*(d + c*Cos[e + f*x])))*Sec[e + f*x]*Sq
rt[a*(1 + Sec[e + f*x])]*Sqrt[3 - 2*Sqrt[2] - Tan[(e + f*x)/4]^2])/(c*(c + d)*f*(c + d*Sec[e + f*x])*((Sqrt[2]
*Sqrt[(-1 + Sqrt[2] - (-2 + Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*(c*EllipticF[ArcSin[Tan[(e + f*
x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + 2*(c + d)*EllipticPi[-3 + 2*Sqrt[2], -ArcSin[Tan[(e + f*x)/4]/S
qrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - d*(EllipticPi[-(((-3 + 2*Sqrt[2])*(c + d))/(3*c + 2*Sqrt[2]*Sqrt[c*(c
- d)] - d)), -ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + EllipticPi[((-3 + 2*Sqrt[2])*(c
+ d))/(-3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] + d), -ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]]
))*Sqrt[Sec[e + f*x]]*Tan[(e + f*x)/4])/(c*(c + d)*Sqrt[3 - 2*Sqrt[2] - Tan[(e + f*x)/4]^2]) + (2*Sqrt[2]*Cos[
(e + f*x)/4]*Sqrt[(-1 + Sqrt[2] - (-2 + Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*(c*EllipticF[ArcSin
[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + 2*(c + d)*EllipticPi[-3 + 2*Sqrt[2], -ArcSin[Tan[(e
+ f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - d*(EllipticPi[-(((-3 + 2*Sqrt[2])*(c + d))/(3*c + 2*Sqrt[2
]*Sqrt[c*(c - d)] - d)), -ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + EllipticPi[((-3 + 2
*Sqrt[2])*(c + d))/(-3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] + d), -ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 -
12*Sqrt[2]]))*Sqrt[Sec[e + f*x]]*Sin[(e + f*x)/4]*Sqrt[3 - 2*Sqrt[2] - Tan[(e + f*x)/4]^2])/(c*(c + d)) - (2*
Sqrt[2]*Cos[(e + f*x)/4]^2*(c*EllipticF[ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + 2*(c
+ d)*EllipticPi[-3 + 2*Sqrt[2], -ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - d*(EllipticP
i[-(((-3 + 2*Sqrt[2])*(c + d))/(3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] - d)), -ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqr
t[2]]], 17 - 12*Sqrt[2]] + EllipticPi[((-3 + 2*Sqrt[2])*(c + d))/(-3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] + d), -ArcS
in[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]]))*Sqrt[Sec[e + f*x]]*(((-2 + Sqrt[2])*Sin[(e + f*x)
/2])/(2*(1 + Cos[(e + f*x)/2])) + ((-1 + Sqrt[2] - (-2 + Sqrt[2])*Cos[(e + f*x)/2])*Sin[(e + f*x)/2])/(2*(1 +
Cos[(e + f*x)/2])^2))*Sqrt[3 - 2*Sqrt[2] - Tan[(e + f*x)/4]^2])/(c*(c + d)*Sqrt[(-1 + Sqrt[2] - (-2 + Sqrt[2])
*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]) - (2*Sqrt[2]*Cos[(e + f*x)/4]^2*Sqrt[(-1 + Sqrt[2] - (-2 + Sqrt[2]
)*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*(c*EllipticF[ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12
*Sqrt[2]] + 2*(c + d)*EllipticPi[-3 + 2*Sqrt[2], -ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2
]] - d*(EllipticPi[-(((-3 + 2*Sqrt[2])*(c + d))/(3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] - d)), -ArcSin[Tan[(e + f*x)/
4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + EllipticPi[((-3 + 2*Sqrt[2])*(c + d))/(-3*c + 2*Sqrt[2]*Sqrt[c*(c
- d)] + d), -ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]]))*Sec[e + f*x]^(3/2)*Sin[e + f*x]*
Sqrt[3 - 2*Sqrt[2] - Tan[(e + f*x)/4]^2])/(c*(c + d)) - (4*Sqrt[2]*Cos[(e + f*x)/4]^2*Sqrt[(-1 + Sqrt[2] - (-2
+ Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*Sqrt[Sec[e + f*x]]*Sqrt[3 - 2*Sqrt[2] - Tan[(e + f*x)/4]
^2]*((c*Sec[(e + f*x)/4]^2)/(4*Sqrt[3 - 2*Sqrt[2]]*Sqrt[1 - Tan[(e + f*x)/4]^2/(3 - 2*Sqrt[2])]*Sqrt[1 - ((17
- 12*Sqrt[2])*Tan[(e + f*x)/4]^2)/(3 - 2*Sqrt[2])]) - ((c + d)*Sec[(e + f*x)/4]^2)/(2*Sqrt[3 - 2*Sqrt[2]]*Sqrt
[1 - Tan[(e + f*x)/4]^2/(3 - 2*Sqrt[2])]*Sqrt[1 - ((17 - 12*Sqrt[2])*Tan[(e + f*x)/4]^2)/(3 - 2*Sqrt[2])]*(1 -
((-3 + 2*Sqrt[2])*Tan[(e + f*x)/4]^2)/(3 - 2*Sqrt[2]))) - d*(-Sec[(e + f*x)/4]^2/(4*Sqrt[3 - 2*Sqrt[2]]*Sqrt[
1 - Tan[(e + f*x)/4]^2/(3 - 2*Sqrt[2])]*Sqrt[1 - ((17 - 12*Sqrt[2])*Tan[(e + f*x)/4]^2)/(3 - 2*Sqrt[2])]*(1 +
((-3 + 2*Sqrt[2])*(c + d)*Tan[(e + f*x)/4]^2)/((3 - 2*Sqrt[2])*(3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] - d)))) - Sec[
(e + f*x)/4]^2/(4*Sqrt[3 - 2*Sqrt[2]]*Sqrt[1 - Tan[(e + f*x)/4]^2/(3 - 2*Sqrt[2])]*Sqrt[1 - ((17 - 12*Sqrt[2])
*Tan[(e + f*x)/4]^2)/(3 - 2*Sqrt[2])]*(1 - ((-3 + 2*Sqrt[2])*(c + d)*Tan[(e + f*x)/4]^2)/((3 - 2*Sqrt[2])*(-3*
c + 2*Sqrt[2]*Sqrt[c*(c - d)] + d)))))))/(c*(c + d))))

________________________________________________________________________________________

Maple [B]  time = 0.288, size = 501, normalized size = 4.8 \begin{align*} -{\frac{\sqrt{2}}{2\,cf} \left ( 2\,\sqrt{{\frac{d}{c-d}}}\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) +d\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\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((a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x)

[Out]

-1/2/f*2^(1/2)/(d/(c-d))^(1/2)/((c+d)*(c-d))^(1/2)/c*(2*(d/(c-d))^(1/2)*((c+d)*(c-d))^(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*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*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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.95385, size = 1704, 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((a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, algorithm="fricas")

[Out]

[(sqrt(-a*d/(c + d))*log((2*(c + d)*sqrt(-a*d/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*si
n(f*x + e) + (a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c + a*d)*cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*
x + e) + d)) + sqrt(-a)*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)))/(c*f), -(2*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a
)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - sqrt(-a*d/(c + d))*log((2*(c + d)*sqrt(-a*d/(c + d))*sq
rt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c +
a*d)*cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)))/(c*f), (2*sqrt(a*d/(c + d))*arctan((c + d)*
sqrt(a*d/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(a*d*sin(f*x + e))) + sqrt(-a)*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)))/(c*f), -2*(sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sq
rt(a)*sin(f*x + e))) - sqrt(a*d/(c + d))*arctan((c + d)*sqrt(a*d/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*x +
e))*cos(f*x + e)/(a*d*sin(f*x + e))))/(c*f)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out