\(\int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx\) [499]

Optimal result

Integrand size = 33, antiderivative size = 83 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=-\frac {(A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}+\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))} \]

[Out]

-(A-B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+(A+B)*(cos(1/
2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+(A-B)*sin(d*x+c)*cos(d*x+c)
^(1/2)/d/(a+a*cos(d*x+c))

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3033, 3057, 2827, 2720, 2719} \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\frac {(A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {(A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)} \]

[In]

Int[(A + B*Sec[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])),x]

[Out]

-(((A - B)*EllipticE[(c + d*x)/2, 2])/(a*d)) + ((A + B)*EllipticF[(c + d*x)/2, 2])/(a*d) + ((A - B)*Sqrt[Cos[c
+ d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3033

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {B+A \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))} \, dx \\ & = \frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{2} a (A+B)-\frac {1}{2} a (A-B) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{a^2} \\ & = \frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(A-B) \int \sqrt {\cos (c+d x)} \, dx}{2 a}+\frac {(A+B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a} \\ & = -\frac {(A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}+\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 7.84 (sec) , antiderivative size = 968, normalized size of antiderivative = 11.66 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\frac {\cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} (A+B \sec (c+d x)) \left (-\frac {2 (-A+B) \csc (c)}{d}-\frac {2 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-A \sin \left (\frac {d x}{2}\right )+B \sin \left (\frac {d x}{2}\right )\right )}{d}\right )}{(B+A \cos (c+d x)) (a+a \sec (c+d x))}-\frac {A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) (A+B \sec (c+d x)) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))}-\frac {B \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) (A+B \sec (c+d x)) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))}+\frac {A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) (A+B \sec (c+d x)) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{2 d (B+A \cos (c+d x)) (a+a \sec (c+d x))}-\frac {B \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) (A+B \sec (c+d x)) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{2 d (B+A \cos (c+d x)) (a+a \sec (c+d x))} \]

[In]

Integrate[(A + B*Sec[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])),x]

[Out]

(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x])*((-2*(-A + B)*Csc[c])/d - (2*Sec[c/2]*Sec[c/2 +
(d*x)/2]*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/d))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])) - (A*Cos[c/2 +
(d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + B*Sec[c +
d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x -
ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[
c + d*x])) - (B*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2
]*Sec[c/2]*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 +
Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(B + A*Cos[c + d*x])*Sqrt
[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) + (A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x])*((Hyperg
eometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[
d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[
c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + Ar
cTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]
^2]]))/(2*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])) - (B*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec[c/2]*(A + B*Sec[
c + d*x])*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[
c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[
c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos
[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]
]]*Sqrt[1 + Tan[c]^2]]))/(2*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x]))

Maple [A] (verified)

Time = 6.29 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.93

method	result	size
default	\(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+\left (2 A -2 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-A +B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(243\)

[In]

int((A+B*sec(d*x+c))/(a+a*sec(d*x+c))/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+
B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+(2*A-2*B)*sin(1/2*d*x+1/2*c)^
4+(-A+B)*sin(1/2*d*x+1/2*c)^2)/a/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1
/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.86 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\frac {2 \, {\left (A - B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/2*(2*(A - B)*sqrt(cos(d*x + c))*sin(d*x + c) + (sqrt(2)*(-I*A - I*B)*cos(d*x + c) + sqrt(2)*(-I*A - I*B))*we
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A + I*B)*cos(d*x + c) + sqrt(2)*(I*A + I
*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + (sqrt(2)*(-I*A + I*B)*cos(d*x + c) + sqrt(2)*
(-I*A + I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + (sqrt(2)*(I*
A - I*B)*cos(d*x + c) + sqrt(2)*(I*A - I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) -
I*sin(d*x + c))))/(a*d*cos(d*x + c) + a*d)

Sympy [F]

\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\frac {\int \frac {A}{\sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )} + \sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )} + \sqrt {\cos {\left (c + d x \right )}}}\, dx}{a} \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))/cos(d*x+c)**(1/2),x)

[Out]

(Integral(A/(sqrt(cos(c + d*x))*sec(c + d*x) + sqrt(cos(c + d*x))), x) + Integral(B*sec(c + d*x)/(sqrt(cos(c +
d*x))*sec(c + d*x) + sqrt(cos(c + d*x))), x))/a

Maxima [F]

\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {\cos \left (d x + c\right )}} \,d x } \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*sqrt(cos(d*x + c))), x)

Giac [F]

\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {\cos \left (d x + c\right )}} \,d x } \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*sqrt(cos(d*x + c))), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\sqrt {\cos \left (c+d\,x\right )}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )} \,d x \]

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + a/cos(c + d*x))),x)

[Out]

int((A + B/cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + a/cos(c + d*x))), x)