3.213 \(\int \frac{A+B \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=177 \[ -\frac{(5 A-2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}-\frac{(5 A-2 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a^2 d (\sec (c+d x)+1)}+\frac{(4 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

((4*A - B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) - ((5*A - 2*B)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((5*A - 2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x]
)/(3*a^2*d*(1 + Sec[c + d*x])) - ((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.325614, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4020, 3787, 3771, 2639, 2641} \[ -\frac{(5 A-2 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a^2 d (\sec (c+d x)+1)}-\frac{(5 A-2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(4 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2),x]

[Out]

((4*A - B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) - ((5*A - 2*B)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((5*A - 2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x]
)/(3*a^2*d*(1 + Sec[c + d*x])) - ((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

Rule 4020

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

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

Rule 2641

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

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a (7 A-B)-\frac{3}{2} a (A-B) \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(5 A-2 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{3}{2} a^2 (4 A-B)-\frac{1}{2} a^2 (5 A-2 B) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{3 a^4}\\ &=-\frac{(5 A-2 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{(5 A-2 B) \int \sqrt{\sec (c+d x)} \, dx}{6 a^2}+\frac{(4 A-B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(5 A-2 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\left ((5 A-2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}+\frac{\left ((4 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}\\ &=\frac{(4 A-B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}-\frac{(5 A-2 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}-\frac{(5 A-2 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 6.75206, size = 854, normalized size = 4.82 \[ -\frac{4 \sqrt{2} A e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac{\sqrt{2} B e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}-\frac{10 A \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sec ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac{4 B \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sec ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac{\sec ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)) \left (\frac{2 \sec \left (\frac{c}{2}\right ) \left (B \sin \left (\frac{d x}{2}\right )-A \sin \left (\frac{d x}{2}\right )\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{2 (B-A) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{4 \sec \left (\frac{c}{2}\right ) \left (4 B \sin \left (\frac{d x}{2}\right )-7 A \sin \left (\frac{d x}{2}\right )\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{2 (\cos (2 c) A+3 A-B) \cos (d x) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right )}{d}+\frac{8 A \cos (c) \sin (d x)}{d}-\frac{4 (4 B-7 A) \tan \left (\frac{c}{2}\right )}{3 d}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{(B+A \cos (c+d x)) (\sec (c+d x) a+a)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x])/(Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2),x]

[Out]

(-4*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]
^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7
/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x]))/(3*d*E^(I*d*x)*(B + A*Cos[c + d*x])*(a
+ a*Sec[c + d*x])^2) + (Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))
]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeom
etric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x]))/(3*d*E^(I*d*x)*(B +
A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) - (10*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c
+ d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x])*Sin[c])/(3*d*(B + A*Cos[c + d*x])*(a + a*Sec[c
+ d*x])^2) + (4*B*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c +
d*x]^(3/2)*(A + B*Sec[c + d*x])*Sin[c])/(3*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) + (Cos[c/2 + (d*x)/2
]^4*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x])*((-2*(3*A - B + A*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2])/d + (2*Sec
[c/2]*Sec[c/2 + (d*x)/2]^3*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(3*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-7*A*
Sin[(d*x)/2] + 4*B*Sin[(d*x)/2]))/(3*d) + (8*A*Cos[c]*Sin[d*x])/d - (4*(-7*A + 4*B)*Tan[c/2])/(3*d) + (2*(-A +
B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2)

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Maple [A]  time = 2.047, size = 421, normalized size = 2.4 \begin{align*}{\frac{1}{6\,{a}^{2}d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 24\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+10\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +24\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -12\,B \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-4\,B \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -6\,B \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -38\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+20\,B \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+15\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-9\,B \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-A+B \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/a^2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*A*cos(1/2*d*x+1/2*c)^6+10*A*cos(1/2*d*x+1/
2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+24
*A*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1
/2*c),2^(1/2))-12*B*cos(1/2*d*x+1/2*c)^6-4*B*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x
+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-6*B*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)
*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-38*A*cos(1/2*d*x+1/2*c)^4+20*B*cos(1/
2*d*x+1/2*c)^4+15*A*cos(1/2*d*x+1/2*c)^2-9*B*cos(1/2*d*x+1/2*c)^2-A+B)/cos(1/2*d*x+1/2*c)^3/(-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

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Maxima [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+B*sec(d*x+c))/(a+a*sec(d*x+c))^2/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{a^{2} \sec \left (d x + c\right )^{3} + 2 \, a^{2} \sec \left (d x + c\right )^{2} + a^{2} \sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(a^2*sec(d*x + c)^3 + 2*a^2*sec(d*x + c)^2 + a^2*sec(d*x + c)
), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A}{\sec ^{\frac{5}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac{3}{2}}{\left (c + d x \right )} + \sqrt{\sec{\left (c + d x \right )}}}\, dx + \int \frac{B \sec{\left (c + d x \right )}}{\sec ^{\frac{5}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac{3}{2}}{\left (c + d x \right )} + \sqrt{\sec{\left (c + d x \right )}}}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A/(sec(c + d*x)**(5/2) + 2*sec(c + d*x)**(3/2) + sqrt(sec(c + d*x))), x) + Integral(B*sec(c + d*x)/(
sec(c + d*x)**(5/2) + 2*sec(c + d*x)**(3/2) + sqrt(sec(c + d*x))), x))/a**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^2*sqrt(sec(d*x + c))), x)