\(\int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx\) [583]

Optimal result

Integrand size = 33, antiderivative size = 203 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\frac {(A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-A b^2-a b B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (3 a^2 A b-A b^3-a^3 B-a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 (a-b) (a+b)^2 d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \cos (c+d x))} \] Output:

(A*b-B*a)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/(a^2-b^2)/d+(2*A*a^2-A*b 
^2-B*a*b)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^2/(a^2-b^2)/d-(3*A*a^2* 
b-A*b^3-B*a^3-B*a*b^2)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^ 
2/(a-b)/(a+b)^2/d-(A*b-B*a)*cos(d*x+c)^(1/2)*sin(d*x+c)/(a^2-b^2)/d/(b+a*c 
os(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 1.73 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.28 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\frac {\frac {4 (-A b+a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (b+a \cos (c+d x))}-\frac {\frac {2 (-A b+a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {(4 a A-4 b B) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{a}+\frac {2 (A b-a B) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(-a+b) (a+b)}}{4 d} \] Input:

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

Output:

((4*(-(A*b) + a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(b + a*Co 
s[c + d*x])) - ((2*(-(A*b) + a*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2 
])/(a + b) + ((4*a*A - 4*b*B)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*Elliptic 
Pi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)))/a + (2*(A*b - a*B)*(-2*a*b*El 
lipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqr 
t[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c 
 + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[Sin[c + d*x]^2]))/((-a + b)*(a + 
 b)))/(4*d)
 

Rubi [A] (verified)

Time = 1.33 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.97, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3433, 3042, 3478, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 3433

\(\displaystyle \int \frac {\sqrt {\cos (c+d x)} (A \cos (c+d x)+B)}{(a \cos (c+d x)+b)^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2}dx\)

\(\Big \downarrow \) 3478

\(\displaystyle \frac {\int -\frac {-\left ((A b-a B) \cos ^2(c+d x)\right )-2 (a A-b B) \cos (c+d x)+A b-a B}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a^2-b^2}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {-\left ((A b-a B) \cos ^2(c+d x)\right )-2 (a A-b B) \cos (c+d x)+A b-a B}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {(a B-A b) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 (a A-b B) \sin \left (c+d x+\frac {\pi }{2}\right )+A b-a B}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3538

\(\displaystyle -\frac {-\frac {\int -\frac {a (A b-a B)-\left (2 A a^2-b B a-A b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {(A b-a B) \int \sqrt {\cos (c+d x)}dx}{a}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\int \frac {a (A b-a B)-\left (2 A a^2-b B a-A b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {(A b-a B) \int \sqrt {\cos (c+d x)}dx}{a}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {a (A b-a B)+\left (-2 A a^2+b B a+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {(A b-a B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {\frac {\int \frac {a (A b-a B)+\left (-2 A a^2+b B a+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3481

\(\displaystyle -\frac {\frac {\frac {\left (a^3 (-B)+3 a^2 A b-a b^2 B-A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {\left (2 a^2 A-a b B-A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}}{a}-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {\left (a^3 (-B)+3 a^2 A b-a b^2 B-A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {\left (2 a^2 A-a b B-A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{a}-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {\frac {\frac {\left (a^3 (-B)+3 a^2 A b-a b^2 B-A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {2 \left (2 a^2 A-a b B-A b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}}{a}-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3284

\(\displaystyle -\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {\frac {\frac {2 \left (a^3 (-B)+3 a^2 A b-a b^2 B-A b^3\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}-\frac {2 \left (2 a^2 A-a b B-A b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}}{a}-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 \left (a^2-b^2\right )}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

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 3120

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 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3433

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[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] && IntegerQ[m] && IntegerQ[n]
 

rule 3478

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

rule 3481

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

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(802\) vs. \(2(206)=412\).

Time = 4.75 (sec) , antiderivative size = 803, normalized size of antiderivative = 3.96

method	result	size
default	\(\text {Expression too large to display}\)	\(803\)

Input:

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

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A/a^2*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1 
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+ 
2/a*(2*A*b-B*a)/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2 
*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt 
icPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2*b*(A*b-B*a)/a^2*(a^2/b/(a^2-b 
^2)*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 
)/(2*a*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)* 
(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2 
*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*a/b/(a^2-b^2)*(sin( 
1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+ 
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)) 
-1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1 
)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos 
(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1 
/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/ 
2*b/(a^2-b^2)/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2 
*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt 
icPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int \frac {A + B \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2} \sqrt {\cos {\left (c + d x \right )}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right ) \sec \left (d x +c \right ) b +\cos \left (d x +c \right ) a}d x \] Input:

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

Output:

int(sqrt(cos(c + d*x))/(cos(c + d*x)*sec(c + d*x)*b + cos(c + d*x)*a),x)