\(\int \frac {\cos (c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\) [375]

Optimal result

Integrand size = 31, antiderivative size = 348 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {A (a-b) \sqrt {a+b} \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a b d}+\frac {A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a d}+\frac {\sqrt {a+b} (A b-2 a B) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 d}+\frac {A \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{a d} \] Output:

A*(a-b)*(a+b)^(1/2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2 
),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/( 
a-b))^(1/2)/a/b/d+A*(a+b)^(1/2)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2 
)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+s 
ec(d*x+c))/(a-b))^(1/2)/a/d+(a+b)^(1/2)*(A*b-2*B*a)*cot(d*x+c)*EllipticPi( 
(a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(b*(1-sec( 
d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d+A*(a+b*sec(d*x+ 
c))^(1/2)*sin(d*x+c)/a/d
 

Mathematica [A] (warning: unable to verify)

Time = 12.49 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.20 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (2 A (a+b) \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-4 a B \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-4 A b \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+8 a B \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+a A \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {3}{2} (c+d x)\right )-a A \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \tan \left (\frac {1}{2} (c+d x)\right )+2 A b \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 a d \sqrt {\frac {\cos (c+d x)}{(1+\cos (c+d x))^2}} \sqrt {\frac {1}{1+\cos (c+d x)}} \sqrt {a+b \sec (c+d x)}} \] Input:

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

Output:

(Sec[(c + d*x)/2]^2*(2*A*(a + b)*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + C 
os[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 4*a* 
B*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin 
[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 4*A*b*Sqrt[(b + a*Cos[c + d*x])/((a 
 + b)*(1 + Cos[c + d*x]))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b 
)/(a + b)] + 8*a*B*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] 
*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + a*A*Sqrt[Cos[ 
c + d*x]/(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sin[(3*(c + d*x))/2] - a*A*S 
qrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Tan[(c + d*x)/2] + 2*A*b*Sqrt[Cos[c + 
 d*x]/(1 + Cos[c + d*x])]*Tan[(c + d*x)/2]))/(4*a*d*Sqrt[Cos[c + d*x]/(1 + 
 Cos[c + d*x])^2]*Sqrt[(1 + Cos[c + d*x])^(-1)]*Sqrt[a + b*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 1.26 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4522, 27, 3042, 4547, 3042, 4409, 3042, 4271, 4319, 4492}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4522

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4547

\(\displaystyle \frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {A b-A \sec (c+d x) b-2 a B}{\sqrt {a+b \sec (c+d x)}}dx+A b \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {\int \frac {A b-A \csc \left (c+d x+\frac {\pi }{2}\right ) b-2 a B}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+A b \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {(A b-2 a B) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+A b \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-A b \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {(A b-2 a B) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-A b \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+A b \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {-A b \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+A b \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} (A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{2 a}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{a d}-\frac {A b \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} (A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{2 a}\)

\(\Big \downarrow \) 4492

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

Input:

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

Output:

-1/2*((-2*A*(a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*S 
ec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a 
 + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) - (2*A*Sqrt[a + b]*C 
ot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b 
)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x 
]))/(a - b))])/d - (2*Sqrt[a + b]*(A*b - 2*a*B)*Cot[c + d*x]*EllipticPi[(a 
 + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sq 
rt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))] 
)/(a*d))/a + (A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(a*d)
 

Defintions of rubi rules used

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 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

rule 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4409

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

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4522

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

rule 4547

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]* 
(b_.) + (a_)], x_Symbol] :> Int[(A - C*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x 
]], x] + Simp[C   Int[Csc[e + f*x]*((1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f 
*x]]), x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 8.79 (sec) , antiderivative size = 580, normalized size of antiderivative = 1.67

method	result	size
default	\(\frac {\left (\left (2 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+2\right ) A \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (1+\cos \left (d x +c \right )\right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, b \operatorname {EllipticPi}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), -1, \sqrt {\frac {a -b}{a +b}}\right )+\left (-4 \cos \left (d x +c \right )^{2}-8 \cos \left (d x +c \right )-4\right ) B \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (1+\cos \left (d x +c \right )\right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a \operatorname {EllipticPi}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), -1, \sqrt {\frac {a -b}{a +b}}\right )+\left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) A \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (1+\cos \left (d x +c \right )\right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a \operatorname {EllipticE}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right )+\left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) A \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (1+\cos \left (d x +c \right )\right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, b \operatorname {EllipticE}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right )+\left (2 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+2\right ) B \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (1+\cos \left (d x +c \right )\right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a \operatorname {EllipticF}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right )+A a \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+A b \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a +b \sec \left (d x +c \right )}}{d a \left (\cos \left (d x +c \right )^{2} a +a \cos \left (d x +c \right )+b \cos \left (d x +c \right )+b \right )}\)	\(580\)

Input:

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

Output:

1/d/a*((2*cos(d*x+c)^2+4*cos(d*x+c)+2)*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos( 
d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*b*EllipticPi(-csc(d*x+c)+ 
cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(-4*cos(d*x+c)^2-8*cos(d*x+c)-4)*B*(1/( 
a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1 
/2)*a*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(-cos(d*x+ 
c)^2-2*cos(d*x+c)-1)*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(co 
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/( 
a+b))^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*A*(1/(a+b)*(b+a*cos(d*x+c))/(1 
+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*b*EllipticE(-csc(d*x 
+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2)*B*(1/( 
a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1 
/2)*a*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+A*a*cos(d*x+c) 
^2*sin(d*x+c)+A*b*cos(d*x+c)*sin(d*x+c))*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c 
)^2*a+a*cos(d*x+c)+b*cos(d*x+c)+b)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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