\(\int \frac {\cos ^4(c+d x) (A+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [191]

Optimal result

Integrand size = 35, antiderivative size = 243 \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {(107 A+112 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 \sqrt {a} d}-\frac {\sqrt {2} (A+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {(21 A+16 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {(43 A+48 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}-\frac {A \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {A \cos ^3(c+d x) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}} \] Output:

1/64*(107*A+112*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(1/ 
2)/d-2^(1/2)*(A+C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^ 
(1/2))/a^(1/2)/d-1/64*(21*A+16*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/96 
*(43*A+48*C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/24*A*cos(d*x 
+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/4*A*cos(d*x+c)^3*sin(d*x+c)/d/ 
(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.66 \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left ((321 A+336 C) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )-192 \sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) \left (-63 A-48 C+(86 A+96 C) \cos (c+d x)-8 A \cos ^2(c+d x)+48 A \cos ^3(c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{192 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[(Cos[c + d*x]^4*(A + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]] 
,x]
 

Output:

(((321*A + 336*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] - 192*Sqrt[2]*(A + C)*Ar 
cTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + Cos[c + d*x]*(-63*A - 48*C + (86*A 
 + 96*C)*Cos[c + d*x] - 8*A*Cos[c + d*x]^2 + 48*A*Cos[c + d*x]^3)*Sqrt[1 - 
 Sec[c + d*x]])*Tan[c + d*x])/(192*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Se 
c[c + d*x])])
 

Rubi [A] (verified)

Time = 1.76 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.14, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.543, Rules used = {3042, 4575, 27, 3042, 4510, 27, 3042, 4510, 27, 3042, 4510, 27, 3042, 4408, 3042, 4261, 216, 4282, 216}

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.

\(\displaystyle \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a \sec (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\)

\(\Big \downarrow \) 4575

\(\displaystyle \frac {\int -\frac {\cos ^3(c+d x) (a A-a (7 A+8 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {\cos ^3(c+d x) (a A-a (7 A+8 C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a A-a (7 A+8 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{8 a}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {\cos ^2(c+d x) \left (a^2 (43 A+48 C)-5 a^2 A \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {\cos ^2(c+d x) \left (a^2 (43 A+48 C)-5 a^2 A \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{6 a}}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (43 A+48 C)-5 a^2 A \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{6 a}}{8 a}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {3 \cos (c+d x) \left (a^3 (21 A+16 C)-a^3 (43 A+48 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}}{8 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \int \frac {\cos (c+d x) \left (a^3 (21 A+16 C)-a^3 (43 A+48 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \int \frac {a^3 (21 A+16 C)-a^3 (43 A+48 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {\int -\frac {a^4 (107 A+112 C)-a^4 (21 A+16 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {a^3 (21 A+16 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^4 (107 A+112 C)-a^4 (21 A+16 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^4 (107 A+112 C)-a^4 (21 A+16 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 4408

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a^3 (107 A+112 C) \int \sqrt {\sec (c+d x) a+a}dx-128 a^4 (A+C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a^3 (107 A+112 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-128 a^4 (A+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-128 a^4 (A+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^4 (107 A+112 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{7/2} (107 A+112 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-128 a^4 (A+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {256 a^4 (A+C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^{7/2} (107 A+112 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{7/2} (107 A+112 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {128 \sqrt {2} a^{7/2} (A+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

Input:

Int[(Cos[c + d*x]^4*(A + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]
 

Output:

(A*Cos[c + d*x]^3*Sin[c + d*x])/(4*d*Sqrt[a + a*Sec[c + d*x]]) - ((a*A*Cos 
[c + d*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) - ((a^2*(43*A + 4 
8*C)*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) - (3*(-1/2* 
((2*a^(7/2)*(107*A + 112*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c 
 + d*x]]])/d - (128*Sqrt[2]*a^(7/2)*(A + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/ 
(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a + (a^3*(21*A + 16*C)*Sin[c + d*x 
])/(d*Sqrt[a + a*Sec[c + d*x]])))/(4*a))/(6*a))/(8*a)
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042

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

rule 4261

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

rule 4282

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

rule 4408

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

rule 4510

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*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d 
*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* 
n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 
 m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
 

rule 4575

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

Maple [A] (warning: unable to verify)

Time = 0.56 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.69

method	result	size
default	\(-\frac {\left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-96 \cos \left (d x +c \right )+48\right ) C +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-48 \cos \left (d x +c \right )^{3}+8 \cos \left (d x +c \right )^{2}-86 \cos \left (d x +c \right )+63\right ) A +321 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right ) A +336 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right ) C +192 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) A +192 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{192 d a \left (\cos \left (d x +c \right )+1\right )}\)	\(410\)

Input:

int(cos(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-1/192/d/a*(sin(d*x+c)*cos(d*x+c)*(-96*cos(d*x+c)+48)*C+sin(d*x+c)*cos(d*x 
+c)*(-48*cos(d*x+c)^3+8*cos(d*x+c)^2-86*cos(d*x+c)+63)*A+321*(cos(d*x+c)+1 
)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+ 
c))/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2))*A+336*(co 
s(d*x+c)+1)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c 
)+cot(d*x+c))/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2)) 
*C+192*2^(1/2)*(cos(d*x+c)+1)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*co 
s(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))*A+192*2^(1/2)*(cos(d 
*x+c)+1)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+ 
1))^(1/2)-cot(d*x+c)+csc(d*x+c))*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1 
)
 

Fricas [A] (verification not implemented)

Time = 1.67 (sec) , antiderivative size = 558, normalized size of antiderivative = 2.30 \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algori 
thm="fricas")
 

Output:

[1/384*(192*sqrt(2)*((A + C)*a*cos(d*x + c) + (A + C)*a)*sqrt(-1/a)*log((2 
*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*s 
in(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) - 1)/(cos(d*x + c)^2 + 2*c 
os(d*x + c) + 1)) - 3*((107*A + 112*C)*cos(d*x + c) + 107*A + 112*C)*sqrt( 
-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
 + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) 
 + 2*(48*A*cos(d*x + c)^4 - 8*A*cos(d*x + c)^3 + 2*(43*A + 48*C)*cos(d*x + 
 c)^2 - 3*(21*A + 16*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + 
c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d), -1/192*(3*((107*A + 112*C)*cos 
(d*x + c) + 107*A + 112*C)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d* 
x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (48*A*cos(d*x + c)^4 - 8*A* 
cos(d*x + c)^3 + 2*(43*A + 48*C)*cos(d*x + c)^2 - 3*(21*A + 16*C)*cos(d*x 
+ c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 192*sqrt(2)*( 
(A + C)*a*cos(d*x + c) + (A + C)*a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + 
a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d* 
x + c) + a*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**4*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{4}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(cos(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algori 
thm="maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + A)*cos(d*x + c)^4/sqrt(a*sec(d*x + c) + a), 
x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1012 vs. \(2 (210) = 420\).

Time = 1.16 (sec) , antiderivative size = 1012, normalized size of antiderivative = 4.16 \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algori 
thm="giac")
 

Output:

1/384*(192*sqrt(2)*(A + C)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*ta 
n(1/2*d*x + 1/2*c)^2 + a))^2)/(sqrt(-a)*sgn(cos(d*x + c))) + 3*(107*A + 11 
2*C)*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c) 
^2 + a))^2 - a*(2*sqrt(2) + 3)))/(sqrt(-a)*sgn(cos(d*x + c))) - 3*(107*A + 
 112*C)*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2 
*c)^2 + a))^2 + a*(2*sqrt(2) - 3)))/(sqrt(-a)*sgn(cos(d*x + c))) + 4*sqrt( 
2)*(1599*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + 
 a))^14*A*sqrt(-a) + 816*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2* 
d*x + 1/2*c)^2 + a))^14*C*sqrt(-a) - 18219*(sqrt(-a)*tan(1/2*d*x + 1/2*c) 
- sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*sqrt(-a)*a - 12528*(sqrt(-a)*t 
an(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*C*sqrt(-a)*a 
 + 91467*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + 
 a))^10*A*sqrt(-a)*a^2 + 64752*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*ta 
n(1/2*d*x + 1/2*c)^2 + a))^10*C*sqrt(-a)*a^2 - 177735*(sqrt(-a)*tan(1/2*d* 
x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*sqrt(-a)*a^3 - 12484 
8*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8* 
C*sqrt(-a)*a^3 + 100413*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d 
*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^4 + 70032*(sqrt(-a)*tan(1/2*d*x + 1/2*c 
) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a)*a^4 - 26881*(sqrt(-a 
)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \] Input:

int((cos(c + d*x)^4*(A + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(1/2),x)
 

Output:

int((cos(c + d*x)^4*(A + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4}}{\sec \left (d x +c \right )+1}d x \right ) a \right )}{a} \] Input:

int(cos(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)
 

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*cos(c + d*x)**4*sec(c + d*x)**2)/(se 
c(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x) + 1)*cos(c + d*x)**4)/(sec(c 
 + d*x) + 1),x)*a))/a