\(\int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [151]

Optimal result

Integrand size = 33, antiderivative size = 206 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {(9 A-14 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 \sqrt {a} d}+\frac {\sqrt {2} (A-B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {(7 A-2 B) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}-\frac {(A-6 B) \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}} \] Output:

-1/8*(9*A-14*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(1/2)/ 
d+2^(1/2)*(A-B)*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+1/8*(7*A-2*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/12*(A-6* 
B)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/3*A*cos(d*x+c)^2*sin(d 
*x+c)/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left ((-27 A+42 B) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+24 \sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) \left (21 A-6 B-2 (A-6 B) \cos (c+d x)+8 A \cos ^2(c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[(Cos[c + d*x]^3*(A + B*Sec[c + d*x]))/Sqrt[a + a*Sec[c + d*x]],x 
]
 

Output:

(((-27*A + 42*B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 24*Sqrt[2]*(A - B)*ArcT 
anh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + Cos[c + d*x]*(21*A - 6*B - 2*(A - 6* 
B)*Cos[c + d*x] + 8*A*Cos[c + d*x]^2)*Sqrt[1 - Sec[c + d*x]])*Tan[c + d*x] 
)/(24*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.37 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.12, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {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.

\(\displaystyle \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a \sec (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\int -\frac {\cos ^2(c+d x) (a (A-6 B)-5 a A \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {\cos ^2(c+d x) (a (A-6 B)-5 a A \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{6 a}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {3 \cos (c+d x) \left (a^2 (7 A-2 B)-a^2 (A-6 B) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \int \frac {\cos (c+d x) \left (a^2 (7 A-2 B)-a^2 (A-6 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a}}{6 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \int \frac {a^2 (7 A-2 B)-a^2 (A-6 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}}{6 a}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {\int -\frac {a^3 (9 A-14 B)-a^3 (7 A-2 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {a^2 (7 A-2 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (9 A-14 B)-a^3 (7 A-2 B) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (9 A-14 B)-a^3 (7 A-2 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 4408

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (9 A-14 B) \int \sqrt {\sec (c+d x) a+a}dx-16 a^3 (A-B) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (9 A-14 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-16 a^3 (A-B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-16 a^3 (A-B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^3 (9 A-14 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{5/2} (9 A-14 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-16 a^3 (A-B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {32 a^3 (A-B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^{5/2} (9 A-14 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{5/2} (9 A-14 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {16 \sqrt {2} a^{5/2} (A-B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\)

Input:

Int[(Cos[c + d*x]^3*(A + B*Sec[c + d*x]))/Sqrt[a + a*Sec[c + d*x]],x]
 

Output:

(A*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) - ((a*(A - 
6*B)*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) - (3*(-1/2* 
((2*a^(5/2)*(9*A - 14*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + 
d*x]]])/d - (16*Sqrt[2]*a^(5/2)*(A - B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqr 
t[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a + (a^2*(7*A - 2*B)*Sin[c + d*x])/(d* 
Sqrt[a + a*Sec[c + d*x]])))/(4*a))/(6*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]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(403\) vs. \(2(177)=354\).

Time = 1.70 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.96

method	result	size
default	\(\frac {\left (\left (27 \cos \left (d x +c \right )+27\right ) A \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\left (-42 \cos \left (d x +c \right )-42\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\left (24 \cos \left (d x +c \right )+24\right ) \sqrt {2}\, A \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (-24 \cos \left (d x +c \right )-24\right ) \sqrt {2}\, B \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (8 \cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )+21\right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (12 \cos \left (d x +c \right )-6\right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{24 d a \left (1+\cos \left (d x +c \right )\right )}\)	\(404\)

Input:

int(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

1/24/d/a*((27*cos(d*x+c)+27)*A*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh( 
2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(cot(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+csc 
(d*x+c)^2-1)^(1/2))+(-42*cos(d*x+c)-42)*B*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/ 
2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(cot(d*x+c)^2-2*csc(d*x+c)*cot 
(d*x+c)+csc(d*x+c)^2-1)^(1/2))+(24*cos(d*x+c)+24)*2^(1/2)*A*(-cos(d*x+c)/( 
1+cos(d*x+c)))^(1/2)*ln((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-cot(d*x+c)+cs 
c(d*x+c))+(-24*cos(d*x+c)-24)*2^(1/2)*B*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2) 
*ln((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-cot(d*x+c)+csc(d*x+c))+sin(d*x+c) 
*cos(d*x+c)*(8*cos(d*x+c)^2-2*cos(d*x+c)+21)*A+sin(d*x+c)*cos(d*x+c)*(12*c 
os(d*x+c)-6)*B)*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 539, normalized size of antiderivative = 2.62 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algorith 
m="fricas")
 

Output:

[-1/48*(24*sqrt(2)*((A - B)*a*cos(d*x + c) + (A - B)*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)*si 
n(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) - 1)/(cos(d*x + c)^2 + 2*co 
s(d*x + c) + 1)) - 3*((9*A - 14*B)*cos(d*x + c) + 9*A - 14*B)*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*(8 
*A*cos(d*x + c)^3 - 2*(A - 6*B)*cos(d*x + c)^2 + 3*(7*A - 2*B)*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), 1/24*(3*((9*A - 14*B)*cos(d*x + c) + 9*A - 14*B)*sqrt(a)*arctan(s 
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)) 
) + (8*A*cos(d*x + c)^3 - 2*(A - 6*B)*cos(d*x + c)^2 + 3*(7*A - 2*B)*cos(d 
*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 24*sqrt(2) 
*((A - B)*a*cos(d*x + c) + (A - B)*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]

\[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \] Input:

integrate(cos(d*x+c)**3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(1/2),x)
 

Output:

Integral((A + B*sec(c + d*x))*cos(c + d*x)**3/sqrt(a*(sec(c + d*x) + 1)), 
x)
 

Maxima [F]

\[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algorith 
m="maxima")
 

Output:

integrate((B*sec(d*x + c) + A)*cos(d*x + c)^3/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. 818 vs. \(2 (177) = 354\).

Time = 0.88 (sec) , antiderivative size = 818, normalized size of antiderivative = 3.97 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algorith 
m="giac")
 

Output:

-1/48*(24*sqrt(2)*(A - B)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan 
(1/2*d*x + 1/2*c)^2 + a))^2)/(sqrt(-a)*sgn(cos(d*x + c))) + 3*(9*A - 14*B) 
*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*(9*A - 14*B) 
*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)*(165 
*(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) - 102*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/ 
2*c)^2 + a))^10*B*sqrt(-a) - 1323*(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 + 954*(sqrt(-a)*tan(1/2*d*x + 
 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*B*sqrt(-a)*a + 3906*(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^2 - 2268*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2* 
c)^2 + a))^6*B*sqrt(-a)*a^2 - 2118*(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(-a)*a^3 + 1044*(sqrt(-a)*tan(1/2*d 
*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a^3 + 393* 
(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A* 
sqrt(-a)*a^4 - 222*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 
1/2*c)^2 + a))^2*B*sqrt(-a)*a^4 - 31*A*sqrt(-a)*a^5 + 18*B*sqrt(-a)*a^5)/( 
((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \] Input:

int((cos(c + d*x)^3*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int((cos(c + d*x)^3*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3}}{\sec \left (d x +c \right )+1}d x \right ) a \right )}{a} \] Input:

int(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x)
 

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x))/(sec(c 
 + d*x) + 1),x)*b + int((sqrt(sec(c + d*x) + 1)*cos(c + d*x)**3)/(sec(c + 
d*x) + 1),x)*a))/a