\(\int \frac {\cos ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [391]

Optimal result

Integrand size = 42, antiderivative size = 165 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {(7 B-4 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} (B-C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {(B-4 C) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {B \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*(7*B-4*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(1/2)/d- 
2^(1/2)*(B-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-1/4*(B-4*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/2*B*cos(d*x+ 
c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left ((7 B-4 C) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )-4 \sqrt {2} (B-C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) (-B+4 C+2 B \cos (c+d x)) \sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[(Cos[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a* 
Sec[c + d*x]],x]
 

Output:

(((7*B - 4*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] - 4*Sqrt[2]*(B - C)*ArcTanh[ 
Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + Cos[c + d*x]*(-B + 4*C + 2*B*Cos[c + d*x 
])*Sqrt[1 - Sec[c + d*x]])*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[ 
a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.16 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 4560, 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.

\(\displaystyle \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a \sec (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4408

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4261

\(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (B-4 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-8 a^2 (B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^2 (7 B-4 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}}{4 a}\)

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 4282

\(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (B-4 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {16 a^2 (B-C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^{3/2} (7 B-4 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {B \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (B-4 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{3/2} (7 B-4 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} (B-C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\)

Input:

Int[(Cos[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c 
+ d*x]],x]
 

Output:

(B*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) - (-1/2*((2*a 
^(3/2)*(7*B - 4*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]] 
)/d - (8*Sqrt[2]*a^(3/2)*(B - C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sq 
rt[a + a*Sec[c + d*x]])])/d)/a + (a*(B - 4*C)*Sin[c + d*x])/(d*Sqrt[a + a* 
Sec[c + d*x]]))/(4*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 4560

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(384\) vs. \(2(140)=280\).

Time = 5.73 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.33

method	result	size
default	\(-\frac {\left (\left (-7 \cos \left (d x +c \right )-7\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (4 \cos \left (d x +c \right )+4\right ) C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (4 \cos \left (d x +c \right )+4\right ) \sqrt {2}\, B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (-4 \cos \left (d x +c \right )-4\right ) \sqrt {2}\, C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (1-2 \cos \left (d x +c \right )\right ) B -4 C \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d a \left (\cos \left (d x +c \right )+1\right )}\)	\(385\)

Input:

int(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,me 
thod=_RETURNVERBOSE)
 

Output:

-1/4/d/a*((-7*cos(d*x+c)-7)*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2 
^(1/2)/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2)*(csc(d* 
x+c)-cot(d*x+c)))+(4*cos(d*x+c)+4)*C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ar 
ctanh(2^(1/2)/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2)* 
(csc(d*x+c)-cot(d*x+c)))+(4*cos(d*x+c)+4)*2^(1/2)*B*(-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) 
)+(-4*cos(d*x+c)-4)*2^(1/2)*C*(-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))+sin(d*x+c)*cos(d*x+c 
)*(1-2*cos(d*x+c))*B-4*C*cos(d*x+c)*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)/( 
cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 1.16 (sec) , antiderivative size = 506, normalized size of antiderivative = 3.07 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [-\frac {4 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right ) + {\left (B - C\right )} a\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - {\left ({\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right ) + 7 \, B - 4 \, C\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left (2 \, B \cos \left (d x + c\right )^{2} - {\left (B - 4 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac {{\left ({\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right ) + 7 \, B - 4 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (2 \, B \cos \left (d x + c\right )^{2} - {\left (B - 4 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - \frac {4 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right ) + {\left (B - C\right )} a\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{4 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \] Input:

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

Output:

[-1/8*(4*sqrt(2)*((B - C)*a*cos(d*x + c) + (B - C)*a)*sqrt(-1/a)*log(-(2*s 
qrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin 
(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1)/(cos(d*x + c)^2 + 2*cos 
(d*x + c) + 1)) - ((7*B - 4*C)*cos(d*x + c) + 7*B - 4*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*(2*B*co 
s(d*x + c)^2 - (B - 4*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), -1/4*(((7*B - 4*C)*cos(d*x + 
c) + 7*B - 4*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))) - (2*B*cos(d*x + c)^2 - (B - 4*C)*cos(d* 
x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 4*sqrt(2)*( 
(B - C)*a*cos(d*x + c) + (B - 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]

\[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \cos \left (d x + c\right )^{3}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \] Input:

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

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*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. 621 vs. \(2 (140) = 280\).

Time = 1.13 (sec) , antiderivative size = 621, normalized size of antiderivative = 3.76 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/8*(4*sqrt(2)*(B - C)*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))) + (7*B - 4*C)*log(a 
bs((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))) - (7*B - 4*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)*(17*(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) 
- 12*(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) - 57*(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 + 76*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(- 
a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*sqrt(-a)*a + 19*(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^2 - 36*(sqrt 
(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*sqrt( 
-a)*a^2 - 3*B*sqrt(-a)*a^3 + 4*C*sqrt(-a)*a^3)/(((sqrt(-a)*tan(1/2*d*x + 1 
/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 
 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^2*sgn(cos(d*x + 
c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) b \right )}{a} \] Input:

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