\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [1162]

Optimal result

Integrand size = 37, antiderivative size = 221 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {(11 A+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(19 A+3 C) \sin (c+d x)}{6 a d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(7 A+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*(11*A+3*C)*arctanh(1/2*a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)*2^(1/2)/(a+ 
a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/a^(3/2)/d-1 
/2*(A+C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)-1/6*(19*A+3* 
C)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+1/6*(7*A+3*C)*co 
s(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\sqrt {\cos (c+d x)} \left (3 \sqrt {2} (11 A+3 C) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)+\sqrt {1-\sec (c+d x)} \left (-4 A+12 A \sec (c+d x)+(19 A+3 C) \sec ^2(c+d x)\right )\right ) \sin (c+d x)}{6 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

-1/6*(Sqrt[Cos[c + d*x]]*(3*Sqrt[2]*(11*A + 3*C)*ArcTan[(Sqrt[2]*Sqrt[Sec[ 
c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^2*Sec[c + d*x]^(5/2) + 
 Sqrt[1 - Sec[c + d*x]]*(-4*A + 12*A*Sec[c + d*x] + (19*A + 3*C)*Sec[c + d 
*x]^2))*Sin[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/ 
2))
 

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 4753, 3042, 4573, 27, 3042, 4510, 27, 3042, 4501, 3042, 4295, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sec ^2(c+d x)+A}{\sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int -\frac {a^2 (19 A+3 C)-2 a^2 (7 A+3 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (19 A+3 C)-2 a^2 (7 A+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (19 A+3 C)-2 a^2 (7 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 4501

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^2 (11 A+3 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^2 (11 A+3 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {6 a^2 (11 A+3 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 \sqrt {2} a^{3/2} (11 A+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A + C)*Sin[c + d*x])/(d*Sqrt 
[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)) + ((2*a*(7*A + 3*C)*Sin[c + d*x 
])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((-3*Sqrt[2]*a^(3/2 
)*(11*A + 3*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]* 
Sqrt[a + a*Sec[c + d*x]])])/d + (2*a^2*(19*A + 3*C)*Sqrt[Sec[c + d*x]]*Sin 
[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(4*a^2))
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 4295

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

rule 4501

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

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 4573

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

rule 4753

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a 
+ b*x])^m*(c*Sec[a + b*x])^m   Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 

Maple [A] (verified)

Time = 3.84 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.47

method	result	size
default	\(\frac {\sqrt {2}\, \left (\frac {A \left (16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )+33 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-33 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-40 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}}\, a}+\frac {C \left (3 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}}\, a}\right )}{4}\)	\(324\)

Input:

int(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x,method=_R 
ETURNVERBOSE)
 

Output:

1/4*2^(1/2)*(1/3*A/d/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/(a/(2*cos(1/2*d*x+1/ 
2*c)^2-1)*cos(1/2*d*x+1/2*c)^2)^(1/2)/a*(16*cos(1/2*d*x+1/2*c)^3*sin(1/2*d 
*x+1/2*c)+33*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)+1)*cos(1/2*d*x+1/2* 
c)-33*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)-1)*cos(1/2*d*x+1/2*c)-40*c 
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)-3*tan(1/2*d*x+1/2*c))+C/d/(2*cos(1/2* 
d*x+1/2*c)^2-1)^(1/2)/(a/(2*cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x+1/2*c)^2)^ 
(1/2)/a*(3*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)+1)*cos(1/2*d*x+1/2*c) 
-3*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)-1)*cos(1/2*d*x+1/2*c)-tan(1/2 
*d*x+1/2*c)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 442, normalized size of antiderivative = 2.00 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left ({\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 11 \, A + 3 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (4 \, A \cos \left (d x + c\right )^{2} - 12 \, A \cos \left (d x + c\right ) - 19 \, A - 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac {3 \, \sqrt {2} {\left ({\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 11 \, A + 3 \, C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right ) + a\right )}}\right ) - 2 \, {\left (4 \, A \cos \left (d x + c\right )^{2} - 12 \, A \cos \left (d x + c\right ) - 19 \, A - 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \] Input:

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, al 
gorithm="fricas")
 

Output:

[1/24*(3*sqrt(2)*((11*A + 3*C)*cos(d*x + c)^2 + 2*(11*A + 3*C)*cos(d*x + c 
) + 11*A + 3*C)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a 
*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos 
(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(4*A*cos(d*x + 
 c)^2 - 12*A*cos(d*x + c) - 19*A - 3*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
+ c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos 
(d*x + c) + a^2*d), -1/12*(3*sqrt(2)*((11*A + 3*C)*cos(d*x + c)^2 + 2*(11* 
A + 3*C)*cos(d*x + c) + 11*A + 3*C)*sqrt(-a)*arctan(1/2*sqrt(2)*sqrt(-a)*s 
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a* 
cos(d*x + c) + a)) - 2*(4*A*cos(d*x + c)^2 - 12*A*cos(d*x + c) - 19*A - 3* 
C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) 
)/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34992 vs. \(2 (186) = 372\).

Time = 0.65 (sec) , antiderivative size = 34992, normalized size of antiderivative = 158.33 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, al 
gorithm="maxima")
 

Output:

1/12*((4*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin 
(3/2*d*x + 3/2*c) - 9*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*ar 
ctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(7/3*arctan2(sin(3/ 
2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^4 + 64*(cos(3*d*x + 3*c)^2*sin(3/2* 
d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) - 9*(cos(3*d*x + 3* 
c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d 
*x + 3/2*c))))*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)) 
)^4 + 4*sin(3/2*d*x + 3/2*c)^5 + 4*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c 
) + sin(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) - 9*(cos(3*d*x + 3*c)^2 + sin( 
3*d*x + 3*c)^2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c) 
)))*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^4 + 64*(c 
os(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin(3/2*d*x + 
3/2*c) - 9*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*arctan2(sin(3 
/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin(5/3*arctan2(sin(3/2*d*x + 3/2 
*c), cos(3/2*d*x + 3/2*c)))^4 + 4*(2*cos(3*d*x + 3*c)^2*cos(3/2*d*x + 3/2* 
c)*sin(3/2*d*x + 3/2*c) + 2*cos(3/2*d*x + 3/2*c)*sin(3*d*x + 3*c)^2*sin(3/ 
2*d*x + 3/2*c) + 2*cos(3*d*x + 3*c)*cos(3/2*d*x + 3/2*c)*sin(3/2*d*x + 3/2 
*c) + 8*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin( 
3/2*d*x + 3/2*c) - 9*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*arc 
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(5/3*arctan2(sin(...
 

Giac [A] (verification not implemented)

Time = 173.05 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {{\left ({\left (\frac {3 \, {\left (\sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a} + \frac {2 \, {\left (23 \, \sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {3 \, {\left (9 \, \sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}} + \frac {3 \, {\left (11 \, \sqrt {2} A + 3 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{12 \, d} \] Input:

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, al 
gorithm="giac")
 

Output:

-1/12*(((3*(sqrt(2)*A*a*sgn(cos(d*x + c)) + sqrt(2)*C*a*sgn(cos(d*x + c))) 
*tan(1/2*d*x + 1/2*c)^2/a + 2*(23*sqrt(2)*A*a*sgn(cos(d*x + c)) + 3*sqrt(2 
)*C*a*sgn(cos(d*x + c)))/a)*tan(1/2*d*x + 1/2*c)^2 + 3*(9*sqrt(2)*A*a*sgn( 
cos(d*x + c)) + sqrt(2)*C*a*sgn(cos(d*x + c)))/a)*tan(1/2*d*x + 1/2*c)/(a* 
tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) + 3*(11*sqrt(2)*A + 3*sqrt(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)))/(a^( 
3/2)*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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