\(\int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1257]

Optimal result

Integrand size = 45, antiderivative size = 197 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3/2} (2 B+3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {a^2 (8 A+6 B-3 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {a (2 A-3 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {2 A \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \] Output:

a^(3/2)*(2*B+3*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d 
*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/3*a^2*(8*A+6*B-3*C)*sin(d*x+c)/d/cos(d*x+ 
c)^(1/2)/(a+a*sec(d*x+c))^(1/2)-1/3*a*(2*A-3*C)*(a+a*sec(d*x+c))^(1/2)*sin 
(d*x+c)/d/cos(d*x+c)^(1/2)+2/3*A*cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(3/2)*s 
in(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 6.88 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.62 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \sqrt {\cos (c+d x)} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \left (3 \sqrt {2} (2 B+3 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 (A+3 C+2 (5 A+3 B) \cos (c+d x)+A \cos (2 (c+d x))) \sec (c+d x) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d} \] Input:

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

Output:

(a*Sqrt[Cos[c + d*x]]*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[ 
2]*(2*B + 3*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] + 2*(A + 3*C + 2*(5*A + 3 
*B)*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[c + d*x]*Sin[(c + d*x)/2]))/(6* 
d)
 

Rubi [A] (verified)

Time = 1.37 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 4753, 3042, 4574, 27, 3042, 4506, 27, 3042, 4503, 3042, 4288, 222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

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

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A+6 B-3 C) a^2+3 (2 B+3 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((8 A+6 B-3 C) a^2+3 (2 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 4503

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \left (3 a^2 (2 B+3 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \left (3 a^2 (2 B+3 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 4288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \left (\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^2 (2 B+3 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \left (\frac {6 a^{5/2} (2 B+3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Sec[c + d*x])^(3/2)*Sin 
[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (-((a^2*(2*A - 3*C)*Sqrt[Sec[c + d*x 
]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d) + ((6*a^(5/2)*(2*B + 3*C)*Arc 
Sinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^3*(8*A + 6 
*B - 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/2 
)/(3*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 222

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

rule 3042

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

rule 4288

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

rule 4503

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

rule 4506

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

rule 4574

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

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 [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(443\) vs. \(2(169)=338\).

Time = 4.82 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.25

method	result	size
default	\(2 \sqrt {2}\, \left (\frac {2 A \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2\right ) \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}}\, \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d}-\frac {C \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 \left (\left (-4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}\right ) \left (12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+3 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}\right ) \left (12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+3 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{4 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{\frac {3}{2}}}-\frac {B \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}\right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \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}}\, \sec \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}\right )\)	\(444\)

Input:

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

Output:

2*2^(1/2)*(2/3*A/d*(cos(1/2*d*x+1/2*c)^2+2)*(a/(2*cos(1/2*d*x+1/2*c)^2-1)* 
cos(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)*a*tan(1/2*d*x 
+1/2*c)-1/4*C/d*(a/(2*cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x+1/2*c)^2)^(1/2)* 
a/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)*((-4*cos(1/2*d*x+1/2*c)^2+2)*tan(1/2*d* 
x+1/2*c)+2^(1/2)*arctanh(1/2*2^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c 
)+1))*(12*cos(1/2*d*x+1/2*c)^3-12*cos(1/2*d*x+1/2*c)+3*sec(1/2*d*x+1/2*c)) 
+2^(1/2)*arctanh(1/2*2^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c)-1))*(1 
2*cos(1/2*d*x+1/2*c)^3-12*cos(1/2*d*x+1/2*c)+3*sec(1/2*d*x+1/2*c)))-1/2*B/ 
d*(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)*(2^(1/2)*arctanh(1/2*2^(1/2)*(cot(1/2*d 
*x+1/2*c)-csc(1/2*d*x+1/2*c)+1))+2^(1/2)*arctanh(1/2*2^(1/2)*(cot(1/2*d*x+ 
1/2*c)-csc(1/2*d*x+1/2*c)-1))-2*sin(1/2*d*x+1/2*c))*a*(a/(2*cos(1/2*d*x+1/ 
2*c)^2-1)*cos(1/2*d*x+1/2*c)^2)^(1/2)*sec(1/2*d*x+1/2*c))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.18 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {4 \, {\left (2 \, A a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {2 \, {\left (2 \, A a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \] Input:

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

Output:

[1/12*(4*(2*A*a*cos(d*x + c)^2 + 2*(5*A + 3*B)*a*cos(d*x + c) + 3*C*a)*sqr 
t((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*( 
(2*B + 3*C)*a*cos(d*x + c)^2 + (2*B + 3*C)*a*cos(d*x + c))*sqrt(a)*log((a* 
cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d* 
x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(c 
os(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), 1/6 
*(2*(2*A*a*cos(d*x + c)^2 + 2*(5*A + 3*B)*a*cos(d*x + c) + 3*C*a)*sqrt((a* 
cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((2*B 
+ 3*C)*a*cos(d*x + c)^2 + (2*B + 3*C)*a*cos(d*x + c))*sqrt(-a)*arctan(2*sq 
rt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x 
+ c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)))/(d*cos(d*x + c)^2 + d*cos 
(d*x + c))]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1895 vs. \(2 (169) = 338\).

Time = 0.35 (sec) , antiderivative size = 1895, normalized size of antiderivative = 9.62 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

1/60*(20*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c 
))*A*sqrt(a) + 6*(2*sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 40*sqrt(2)*a*sin(1/2* 
d*x + 1/2*c) - 2*sqrt(2)*a*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d 
*x + 3/2*c))) - 20*sqrt(2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2 
*d*x + 3/2*c))) + 5*a*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* 
d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3 
/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 
3/2*c))) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3 
/2*c))) + 2) - 5*a*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x 
 + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* 
c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2 
*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* 
c))) + 2) + 5*a*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 
3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)) 
)^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c) 
)) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)) 
) + 2) - 5*a*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2 
*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 
 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) 
- 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 414 vs. \(2 (169) = 338\).

Time = 0.59 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.10 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/6*(3*(2*B*a^(3/2)*sgn(cos(d*x + c)) + 3*C*a^(3/2)*sgn(cos(d*x + 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))) - 3*(2*B*a^(3/2)*sgn(cos(d*x + c)) + 3*C*a^(3/2)*sg 
n(cos(d*x + 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))) + 4*(6*sqrt(2)*A*a^3*sgn(cos(d* 
x + c)) + 3*sqrt(2)*B*a^3*sgn(cos(d*x + c)) + (4*sqrt(2)*A*a^3*sgn(cos(d*x 
 + c)) + 3*sqrt(2)*B*a^3*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/ 
2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) + 12*(3*sqrt(2)*(sqrt( 
a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^(5/2)* 
sgn(cos(d*x + c)) - sqrt(2)*C*a^(7/2)*sgn(cos(d*x + c)))/((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))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a*(int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)*sec( 
c + d*x)**3,x)*c + int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d 
*x)*sec(c + d*x)**2,x)*b + int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*c 
os(c + d*x)*sec(c + d*x)**2,x)*c + int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + 
 d*x))*cos(c + d*x)*sec(c + d*x),x)*a + int(sqrt(sec(c + d*x) + 1)*sqrt(co 
s(c + d*x))*cos(c + d*x)*sec(c + d*x),x)*b + int(sqrt(sec(c + d*x) + 1)*sq 
rt(cos(c + d*x))*cos(c + d*x),x)*a)