\(\int \frac {(a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [1141]

Optimal result

Integrand size = 37, antiderivative size = 191 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {a^{3/2} (24 A+11 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{8 d}+\frac {a^2 (24 A+19 C) \sin (c+d x)}{24 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a C \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \] Output:

1/8*a^(3/2)*(24*A+11*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+1/24*a^2*(24*A+19*C)*sin(d*x+c)/d/cos 
(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1/4*a*C*(a+a*sec(d*x+c))^(1/2)*sin(d* 
x+c)/d/cos(d*x+c)^(3/2)+1/3*C*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+ 
c)^(3/2)
 

Mathematica [A] (warning: unable to verify)

Time = 3.06 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.03 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {a^2 \sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left (33 C \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-72 A \arcsin \left (\sqrt {\sec (c+d x)}\right )+22 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+24 A \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+33 C \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \sin (c+d x)}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^2*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*(33*C*ArcSin[Sqrt[1 - Sec[c + d 
*x]]] - 72*A*ArcSin[Sqrt[Sec[c + d*x]]] + 22*C*Sqrt[1 - Sec[c + d*x]]*Sec[ 
c + d*x]^(3/2) + 8*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2) + 24*A*Sqrt 
[-((-1 + Sec[c + d*x])*Sec[c + d*x])] + 33*C*Sqrt[-((-1 + Sec[c + d*x])*Se 
c[c + d*x])])*Sin[c + d*x])/(24*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c 
 + d*x])])
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.06, 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, 4577, 27, 3042, 4506, 27, 3042, 4504, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4577

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \int \frac {1}{2} \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a} \left ((24 A+7 C) a^2+(24 A+19 C) \sec (c+d x) a^2\right )dx+\frac {3 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a} \left ((24 A+7 C) a^2+(24 A+19 C) \sec (c+d x) a^2\right )dx+\frac {3 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((24 A+7 C) a^2+(24 A+19 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {3 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\)

\(\Big \downarrow \) 4504

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {3}{2} a^2 (24 A+11 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (24 A+19 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {3}{2} a^2 (24 A+11 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^3 (24 A+19 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\)

\(\Big \downarrow \) 4288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {a^3 (24 A+19 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 a^2 (24 A+11 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {3 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {3 a^{5/2} (24 A+11 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^3 (24 A+19 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sec[c + d*x]^(3/2)*(a + a*Sec[c 
+ d*x])^(3/2)*Sin[c + d*x])/(3*d) + ((3*a^2*C*Sec[c + d*x]^(3/2)*Sqrt[a + 
a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + ((3*a^(5/2)*(24*A + 11*C)*ArcSinh[(S 
qrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a^3*(24*A + 19*C)*Sec 
[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/4)/(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 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 4504

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C 
ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] 
 + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1))   Int[Sqrt[a + b*Csc[e + f* 
x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, 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 4577

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C) 
*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*(m + n + 1))), 
x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n 
*Simp[A*b*(m + n + 1) + b*C*n + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, 
 b, d, e, f, A, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && 
!LtQ[n, -2^(-1)] && NeQ[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. \(520\) vs. \(2(161)=322\).

Time = 3.94 (sec) , antiderivative size = 521, normalized size of antiderivative = 2.73

method	result	size
default	\(2 \sqrt {2}\, \left (-\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 (-528 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+616 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-252 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+38\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 (528 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-1056 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+792 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-264 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+33 \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 (528 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-1056 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+792 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-264 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+33 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{96 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{\frac {7}{2}}}-\frac {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}}\, 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}}}\right )\)	\(521\)

Input:

int((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x,method=_R 
ETURNVERBOSE)
 

Output:

2*2^(1/2)*(-1/96*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)^(7/2)*((-528*cos(1/2*d*x+1/2*c)^6+616*co 
s(1/2*d*x+1/2*c)^4-252*cos(1/2*d*x+1/2*c)^2+38)*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))*(528*cos(1 
/2*d*x+1/2*c)^7-1056*cos(1/2*d*x+1/2*c)^5+792*cos(1/2*d*x+1/2*c)^3-264*cos 
(1/2*d*x+1/2*c)+33*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))*(528*cos(1/2*d*x+1/2*c)^7-1056*cos(1/2 
*d*x+1/2*c)^5+792*cos(1/2*d*x+1/2*c)^3-264*cos(1/2*d*x+1/2*c)+33*sec(1/2*d 
*x+1/2*c)))-1/4*A/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) 
)*(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))))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.29 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\left [\frac {4 \, {\left (3 \, {\left (8 \, A + 11 \, C\right )} a \cos \left (d x + c\right )^{2} + 22 \, C a \cos \left (d x + c\right ) + 8 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (24 \, A + 11 \, C\right )} a \cos \left (d x + c\right )^{4} + {\left (24 \, A + 11 \, C\right )} a \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, \frac {2 \, {\left (3 \, {\left (8 \, A + 11 \, C\right )} a \cos \left (d x + c\right )^{2} + 22 \, C a \cos \left (d x + c\right ) + 8 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (24 \, A + 11 \, C\right )} a \cos \left (d x + c\right )^{4} + {\left (24 \, A + 11 \, C\right )} a \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \] Input:

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

Output:

[1/96*(4*(3*(8*A + 11*C)*a*cos(d*x + c)^2 + 22*C*a*cos(d*x + c) + 8*C*a)*s 
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3 
*((24*A + 11*C)*a*cos(d*x + c)^4 + (24*A + 11*C)*a*cos(d*x + c)^3)*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)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x + c)^4 + d*cos(d*x + 
c)^3), 1/48*(2*(3*(8*A + 11*C)*a*cos(d*x + c)^2 + 22*C*a*cos(d*x + c) + 8* 
C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + 
c) + 3*((24*A + 11*C)*a*cos(d*x + c)^4 + (24*A + 11*C)*a*cos(d*x + c)^3)*s 
qrt(-a)*arctan(2*sqrt(-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)^4 + d*cos(d*x + c)^3)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3506 vs. \(2 (161) = 322\).

Time = 0.35 (sec) , antiderivative size = 3506, normalized size of antiderivative = 18.36 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/96*(24*(3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2 
*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*lo 
g(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2* 
d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(2*cos(1/2*d*x + 
 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2* 
sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin 
(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d 
*x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 + 3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 
 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin 
(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 
1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) 
 + 2) + a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt 
(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*c 
os(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 
 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(2*d*x + 2*c)^2 + 4*sqrt 
(2)*a*sin(3/2*d*x + 3/2*c) - 4*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 2*(2*sqrt( 
2)*a*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 3*a*log(2*c 
os(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 
 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/ 
2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 783 vs. \(2 (161) = 322\).

Time = 0.72 (sec) , antiderivative size = 783, normalized size of antiderivative = 4.10 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/48*(3*(24*A*a^(3/2)*sgn(cos(d*x + c)) + 11*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*(24*A*a^(3/2)*sgn(cos(d*x + c)) + 11*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))) + 4*(72*sqrt(2)*(sqrt(a)*t 
an(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*a^(5/2)*sgn 
(cos(d*x + c)) + 33*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2 
*d*x + 1/2*c)^2 + a))^10*C*a^(5/2)*sgn(cos(d*x + c)) - 888*sqrt(2)*(sqrt(a 
)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*a^(7/2)*s 
gn(cos(d*x + c)) - 303*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan( 
1/2*d*x + 1/2*c)^2 + a))^8*C*a^(7/2)*sgn(cos(d*x + c)) + 3024*sqrt(2)*(sqr 
t(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(9/2 
)*sgn(cos(d*x + c)) + 2394*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a* 
tan(1/2*d*x + 1/2*c)^2 + a))^6*C*a^(9/2)*sgn(cos(d*x + c)) - 1776*sqrt(2)* 
(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*a^ 
(11/2)*sgn(cos(d*x + c)) - 1806*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sq 
rt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*a^(11/2)*sgn(cos(d*x + c)) + 360*sqr 
t(2)*(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^(13/2)*sgn(cos(d*x + c)) + 309*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^(13/2)*sgn(cos(d*x + c)) - ...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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