\(\int \frac {(a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [262]

Optimal result

Integrand size = 37, antiderivative size = 169 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {3 a^{3/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {a^2 (8 A-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}-\frac {a (2 A-3 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 1.18 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.71 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^2 \left (-9 C \arcsin \left (\sqrt {\sec (c+d x)}\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+\sqrt {1-\sec (c+d x)} (2 A \sin (c+d x)+(10 A+3 C \sec (c+d x)) \tan (c+d x))\right )}{3 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^2*(-9*C*ArcSin[Sqrt[Sec[c + d*x]]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + Sq 
rt[1 - Sec[c + d*x]]*(2*A*Sin[c + d*x] + (10*A + 3*C*Sec[c + d*x])*Tan[c + 
 d*x])))/(3*d*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*Sqrt[a*(1 + Sec[c 
+ d*x])])
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 4575, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4575

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A-3 C) a^2+9 C \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A-3 C) a^2+9 C \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((8 A-3 C) a^2+9 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {\frac {1}{2} \left (9 a^2 C \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^3 (8 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (9 a^2 C \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^3 (8 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 a^3 (8 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {18 a^2 C \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {1}{2} \left (\frac {18 a^{5/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (8 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

Input:

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

Output:

(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) + ((18*a^(5/2)*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d 
*x]]])/d + (2*a^3*(8*A - 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 4575

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*Co 
t[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 
*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 
 C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || 
 EqQ[m + n + 1, 0])
 

Maple [A] (warning: unable to verify)

Time = 1.33 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.16

method	result	size
default	\(\frac {a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-9-9 \sec \left (d x +c \right )\right )+\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-9-9 \sec \left (d x +c \right )\right )+A \left (8 \sin \left (d x +c \right )+40 \tan \left (d x +c \right )\right )+12 C \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{12 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(196\)
parts	\(\frac {A \left (2 \sin \left (d x +c \right )+10 \tan \left (d x +c \right )\right ) a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (3 \cos \left (d x +c \right )+3\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {C a \sqrt {\sec \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}-3 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )-3 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )\right )}{2 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(215\)

Input:

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

Output:

1/12/d*a*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*(2^(1/2) 
*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2/(-1/(cos(d*x+c)+1))^(1/2)*(cot(d*x 
+c)-csc(d*x+c)+1))*(-9-9*sec(d*x+c))+2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2)*a 
rctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))*(-9-9*sec(d 
*x+c))+A*(8*sin(d*x+c)+40*tan(d*x+c))+12*C*sec(d*x+c)*tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 365, normalized size of antiderivative = 2.16 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {9 \, {\left (C a \cos \left (d x + c\right ) + C a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (2 \, A a \cos \left (d x + c\right )^{2} + 10 \, A a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{12 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {9 \, {\left (C a \cos \left (d x + c\right ) + C a\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (2 \, A a \cos \left (d x + c\right )^{2} + 10 \, A a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

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

Output:

[1/12*(9*(C*a*cos(d*x + c) + C*a)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos( 
d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + 
 c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c 
)^3 + cos(d*x + c)^2)) + 4*(2*A*a*cos(d*x + c)^2 + 10*A*a*cos(d*x + c) + 3 
*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + 
c)))/(d*cos(d*x + c) + d), 1/6*(9*(C*a*cos(d*x + c) + C*a)*sqrt(-a)*arctan 
(1/2*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/ 
cos(d*x + c))/(a*sqrt(cos(d*x + c))*sin(d*x + c))) + 2*(2*A*a*cos(d*x + c) 
^2 + 10*A*a*cos(d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* 
sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)]
 

Sympy [F(-1)]

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

integrate((a+a*sec(d*x+c))**(3/2)*(A+C*sec(d*x+c)**2)/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. 1183 vs. \(2 (145) = 290\).

Time = 0.25 (sec) , antiderivative size = 1183, normalized size of antiderivative = 7.00 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/12*(4*(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) + 3*(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)*c 
os(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)*si 
n(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*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))*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*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) - 3*a*log(2*cos(1/...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (145) = 290\).

Time = 1.93 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.98 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {3 \, \sqrt {2} C a^{\frac {7}{2}} {\left (\frac {3 \, \sqrt {2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{a {\left | a \right |}} + \frac {8 \, {\left (3 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )}}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )} a}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \frac {16 \, {\left (2 \, \sqrt {2} A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, \sqrt {2} A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}}}{12 \, d} \] Input:

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

Output:

1/12*(3*sqrt(2)*C*a^(7/2)*(3*sqrt(2)*log(abs(2*(sqrt(a)*tan(1/2*d*x + 1/2* 
c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2 
*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4 
*sqrt(2)*abs(a) - 6*a))/(a*abs(a)) + 8*(3*(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)*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)*a))*sgn(cos(d*x + c)) + 
16*(2*sqrt(2)*A*a^3*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 + 3*sqrt(2)*A 
*a^3*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a 
)^(3/2))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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