\(\int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx\) [286]

Optimal result

Integrand size = 37, antiderivative size = 152 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(7 A-C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(5 A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}} \] Output:

-1/4*(7*A-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))*2^(1/2)/a^(3/2)/d-1/2*(A+C)*sec(d*x+c)^(1/2)*sin(d*x+c) 
/d/(a+a*sec(d*x+c))^(3/2)+1/2*(5*A+C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a 
*sec(d*x+c))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(383\) vs. \(2(152)=304\).

Time = 1.30 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.52 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\frac {10 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+2 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+8 A \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sin (c+d x)+7 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-\sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)+7 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-\sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)+2 C \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) (1+\sec (c+d x)) \tan (c+d x)+2 C \arcsin \left (\sqrt {\sec (c+d x)}\right ) (1+\sec (c+d x)) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

(10*A*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + 2*C*Sqrt[1 
- Sec[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + 8*A*Sqrt[-((-1 + Sec[c + 
 d*x])*Sec[c + d*x])]*Sin[c + d*x] + 7*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[ 
c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - Sqrt[2]*C*ArcTan[(Sqrt[2 
]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] + 7*Sqrt[2]*A*A 
rcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Ta 
n[c + d*x] - Sqrt[2]*C*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c 
+ d*x]]]*Sec[c + d*x]*Tan[c + d*x] + 2*C*ArcSin[Sqrt[1 - Sec[c + d*x]]]*(1 
 + Sec[c + d*x])*Tan[c + d*x] + 2*C*ArcSin[Sqrt[Sec[c + d*x]]]*(1 + Sec[c 
+ d*x])*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^( 
3/2))
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {3042, 4573, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

\(\displaystyle -\frac {\int -\frac {a (5 A+C)-2 a (A-C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (5 A+C)-2 a (A-C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (5 A+C)-2 a (A-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4501

\(\displaystyle \frac {\frac {2 a (5 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a (7 A-C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (5 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a (7 A-C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {2 a (7 A-C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a (5 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 a (5 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {\sqrt {2} \sqrt {a} (7 A-C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

Input:

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

Output:

-1/2*((A + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/ 
2)) + (-((Sqrt[2]*Sqrt[a]*(7*A - C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Si 
n[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d) + (2*a*(5*A + C)*Sqrt[ 
Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(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 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)]
 

Maple [A] (verified)

Time = 1.97 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.41

method	result	size
default	\(\frac {\left (-\frac {\left (-\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) C}{4}-\frac {\left (-\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-9 \csc \left (d x +c \right )+9 \cot \left (d x +c \right )\right ) A}{4}+\frac {7 A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}}{4}-\frac {C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )}{4}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{2} \sqrt {\sec \left (d x +c \right )}}\)	\(215\)
parts	\(\frac {A \left (\frac {\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{4}+\frac {7 \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}}{4}+\frac {9 \csc \left (d x +c \right )}{4}-\frac {9 \cot \left (d x +c \right )}{4}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{2} \sqrt {\sec \left (d x +c \right )}}+\frac {C \left (\left (\cos \left (d x +c \right )+1\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sqrt {2}\, \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \cos \left (d x +c \right )^{2}}{4 d \,a^{2} \left (\cos \left (d x +c \right )+1\right )^{2} \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(250\)

Input:

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

Output:

1/d*(-1/4*(-(1-cos(d*x+c))^3*csc(d*x+c)^3-csc(d*x+c)+cot(d*x+c))*C-1/4*(-( 
1-cos(d*x+c))^3*csc(d*x+c)^3-9*csc(d*x+c)+9*cot(d*x+c))*A+7/4*A*arctan(1/2 
*2^(1/2)/(-1/(cos(d*x+c)+1))^(1/2)*(-csc(d*x+c)+cot(d*x+c)))*(-2/(cos(d*x+ 
c)+1))^(1/2)-1/4*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)/(-1/(cos(d 
*x+c)+1))^(1/2)*(-csc(d*x+c)+cot(d*x+c))))/a^2*(a*(1+sec(d*x+c)))^(1/2)/se 
c(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.80 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {\sqrt {2} {\left ({\left (7 \, A - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - C\right )} \cos \left (d x + c\right ) + 7 \, A - C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - \frac {4 \, {\left (4 \, A \cos \left (d x + c\right )^{2} + {\left (5 \, A + C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac {\sqrt {2} {\left ({\left (7 \, A - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - C\right )} \cos \left (d x + c\right ) + 7 \, A - C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (4 \, A \cos \left (d x + c\right )^{2} + {\left (5 \, A + C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \] Input:

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

Output:

[-1/8*(sqrt(2)*((7*A - C)*cos(d*x + c)^2 + 2*(7*A - C)*cos(d*x + c) + 7*A 
- 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 + (5* 
A + C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/ 
sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), 
 1/4*(sqrt(2)*((7*A - C)*cos(d*x + c)^2 + 2*(7*A - C)*cos(d*x + c) + 7*A - 
 C)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c 
))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + 2*(4*A*cos(d*x + c)^2 + (5*A + C 
)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt( 
cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 22898 vs. \(2 (127) = 254\).

Time = 0.83 (sec) , antiderivative size = 22898, normalized size of antiderivative = 150.64 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*((4*(7*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/ 
2*d*x + 1/2*c) + 1) - 7*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^ 
2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 8*sin(1/2*d*x + 1/2*c))*cos(3/2*d*x + 3/ 
2*c)^4 + 63*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1 
/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 
 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(1/2*d*x + 1/2*c)^4 + 4*(7*log(cos(1/2* 
d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 7* 
log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2* 
c) + 1) - 8*sin(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c)^4 + 70*(log(cos(1/2 
*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - l 
og(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c 
) + 1))*cos(1/2*d*x + 1/2*c)^2*sin(1/2*d*x + 1/2*c)^2 + 7*(log(cos(1/2*d*x 
 + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(c 
os(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 
1))*sin(1/2*d*x + 1/2*c)^4 - 8*sin(1/2*d*x + 1/2*c)^5 + 28*(7*(log(cos(1/2 
*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - l 
og(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c 
) + 1))*cos(1/2*d*x + 1/2*c) - 8*cos(1/2*d*x + 1/2*c)*sin(1/2*d*x + 1/2*c) 
)*cos(3/2*d*x + 3/2*c)^3 + 4*(21*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x 
 + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 ...
 

Giac [A] (verification not implemented)

Time = 2.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.91 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {{\left (\frac {\sqrt {2} {\left (A a^{2} + C a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3}} + \frac {\sqrt {2} {\left (9 \, A a^{2} + C a^{2}\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} + \frac {\sqrt {2} {\left (7 \, A - C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}}}}{4 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

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

Output:

1/4*((sqrt(2)*(A*a^2 + C*a^2)*tan(1/2*d*x + 1/2*c)^2/a^3 + sqrt(2)*(9*A*a^ 
2 + C*a^2)/a^3)*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) + 
sqrt(2)*(7*A - 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))/(d*sgn(cos(d*x + c)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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