\(\int \frac {\sqrt {\cos (c+d x)} (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [1163]

Optimal result

Integrand size = 37, antiderivative size = 172 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(7 A-C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(5 A+C) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \] 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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/a^(3/2)/d-1/2 
*(A+C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(3/2)+1/2*(5*A+C)*si 
n(d*x+c)/a/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 2.05 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.97 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {\sec (c+d x)} \left (7 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )-\sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )+\frac {8 A \sqrt {1-\sec (c+d x)}}{\sqrt {\sec (c+d x)}}+7 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x)-\sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x)+10 A \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+2 C \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+2 C \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) (1+\sec (c+d x))+2 C \arcsin \left (\sqrt {\sec (c+d x)}\right ) (1+\sec (c+d x))\right ) \sin (c+d x)}{4 d \sqrt {-1+\cos (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

(Sqrt[Sec[c + d*x]]*(7*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[ 
1 - Sec[c + d*x]]] - Sqrt[2]*C*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 
- Sec[c + d*x]]] + (8*A*Sqrt[1 - Sec[c + d*x]])/Sqrt[Sec[c + d*x]] + 7*Sqr 
t[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + 
 d*x] - Sqrt[2]*C*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x 
]]]*Sec[c + d*x] + 10*A*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] + 2*C*Sq 
rt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] + 2*C*ArcSin[Sqrt[1 - Sec[c + d*x] 
]]*(1 + Sec[c + d*x]) + 2*C*ArcSin[Sqrt[Sec[c + d*x]]]*(1 + Sec[c + d*x])) 
*Sin[c + d*x])/(4*d*Sqrt[-1 + Cos[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {3042, 4753, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4501

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4295

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

\(\Big \downarrow \) 219

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

Input:

Int[(Sqrt[Cos[c + d*x]]*(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)*Sqrt[Sec[c + d*x]]*Si 
n[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]]*Sin[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)]
 

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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(298\) vs. \(2(143)=286\).

Time = 3.54 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.74

method	result	size
default	\(\frac {\sqrt {2}\, \left (-\frac {A \left (7 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-7 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}}\, a}+\frac {C \left (\ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \sqrt {\frac {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}}\, \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, a}\right )}{4}\)	\(299\)

Input:

int(cos(d*x+c)^(1/2)*(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)*(-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*(7*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)-7*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)-8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)-tan(1/2*d*x+ 
1/2*c))+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)/(2*c 
os(1/2*d*x+1/2*c)^2-1)^(1/2)/a*(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)-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 = 0.09 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.41 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {\sqrt {2} {\left ({\left (7 \, A - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - C\right )} \cos \left (d x + c\right ) + 7 \, A - C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (4 \, A \cos \left (d x + c\right ) + 5 \, A + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac {\sqrt {2} {\left ({\left (7 \, A - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - C\right )} \cos \left (d x + c\right ) + 7 \, A - C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right ) + a\right )}}\right ) + 2 \, {\left (4 \, A \cos \left (d x + c\right ) + 5 \, A + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \] Input:

integrate(cos(d*x+c)^(1/2)*(A+C*sec(d*x+c)^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) + 5*A + 
 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), 1/4*(sqrt(2)*((7 
*A - C)*cos(d*x + c)^2 + 2*(7*A - C)*cos(d*x + c) + 7*A - C)*sqrt(-a)*arct 
an(1/2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d 
*x + c))*sin(d*x + c)/(a*cos(d*x + c) + a)) + 2*(4*A*cos(d*x + c) + 5*A + 
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]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8312 vs. \(2 (143) = 286\).

Time = 0.36 (sec) , antiderivative size = 8312, normalized size of antiderivative = 48.33 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(1/2)*(A+C*sec(d*x+c)^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 = 167.69 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {{\left (\frac {\sqrt {2} {\left (A a^{2} + C a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {\sqrt {2} {\left (9 \, A a^{2} + C a^{2}\right )}}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} + \frac {\sqrt {2} {\left (7 \, A - C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{4 \, d} \] Input:

integrate(cos(d*x+c)^(1/2)*(A+C*sec(d*x+c)^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*sgn(cos(d*x + c) 
)) + sqrt(2)*(9*A*a^2 + C*a^2)/(a^3*sgn(cos(d*x + c))))*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)*sgn( 
cos(d*x + c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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