\(\int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [1252]

Optimal result

Integrand size = 45, antiderivative size = 247 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {a} (48 A+40 B+35 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 d}+\frac {a (8 B+C) \sin (c+d x)}{24 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (48 A+40 B+35 C) \sin (c+d x)}{96 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (48 A+40 B+35 C) \sin (c+d x)}{64 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {C \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x)} \] Output:

1/64*a^(1/2)*(48*A+40*B+35*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*(8*B+C)*sin(d*x+c)/d/cos 
(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2)+1/96*a*(48*A+40*B+35*C)*sin(d*x+c)/d/ 
cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+1/64*a*(48*A+40*B+35*C)*sin(d*x+c) 
/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1/4*C*(a+a*sec(d*x+c))^(1/2)*si 
n(d*x+c)/d/cos(d*x+c)^(7/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 1.55 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a \sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left (3 (48 A+40 B+35 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+2 (48 A+40 B+35 C) \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 (8 B+7 C) \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+48 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {7}{2}}(c+d x)+3 (48 A+40 B+35 C) \sqrt {(-1+\cos (c+d x)) \sec ^2(c+d x)}\right ) \sin (c+d x)}{192 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2 
))/Cos[c + d*x]^(5/2),x]
 

Output:

(a*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*(3*(48*A + 40*B + 35*C)*ArcSin[Sq 
rt[1 - Sec[c + d*x]]] + 2*(48*A + 40*B + 35*C)*Sqrt[1 - Sec[c + d*x]]*Sec[ 
c + d*x]^(3/2) + 8*(8*B + 7*C)*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2) + 
 48*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(7/2) + 3*(48*A + 40*B + 35*C)*S 
qrt[(-1 + Cos[c + d*x])*Sec[c + d*x]^2])*Sin[c + d*x])/(192*d*Sqrt[1 - Sec 
[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {3042, 4753, 3042, 4576, 27, 3042, 4504, 3042, 4290, 3042, 4290, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4576

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {1}{2} \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (a (8 A+5 C)+a (8 B+C) \sec (c+d x))dx}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (a (8 A+5 C)+a (8 B+C) \sec (c+d x))dx}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (8 A+5 C)+a (8 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 4504

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (8 B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 (8 B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 4290

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 4290

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 4288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (8 B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

Input:

Int[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Cos 
[c + d*x]^(5/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sec[c + d*x]^(7/2)*Sqrt[a + a*Se 
c[c + d*x]]*Sin[c + d*x])/(4*d) + ((a^2*(8*B + C)*Sec[c + d*x]^(7/2)*Sin[c 
 + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(48*A + 40*B + 35*C)*((a*Sec[ 
c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (3*((Sqrt[a] 
*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a*Sec[c + 
d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4))/6)/(8*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 4290

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

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 4576

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[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs 
c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Cs 
c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* 
B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 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. \(463\) vs. \(2(211)=422\).

Time = 3.20 (sec) , antiderivative size = 464, normalized size of antiderivative = 1.88

method	result	size
default	\(\frac {\left (144 A \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+120 B \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+105 C \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+144 A \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+120 B \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+105 C \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \left (144 \cos \left (d x +c \right )+96\right ) \sqrt {2}\, A \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (120 \cos \left (d x +c \right )^{2}+80 \cos \left (d x +c \right )+64\right ) \sqrt {2}\, B \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right ) \left (105 \cos \left (d x +c \right )^{3}+70 \cos \left (d x +c \right )^{2}+56 \cos \left (d x +c \right )+48\right ) \sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{384 d \cos \left (d x +c \right )^{\frac {7}{2}} \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(464\)

Input:

int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2 
),x,method=_RETURNVERBOSE)
 

Output:

1/384/d*(144*A*cos(d*x+c)^4*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(cos( 
d*x+c)+1))^(1/2))+120*B*cos(d*x+c)^4*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/ 
(-1/(cos(d*x+c)+1))^(1/2))+105*C*cos(d*x+c)^4*arctan(1/2*(cot(d*x+c)-csc(d 
*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))+144*A*cos(d*x+c)^4*arctan(1/2/(-1/(cos 
(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1))+120*B*cos(d*x+c)^4*arctan(1/2 
/(-1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1))+105*C*cos(d*x+c)^4*a 
rctan(1/2/(-1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1))+sin(d*x+c)* 
cos(d*x+c)^2*(144*cos(d*x+c)+96)*2^(1/2)*A*(-2/(cos(d*x+c)+1))^(1/2)+sin(d 
*x+c)*cos(d*x+c)*(120*cos(d*x+c)^2+80*cos(d*x+c)+64)*2^(1/2)*B*(-2/(cos(d* 
x+c)+1))^(1/2)+sin(d*x+c)*(105*cos(d*x+c)^3+70*cos(d*x+c)^2+56*cos(d*x+c)+ 
48)*2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2))*(a*(1+sec(d*x+c)))^(1/2)/cos(d*x+ 
c)^(7/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 497, normalized size of antiderivative = 2.01 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\left [\frac {4 \, {\left (3 \, {\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 48 \, 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 ) + 3 \, {\left ({\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{4}\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 )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}}, \frac {2 \, {\left (3 \, {\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 48 \, 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 ) + 3 \, {\left ({\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{4}\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 )}{384 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}}\right ] \] Input:

integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c 
)^(5/2),x, algorithm="fricas")
 

Output:

[1/768*(4*(3*(48*A + 40*B + 35*C)*cos(d*x + c)^3 + 2*(48*A + 40*B + 35*C)* 
cos(d*x + c)^2 + 8*(8*B + 7*C)*cos(d*x + c) + 48*C)*sqrt((a*cos(d*x + c) + 
 a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((48*A + 40*B + 35*C 
)*cos(d*x + c)^5 + (48*A + 40*B + 35*C)*cos(d*x + c)^4)*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)^5 + d*cos(d*x + c)^4), 1/38 
4*(2*(3*(48*A + 40*B + 35*C)*cos(d*x + c)^3 + 2*(48*A + 40*B + 35*C)*cos(d 
*x + c)^2 + 8*(8*B + 7*C)*cos(d*x + c) + 48*C)*sqrt((a*cos(d*x + c) + a)/c 
os(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((48*A + 40*B + 35*C)*cos 
(d*x + c)^5 + (48*A + 40*B + 35*C)*cos(d*x + c)^4)*sqrt(-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)^5 + d*cos(d* 
x + c)^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x 
+c)**(5/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6492 vs. \(2 (211) = 422\).

Time = 0.65 (sec) , antiderivative size = 6492, normalized size of antiderivative = 26.28 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c 
)^(5/2),x, algorithm="maxima")
 

Output:

-1/768*(48*(12*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos 
(7/2*arctan2(sin(d*x + c), cos(d*x + c))) + 4*(sqrt(2)*sin(4*d*x + 4*c) + 
2*sqrt(2)*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(d*x + c), cos(d*x + c))) - 
 4*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(3/2*arctan2 
(sin(d*x + c), cos(d*x + c))) - 12*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*s 
in(2*d*x + 2*c))*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 3*(2*(2*co 
s(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 
2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin( 
2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c 
), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2 
*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2* 
arctan2(sin(d*x + c), cos(d*x + c))) + 2) + 3*(2*(2*cos(2*d*x + 2*c) + 1)* 
cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 
 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*c 
os(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 
+ 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arc 
tan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c) 
, cos(d*x + c))) + 2) - 3*(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + c 
os(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d* 
x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) +...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1317 vs. \(2 (211) = 422\).

Time = 0.74 (sec) , antiderivative size = 1317, normalized size of antiderivative = 5.33 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c 
)^(5/2),x, algorithm="giac")
 

Output:

1/384*(3*(48*A*sqrt(a)*sgn(cos(d*x + c)) + 40*B*sqrt(a)*sgn(cos(d*x + c)) 
+ 35*C*sqrt(a)*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*(48*A*sqrt 
(a)*sgn(cos(d*x + c)) + 40*B*sqrt(a)*sgn(cos(d*x + c)) + 35*C*sqrt(a)*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*sqrt(2)*(240*(sqrt(a)*tan(1/2 
*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*A*a^(3/2)*sgn(cos(d 
*x + c)) - 504*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c) 
^2 + a))^14*B*a^(3/2)*sgn(cos(d*x + c)) + 279*(sqrt(a)*tan(1/2*d*x + 1/2*c 
) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*C*a^(3/2)*sgn(cos(d*x + c)) - 1 
968*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12 
*A*a^(5/2)*sgn(cos(d*x + c)) + 5976*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a 
*tan(1/2*d*x + 1/2*c)^2 + a))^12*B*a^(5/2)*sgn(cos(d*x + c)) + 285*(sqrt(a 
)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12*C*a^(5/2)* 
sgn(cos(d*x + c)) - 2640*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d* 
x + 1/2*c)^2 + a))^10*A*a^(7/2)*sgn(cos(d*x + c)) - 31320*(sqrt(a)*tan(1/2 
*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*B*a^(7/2)*sgn(cos(d 
*x + c)) - 4605*(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^(7/2)*sgn(cos(d*x + c)) + 41616*(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^(9/2)*sgn(cos(d*x + c)...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(((a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/c 
os(c + d*x)^(5/2),x)
 

Output:

int(((a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/c 
os(c + d*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{3}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{3}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a \right ) \] Input:

int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2 
),x)
 

Output:

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