\(\int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1268]

Optimal result

Integrand size = 45, antiderivative size = 253 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (40 A+38 B+25 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{8 d}+\frac {a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (24 A+42 B+31 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {a (6 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d \sqrt {\cos (c+d x)}}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}} \] Output:

1/8*a^(5/2)*(40*A+38*B+25*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^3*(24*A-54*B-49*C)*sin(d* 
x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+1/24*a^2*(24*A+42*B+31*C)*( 
a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+1/12*a*(6*B+5*C)*(a+a* 
sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+1/3*C*(a+a*sec(d*x+c))^(5/ 
2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 7.28 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.70 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \left (3 (40 A+18 B+25 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \sqrt {\sec (c+d x)} \sin (c+d x)-60 B \arcsin \left (\sqrt {\sec (c+d x)}\right ) \sqrt {\sec (c+d x)} \sin (c+d x)+\sqrt {1-\sec (c+d x)} \left (48 A \sin (c+d x)+\left (24 A+66 B+75 C+2 (6 B+17 C) \sec (c+d x)+8 C \sec ^2(c+d x)\right ) \tan (c+d x)\right )\right )}{24 d \sqrt {-1+\cos (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^3*(3*(40*A + 18*B + 25*C)*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Sqrt[Sec[c + d 
*x]]*Sin[c + d*x] - 60*B*ArcSin[Sqrt[Sec[c + d*x]]]*Sqrt[Sec[c + d*x]]*Sin 
[c + d*x] + Sqrt[1 - Sec[c + d*x]]*(48*A*Sin[c + d*x] + (24*A + 66*B + 75* 
C + 2*(6*B + 17*C)*Sec[c + d*x] + 8*C*Sec[c + d*x]^2)*Tan[c + d*x])))/(24* 
d*Sqrt[-1 + Cos[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.74 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.06, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.356, Rules used = {3042, 4753, 3042, 4576, 27, 3042, 4506, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4576

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\sec (c+d x) a+a)^{5/2} (a (6 A-C)+a (6 B+5 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx}{3 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\sec (c+d x) a+a)^{5/2} (a (6 A-C)+a (6 B+5 C) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (6 A-C)+a (6 B+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 4506

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 (8 A-2 B-3 C) a^2+(24 A+42 B+31 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 (8 A-2 B-3 C) a^2+(24 A+42 B+31 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 (8 A-2 B-3 C) a^2+(24 A+42 B+31 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 4506

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\int \frac {\sqrt {\sec (c+d x) a+a} \left ((24 A-54 B-49 C) a^3+3 (40 A+38 B+25 C) \sec (c+d x) a^3\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((24 A-54 B-49 C) a^3+3 (40 A+38 B+25 C) \sec (c+d x) a^3\right )}{\sqrt {\sec (c+d x)}}dx+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((24 A-54 B-49 C) a^3+3 (40 A+38 B+25 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 4503

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {1}{2} \left (3 a^3 (40 A+38 B+25 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {1}{2} \left (3 a^3 (40 A+38 B+25 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 4288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {1}{2} \left (\frac {2 a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^3 (40 A+38 B+25 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (6 B+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}+\frac {1}{4} \left (\frac {a^3 (24 A+42 B+31 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}+\frac {1}{2} \left (\frac {6 a^{7/2} (40 A+38 B+25 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^4 (24 A-54 B-49 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )\right )}{6 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c 
+ d*x])^(5/2)*Sin[c + d*x])/(3*d) + ((a^2*(6*B + 5*C)*Sqrt[Sec[c + d*x]]*( 
a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d) + ((a^3*(24*A + 42*B + 31*C) 
*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d + ((6*a^(7/2) 
*(40*A + 38*B + 25*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d* 
x]]])/d + (2*a^4*(24*A - 54*B - 49*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d* 
Sqrt[a + a*Sec[c + d*x]]))/2)/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 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 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. \(796\) vs. \(2(217)=434\).

Time = 4.98 (sec) , antiderivative size = 797, normalized size of antiderivative = 3.15

method	result	size
default	\(\text {Expression too large to display}\)	\(797\)

Input:

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

Output:

4*2^(1/2)*(-1/8*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/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)*((-16*cos(1/2*d*x+1/2*c)^4+12*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))*(20*cos(1/2*d*x+1/2*c)^3-20*cos(1/2*d 
*x+1/2*c)+5*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))*(20*cos(1/2*d*x+1/2*c)^3-20*cos(1/2*d*x+1/2*c 
)+5*sec(1/2*d*x+1/2*c)))-1/192*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/(2*cos(1/2*d*x+1/2*c)^2-1)^(7/2)*((-1200*cos(1/2*d* 
x+1/2*c)^6+1528*cos(1/2*d*x+1/2*c)^4-660*cos(1/2*d*x+1/2*c)^2+98)*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))*(1200*cos(1/2*d*x+1/2*c)^7-2400*cos(1/2*d*x+1/2*c)^5+1800*cos(1/2* 
d*x+1/2*c)^3-600*cos(1/2*d*x+1/2*c)+75*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))*(1200*cos(1/2*d*x+ 
1/2*c)^7-2400*cos(1/2*d*x+1/2*c)^5+1800*cos(1/2*d*x+1/2*c)^3-600*cos(1/2*d 
*x+1/2*c)+75*sec(1/2*d*x+1/2*c)))-1/32*B/d*(a/(2*cos(1/2*d*x+1/2*c)^2-1)*c 
os(1/2*d*x+1/2*c)^2)^(1/2)*a^2/(2*cos(1/2*d*x+1/2*c)^2-1)^(5/2)*((-88*cos( 
1/2*d*x+1/2*c)^4+80*cos(1/2*d*x+1/2*c)^2-18)*tan(1/2*d*x+1/2*c)+2^(1/2)*ar 
ctanh(1/2*2^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c)+1))*(152*cos(1/2* 
d*x+1/2*c)^5-228*cos(1/2*d*x+1/2*c)^3+114*cos(1/2*d*x+1/2*c)-19*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...
 

Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 515, normalized size of antiderivative = 2.04 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {4 \, {\left (48 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, A + 22 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, B + 17 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\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 (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \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 (48 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, A + 22 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, B + 17 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\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 (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \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(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d* 
x+c)^2),x, algorithm="fricas")
 

Output:

[1/96*(4*(48*A*a^2*cos(d*x + c)^3 + 3*(8*A + 22*B + 25*C)*a^2*cos(d*x + c) 
^2 + 2*(6*B + 17*C)*a^2*cos(d*x + c) + 8*C*a^2)*sqrt((a*cos(d*x + c) + a)/ 
cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((40*A + 38*B + 25*C)*a^ 
2*cos(d*x + c)^4 + (40*A + 38*B + 25*C)*a^2*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*(48*A*a^2*cos(d*x + c)^3 + 3*(8*A + 22*B + 25*C)*a^2*cos(d*x + c)^ 
2 + 2*(6*B + 17*C)*a^2*cos(d*x + c) + 8*C*a^2)*sqrt((a*cos(d*x + c) + a)/c 
os(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((40*A + 38*B + 25*C)*a^2 
*cos(d*x + c)^4 + (40*A + 38*B + 25*C)*a^2*cos(d*x + c)^3)*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)^4 + 
d*cos(d*x + c)^3)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17788 vs. \(2 (217) = 434\).

Time = 3.22 (sec) , antiderivative size = 17788, normalized size of antiderivative = 70.31 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

1/96*(24*(8*a^2*cos(1/2*d*x + 1/2*c)^4*sin(1/2*d*x + 1/2*c) + 16*a^2*cos(1 
/2*d*x + 1/2*c)^2*sin(1/2*d*x + 1/2*c)^3 + 8*a^2*sin(1/2*d*x + 1/2*c)^5 + 
5*(sqrt(2)*a^2*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) - sqrt 
(2)*a^2*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) + sqrt(2)*a^2 
*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) - sqrt(2)*a^2*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(1/2*d*x + 1/2*c)^4 + 1 
0*(sqrt(2)*a^2*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) - sqrt 
(2)*a^2*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) + sqrt(2)*a^2 
*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) - sqrt(2)*a^2*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(1/2*d*x + 1/2*c)^2*sin 
(1/2*d*x + 1/2*c)^2 + 5*(sqrt(2)*a^2*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...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1122 vs. \(2 (217) = 434\).

Time = 0.83 (sec) , antiderivative size = 1122, normalized size of antiderivative = 4.43 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

1/48*(96*sqrt(2)*A*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) + 3*(40*A*a^(5/2)*sgn(cos(d*x + c)) + 38*B*a^(5/2)* 
sgn(cos(d*x + c)) + 25*C*a^(5/2)*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*(40*A*a^(5/2)*sgn(cos(d*x + c)) + 38*B*a^(5/2)*sgn(cos(d*x + c)) + 
 25*C*a^(5/2)*sgn(cos(d*x + c)))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - s 
qrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*(72*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*A* 
a^(7/2)*sgn(cos(d*x + c)) + 114*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))^10*B*a^(7/2)*sgn(cos(d*x + c)) + 75*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^(7/2)*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^(9/2)*sgn(cos(d*x + c)) - 1710* 
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*B*a^(9/2)*sgn(cos(d*x + c)) - 1125*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^(9/2)*sgn(cos(d*x + c)) + 3 
024*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*A*a^(11/2)*sgn(cos(d*x + c)) + 6804*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*B*a^(11/2)*sgn(cos(d*x + c 
)) + 6174*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + ...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) b +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}d x \right ) a \right ) \] Input:

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

Output:

sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)** 
4,x)*c + int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**3,x)* 
b + 2*int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**3,x)*c + 
 int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**2,x)*a + 2*in 
t(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**2,x)*b + int(sqr 
t(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**2,x)*c + 2*int(sqrt(s 
ec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x),x)*a + int(sqrt(sec(c + d 
*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x),x)*b + int(sqrt(sec(c + d*x) + 1) 
*sqrt(cos(c + d*x)),x)*a)