\(\int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [1148]

Optimal result

Integrand size = 37, antiderivative size = 230 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {5 a^{5/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {a^3 (64 A+15 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (16 A-15 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 a A \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d} \] Output:

5*a^(5/2)*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+1/15*a^3*(64*A+15*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2 
)/(a+a*sec(d*x+c))^(1/2)-1/15*a^2*(16*A-15*C)*(a+a*sec(d*x+c))^(1/2)*sin(d 
*x+c)/d/cos(d*x+c)^(1/2)+2/3*a*A*cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(3/2)*s 
in(d*x+c)/d+2/5*A*cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d
 

Mathematica [A] (warning: unable to verify)

Time = 1.35 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.57 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \left (75 C \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \sqrt {\sec (c+d x)} \sin (c+d x)+\sqrt {1-\sec (c+d x)} \left (\left (86 A+30 C+6 A \cos ^2(c+d x)\right ) \sin (c+d x)+14 A \sin (2 (c+d x))+15 C \tan (c+d x)\right )\right )}{15 d \sqrt {-1+\cos (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^3*(75*C*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x] 
+ Sqrt[1 - Sec[c + d*x]]*((86*A + 30*C + 6*A*Cos[c + d*x]^2)*Sin[c + d*x] 
+ 14*A*Sin[2*(c + d*x)] + 15*C*Tan[c + d*x])))/(15*d*Sqrt[-1 + Cos[c + d*x 
]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.64 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.07, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.432, Rules used = {3042, 4753, 3042, 4575, 27, 3042, 4505, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4575

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{3} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 a^2 (8 A+5 C)-a^2 (16 A-15 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 a^2 (8 A+5 C)-a^2 (16 A-15 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}}dx+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a^2 (8 A+5 C)-a^2 (16 A-15 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4506

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\int \frac {\sqrt {\sec (c+d x) a+a} \left ((64 A+15 C) a^3+75 C \sec (c+d x) a^3\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((64 A+15 C) a^3+75 C \sec (c+d x) a^3\right )}{\sqrt {\sec (c+d x)}}dx-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((64 A+15 C) a^3+75 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4503

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {1}{2} \left (75 a^3 C \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^4 (64 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {1}{2} \left (75 a^3 C \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^4 (64 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {2 a^4 (64 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {150 a^3 C \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \left (\frac {1}{2} \left (\frac {150 a^{7/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^4 (64 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^3 (16 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}\right )}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Sec[c + d*x])^(5/2)*Sin 
[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ((10*a^2*A*(a + a*Sec[c + d*x])^(3/2 
)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (-((a^3*(16*A - 15*C)*Sqrt[Sec[ 
c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d) + ((150*a^(7/2)*C*ArcS 
inh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^4*(64*A + 1 
5*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/2)/3)/ 
(5*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 4505

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot 
[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim 
p[b/(a*d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim 
p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] 
/; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 
] && GtQ[m, 1/2] && LtQ[n, -1]
 

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])
 

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 [A] (warning: unable to verify)

Time = 1.83 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.02

\[\frac {\left (\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-75 \cos \left (d x +c \right )^{2}-75 \cos \left (d x +c \right )\right )+\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-75 \cos \left (d x +c \right )^{2}-75 \cos \left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (24 \cos \left (d x +c \right )^{2}+112 \cos \left (d x +c \right )+344\right ) A +\left (120 \cos \left (d x +c \right )+60\right ) \sin \left (d x +c \right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, a^{2}}{60 d \sqrt {\cos \left (d x +c \right )}\, \left (\cos \left (d x +c \right )+1\right )}\]

Input:

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

Output:

1/60/d*(2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(cot(d*x+c)-csc(d*x 
+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))*(-75*cos(d*x+c)^2-75*cos(d*x+c))+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))*(-75*cos(d*x+c)^2-75*cos(d*x+c))+sin(d*x+c)*cos(d*x+c)* 
(24*cos(d*x+c)^2+112*cos(d*x+c)+344)*A+(120*cos(d*x+c)+60)*sin(d*x+c)*C)*( 
a*(1+sec(d*x+c)))^(1/2)*a^2/cos(d*x+c)^(1/2)/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 453, normalized size of antiderivative = 1.97 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {4 \, {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 28 \, A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (43 \, A + 15 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 75 \, {\left (C a^{2} \cos \left (d x + c\right )^{2} + C a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{60 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {2 \, {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 28 \, A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (43 \, A + 15 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 75 \, {\left (C a^{2} \cos \left (d x + c\right )^{2} + C a^{2} \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{30 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \] Input:

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

Output:

[1/60*(4*(6*A*a^2*cos(d*x + c)^3 + 28*A*a^2*cos(d*x + c)^2 + 2*(43*A + 15* 
C)*a^2*cos(d*x + c) + 15*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sq 
rt(cos(d*x + c))*sin(d*x + c) + 75*(C*a^2*cos(d*x + c)^2 + C*a^2*cos(d*x + 
 c))*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/c 
os(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)^2 + 
d*cos(d*x + c)), 1/30*(2*(6*A*a^2*cos(d*x + c)^3 + 28*A*a^2*cos(d*x + c)^2 
 + 2*(43*A + 15*C)*a^2*cos(d*x + c) + 15*C*a^2)*sqrt((a*cos(d*x + c) + a)/ 
cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 75*(C*a^2*cos(d*x + c)^2 + 
 C*a^2*cos(d*x + c))*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)^2 + d*cos(d*x + c))]
 

Sympy [F(-1)]

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

integrate(cos(d*x+c)**(5/2)*(a+a*sec(d*x+c))**(5/2)*(A+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. 8175 vs. \(2 (196) = 392\).

Time = 0.48 (sec) , antiderivative size = 8175, normalized size of antiderivative = 35.54 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

1/1260*(42*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d* 
x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*A*sqrt(a) - 5*(1449*sqr 
t(2)*a^2*cos(5/2*d*x + 5/2*c)^3*sin(2*d*x + 2*c) - 1260*sqrt(2)*a^2*sin(1/ 
2*d*x + 1/2*c)^3 - 1449*(sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*sin(5 
/2*d*x + 5/2*c)^3 + 21*(25*sqrt(2)*a^2*cos(2*d*x + 2*c)^2*sin(3/2*d*x + 3/ 
2*c) + 25*sqrt(2)*a^2*sin(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) - 60*sqrt(2) 
*a^2*sin(1/2*d*x + 1/2*c) + 5*(5*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) - 12*sqr 
t(2)*a^2*sin(1/2*d*x + 1/2*c))*cos(2*d*x + 2*c) + (25*sqrt(2)*a^2*cos(3/2* 
d*x + 3/2*c) + 198*sqrt(2)*a^2*cos(1/2*d*x + 1/2*c))*sin(2*d*x + 2*c))*cos 
(5/2*d*x + 5/2*c)^2 - 21*(12*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) - 25*(sqrt(2 
)*a^2*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*a^2*sin(1/2*d*x + 1/2*c)^2)*sin(3/2 
*d*x + 3/2*c))*cos(2*d*x + 2*c)^2 + 21*(25*sqrt(2)*a^2*cos(2*d*x + 2*c)^2* 
sin(3/2*d*x + 3/2*c) + 25*sqrt(2)*a^2*sin(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2 
*c) + 69*sqrt(2)*a^2*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) - 198*sqrt(2)*a 
^2*sin(1/2*d*x + 1/2*c) + (25*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) - 198*sqrt( 
2)*a^2*sin(1/2*d*x + 1/2*c))*cos(2*d*x + 2*c) + 5*(5*sqrt(2)*a^2*cos(3/2*d 
*x + 3/2*c) + 12*sqrt(2)*a^2*cos(1/2*d*x + 1/2*c))*sin(2*d*x + 2*c))*sin(5 
/2*d*x + 5/2*c)^2 - 21*(12*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) - 25*(sqrt(2)* 
a^2*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*a^2*sin(1/2*d*x + 1/2*c)^2)*sin(3/2*d 
*x + 3/2*c))*sin(2*d*x + 2*c)^2 - 35*(sqrt(2)*a^2*cos(5/2*d*x + 5/2*c)^...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (196) = 392\).

Time = 0.72 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.86 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/30*(75*C*a^(5/2)*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)))*sgn(cos(d*x + c)) - 75*C*a^(5 
/2)*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)))*sgn(cos(d*x + c)) + 60*(3*sqrt(2)*(sqrt(a)*t 
an(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^(7/2)*sgn( 
cos(d*x + c)) - sqrt(2)*C*a^(9/2)*sgn(cos(d*x + c)))/((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) + 4*(60*sqrt(2) 
*A*a^5*sgn(cos(d*x + c)) + 15*sqrt(2)*C*a^5*sgn(cos(d*x + c)) + (80*sqrt(2 
)*A*a^5*sgn(cos(d*x + c)) + 30*sqrt(2)*C*a^5*sgn(cos(d*x + c)) + (32*sqrt( 
2)*A*a^5*sgn(cos(d*x + c)) + 15*sqrt(2)*C*a^5*sgn(cos(d*x + c)))*tan(1/2*d 
*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x 
 + 1/2*c)^2 + a)^(5/2))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)** 
2*sec(c + d*x)**4,x)*c + 2*int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*c 
os(c + d*x)**2*sec(c + d*x)**3,x)*c + int(sqrt(sec(c + d*x) + 1)*sqrt(cos( 
c + d*x))*cos(c + d*x)**2*sec(c + d*x)**2,x)*a + int(sqrt(sec(c + d*x) + 1 
)*sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**2,x)*c + 2*int(sqrt(sec 
(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x),x)*a + int( 
sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*a)