\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {11}{2}}(c+d x) \, dx\) [1329]

Optimal result

Integrand size = 45, antiderivative size = 234 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\frac {2 a^3 (584 A+690 B+903 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (8 A+10 B+11 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (64 A+90 B+63 C) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 a (5 A+9 B) (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d} \] Output:

2/315*a^3*(584*A+690*B+903*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c 
))^(1/2)+2/15*a^3*(8*A+10*B+11*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d 
*x+c))^(1/2)+2/315*a^2*(64*A+90*B+63*C)*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^ 
(5/2)*sin(d*x+c)/d+2/63*a*(5*A+9*B)*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(7/2 
)*sin(d*x+c)/d+2/9*A*(a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(9/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.68 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (2908 A+2790 B+2961 C+2 (1396 A+1215 B+882 C) \cos (c+d x)+4 (803 A+870 B+966 C) \cos (2 (c+d x))+584 A \cos (3 (c+d x))+690 B \cos (3 (c+d x))+588 C \cos (3 (c+d x))+584 A \cos (4 (c+d x))+690 B \cos (4 (c+d x))+903 C \cos (4 (c+d x))) \sec ^{\frac {9}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{1260 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 
2)*Sec[c + d*x]^(11/2),x]
 

Output:

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(2908*A + 2790*B + 2961*C + 2*(1396*A + 12 
15*B + 882*C)*Cos[c + d*x] + 4*(803*A + 870*B + 966*C)*Cos[2*(c + d*x)] + 
584*A*Cos[3*(c + d*x)] + 690*B*Cos[3*(c + d*x)] + 588*C*Cos[3*(c + d*x)] + 
 584*A*Cos[4*(c + d*x)] + 690*B*Cos[4*(c + d*x)] + 903*C*Cos[4*(c + d*x)]) 
*Sec[c + d*x]^(9/2)*Tan[(c + d*x)/2])/(1260*d)
 

Rubi [A] (verified)

Time = 1.74 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.16, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4709, 3042, 3522, 27, 3042, 3454, 27, 3042, 3454, 27, 3042, 3459, 3042, 3250}

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.

\(\displaystyle \int \sec ^{\frac {11}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (c+d x)^{11/2} (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4709

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^{5/2} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {11}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx\)

\(\Big \downarrow \) 3522

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^{5/2} (a (5 A+9 B)+a (2 A+9 C) \cos (c+d x))}{2 \cos ^{\frac {9}{2}}(c+d x)}dx}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^{5/2} (a (5 A+9 B)+a (2 A+9 C) \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)}dx}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+9 B)+a (2 A+9 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3454

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{7} \int \frac {(\cos (c+d x) a+a)^{3/2} \left ((64 A+90 B+63 C) a^2+3 (8 A+6 B+21 C) \cos (c+d x) a^2\right )}{2 \cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \int \frac {(\cos (c+d x) a+a)^{3/2} \left ((64 A+90 B+63 C) a^2+3 (8 A+6 B+21 C) \cos (c+d x) a^2\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((64 A+90 B+63 C) a^2+3 (8 A+6 B+21 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a^2 (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3454

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \left (\frac {2}{5} \int \frac {\sqrt {\cos (c+d x) a+a} \left (63 (8 A+10 B+11 C) a^3+(248 A+270 B+441 C) \cos (c+d x) a^3\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^3 (64 A+90 B+63 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \left (\frac {1}{5} \int \frac {\sqrt {\cos (c+d x) a+a} \left (63 (8 A+10 B+11 C) a^3+(248 A+270 B+441 C) \cos (c+d x) a^3\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^3 (64 A+90 B+63 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \left (\frac {1}{5} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (63 (8 A+10 B+11 C) a^3+(248 A+270 B+441 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^3 (64 A+90 B+63 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3459

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \left (\frac {1}{5} \left (a^3 (584 A+690 B+903 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {42 a^4 (8 A+10 B+11 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (64 A+90 B+63 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \left (\frac {1}{5} \left (a^3 (584 A+690 B+903 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {42 a^4 (8 A+10 B+11 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (64 A+90 B+63 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3250

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a^2 (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (\frac {1}{5} \left (\frac {42 a^4 (8 A+10 B+11 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 a^4 (584 A+690 B+903 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (64 A+90 B+63 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )}{9 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )\)

Input:

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec 
[c + d*x]^(11/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^(5/2)*Sin 
[c + d*x])/(9*d*Cos[c + d*x]^(9/2)) + ((2*a^2*(5*A + 9*B)*(a + a*Cos[c + d 
*x])^(3/2)*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((2*a^3*(64*A + 90*B + 
 63*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ( 
(42*a^4*(8*A + 10*B + 11*C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*Sqrt[a + a 
*Cos[c + d*x]]) + (2*a^4*(584*A + 690*B + 903*C)*Sin[c + d*x])/(d*Sqrt[Cos 
[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/5)/7)/(9*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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3250

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq 
rt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, 
 e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3454

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

rule 3459

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) 
*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* 
c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d))   Int[Sqrt[a + b*Sin[e + f*x] 
]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1]
 

rule 3522

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - 
d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^m* 
(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( 
a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* 
(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, 
 x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ 
[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
 

rule 4709

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.69

\[\frac {2 a^{2} \sin \left (d x +c \right ) \left (\left (584 \cos \left (d x +c \right )^{4}+292 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+130 \cos \left (d x +c \right )+35\right ) A +\cos \left (d x +c \right ) \left (690 \cos \left (d x +c \right )^{3}+345 \cos \left (d x +c \right )^{2}+180 \cos \left (d x +c \right )+45\right ) B +\cos \left (d x +c \right )^{2} \left (903 \cos \left (d x +c \right )^{2}+294 \cos \left (d x +c \right )+63\right ) C \right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{\frac {11}{2}}}{315 d \left (1+\cos \left (d x +c \right )\right )}\]

Input:

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(11/ 
2),x)
 

Output:

2/315/d*a^2*sin(d*x+c)*((584*cos(d*x+c)^4+292*cos(d*x+c)^3+219*cos(d*x+c)^ 
2+130*cos(d*x+c)+35)*A+cos(d*x+c)*(690*cos(d*x+c)^3+345*cos(d*x+c)^2+180*c 
os(d*x+c)+45)*B+cos(d*x+c)^2*(903*cos(d*x+c)^2+294*cos(d*x+c)+63)*C)*((1+c 
os(d*x+c))*a)^(1/2)*cos(d*x+c)*sec(d*x+c)^(11/2)/(1+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.61 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\frac {2 \, {\left ({\left (584 \, A + 690 \, B + 903 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (292 \, A + 345 \, B + 294 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (73 \, A + 60 \, B + 21 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (26 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right ) + 35 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c 
)^(11/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

2/315*((584*A + 690*B + 903*C)*a^2*cos(d*x + c)^4 + (292*A + 345*B + 294*C 
)*a^2*cos(d*x + c)^3 + 3*(73*A + 60*B + 21*C)*a^2*cos(d*x + c)^2 + 5*(26*A 
 + 9*B)*a^2*cos(d*x + c) + 35*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) 
/((d*cos(d*x + c)^5 + d*cos(d*x + c)^4)*sqrt(cos(d*x + c)))
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x 
+c)**(11/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 866 vs. \(2 (204) = 408\).

Time = 0.23 (sec) , antiderivative size = 866, normalized size of antiderivative = 3.70 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c 
)^(11/2),x, algorithm="maxima")
 

Output:

8/315*((315*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 945*sqrt(2)* 
a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 1449*sqrt(2)*a^(5/2)*sin(d*x 
 + c)^5/(cos(d*x + c) + 1)^5 - 1287*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d* 
x + c) + 1)^7 + 572*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 
104*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)*A*(sin(d*x + c) 
^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/ 
2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(3*sin(d*x + c)^2/(cos(d* 
x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(co 
s(d*x + c) + 1)^6 + 1)) + 15*(21*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c 
) + 1) - 77*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 119*sqrt 
(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 99*sqrt(2)*a^(5/2)*sin(d 
*x + c)^7/(cos(d*x + c) + 1)^7 + 44*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d* 
x + c) + 1)^9 - 8*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)*B 
*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) 
+ 1) + 1)^(11/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(3*sin(d*x 
+ c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin( 
d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)) + 21*(15*sqrt(2)*a^(5/2)*sin(d*x + c 
)/(cos(d*x + c) + 1) - 65*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1 
)^3 + 113*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 99*sqrt(2) 
*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 44*sqrt(2)*a^(5/2)*sin(d...
 

Giac [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c 
)^(11/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 3.87 (sec) , antiderivative size = 749, normalized size of antiderivative = 3.20 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\text {Too large to display} \] Input:

int((1/cos(c + d*x))^(11/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) 
 + C*cos(c + d*x)^2),x)
 

Output:

((1/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a^2*(a + a*(e 
xp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(584*A + 690*B + 903* 
C)*2i)/(315*d) - (C*a^2*exp(c*1i + d*x*1i)*(a + a*(exp(- c*1i - d*x*1i)/2 
+ exp(c*1i + d*x*1i)/2))^(1/2)*2i)/d + (C*a^2*exp(c*8i + d*x*8i)*(a + a*(e 
xp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*2i)/d - (a^2*exp(c*3i 
 + d*x*3i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*( 
2*A + 5*B + 10*C)*4i)/(3*d) + (a^2*exp(c*6i + d*x*6i)*(a + a*(exp(- c*1i - 
 d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(2*A + 5*B + 10*C)*4i)/(3*d) + ( 
a^2*exp(c*4i + d*x*4i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i) 
/2))^(1/2)*(24*A + 25*B + 33*C)*4i)/(5*d) - (a^2*exp(c*5i + d*x*5i)*(a + a 
*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(24*A + 25*B + 33* 
C)*4i)/(5*d) + (a^2*exp(c*2i + d*x*2i)*(a + a*(exp(- c*1i - d*x*1i)/2 + ex 
p(c*1i + d*x*1i)/2))^(1/2)*(146*A + 155*B + 182*C)*4i)/(35*d) - (a^2*exp(c 
*7i + d*x*7i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2 
)*(146*A + 155*B + 182*C)*4i)/(35*d) - (a^2*exp(c*9i + d*x*9i)*(a + a*(exp 
(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(584*A + 690*B + 903*C) 
*2i)/(315*d)))/(exp(c*1i + d*x*1i) + 4*exp(c*2i + d*x*2i) + 4*exp(c*3i + d 
*x*3i) + 6*exp(c*4i + d*x*4i) + 6*exp(c*5i + d*x*5i) + 4*exp(c*6i + d*x*6i 
) + 4*exp(c*7i + d*x*7i) + exp(c*8i + d*x*8i) + exp(c*9i + d*x*9i) + 1)
 

Reduce [F]

\[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x) \, dx=\sqrt {a}\, a^{2} \left (2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{5}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{5}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{5}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{5}d x \right ) b +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{5}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{5}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{5}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{5}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}d x \right ) a \right ) \] Input:

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(11/ 
2),x)
 

Output:

sqrt(a)*a**2*(2*int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x) 
*sec(c + d*x)**5,x)*a + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos( 
c + d*x)*sec(c + d*x)**5,x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 
 1)*cos(c + d*x)**4*sec(c + d*x)**5,x)*c + int(sqrt(sec(c + d*x))*sqrt(cos 
(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**5,x)*b + 2*int(sqrt(sec(c + d 
*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**5,x)*c + int(sqr 
t(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**5,x)* 
a + 2*int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c 
+ d*x)**5,x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x 
)**2*sec(c + d*x)**5,x)*c + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)* 
sec(c + d*x)**5,x)*a)