\(\int (a+a \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx\) [1225]

Optimal result

Integrand size = 37, antiderivative size = 230 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {5 a^{5/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}-\frac {a^3 (64 A+15 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 a^2 (8 A+5 C) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \]

[Out]

2/3*a*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*A*(a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(5/2)*sin
(d*x+c)/d-1/15*a^3*(64*A+15*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+5*a^(5/2)*C*arcsin(sin(d*x
+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/5*a^2*(8*A+5*C)*sin(d*x+c)*(a+a*cos(
d*x+c))^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4306, 3123, 3054, 3060, 2853, 222} \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {5 a^{5/2} C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {a^3 (64 A+15 C) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (8 A+5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}{5 d}+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d} \]

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]

[Out]

(5*a^(5/2)*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d
- (a^3*(64*A + 15*C)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*a^2*(8*A + 5*C)*Sqr
t[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^
(3/2)*Sin[c + d*x])/(3*d) + (2*A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d)

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 2853

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 3054

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((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]*(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] - Dist[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 3060

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

Rule 3123

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Si
n[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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 4306

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

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{5/2} \left (\frac {5 a A}{2}-\frac {1}{2} a (2 A-5 C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{5 a} \\ & = \frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{3/2} \left (\frac {3}{4} a^2 (8 A+5 C)-\frac {1}{4} a^2 (16 A-15 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a} \\ & = \frac {2 a^2 (8 A+5 C) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{8} a^3 (32 A+45 C)-\frac {1}{8} a^3 (64 A+15 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{15 a} \\ & = -\frac {a^3 (64 A+15 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 a^2 (8 A+5 C) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{2} \left (5 a^2 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {a^3 (64 A+15 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 a^2 (8 A+5 C) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac {\left (5 a^2 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d} \\ & = \frac {5 a^{5/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (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 {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 a^2 (8 A+5 C) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.61 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (300 \sqrt {2} C \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {5}{2}}(c+d x)+2 (196 A+60 C+(112 A+45 C) \cos (c+d x)+4 (43 A+15 C) \cos (2 (c+d x))+15 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{120 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^(5/2)*(300*Sqrt[2]*C*ArcSin[Sqrt[2]*Sin[(c + d*x
)/2]]*Cos[c + d*x]^(5/2) + 2*(196*A + 60*C + (112*A + 45*C)*Cos[c + d*x] + 4*(43*A + 15*C)*Cos[2*(c + d*x)] +
15*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(120*d)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.03

\[\frac {a^{2} \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (75 C \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+75 C \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+15 C \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+86 A \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+30 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+28 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 A \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{15 d \left (1+\cos \left (d x +c \right )\right )}\]

[In]

int((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x)

[Out]

1/15*a^2/d*sec(d*x+c)^(7/2)*((1+cos(d*x+c))*a)^(1/2)/(1+cos(d*x+c))*(75*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))+75*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))+15*C*cos(d*x+c)^4*sin(d*x+c)+86*A*cos(d*x+c)^3*sin(
d*x+c)+30*C*cos(d*x+c)^3*sin(d*x+c)+28*A*cos(d*x+c)^2*sin(d*x+c)+6*A*cos(d*x+c)*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.75 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {75 \, {\left (C a^{2} \cos \left (d x + c\right )^{3} + C a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (15 \, C a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (43 \, A + 15 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 28 \, A a^{2} \cos \left (d x + c\right ) + 6 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/15*(75*(C*a^2*cos(d*x + c)^3 + C*a^2*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x +
c))/(sqrt(a)*sin(d*x + c))) - (15*C*a^2*cos(d*x + c)^3 + 2*(43*A + 15*C)*a^2*cos(d*x + c)^2 + 28*A*a^2*cos(d*
x + c) + 6*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)
^2)

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1673 vs. \(2 (196) = 392\).

Time = 0.55 (sec) , antiderivative size = 1673, normalized size of antiderivative = 7.27 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Too large to display} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/60*(5*(2*(5*a^2*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 3*(a^2*cos(2*d*x + 2*c)^2*sin(d*x
+ c) + a^2*sin(2*d*x + 2*c)^2*sin(d*x + c) + 2*a^2*cos(2*d*x + 2*c)*sin(d*x + c) + a^2*sin(d*x + c))*cos(1/2*
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 3*((a^2*cos(d*x + c) - a^2)*cos(2*d*x + 2*c)^2 + a^2*cos(d*
x + c) + (a^2*cos(d*x + c) - a^2)*sin(2*d*x + 2*c)^2 - a^2 + 2*(a^2*cos(d*x + c) - a^2)*cos(2*d*x + 2*c))*sin(
1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d
*x + 2*c) + 1)*sqrt(a) + 15*((a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*
arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(
sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1
))) + 1) - (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2(-(cos(2*d*
x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x
+ 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) - (a^2*co
s(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*
d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*
d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c) + 1)) + 1) + (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2((
cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(s
in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) +
1)^(1/4)*sqrt(a) + 2*((12*a^2*sin(5*d*x + 5*c) + 15*a^2*sin(4*d*x + 4*c) + 24*a^2*sin(3*d*x + 3*c) + 35*a^2*si
n(2*d*x + 2*c) + 12*a^2*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (12*a^2*cos(5
*d*x + 5*c) + 15*a^2*cos(4*d*x + 4*c) + 24*a^2*cos(3*d*x + 3*c) + 35*a^2*cos(2*d*x + 2*c) + 12*a^2*cos(d*x + c
) + 20*a^2)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 27*(a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*
d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a)
)*C/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(5/4) + 32*(15*sqrt(2)*a^(5/2)*sin(d*x
+ c)/(cos(d*x + c) + 1) - 35*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 28*sqrt(2)*a^(5/2)*sin(d*x
+ c)^5/(cos(d*x + c) + 1)^5 - 8*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)*A/((sin(d*x + c)/(cos(d*x
+ c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)))/d

Giac [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]

[In]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^(5/2),x)

[Out]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^(5/2), x)