\(\int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx\) [465]

Optimal result

Integrand size = 33, antiderivative size = 199 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {4 a^2 (4 A+5 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (A+2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^2 (4 A+5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d} \]

[Out]

2/15*a^2*(7*A+5*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*A*sec(d*x+c)^(3/2)*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d+4/5*
a^2*(4*A+5*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4/5*a^2*(4*A+5*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*
EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+4/3*a^2*(A+2*B)*(cos(1/2*d*x+1/2*c)^
2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3039, 4103, 4082, 3872, 3856, 2720, 3853, 2719} \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 a^2 (7 A+5 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {4 a^2 (4 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (A+2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {4 a^2 (4 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d} \]

[In]

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

[Out]

(-4*a^2*(4*A + 5*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(A + 2*B)*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (4*a^2*(4*A + 5*B)*Sqrt[Sec[c + d*x]]
*Sin[c + d*x])/(5*d) + (2*a^2*(7*A + 5*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*A*Sec[c + d*x]^(3/2)*(a
^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(5*d)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3039

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4082

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

Rule 4103

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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 (B+A \sec (c+d x)) \, dx \\ & = \frac {2 A \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x)) \left (\frac {1}{2} a (A+5 B)+\frac {1}{2} a (7 A+5 B) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {4}{15} \int \sqrt {\sec (c+d x)} \left (\frac {5}{2} a^2 (A+2 B)+\frac {3}{2} a^2 (4 A+5 B) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {1}{3} \left (2 a^2 (A+2 B)\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{5} \left (2 a^2 (4 A+5 B)\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {4 a^2 (4 A+5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}-\frac {1}{5} \left (2 a^2 (4 A+5 B)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (2 a^2 (A+2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {4 a^2 (A+2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^2 (4 A+5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}-\frac {1}{5} \left (2 a^2 (4 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {4 a^2 (4 A+5 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (A+2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^2 (4 A+5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.55 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.50 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {a^2 e^{-i c} \left (-1+e^{2 i c}\right ) (1+\cos (c+d x))^2 \csc (c) \left (10 A+5 B-18 A e^{i (c+d x)}-30 B e^{i (c+d x)}-54 A e^{3 i (c+d x)}-60 B e^{3 i (c+d x)}-10 A e^{4 i (c+d x)}-5 B e^{4 i (c+d x)}-24 A e^{5 i (c+d x)}-30 B e^{5 i (c+d x)}-10 i (A+2 B) \left (1+e^{2 i (c+d x)}\right )^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 (4 A+5 B) e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)}}{60 d \left (1+e^{2 i (c+d x)}\right )^2} \]

[In]

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

[Out]

(a^2*(-1 + E^((2*I)*c))*(1 + Cos[c + d*x])^2*Csc[c]*(10*A + 5*B - 18*A*E^(I*(c + d*x)) - 30*B*E^(I*(c + d*x))
- 54*A*E^((3*I)*(c + d*x)) - 60*B*E^((3*I)*(c + d*x)) - 10*A*E^((4*I)*(c + d*x)) - 5*B*E^((4*I)*(c + d*x)) - 2
4*A*E^((5*I)*(c + d*x)) - 30*B*E^((5*I)*(c + d*x)) - (10*I)*(A + 2*B)*(1 + E^((2*I)*(c + d*x)))^2*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(4*A + 5*B)*E^(I*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometri
c2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[(c + d*x)/2]^4*Sqrt[Sec[c + d*x]])/(60*d*E^(I*c)*(1 + E^((2*I)*
(c + d*x)))^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(713\) vs. \(2(227)=454\).

Time = 63.49 (sec) , antiderivative size = 714, normalized size of antiderivative = 3.59

method	result	size
default	\(\text {Expression too large to display}\)	\(714\)
parts	\(\text {Expression too large to display}\)	\(916\)

[In]

int((a+cos(d*x+c)*a)^2*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-8*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(1/4*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1
/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(
1/2))+1/20*A/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(2
4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(2*sin(1
/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)
^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(1/2*A+1/4*B)*(-1/6
*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2
)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+(1/4*A+1/2*B)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2
*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.20 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (4 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (4 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (6 \, {\left (4 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 3 \, A a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

-2/15*(5*I*sqrt(2)*(A + 2*B)*a^2*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*
I*sqrt(2)*(A + 2*B)*a^2*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)
*(4*A + 5*B)*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c
))) - 3*I*sqrt(2)*(4*A + 5*B)*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c
) - I*sin(d*x + c))) - (6*(4*A + 5*B)*a^2*cos(d*x + c)^2 + 5*(2*A + B)*a^2*cos(d*x + c) + 3*A*a^2)*sin(d*x + c
)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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