\(\int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\) [1276]

Optimal result

Integrand size = 43, antiderivative size = 212 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {4 a^2 (A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {4 a^2 (2 A+3 B+2 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 a^2 (5 A+3 B-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 (4 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \]

[Out]

2/3*A*(a+a*cos(d*x+c))^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d-2/3*a^2*(5*A+3*B-C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/3*(
4*A+3*B)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4*a^2*(A-C)*(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*(2*A+3*B+2*C)*(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 = 0.73 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4306, 3122, 3054, 3047, 3102, 2827, 2720, 2719} \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {2 a^2 (5 A+3 B-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {4 a^2 (2 A+3 B+2 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 (4 A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{3 d}-\frac {4 a^2 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[In]

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

[Out]

(-4*a^2*(A - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (4*a^2*(2*A + 3*B + 2*C)*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (2*a^2*(5*A + 3*B - C)*Sin[c + d*x])/
(3*d*Sqrt[Sec[c + d*x]]) + (2*(4*A + 3*B)*(a^2 + a^2*Cos[c + d*x])*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d) + (2
*A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3047

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

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 3102

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

Rule 3122

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(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] + 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
- 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 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))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^2 \left (\frac {1}{2} a (4 A+3 B)-\frac {3}{2} a (A-C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{3 a} \\ & = \frac {2 (4 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x)) \left (\frac {3}{4} a^2 (3 A+3 B+C)-\frac {3}{4} a^2 (5 A+3 B-C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{3 a} \\ & = \frac {2 (4 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{4} a^3 (3 A+3 B+C)+\left (-\frac {3}{4} a^3 (5 A+3 B-C)+\frac {3}{4} a^3 (3 A+3 B+C)\right ) \cos (c+d x)-\frac {3}{4} a^3 (5 A+3 B-C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{3 a} \\ & = -\frac {2 a^2 (5 A+3 B-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 (4 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{4} a^3 (2 A+3 B+2 C)-\frac {9}{4} a^3 (A-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{9 a} \\ & = -\frac {2 a^2 (5 A+3 B-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 (4 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}-\left (2 a^2 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (2 a^2 (2 A+3 B+2 C) \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-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {4 a^2 (2 A+3 B+2 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 a^2 (5 A+3 B-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 (4 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.56 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {a^2 \sec ^{\frac {3}{2}}(c+d x) \left (-24 (A-C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+8 (2 A+3 B+2 C) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 (2 A+C+6 (2 A+B) \cos (c+d x)+C \cos (2 (c+d x))) \sin (c+d x)\right )}{6 d} \]

[In]

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

[Out]

(a^2*Sec[c + d*x]^(3/2)*(-24*(A - C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 8*(2*A + 3*B + 2*C)*Cos[c
+ d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 2*(2*A + C + 6*(2*A + B)*Cos[c + d*x] + C*Cos[2*(c + d*x)])*Sin[c + d
*x]))/(6*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(800\) vs. \(2(242)=484\).

Time = 158.22 (sec) , antiderivative size = 801, normalized size of antiderivative = 3.78

method	result	size
default	\(\text {Expression too large to display}\)	\(801\)
parts	\(\text {Expression too large to display}\)	\(886\)

[In]

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

[Out]

-4/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c
)^2+1)/sin(1/2*d*x+1/2*c)^3*(-4*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+12*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+
1/2*c)^4-4*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))*sin(1/2*d*x+1/2*c)^2-6*A*(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+6*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-6*B*(2*sin(1/2*d*x+1/2*c)^2
-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+4*C*cos(1/2*
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-4*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)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+6*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-7*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A+2*A*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*A*(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))-3*B*cos(1/2*d*x+
1/2*c)*sin(1/2*d*x+1/2*c)^2+3*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(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)*C+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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*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)))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+
2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.08 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {2 \, {\left (i \, \sqrt {2} {\left (2 \, A + 3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} {\left (2 \, A + 3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (C a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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