\(\int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) [581]

Optimal result

Integrand size = 23, antiderivative size = 194 \[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {2 b \left (9 a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a \left (5 a^2+21 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {32 a^2 b \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a \left (5 a^2+21 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (9 a^2+5 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

-2/5*b*(9*a^2+5*b^2)*(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))/d+2
/21*a*(5*a^2+21*b^2)*(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))/d+3
2/35*a^2*b*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/21*a*(5*a^2+21*b^2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/7*a^2*(a+b*cos(
d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/5*b*(9*a^2+5*b^2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2871, 3100, 2827, 2716, 2720, 2719} \[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a \left (5 a^2+21 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {2 b \left (9 a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a \left (5 a^2+21 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (9 a^2+5 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {32 a^2 b \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

Int[(a + b*Cos[c + d*x])^3/Cos[c + d*x]^(9/2),x]

[Out]

(-2*b*(9*a^2 + 5*b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*a*(5*a^2 + 21*b^2)*EllipticF[(c + d*x)/2, 2])/(21*
d) + (32*a^2*b*Sin[c + d*x])/(35*d*Cos[c + d*x]^(5/2)) + (2*a*(5*a^2 + 21*b^2)*Sin[c + d*x])/(21*d*Cos[c + d*x
]^(3/2)) + (2*b*(9*a^2 + 5*b^2)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*a^2*(a + b*Cos[c + d*x])*Sin[c + d
*x])/(7*d*Cos[c + d*x]^(7/2))

Rule 2716

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

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 2871

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/
(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e
+ f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 +
c*d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 +
d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3100

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {8 a^2 b+\frac {1}{2} a \left (5 a^2+21 b^2\right ) \cos (c+d x)+\frac {1}{2} b \left (3 a^2+7 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {32 a^2 b \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4}{35} \int \frac {\frac {5}{4} a \left (5 a^2+21 b^2\right )+\frac {7}{4} b \left (9 a^2+5 b^2\right ) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {32 a^2 b \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{5} \left (b \left (9 a^2+5 b^2\right )\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{7} \left (a \left (5 a^2+21 b^2\right )\right ) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {32 a^2 b \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a \left (5 a^2+21 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (9 a^2+5 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {1}{5} \left (b \left (9 a^2+5 b^2\right )\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (a \left (5 a^2+21 b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {2 b \left (9 a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a \left (5 a^2+21 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {32 a^2 b \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a \left (5 a^2+21 b^2\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (9 a^2+5 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {-42 b \left (9 a^2+5 b^2\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 a \left (5 a^2+21 b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+126 a^2 b \sin (c+d x)+378 a^2 b \cos ^2(c+d x) \sin (c+d x)+210 b^3 \cos ^2(c+d x) \sin (c+d x)+25 a^3 \sin (2 (c+d x))+105 a b^2 \sin (2 (c+d x))+30 a^3 \tan (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

Integrate[(a + b*Cos[c + d*x])^3/Cos[c + d*x]^(9/2),x]

[Out]

(-42*b*(9*a^2 + 5*b^2)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 10*a*(5*a^2 + 21*b^2)*Cos[c + d*x]^(5/2)
*EllipticF[(c + d*x)/2, 2] + 126*a^2*b*Sin[c + d*x] + 378*a^2*b*Cos[c + d*x]^2*Sin[c + d*x] + 210*b^3*Cos[c +
d*x]^2*Sin[c + d*x] + 25*a^3*Sin[2*(c + d*x)] + 105*a*b^2*Sin[2*(c + d*x)] + 30*a^3*Tan[c + d*x])/(105*d*Cos[c
+ d*x]^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(819\) vs. \(2(226)=452\).

Time = 14.69 (sec) , antiderivative size = 820, normalized size of antiderivative = 4.23

method	result	size
default	\(\text {Expression too large to display}\)	\(820\)
parts	\(\text {Expression too large to display}\)	\(1008\)

[In]

int((a+cos(d*x+c)*b)^3/cos(d*x+c)^(9/2),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a^3*(-1/56*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)^4-5/42*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+5/21*(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)))+
2*b^3/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)*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2)))+6/5*a^2*b/sin(1/2*d*x+1/2*c)^2/(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)*(24*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)*EllipticE(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*co
s(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)*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)+6*a*b^2*(-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))))/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 = 0.11 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (5 i \, a^{3} + 21 i \, a b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-5 i \, a^{3} - 21 i \, a b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (9 i \, a^{2} b + 5 i \, b^{3}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (-9 i \, a^{2} b - 5 i \, b^{3}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (63 \, a^{2} b \cos \left (d x + c\right ) + 21 \, {\left (9 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 15 \, a^{3} + 5 \, {\left (5 \, a^{3} + 21 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate((a+b*cos(d*x+c))^3/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

-1/105*(5*sqrt(2)*(5*I*a^3 + 21*I*a*b^2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c)) + 5*sqrt(2)*(-5*I*a^3 - 21*I*a*b^2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c
)) + 21*sqrt(2)*(9*I*a^2*b + 5*I*b^3)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x
+ c) + I*sin(d*x + c))) + 21*sqrt(2)*(-9*I*a^2*b - 5*I*b^3)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrass
PInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(63*a^2*b*cos(d*x + c) + 21*(9*a^2*b + 5*b^3)*cos(d*x + c)
^3 + 15*a^3 + 5*(5*a^3 + 21*a*b^2)*cos(d*x + c)^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))**3/cos(d*x+c)**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^3/cos(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^3/cos(d*x + c)^(9/2), x)

Giac [F]

\[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^3/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^3/cos(d*x + c)^(9/2), x)

Mupad [B] (verification not implemented)

Time = 16.46 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.76 \[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\frac {2\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7}+2\,b^3\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+\frac {6\,a^2\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5}+2\,a\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \]

[In]

int((a + b*cos(c + d*x))^3/cos(c + d*x)^(9/2),x)

[Out]

((2*a^3*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/7 + 2*b^3*cos(c + d*x)^3*sin(c + d*x)*hyper
geom([-1/4, 1/2], 3/4, cos(c + d*x)^2) + (6*a^2*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c
+ d*x)^2))/5 + 2*a*b^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos(c + d*
x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2))