Integrand size = 23, antiderivative size = 149 \[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {6 a \left (a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \left (a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {8 a^2 b \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 a \left (a^2+5 b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a^2 (a+b \cos (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
-6/5*a*(a^2+5*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti cE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2*b*(a^2+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+8/5*a^2*b*sin (d*x+c)/d/cos(d*x+c)^(3/2)+2/5*a^2*(a+b*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c )^(5/2)+6/5*a*(a^2+5*b^2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
Time = 0.94 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {-6 a \left (a^2+5 b^2\right ) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 b \left (a^2+b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+10 a^2 b \sin (c+d x)+3 \left (a^3+5 a b^2\right ) \sin (2 (c+d x))+2 a^3 \tan (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)} \]
(-6*a*(a^2 + 5*b^2)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*b*(a ^2 + b^2)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 10*a^2*b*Sin[c + d*x] + 3*(a^3 + 5*a*b^2)*Sin[2*(c + d*x)] + 2*a^3*Tan[c + d*x])/(5*d*Cos[c + d*x]^(3/2))
Time = 0.78 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.95, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3271, 27, 3042, 3500, 27, 3042, 3227, 3042, 3116, 3042, 3119, 3120}
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.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3271 |
| \(\displaystyle \frac {2}{5} \int \frac {12 b a^2+3 \left (a^2+5 b^2\right ) \cos (c+d x) a+b \left (a^2+5 b^2\right ) \cos ^2(c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {12 b a^2+3 \left (a^2+5 b^2\right ) \cos (c+d x) a+b \left (a^2+5 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {12 b a^2+3 \left (a^2+5 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+b \left (a^2+5 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {3 \left (3 a \left (a^2+5 b^2\right )+5 b \left (a^2+b^2\right ) \cos (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {8 a^2 b \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {3 a \left (a^2+5 b^2\right )+5 b \left (a^2+b^2\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {8 a^2 b \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {3 a \left (a^2+5 b^2\right )+5 b \left (a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {8 a^2 b \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (3 a \left (a^2+5 b^2\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+5 b \left (a^2+b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {8 a^2 b \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (3 a \left (a^2+5 b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+5 b \left (a^2+b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {8 a^2 b \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \frac {1}{5} \left (5 b \left (a^2+b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )+\frac {8 a^2 b \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (5 b \left (a^2+b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {8 a^2 b \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (5 b \left (a^2+b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a^2 b \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {10 b \left (a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+3 a \left (a^2+5 b^2\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a^2 b \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 \sin (c+d x) (a+b \cos (c+d x))}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
(2*a^2*(a + b*Cos[c + d*x])*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((10* b*(a^2 + b^2)*EllipticF[(c + d*x)/2, 2])/d + (8*a^2*b*Sin[c + d*x])/(d*Cos [c + d*x]^(3/2)) + 3*a*(a^2 + 5*b^2)*((-2*EllipticE[(c + d*x)/2, 2])/d + ( 2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/5
3.6.80.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
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]
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]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co s[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] + Simp[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])
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] + Simp[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]
Leaf count of result is larger than twice the leaf count of optimal. \(710\) vs. \(2(187)=374\).
Time = 12.05 (sec) , antiderivative size = 711, normalized size of antiderivative = 4.77
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(711\) |
| parts | \(\text {Expression too large to display}\) | \(783\) |
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b^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))+2/ 5*a^3/(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*(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(co s(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)*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^2*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)))+6*a*b^2/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
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.64 \[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, a^{2} b + i \, b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, a^{2} b - i \, b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (i \, a^{3} + 5 i \, a b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-i \, a^{3} - 5 i \, a b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (5 \, a^{2} b \cos \left (d x + c\right ) + a^{3} + 3 \, {\left (a^{3} + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{5 \, d \cos \left (d x + c\right )^{3}} \]
-1/5*(5*sqrt(2)*(I*a^2*b + I*b^3)*cos(d*x + c)^3*weierstrassPInverse(-4, 0 , cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*a^2*b - I*b^3)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2) *(I*a^3 + 5*I*a*b^2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-I*a^3 - 5*I*a*b^ 2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* x + c) - I*sin(d*x + c))) - 2*(5*a^2*b*cos(d*x + c) + a^3 + 3*(a^3 + 5*a*b ^2)*cos(d*x + c)^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
Time = 15.55 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2\,b^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,a\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(2*b^3*ellipticF(c/2 + (d*x)/2, 2))/d + (2*a^3*sin(c + d*x)*hypergeom([-5/ 4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^( 1/2)) + (6*a*b^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2)) /(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*a^2*b*sin(c + d*x)*hyp ergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(3/2)*(sin(c + d *x)^2)^(1/2))