Integrand size = 31, antiderivative size = 101 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2 a (3 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a (A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:
2/5*a*(3*A+5*B)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*a*(A+B)*Invers eJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/3*a*(A+B)*cos(d*x+c)^(1/2)*sin(d*x+c) /d+2/5*a*A*cos(d*x+c)^(3/2)*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.10 (sec) , antiderivative size = 363, normalized size of antiderivative = 3.59 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-6 (3 A+5 B) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sec (c) \sin (d x+\arctan (\tan (c)))+\left (9 (3 A+5 B) \cos (c-d x-\arctan (\tan (c))) \csc (c) \sec (c)+9 A \cos (c+d x+\arctan (\tan (c))) \csc (c) \sec (c)+15 B \cos (c+d x+\arctan (\tan (c))) \csc (c) \sec (c)-36 A \cos (c+d x) \cot (c) \sqrt {\sec ^2(c)}-60 B \cos (c+d x) \cot (c) \sqrt {\sec ^2(c)}-20 (A+B) \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sqrt {\sec ^2(c)} \sec (d x-\arctan (\cot (c))) \sin (c)+20 A \cos (c+d x) \sqrt {\sec ^2(c)} \sin (c+d x)+20 B \cos (c+d x) \sqrt {\sec ^2(c)} \sin (c+d x)+12 A \cos ^2(c+d x) \sqrt {\sec ^2(c)} \sin (c+d x)\right ) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )}{60 d \sqrt {\cos (c+d x)} \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}} \] Input:
Integrate[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(-6*(3*A + 5*B)*HypergeometricPFQ [{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sec[c]*Sin[d*x + ArcTan [Tan[c]]] + (9*(3*A + 5*B)*Cos[c - d*x - ArcTan[Tan[c]]]*Csc[c]*Sec[c] + 9 *A*Cos[c + d*x + ArcTan[Tan[c]]]*Csc[c]*Sec[c] + 15*B*Cos[c + d*x + ArcTan [Tan[c]]]*Csc[c]*Sec[c] - 36*A*Cos[c + d*x]*Cot[c]*Sqrt[Sec[c]^2] - 60*B*C os[c + d*x]*Cot[c]*Sqrt[Sec[c]^2] - 20*(A + B)*Cos[c + d*x]*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Si n[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Sec[c]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + 20*A*Cos[c + d*x]*Sqrt[Sec[c]^2]*Sin[c + d*x] + 20*B*Cos[c + d*x]*Sqrt[ Sec[c]^2]*Sin[c + d*x] + 12*A*Cos[c + d*x]^2*Sqrt[Sec[c]^2]*Sin[c + d*x])* Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(60*d*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c]^2 ]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])
Time = 0.70 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 3433, 3042, 3447, 3042, 3502, 27, 3042, 3227, 3042, 3115, 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.
| \(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a) (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3433 |
| \(\displaystyle \int \sqrt {\cos (c+d x)} (a \cos (c+d x)+a) (A \cos (c+d x)+B)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int \sqrt {\cos (c+d x)} \left ((a A+a B) \cos (c+d x)+a A \cos ^2(c+d x)+a B\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left ((a A+a B) \sin \left (c+d x+\frac {\pi }{2}\right )+a A \sin \left (c+d x+\frac {\pi }{2}\right )^2+a B\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {\cos (c+d x)} (a (3 A+5 B)+5 a (A+B) \cos (c+d x))dx+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\cos (c+d x)} (a (3 A+5 B)+5 a (A+B) \cos (c+d x))dx+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a (3 A+5 B)+5 a (A+B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (5 a (A+B) \int \cos ^{\frac {3}{2}}(c+d x)dx+a (3 A+5 B) \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (a (3 A+5 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 a (A+B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (a (3 A+5 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 a (A+B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (a (3 A+5 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 a (A+B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 a (3 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 a (3 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+5 a (A+B) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
Input:
Int[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(2*a*A*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + ((2*a*(3*A + 5*B)*Elliptic E[(c + d*x)/2, 2])/d + 5*a*(A + B)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/5
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[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* (d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
Leaf count of result is larger than twice the leaf count of optimal. \(354\) vs. \(2(92)=184\).
Time = 9.79 (sec) , antiderivative size = 355, normalized size of antiderivative = 3.51
| method | result | size |
| default | \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \left (-24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (44 A +20 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-16 A -10 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}-9 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}+5 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}-15 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{15 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(355\) |
Input:
int(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVER BOSE)
Output:
-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(-24*A*cos (1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(44*A+20*B)*sin(1/2*d*x+1/2*c)^4*cos( 1/2*d*x+1/2*c)+(-16*A-10*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*A*El lipticF(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)-9*A*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)+5*B*EllipticF(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)-15*B*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))/(-2*sin(1/2*d*x+1/2*c)^4+sin(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 complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.59 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {-5 i \, \sqrt {2} {\left (A + B\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (A + B\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (3 \, A + 5 \, B\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (3 \, A + 5 \, B\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, A a \cos \left (d x + c\right ) + 5 \, {\left (A + B\right )} a\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d} \] Input:
integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm= "fricas")
Output:
1/15*(-5*I*sqrt(2)*(A + B)*a*weierstrassPInverse(-4, 0, cos(d*x + c) + I*s in(d*x + c)) + 5*I*sqrt(2)*(A + B)*a*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)*(3*A + 5*B)*a*weierstrassZeta(-4, 0, we ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(3* A + 5*B)*a*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*A*a*cos(d*x + c) + 5*(A + B)*a)*sqrt(cos(d*x + c ))*sin(d*x + c))/d
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(5/2)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
Timed out
\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm= "maxima")
Output:
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)*cos(d*x + c)^(5/2), x)
\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm= "giac")
Output:
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)*cos(d*x + c)^(5/2), x)
Time = 0.35 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.27 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2\,A\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,A\,a\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int(cos(c + d*x)^(5/2)*(A + B/cos(c + d*x))*(a + a/cos(c + d*x)),x)
Output:
(2*A*a*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3 *d) + (2*B*a*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2 )))/(3*d) + (2*B*a*ellipticE(c/2 + (d*x)/2, 2))/d - (2*A*a*cos(c + d*x)^(7 /2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))
\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:
int(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
a*(int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**2,x)*b + int(sqrt( cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x),x)*a + int(sqrt(cos(c + d*x))*c os(c + d*x)**2*sec(c + d*x),x)*b + int(sqrt(cos(c + d*x))*cos(c + d*x)**2, x)*a)