Integrand size = 43, antiderivative size = 209 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac {2 \left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^4 d}+\frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^4 (a+b) d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d} \] Output:
2/5*(5*A*b^2-5*B*a*b+a^2*(3*A+5*C))*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/ a^3/d-2/3*(3*A*b^3-B*a^3-3*B*a*b^2+a^2*b*(A+3*C))*InverseJacobiAM(1/2*d*x+ 1/2*c,2^(1/2))/a^4/d+2*b^2*(A*b^2-a*(B*b-C*a))*EllipticPi(sin(1/2*d*x+1/2* c),2*a/(a+b),2^(1/2))/a^4/(a+b)/d-2/3*(A*b-B*a)*cos(d*x+c)^(1/2)*sin(d*x+c )/a^2/d+2/5*A*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d
Time = 2.52 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.30 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\frac {2 a^2 \left (5 A b^2-5 a b B+3 a^2 (3 A+5 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+2 a^2 (4 A b+5 a B) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )+4 a^2 \sqrt {\cos (c+d x)} (-5 A b+5 a B+3 a A \cos (c+d x)) \sin (c+d x)+\frac {6 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{b \sqrt {\sin ^2(c+d x)}}}{30 a^4 d} \] Input:
Integrate[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]
Output:
((2*a^2*(5*A*b^2 - 5*a*b*B + 3*a^2*(3*A + 5*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + 2*a^2*(4*A*b + 5*a*B)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)) + 4*a^2*Sqr t[Cos[c + d*x]]*(-5*A*b + 5*a*B + 3*a*A*Cos[c + d*x])*Sin[c + d*x] + (6*(5 *A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x] )/(b*Sqrt[Sin[c + d*x]^2]))/(30*a^4*d)
Time = 1.84 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.09, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4600, 3042, 3528, 27, 3042, 3528, 27, 3042, 3538, 27, 3042, 3119, 3481, 3042, 3120, 3284}
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.
| \(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^{5/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{a+b \sec (c+d x)}dx\) |
| \(\Big \downarrow \) 4600 |
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{a \cos (c+d x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{a \sin \left (c+d x+\frac {\pi }{2}\right )+b}dx\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {2 \int \frac {\sqrt {\cos (c+d x)} \left (-5 (A b-a B) \cos ^2(c+d x)+a (3 A+5 C) \cos (c+d x)+3 A b\right )}{2 (b+a \cos (c+d x))}dx}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {\cos (c+d x)} \left (-5 (A b-a B) \cos ^2(c+d x)+a (3 A+5 C) \cos (c+d x)+3 A b\right )}{b+a \cos (c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (-5 (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (3 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 A b\right )}{b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {\frac {2 \int -\frac {-3 \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right ) \cos ^2(c+d x)-a (4 A b+5 a B) \cos (c+d x)+5 b (A b-a B)}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {-3 \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right ) \cos ^2(c+d x)-a (4 A b+5 a B) \cos (c+d x)+5 b (A b-a B)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {-3 \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (4 A b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+5 b (A b-a B)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \frac {-\frac {-\frac {3 \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \sqrt {\cos (c+d x)}dx}{a}-\frac {\int -\frac {5 \left (a b (A b-a B)+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\frac {5 \int \frac {a b (A b-a B)+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {3 \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \sqrt {\cos (c+d x)}dx}{a}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {5 \int \frac {a b (A b-a B)+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {3 \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {-\frac {\frac {5 \int \frac {a b (A b-a B)+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {-\frac {\frac {5 \left (\frac {\left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}-\frac {3 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {5 \left (\frac {\left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {3 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {-\frac {\frac {5 \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {-\frac {\frac {5 \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {6 b^2 \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
Input:
Int[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Se c[c + d*x]),x]
Output:
(2*A*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*a*d) + (-1/3*((-6*(5*A*b^2 - 5*a* b*B + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(a*d) + (5*((2*(3*A*b^3 - a^3*B - 3*a*b^2*B + a^2*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(a*d) - (6*b^2*(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/ (a*(a + b)*d)))/a)/a - (10*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3 *a*d))/(5*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], 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[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x _)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) *(x_)]^2), x_Symbol] :> Simp[d^(m + 2) Int[(b + a*Cos[e + f*x])^m*(d*Cos[ e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr eeQ[{a, b, d, e, f, A, B, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(800\) vs. \(2(202)=404\).
Time = 27.37 (sec) , antiderivative size = 801, normalized size of antiderivative = 3.83
Input:
int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x,me thod=_RETURNVERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b^2*(A*b^2- B*a*b+C*a^2)/a^3/(a^2-a*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)*Ellip ticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-2*(A*a^3+A*a^2*b+A*a*b^2+A*b^3 -B*a^3-B*a^2*b-B*a*b^2+C*a^3+C*a^2*b)/a^4*(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))-4/5*A/a/(-2*sin(1/2*d*x+1/ 2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2 *c)-14*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2*cos( 1/2*d*x+1/2*c)-5*(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))+9*(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)^(1/2))+2/ a^3*(3*A*a^2+2*A*a*b+A*b^2-2*B*a^2-B*a*b+C*a^2)*(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))-EllipticE(cos(1/2*d *x+1/2*c),2^(1/2)))-4/3/a^2*(3*A*a+A*b-B*a)*(2*cos(1/2*d*x+1/2*c)*sin(1/2* d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+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)*EllipticF(cos(1/2*d*x+1/2*c),2^ (1/2))-3*(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)^(1/2))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*...
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \sec \left (d x + c\right ) + a} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) ),x, algorithm="fricas")
Output:
integral((C*cos(d*x + c)^2*sec(d*x + c)^2 + B*cos(d*x + c)^2*sec(d*x + c) + A*cos(d*x + c)^2)*sqrt(cos(d*x + c))/(b*sec(d*x + c) + a), x)
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+ c)),x)
Output:
Timed out
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \sec \left (d x + c\right ) + a} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) ),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(5/2)/(b*se c(d*x + c) + a), x)
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \sec \left (d x + c\right ) + a} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) ),x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(5/2)/(b*se c(d*x + c) + a), x)
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \] Input:
int((cos(c + d*x)^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/co s(c + d*x)),x)
Output:
int((cos(c + d*x)^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/co s(c + d*x)), x)
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right ) b +a}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )}{\sec \left (d x +c \right ) b +a}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\sec \left (d x +c \right ) b +a}d x \right ) a \] Input:
int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)
Output:
int((sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**2)/(sec(c + d*x)*b + a),x)*c + int((sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x))/(sec(c + d*x)*b + a),x)*b + int((sqrt(cos(c + d*x))*cos(c + d*x)**2)/(sec(c + d*x)* b + a),x)*a