Integrand size = 45, antiderivative size = 163 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {2} (A-B+C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}-\frac {2 (A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}} \] Output:
2^(1/2)*(A-B+C)*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/(a+ a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(1/2)/d-2/3*(A-3* B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/3*A*sec(d*x+c)^( 3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 6.92 (sec) , antiderivative size = 661, normalized size of antiderivative = 4.06 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx =\text {Too large to display} \] Input:
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/Sqr t[a + a*Cos[c + d*x]],x]
Output:
(2*Cos[c/2 + (d*x)/2]*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2*S in[c/2 + (d*x)/2]^2]*((2*B*Sin[c/2 + (d*x)/2])/(3*(1 - 2*Sin[c/2 + (d*x)/2 ]^2)^(3/2)) - (4*C*Sin[c/2 + (d*x)/2]^3)/(3*(1 - 2*Sin[c/2 + (d*x)/2]^2)^( 3/2)) + (4*B*Sin[c/2 + (d*x)/2])/(3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (( A - B + C)*Csc[c/2 + (d*x)/2]^5*(-12*Cos[(c + d*x)/2]^4*HypergeometricPFQ[ {2, 2, 7/2}, {1, 9/2}, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2) )]*Sin[c/2 + (d*x)/2]^8 - 12*Hypergeometric2F1[2, 7/2, 9/2, -(Sin[c/2 + (d *x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8*(4 - 7*Sin[c/ 2 + (d*x)/2]^2 + 3*Sin[c/2 + (d*x)/2]^4) + 7*Sqrt[-(Sin[c/2 + (d*x)/2]^2/( 1 - 2*Sin[c/2 + (d*x)/2]^2))]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^3*(15 - 20*Sin[ c/2 + (d*x)/2]^2 + 8*Sin[c/2 + (d*x)/2]^4)*((3 - 7*Sin[c/2 + (d*x)/2]^2)*S qrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))] - 3*ArcTanh[Sqrt [-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*(1 - 2*Sin[c/2 + ( d*x)/2]^2))))/(63*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2))))/(d*Sqrt[a*(1 + Cos [c + d*x])])
Time = 0.96 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4709, 3042, 3522, 27, 3042, 3463, 27, 3042, 3261, 218}
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.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a \cos (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^{5/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )}{\sqrt {a \cos (c+d x)+a}}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+B \cos (c+d x)+A}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int -\frac {a (A-3 B)-a (2 A+3 C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a (A-3 B)-a (2 A+3 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a (A-3 B)-a (2 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {3 a^2 (A-B+C)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a (A-3 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-3 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a (A-B+C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{3 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-3 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a (A-B+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {6 a^2 (A-B+C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a (A-3 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-3 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {3 \sqrt {2} \sqrt {a} (A-B+C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{3 a}\right )\) |
Input:
Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/Sqrt[a + a*Cos[c + d*x]],x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sin[c + d*x])/(3*d*Cos[c + d*x ]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((-3*Sqrt[2]*Sqrt[a]*(A - B + C)*ArcTa n[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d* x]])])/d + (2*a*(A - 3*B)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*C os[c + d*x]]))/(3*a))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 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^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] + Simp[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])
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Time = 2.46 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.54
| method | result | size |
| default | \(-\frac {\cos \left (d x +c \right ) \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (A \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )\right )+B \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (-3 \cos \left (d x +c \right )^{2}-3 \cos \left (d x +c \right )\right )+C \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )\right )+\sin \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right ) \sqrt {2}\, A -\frac {3 \sqrt {2}\, \sin \left (2 d x +2 c \right ) B}{2}\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}{3 d a \left (1+\cos \left (d x +c \right )\right )}\) | \(251\) |
| parts | \(-\frac {A \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (\sin \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right ) \sqrt {2}\, \cos \left (d x +c \right )+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \left (3 \cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}\right )\right )}{3 d a \left (1+\cos \left (d x +c \right )\right )}+\frac {B \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right ) \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )+\sqrt {2}\, \sin \left (d x +c \right )\right )}{d a \left (1+\cos \left (d x +c \right )\right )}-\frac {C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right )^{2}}{d a}\) | \(305\) |
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2 ),x,method=_RETURNVERBOSE)
Output:
-1/3/d/a*cos(d*x+c)*2^(1/2)*((1+cos(d*x+c))*a)^(1/2)*(A*arcsin(-csc(d*x+c) +cot(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(3*cos(d*x+c)^2+3*cos(d*x+c ))+B*arcsin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-3* cos(d*x+c)^2-3*cos(d*x+c))+C*arcsin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x+c)/(1 +cos(d*x+c)))^(1/2)*(3*cos(d*x+c)^2+3*cos(d*x+c))+sin(d*x+c)*(-1+cos(d*x+c ))*2^(1/2)*A-3/2*2^(1/2)*sin(2*d*x+2*c)*B)*sec(d*x+c)^(5/2)/(1+cos(d*x+c))
Time = 0.11 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.89 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\frac {3 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{2} + {\left (A - B + C\right )} a \cos \left (d x + c\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}} + \frac {2 \, {\left ({\left (A - 3 \, B\right )} \cos \left (d x + c\right ) - A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c) )^(1/2),x, algorithm="fricas")
Output:
-1/3*(3*sqrt(2)*((A - B + C)*a*cos(d*x + c)^2 + (A - B + C)*a*cos(d*x + c) )*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin( d*x + c)))/sqrt(a) + 2*((A - 3*B)*cos(d*x + c) - A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c) )
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(5/2)/(a+a*cos(d*x+ c))**(1/2),x)
Output:
Timed out
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {a \cos \left (d x + c\right ) + a}} \,d x } \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c) )^(1/2),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(5/2)/sqrt( a*cos(d*x + c) + a), x)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {a \cos \left (d x + c\right ) + a}} \,d x } \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c) )^(1/2),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(5/2)/sqrt( a*cos(d*x + c) + a), x)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int(((1/cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^(1/2),x)
Output:
int(((1/cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^(1/2), x)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )+1}d x \right ) a \right )}{a} \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2 ),x)
Output:
(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec( c + d*x)**2)/(cos(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**2)/(cos(c + d*x) + 1),x)*c + int ((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*sec(c + d*x)**2)/(cos(c + d*x) + 1),x)*a))/a