Integrand size = 35, antiderivative size = 223 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(11 A-7 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(19 A-15 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}} \] Output:
1/4*(11*A-7*B)*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)*2^(1/2)/a^(3/2)/d-1/ 6*(19*A-15*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/2*( A-B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/6*(7*A-3*B)*se c(d*x+c)^(3/2)*sin(d*x+c)/a/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 = 7.09 (sec) , antiderivative size = 981, normalized size of antiderivative = 4.40 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:
Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2))/(a + a*Cos[c + d*x])^( 3/2),x]
Output:
(2*Cos[c/2 + (d*x)/2]^3*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2 *Sin[c/2 + (d*x)/2]^2]*(-1/12*((A - B)*(1 - 2*Sin[c/2 + (d*x)/2]))/((1 + S in[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + ((A - B)*(1 + 2*S in[c/2 + (d*x)/2]))/(12*(1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2] ^2)^(3/2)) - ((A - B)*(5*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[ c/2 + (d*x)/2]^2]] + (1 + Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*Sq rt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/2 + ((A - B)*(5*ArcTan[(1 + 2*Sin[c/2 + (d*x)/2])/ Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (1 - Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 + ( d*x)/2]^2])/(1 + Sin[c/2 + (d*x)/2])))/2 + ((A + 3*B)*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*Hy pergeometric2F1[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)*Sqrt[-(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))))/(126*(1 - 2*Si...
Time = 1.26 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3440, 3042, 3457, 27, 3042, 3463, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(a \cos (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3440 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (7 A-3 B)-4 a (A-B) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (7 A-3 B)-4 a (A-B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (7 A-3 B)-4 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int -\frac {a^2 (19 A-15 B)-2 a^2 (7 A-3 B) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^2 (19 A-15 B)-2 a^2 (7 A-3 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^2 (19 A-15 B)-2 a^2 (7 A-3 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {3 a^3 (11 A-7 B)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^2 (19 A-15 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (19 A-15 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a^2 (11 A-7 B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (19 A-15 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a^2 (11 A-7 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {6 a^3 (11 A-7 B) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (19 A-15 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (19 A-15 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {3 \sqrt {2} a^{3/2} (11 A-7 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
Input:
Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2))/(a + a*Cos[c + d*x])^(3/2),x ]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A - B)*Sin[c + d*x])/(d*Cos[ c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((2*a*(7*A - 3*B)*Sin[c + d*x ])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((-3*Sqrt[2]*a^(3/2 )*(11*A - 7*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*S qrt[a + a*Cos[c + d*x]])])/d + (2*a^2*(19*A - 15*B)*Sin[c + d*x])/(d*Sqrt[ Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/(3*a))/(4*a^2))
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[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p Int[(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g , m, n, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && !(IntegerQ[m] && I ntegerQ[n])
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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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])
Time = 14.53 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.10
| method | result | size |
| default | \(-\frac {\sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (33 \cos \left (d x +c \right )^{4}+66 \cos \left (d x +c \right )^{3}+33 \cos \left (d x +c \right )^{2}\right )+B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-21 \cos \left (d x +c \right )^{4}-42 \cos \left (d x +c \right )^{3}-21 \cos \left (d x +c \right )^{2}\right )+\cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (19 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )-4\right ) \sqrt {2}\, A +\cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \left (-15 \cos \left (d x +c \right )-12\right ) \sqrt {2}\, B \right )}{12 d \left (\cos \left (d x +c \right )+1\right )^{2} a^{2}}\) | \(245\) |
| parts | \(-\frac {A \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (\cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (19 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )-4\right ) \sqrt {2}+\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (33 \cos \left (d x +c \right )^{4}+66 \cos \left (d x +c \right )^{3}+33 \cos \left (d x +c \right )^{2}\right )\right )}{12 d \left (\cos \left (d x +c \right )+1\right )^{2} a^{2}}+\frac {B \cos \left (d x +c \right )^{2} \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (\sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+4\right ) \sqrt {2}+\frac {7 \left (3+\cos \left (2 d x +2 c \right )+4 \cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{2}\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}{4 d \left (\cos \left (d x +c \right )+1\right )^{2} a^{2}}\) | \(275\) |
Input:
int((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x,method=_RET URNVERBOSE)
Output:
-1/12/d*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)*sec(d*x+c)^(5/2)/(cos(d*x+c)+1)^2 *(A*arcsin(cot(d*x+c)-csc(d*x+c))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(33*co s(d*x+c)^4+66*cos(d*x+c)^3+33*cos(d*x+c)^2)+B*arcsin(cot(d*x+c)-csc(d*x+c) )*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-21*cos(d*x+c)^4-42*cos(d*x+c)^3-21*c os(d*x+c)^2)+cos(d*x+c)*sin(d*x+c)*(19*cos(d*x+c)^2+12*cos(d*x+c)-4)*2^(1/ 2)*A+cos(d*x+c)^2*sin(d*x+c)*(-15*cos(d*x+c)-12)*2^(1/2)*B)/a^2
Time = 0.11 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {3 \, \sqrt {2} {\left ({\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (19 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (A - B\right )} \cos \left (d x + c\right ) - 4 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algo rithm="fricas")
Output:
-1/12*(3*sqrt(2)*((11*A - 7*B)*cos(d*x + c)^3 + 2*(11*A - 7*B)*cos(d*x + c )^2 + (11*A - 7*B)*cos(d*x + c))*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c ) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((19*A - 15*B)*cos(d *x + c)^2 + 12*(A - B)*cos(d*x + c) - 4*A)*sqrt(a*cos(d*x + c) + a)*sin(d* x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))
Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(3/2),x)
Output:
Timed out
\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algo rithm="maxima")
Output:
integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^(3/ 2), x)
\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algo rithm="giac")
Output:
integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^(3/ 2), x)
Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:
int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(5/2))/(a + a*cos(c + d*x))^(3/ 2),x)
Output:
int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(5/2))/(a + a*cos(c + d*x))^(3/ 2), x)
\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) a \right )}{a^{2}} \] Input:
int((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),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)**2 + 2*cos(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*sec(c + d*x)**2)/(cos(c + d*x)**2 + 2*cos( c + d*x) + 1),x)*a))/a**2