Integrand size = 35, antiderivative size = 270 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(15 A-11 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}+\frac {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-5 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}} \]
-1/2*(A-B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-1/30*(39*A -35*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+1/10*(9*A-5* B)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/4*(15*A-11*B)* arctan(1/2*sin(d*x+c)*a^(1/2)*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^(3/2)/d*2^(1/2)+1/30*(147*A-95*B) *sin(d*x+c)*sec(d*x+c)^(1/2)/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 = 7.23 (sec) , antiderivative size = 2166, normalized size of antiderivative = 8.02 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Result too large to show} \]
(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]*(-1/20*((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)^(5/2)) + ((A - B)*(1 + 2*S in[c/2 + (d*x)/2]))/(20*(1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2] ^2)^(5/2)) - ((A - B)*(-105*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*S in[c/2 + (d*x)/2]^2]] + (4 + 3*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2 ])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) - (19 + 29*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) - (67*Sqrt[1 - 2*S in[c/2 + (d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/30 + ((A - B)*(-105*ArcTa n[(1 + 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (4 - 3*Si n[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^( 3/2)) - (19 - 29*Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*Sqrt[1 - 2* Sin[c/2 + (d*x)/2]^2]) - (67*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 + Sin[c/ 2 + (d*x)/2])))/30 + ((-A - 3*B)*Csc[c/2 + (d*x)/2]^7*(4725*Sin[c/2 + (d*x )/2]^2 - 48825*Sin[c/2 + (d*x)/2]^4 + 210105*Sin[c/2 + (d*x)/2]^6 - 486630 *Sin[c/2 + (d*x)/2]^8 + 655812*Sin[c/2 + (d*x)/2]^10 - 710*Hypergeometric2 F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c /2 + (d*x)/2]^10 - 40*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 9/2}, {1, 1, 11/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^10 - 518760*Sin[c/2 + (d*x)/2]^12 + 1770*Hypergeometric2F1[...
Time = 1.63 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.07, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 3440, 3042, 3457, 27, 3042, 3463, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \cos (c+d x))}{(a \cos (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3440 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A-5 B)-6 a (A-B) \cos (c+d x)}{2 \cos ^{\frac {7}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A-5 B)-6 a (A-B) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A-5 B)-6 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int -\frac {a^2 (39 A-35 B)-4 a^2 (9 A-5 B) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{5 a}+\frac {2 a (9 A-5 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A-5 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^2 (39 A-35 B)-4 a^2 (9 A-5 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{5 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A-5 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^2 (39 A-35 B)-4 a^2 (9 A-5 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A-5 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {a^3 (147 A-95 B)-2 a^3 (39 A-35 B) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 a^2 (39 A-35 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{5 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A-5 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (39 A-35 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (147 A-95 B)-2 a^3 (39 A-35 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A-5 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (39 A-35 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (147 A-95 B)-2 a^3 (39 A-35 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A-5 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (39 A-35 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {15 a^4 (15 A-11 B)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^3 (147 A-95 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{5 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A-5 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (39 A-35 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (147 A-95 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-15 a^3 (15 A-11 B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A-5 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (39 A-35 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (147 A-95 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-15 a^3 (15 A-11 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A-5 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (39 A-35 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {30 a^4 (15 A-11 B) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^3 (147 A-95 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{5 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A-5 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (39 A-35 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (147 A-95 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {15 \sqrt {2} a^{5/2} (15 A-11 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{3 a}}{5 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A - B)*Sin[c + d*x])/(d*Cos[ c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(3/2)) + ((2*a*(9*A - 5*B)*Sin[c + d*x ])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) - ((2*a^2*(39*A - 35* B)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((-15 *Sqrt[2]*a^(5/2)*(15*A - 11*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt [Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/d + (2*a^3*(147*A - 95*B)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/(3*a))/(5*a))/(4 *a^2))
3.6.30.3.1 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[((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 = 10.38 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.69
| method | result | size |
| default | \(\frac {\left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (225 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-165 B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+147 A \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}+450 A \left (\cos ^{4}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-95 B \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}-330 B \left (\cos ^{4}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+108 A \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+225 A \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-60 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}-165 B \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-12 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+20 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+12 A \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {2}}{60 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{2}}\) | \(457\) |
| parts | \(\frac {A \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \left (75 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )+49 \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+150 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )+36 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+75 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-4 \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {2}}{20 d \left (1+\cos \left (d x +c \right )\right )^{2} a^{2}}-\frac {B \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \left (33 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )+19 \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+66 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )+12 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+33 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-4 \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right ) \sqrt {2}}{12 d \left (1+\cos \left (d x +c \right )\right )^{2} a^{2}}\) | \(489\) |
1/60/a^2/d*sec(d*x+c)^(7/2)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))^2*(225 *A*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^ (1/2)-165*B*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^5*(cos(d*x+c)/(1+cos( d*x+c)))^(1/2)+147*A*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)+450*A*cos(d*x+c)^4*ar csin(cot(d*x+c)-csc(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-95*B*sin(d*x +c)*cos(d*x+c)^4*2^(1/2)-330*B*cos(d*x+c)^4*arcsin(cot(d*x+c)-csc(d*x+c))* (cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+108*A*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)+22 5*A*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d *x+c))-60*B*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)-165*B*cos(d*x+c)^3*(cos(d*x+c) /(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))-12*A*sin(d*x+c)*cos(d *x+c)^2*2^(1/2)+20*B*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)+12*A*sin(d*x+c)*cos(d *x+c)*2^(1/2))*2^(1/2)
Time = 0.29 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {15 \, \sqrt {2} {\left ({\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (147 \, A - 95 \, B\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 12 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]
1/60*(15*sqrt(2)*((15*A - 11*B)*cos(d*x + c)^4 + 2*(15*A - 11*B)*cos(d*x + c)^3 + (15*A - 11*B)*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d* x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((147*A - 95*B) *cos(d*x + c)^3 + 12*(9*A - 5*B)*cos(d*x + c)^2 - 4*(3*A - 5*B)*cos(d*x + c) + 12*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2* d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)
Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]