Integrand size = 33, antiderivative size = 264 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(39 A-20 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \]
1/4*(39*A-20*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2) /d-1/32*(219*A-115*B)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+ c))^(1/2))/a^(5/2)/d*2^(1/2)-1/4*(A-B)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d* x+c))^(5/2)-1/16*(19*A-11*B)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(3 /2)-7/16*(9*A-5*B)*tan(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)+1/16*(31*A-15*B )*sec(d*x+c)*tan(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)
Time = 4.02 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.67 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {-8 (219 A-115 B) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )+32 \sqrt {2} (39 A-20 B) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )+(-158 A+110 B+(-269 A+169 B) \cos (c+d x)+(-190 A+110 B) \cos (2 (c+d x))-63 A \cos (3 (c+d x))+35 B \cos (3 (c+d x))) \sec ^2(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{64 a d (a (1+\cos (c+d x)))^{3/2}} \]
(-8*(219*A - 115*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + 32*Sqrt [2]*(39*A - 20*B)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + ( -158*A + 110*B + (-269*A + 169*B)*Cos[c + d*x] + (-190*A + 110*B)*Cos[2*(c + d*x)] - 63*A*Cos[3*(c + d*x)] + 35*B*Cos[3*(c + d*x)])*Sec[c + d*x]^2*T an[(c + d*x)/2])/(64*a*d*(a*(1 + Cos[c + d*x]))^(3/2))
Time = 1.91 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.07, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 3457, 27, 3042, 3457, 27, 3042, 3463, 27, 3042, 3463, 25, 3042, 3464, 3042, 3128, 219, 3252, 219}
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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \cos (c+d x))}{(a \cos (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {(4 a (3 A-B)-7 a (A-B) \cos (c+d x)) \sec ^3(c+d x)}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(4 a (3 A-B)-7 a (A-B) \cos (c+d x)) \sec ^3(c+d x)}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {4 a (3 A-B)-7 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (4 a^2 (31 A-15 B)-5 a^2 (19 A-11 B) \cos (c+d x)\right ) \sec ^3(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\left (4 a^2 (31 A-15 B)-5 a^2 (19 A-11 B) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {4 a^2 (31 A-15 B)-5 a^2 (19 A-11 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {2 \left (14 a^3 (9 A-5 B)-3 a^3 (31 A-15 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (14 a^3 (9 A-5 B)-3 a^3 (31 A-15 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {14 a^3 (9 A-5 B)-3 a^3 (31 A-15 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {\int -\frac {\left (4 a^4 (39 A-20 B)-7 a^4 (9 A-5 B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {14 a^3 (9 A-5 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{a}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (4 a^4 (39 A-20 B)-7 a^4 (9 A-5 B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {4 a^4 (39 A-20 B)-7 a^4 (9 A-5 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3464 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {4 a^3 (39 A-20 B) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-a^4 (219 A-115 B) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {4 a^3 (39 A-20 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^4 (219 A-115 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {4 a^3 (39 A-20 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^4 (219 A-115 B) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {4 a^3 (39 A-20 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {\sqrt {2} a^{7/2} (219 A-115 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {8 a^4 (39 A-20 B) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {\sqrt {2} a^{7/2} (219 A-115 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {8 a^{7/2} (39 A-20 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{7/2} (219 A-115 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
-1/4*((A - B)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^(5/2)) + (-1/2*(a*(19*A - 11*B)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^ (3/2)) + ((2*a^2*(31*A - 15*B)*Sec[c + d*x]*Tan[c + d*x])/(d*Sqrt[a + a*Co s[c + d*x]]) - (-(((8*a^(7/2)*(39*A - 20*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x]) /Sqrt[a + a*Cos[c + d*x]]])/d - (Sqrt[2]*a^(7/2)*(219*A - 115*B)*ArcTanh[( Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d)/a) + (14*a^3 *(9*A - 5*B)*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/a)/(4*a^2))/(8*a^ 2)
3.2.22.3.1 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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[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*(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])
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A *b - a*B)/(b*c - a*d) Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c - A*d)/(b*c - a*d) Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1437\) vs. \(2(229)=458\).
Time = 7.58 (sec) , antiderivative size = 1438, normalized size of antiderivative = 5.45
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1438\) |
| default | \(\text {Expression too large to display}\) | \(1610\) |
-1/8*A*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(876*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin( 1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^8*a-62 4*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/ 2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^8*a-624 *ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2) *(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^8*a-876*2 ^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2 *c))*cos(1/2*d*x+1/2*c)^6*a+252*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^( 1/2)*cos(1/2*d*x+1/2*c)^6+624*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2 )*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a) )*cos(1/2*d*x+1/2*c)^6*a+624*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)* a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))* cos(1/2*d*x+1/2*c)^6*a+219*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2 )^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*a*cos(1/2*d*x+1/2*c)^4-188*2^(1/2)*(a*sin (1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4-156*ln(-4/(2*cos(1/2 *d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+ 1/2*c)^2)^(1/2)*a^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^4*a-156*ln(4/(2*cos(1/2*d *x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/ 2*c)^2)^(1/2)*a^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a+19*cos(1/2*d*x+1/2*c)^2 *(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*2^(1/2)*a^(1/2)+2*2^(1/2)*(a*sin(1/2*d*...
Time = 0.41 (sec) , antiderivative size = 428, normalized size of antiderivative = 1.62 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (7 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (19 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (5 \, A - 4 \, B\right )} \cos \left (d x + c\right ) - 8 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]
-1/64*(sqrt(2)*((219*A - 115*B)*cos(d*x + c)^5 + 3*(219*A - 115*B)*cos(d*x + c)^4 + 3*(219*A - 115*B)*cos(d*x + c)^3 + (219*A - 115*B)*cos(d*x + c)^ 2)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqr t(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((39*A - 20*B)*cos(d*x + c)^5 + 3*(39*A - 20*B)*cos(d*x + c)^ 4 + 3*(39*A - 20*B)*cos(d*x + c)^3 + (39*A - 20*B)*cos(d*x + c)^2)*sqrt(a) *log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*sqrt(a*cos(d*x + c) + a)*s qrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(7*(9*A - 5*B)*cos(d*x + c)^3 + 5*(19*A - 11*B)*cos(d*x + c)^2 + 4*(5*A - 4*B)*cos(d*x + c) - 8*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)) /(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)
Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]