Integrand size = 35, antiderivative size = 130 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {4 a (4 A+5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (4 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}} \]
2/15*a*(4*A+5*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/5* a*A*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+4/15*a*(4*A+5*B)* sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)
Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.60 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \sqrt {a (1+\cos (c+d x))} (7 A+5 B+(4 A+5 B) \cos (c+d x)+(4 A+5 B) \cos (2 (c+d x))) \sec ^{\frac {5}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{15 d} \]
(2*Sqrt[a*(1 + Cos[c + d*x])]*(7*A + 5*B + (4*A + 5*B)*Cos[c + d*x] + (4*A + 5*B)*Cos[2*(c + d*x)])*Sec[c + d*x]^(5/2)*Tan[(c + d*x)/2])/(15*d)
Time = 0.75 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 3440, 3042, 3459, 3042, 3251, 3042, 3250}
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.
| \(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a} (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3440 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x) a+a} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} (4 A+5 B) \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} (4 A+5 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} (4 A+5 B) \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} (4 A+5 B) \left (\frac {2}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3250 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} (4 A+5 B) \left (\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a*A*Sin[c + d*x])/(5*d*Cos[c + d *x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + ((4*A + 5*B)*((2*a*Sin[c + d*x])/(3* d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*a*Sin[c + d*x])/(3*d*S qrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])))/5)
3.5.95.3.1 Defintions of rubi rules used
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq rt[a + b*Sin[e + f*x]]*Sqrt[c + d*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[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
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[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) *(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d)) Int[Sqrt[a + b*Sin[e + f*x] ]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, -1]
Time = 9.73 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.65
| method | result | size |
| default | \(-\frac {2 \cot \left (d x +c \right ) \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \left (\cos \left (d x +c \right )-1\right ) \left (8 A \left (\cos ^{2}\left (d x +c \right )\right )+10 B \left (\cos ^{2}\left (d x +c \right )\right )+4 A \cos \left (d x +c \right )+5 B \cos \left (d x +c \right )+3 A \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{15 d}\) | \(84\) |
| parts | \(-\frac {2 A \cot \left (d x +c \right ) \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (8 \left (\cos ^{3}\left (d x +c \right )\right )-4 \left (\cos ^{2}\left (d x +c \right )\right )-\cos \left (d x +c \right )-3\right )}{15 d}-\frac {2 B \cos \left (d x +c \right ) \cot \left (d x +c \right ) \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \left (2 \left (\cos ^{2}\left (d x +c \right )\right )-\cos \left (d x +c \right )-1\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{3 d}\) | \(122\) |
-2/15/d*cot(d*x+c)*sec(d*x+c)^(7/2)*(cos(d*x+c)-1)*(8*A*cos(d*x+c)^2+10*B* cos(d*x+c)^2+4*A*cos(d*x+c)+5*B*cos(d*x+c)+3*A)*(a*(1+cos(d*x+c)))^(1/2)
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.66 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \, {\left (2 \, {\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 3 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )}} \]
2/15*(2*(4*A + 5*B)*cos(d*x + c)^2 + (4*A + 5*B)*cos(d*x + c) + 3*A)*sqrt( a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^3 + d*cos(d*x + c)^2)*sq rt(cos(d*x + c)))
Timed out. \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 475 vs. \(2 (112) = 224\).
Time = 0.35 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.65 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \, {\left (\frac {A {\left (\frac {15 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {25 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {17 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {7 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}} + \frac {5 \, B {\left (\frac {3 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {\sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}}\right )}}{15 \, d} \]
2/15*(A*(15*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sqrt(2)*s qrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 17*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 7*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d *x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(3*si n(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)) + 5*B*(3*sqrt(2)*sqrt(a)*sin(d *x + c)/(cos(d*x + c) + 1) - 7*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c ) + 1)^3 + 5*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - sqrt(2) *sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)*(sin(d*x + c)^2/(cos(d*x + c ) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c )/(cos(d*x + c) + 1) + 1)^(7/2)*(3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3 *sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)))/d
Timed out. \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]
Time = 2.82 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.51 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {4\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (14\,A\,\sin \left (c+d\,x\right )+10\,B\,\sin \left (c+d\,x\right )+8\,A\,\sin \left (2\,c+2\,d\,x\right )+18\,A\,\sin \left (3\,c+3\,d\,x\right )+4\,A\,\sin \left (4\,c+4\,d\,x\right )+4\,A\,\sin \left (5\,c+5\,d\,x\right )+10\,B\,\sin \left (2\,c+2\,d\,x\right )+15\,B\,\sin \left (3\,c+3\,d\,x\right )+5\,B\,\sin \left (4\,c+4\,d\,x\right )+5\,B\,\sin \left (5\,c+5\,d\,x\right )\right )}{15\,d\,\left (10\,\cos \left (c+d\,x\right )+8\,\cos \left (2\,c+2\,d\,x\right )+5\,\cos \left (3\,c+3\,d\,x\right )+2\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (5\,c+5\,d\,x\right )+6\right )} \]
(4*(a*(cos(c + d*x) + 1))^(1/2)*(1/cos(c + d*x))^(1/2)*(14*A*sin(c + d*x) + 10*B*sin(c + d*x) + 8*A*sin(2*c + 2*d*x) + 18*A*sin(3*c + 3*d*x) + 4*A*s in(4*c + 4*d*x) + 4*A*sin(5*c + 5*d*x) + 10*B*sin(2*c + 2*d*x) + 15*B*sin( 3*c + 3*d*x) + 5*B*sin(4*c + 4*d*x) + 5*B*sin(5*c + 5*d*x)))/(15*d*(10*cos (c + d*x) + 8*cos(2*c + 2*d*x) + 5*cos(3*c + 3*d*x) + 2*cos(4*c + 4*d*x) + cos(5*c + 5*d*x) + 6))