Integrand size = 35, antiderivative size = 203 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{7/2}} \, dx=\frac {(63 A+13 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{64 \sqrt {2} a^{7/2} d}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {(5 A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}-\frac {(103 A+5 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}} \] Output:
1/128*(63*A+13*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))*2^(1/2)/a^(7/2)/d-1/6*(A-B)*cos(d*x+c)^(1/2)*sin(d* x+c)/d/(a+a*cos(d*x+c))^(7/2)-1/16*(5*A-B)*cos(d*x+c)^(1/2)*sin(d*x+c)/a/d /(a+a*cos(d*x+c))^(5/2)-1/192*(103*A+5*B)*cos(d*x+c)^(1/2)*sin(d*x+c)/a^2/ d/(a+a*cos(d*x+c))^(3/2)
Time = 3.03 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.83 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{7/2}} \, dx=-\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-48 (63 A+13 B) \text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos ^6\left (\frac {1}{2} (c+d x)\right )+\cos (c+d x) (493 A-73 B+(532 A-4 B) \cos (c+d x)+(103 A+5 B) \cos (2 (c+d x))) \sqrt {2-2 \sec (c+d x)}\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{1536 \sqrt {2} a^3 d \sqrt {-1+\cos (c+d x)} \sqrt {a (1+\cos (c+d x))}} \] Input:
Integrate[(A + B*Cos[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(7 /2)),x]
Output:
-1/1536*(Sec[(c + d*x)/2]^4*(-48*(63*A + 13*B)*ArcTanh[Sqrt[-(Sec[c + d*x] *Sin[(c + d*x)/2]^2)]]*Cos[(c + d*x)/2]^6 + Cos[c + d*x]*(493*A - 73*B + ( 532*A - 4*B)*Cos[c + d*x] + (103*A + 5*B)*Cos[2*(c + d*x)])*Sqrt[2 - 2*Sec [c + d*x]])*Tan[(c + d*x)/2])/(Sqrt[2]*a^3*d*Sqrt[-1 + Cos[c + d*x]]*Sqrt[ a*(1 + Cos[c + d*x])])
Time = 1.08 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 3457, 27, 3042, 3457, 27, 3042, 3457, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {a (11 A+B)-4 a (A-B) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{5/2}}dx}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (11 A+B)-4 a (A-B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{5/2}}dx}{12 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (11 A+B)-4 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx}{12 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (73 A+11 B)-6 a^2 (5 A-B) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {3 a (5 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (73 A+11 B)-6 a^2 (5 A-B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {3 a (5 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (73 A+11 B)-6 a^2 (5 A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {3 a (5 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {3 a^3 (63 A+13 B)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a^2 (103 A+5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {3 a (5 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3}{4} a (63 A+13 B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-\frac {a^2 (103 A+5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {3 a (5 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{4} a (63 A+13 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {a^2 (103 A+5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {3 a (5 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {\frac {-\frac {3 a^2 (63 A+13 B) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{2 d}-\frac {a^2 (103 A+5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {3 a (5 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {3 \sqrt {a} (63 A+13 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} d}-\frac {a^2 (103 A+5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {3 a (5 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\) |
Input:
Int[(A + B*Cos[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(7/2)),x ]
Output:
-1/6*((A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(7/ 2)) + ((-3*a*(5*A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((3*Sqrt[a]*(63*A + 13*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/( Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*d) - (a^ 2*(103*A + 5*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x]) ^(3/2)))/(8*a^2))/(12*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[((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])
Time = 13.00 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.42
| method | result | size |
| default | \(-\frac {\left (\sin \left (d x +c \right ) \left (103 \cos \left (d x +c \right )^{2}+266 \cos \left (d x +c \right )+195\right ) \sqrt {2}\, A \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-39\right ) \sqrt {2}\, B \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (189 \cos \left (d x +c \right )^{3}+567 \cos \left (d x +c \right )^{2}+567 \cos \left (d x +c \right )+189\right ) A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+\left (39 \cos \left (d x +c \right )^{3}+117 \cos \left (d x +c \right )^{2}+117 \cos \left (d x +c \right )+39\right ) B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}}{384 d \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+1\right ) a^{4}}\) | \(288\) |
| parts | \(-\frac {A \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\, \left (103 \cos \left (d x +c \right )^{2}+266 \cos \left (d x +c \right )+195\right )+\left (189 \cos \left (d x +c \right )^{4}+756 \cos \left (d x +c \right )^{3}+1134 \cos \left (d x +c \right )^{2}+756 \cos \left (d x +c \right )+189\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{192 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\cos \left (d x +c \right )}\, \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) a^{4}}-\frac {B \left (\sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-39\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (39 \cos \left (d x +c \right )^{3}+117 \cos \left (d x +c \right )^{2}+117 \cos \left (d x +c \right )+39\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}}{384 d \left (\cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{4}}\) | \(375\) |
Input:
int((A+B*cos(d*x+c))/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x,method=_RET URNVERBOSE)
Output:
-1/384/d*(sin(d*x+c)*(103*cos(d*x+c)^2+266*cos(d*x+c)+195)*2^(1/2)*A*(cos( d*x+c)/(cos(d*x+c)+1))^(1/2)+sin(d*x+c)*(5*cos(d*x+c)^2-2*cos(d*x+c)-39)*2 ^(1/2)*B*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+(189*cos(d*x+c)^3+567*cos(d*x+c )^2+567*cos(d*x+c)+189)*A*arcsin(cot(d*x+c)-csc(d*x+c))+(39*cos(d*x+c)^3+1 17*cos(d*x+c)^2+117*cos(d*x+c)+39)*B*arcsin(cot(d*x+c)-csc(d*x+c)))*cos(d* x+c)^(1/2)*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)/(cos(d*x+c)/(cos(d*x+c)+1))^(1 /2)/(cos(d*x+c)^4+4*cos(d*x+c)^3+6*cos(d*x+c)^2+4*cos(d*x+c)+1)/a^4
Time = 0.12 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.31 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{7/2}} \, dx=\frac {3 \, \sqrt {2} {\left ({\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right ) + 63 \, A + 13 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (103 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (133 \, A - B\right )} \cos \left (d x + c\right ) + 195 \, A - 39 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x, algo rithm="fricas")
Output:
1/384*(3*sqrt(2)*((63*A + 13*B)*cos(d*x + c)^4 + 4*(63*A + 13*B)*cos(d*x + c)^3 + 6*(63*A + 13*B)*cos(d*x + c)^2 + 4*(63*A + 13*B)*cos(d*x + c) + 63 *A + 13*B)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqr t(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) - 2*((10 3*A + 5*B)*cos(d*x + c)^2 + 2*(133*A - B)*cos(d*x + c) + 195*A - 39*B)*sqr t(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^4*d*cos(d*x + c) ^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c ) + a^4*d)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)**(1/2)/(a+a*cos(d*x+c))**(7/2),x)
Output:
Timed out
\[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{7/2}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x, algo rithm="maxima")
Output:
integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(7/2)*sqrt(cos(d*x + c))), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x, algo rithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{7/2}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \] Input:
int((A + B*cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + a*cos(c + d*x))^(7/2)),x )
Output:
int((A + B*cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + a*cos(c + d*x))^(7/2)), x)
\[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{7/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5}+4 \cos \left (d x +c \right )^{4}+6 \cos \left (d x +c \right )^{3}+4 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+1}d x \right ) b \right )}{a^{4}} \] Input:
int((A+B*cos(d*x+c))/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x)
Output:
(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x)**5 + 4*cos(c + d*x)**4 + 6*cos(c + d*x)**3 + 4*cos(c + d*x)**2 + cos(c + d*x )),x)*a + int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x)**4 + 4*cos(c + d*x)**3 + 6*cos(c + d*x)**2 + 4*cos(c + d*x) + 1),x)*b))/a**4