Integrand size = 35, antiderivative size = 221 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(7 A+5 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 \sqrt {2} a^{7/2} d}+\frac {(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(A-13 B) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}+\frac {(17 A+67 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \] Output:
1/128*(7*A+5*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^(7/2)/d+1 /6*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(3/2)+1/48*(A-13*B )*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2)+1/192*(17*A+67*B) *sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(561\) vs. \(2(221)=442\).
Time = 6.77 (sec) , antiderivative size = 561, normalized size of antiderivative = 2.54 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {A \cos ^7\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {1}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \left (27-106 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+121 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )-34 \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )+\frac {21 \text {arctanh}\left (\sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \cos ^6\left (\frac {1}{2} (c+d x)\right )}{\sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}}\right )}{24 d (a (1+\cos (c+d x)))^{7/2}}+\frac {B \cos ^7\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {1}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )} \left (15 \arcsin \left (\frac {\sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )}}\right )+\frac {33 \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {1-\sec ^2\left (\frac {1}{2} (c+d x)\right ) \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}{\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )}}-\frac {26 \sin ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {1-\sec ^2\left (\frac {1}{2} (c+d x)\right ) \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}{\cos ^2\left (\frac {1}{2} (c+d x)\right )^{3/2}}+\frac {8 \sin ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {1-\sec ^2\left (\frac {1}{2} (c+d x)\right ) \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}{\cos ^2\left (\frac {1}{2} (c+d x)\right )^{5/2}}\right )}{24 d (a (1+\cos (c+d x)))^{7/2}} \] Input:
Integrate[(A + B*Cos[c + d*x])/((a + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(3 /2)),x]
Output:
(A*Cos[c/2 + (d*x)/2]^7*Sec[(c + d*x)/2]^6*Sin[c/2 + (d*x)/2]*Sqrt[(1 - 2* Sin[c/2 + (d*x)/2]^2)^(-1)]*(27 - 106*Sin[c/2 + (d*x)/2]^2 + 121*Sin[c/2 + (d*x)/2]^4 - 34*Sin[c/2 + (d*x)/2]^6 + (21*ArcTanh[Sqrt[-(Sin[c/2 + (d*x) /2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*Cos[(c + d*x)/2]^6)/Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]))/(24*d*(a*(1 + Cos[c + d*x]))^ (7/2)) + (B*Cos[c/2 + (d*x)/2]^7*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*S qrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*(15*ArcSin[Sin[c/2 + (d*x)/2]/Sqrt[Cos[(c + d*x)/2]^2]] + (33*Sin[c/2 + (d*x)/2]*Sqrt[1 - Sec[(c + d*x)/2]^2*Sin[c/2 + (d*x)/2]^2])/Sqrt[Cos[(c + d*x)/2]^2] - (26*Sin[c/2 + (d*x)/2]^3*Sqrt[1 - Sec[(c + d*x)/2]^2*Sin[c/2 + (d*x)/2]^2])/(Cos[(c + d*x)/2]^2)^(3/2) + (8*Sin[c/2 + (d*x)/2]^5*Sqrt[1 - Sec[(c + d*x)/2]^2*Sin[c/2 + (d*x)/2]^2]) /(Cos[(c + d*x)/2]^2)^(5/2)))/(24*d*(a*(1 + Cos[c + d*x]))^(7/2))
Time = 1.29 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.07, 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, 3456, 27, 3042, 3456, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3440 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(\cos (c+d x) a+a)^{7/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} (3 a (A-B)+2 a (A+5 B) \cos (c+d x))}{2 (\cos (c+d x) a+a)^{5/2}}dx}{6 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} (3 a (A-B)+2 a (A+5 B) \cos (c+d x))}{(\cos (c+d x) a+a)^{5/2}}dx}{12 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 a (A-B)+2 a (A+5 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx}{12 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {(A-13 B) a^2+18 (A+3 B) \cos (c+d x) a^2}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}+\frac {a (A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {(A-13 B) a^2+18 (A+3 B) \cos (c+d x) a^2}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}+\frac {a (A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {(A-13 B) a^2+18 (A+3 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}+\frac {a (A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int \frac {3 a^3 (7 A+5 B)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}+\frac {a^2 (17 A+67 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {a (A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {3}{4} a (7 A+5 B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx+\frac {a^2 (17 A+67 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {a (A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {3}{4} a (7 A+5 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {a^2 (17 A+67 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {a (A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {a^2 (17 A+67 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {3 a^2 (7 A+5 B) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{2 d}}{8 a^2}+\frac {a (A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {a^2 (17 A+67 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac {3 \sqrt {a} (7 A+5 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} d}}{8 a^2}+\frac {a (A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\) |
Input:
Int[(A + B*Cos[c + d*x])/((a + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(3/2)),x ]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((A - B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)) + ((a*(A - 13*B)*Sqrt[Cos[c + d*x] ]*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((3*Sqrt[a]*(7*A + 5*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*(17*A + 67*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[(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[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{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[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[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*(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 = 10.69 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.29
| method | result | size |
| default | \(\frac {\sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (\tan \left (d x +c \right ) \left (17 \cos \left (d x +c \right )^{2}+70 \cos \left (d x +c \right )+21\right ) \sqrt {2}\, A \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\tan \left (d x +c \right ) \left (67 \cos \left (d x +c \right )^{2}+50 \cos \left (d x +c \right )+15\right ) \sqrt {2}\, B \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (-21 \cos \left (d x +c \right )^{2}-63 \cos \left (d x +c \right )-63-21 \sec \left (d x +c \right )\right )+B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (-15 \cos \left (d x +c \right )^{2}-45 \cos \left (d x +c \right )-45-15 \sec \left (d x +c \right )\right )\right )}{384 d \left (\cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{4}}\) | \(284\) |
| parts | \(\frac {A \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (\tan \left (d x +c \right ) \left (17 \cos \left (d x +c \right )^{2}+70 \cos \left (d x +c \right )+21\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (-21 \cos \left (d x +c \right )^{2}-63 \cos \left (d x +c \right )-63-21 \sec \left (d x +c \right )\right )\right )}{384 d \left (\cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{4}}+\frac {B \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (\tan \left (d x +c \right ) \left (67 \cos \left (d x +c \right )^{2}+50 \cos \left (d x +c \right )+15\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (-15 \cos \left (d x +c \right )^{2}-45 \cos \left (d x +c \right )-45-15 \sec \left (d x +c \right )\right )\right )}{384 d \left (\cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{4}}\) | \(374\) |
Input:
int((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(3/2),x,method=_RET URNVERBOSE)
Output:
1/384/d*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)/(cos(d*x+c)^4+4*cos(d*x+c)^3+6*co s(d*x+c)^2+4*cos(d*x+c)+1)/sec(d*x+c)^(3/2)/(cos(d*x+c)/(cos(d*x+c)+1))^(1 /2)*(tan(d*x+c)*(17*cos(d*x+c)^2+70*cos(d*x+c)+21)*2^(1/2)*A*(cos(d*x+c)/( cos(d*x+c)+1))^(1/2)+tan(d*x+c)*(67*cos(d*x+c)^2+50*cos(d*x+c)+15)*2^(1/2) *B*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+A*arcsin(cot(d*x+c)-csc(d*x+c))*(-21* cos(d*x+c)^2-63*cos(d*x+c)-63-21*sec(d*x+c))+B*arcsin(cot(d*x+c)-csc(d*x+c ))*(-15*cos(d*x+c)^2-45*cos(d*x+c)-45-15*sec(d*x+c)))/a^4
Time = 0.13 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.16 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {3 \, \sqrt {2} {\left ({\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 7 \, A + 5 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left ({\left (17 \, A + 67 \, B\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(3/2),x, algo rithm="fricas")
Output:
-1/384*(3*sqrt(2)*((7*A + 5*B)*cos(d*x + c)^4 + 4*(7*A + 5*B)*cos(d*x + c) ^3 + 6*(7*A + 5*B)*cos(d*x + c)^2 + 4*(7*A + 5*B)*cos(d*x + c) + 7*A + 5*B )*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt (a)*sin(d*x + c))) - 2*((17*A + 67*B)*cos(d*x + c)^3 + 10*(7*A + 5*B)*cos( d*x + c)^2 + 3*(7*A + 5*B)*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(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. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))**(7/2)/sec(d*x+c)**(3/2),x)
Output:
Timed out
\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(3/2),x, algo rithm="maxima")
Output:
integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(7/2)*sec(d*x + c)^(3 /2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(3/2),x, algo rithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \] Input:
int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(7/2 )),x)
Output:
int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(7/2 )), x)
\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}+4 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}}{\cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}+4 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )^{2}}d x \right ) a \right )}{a^{4}} \] Input:
int((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2)/sec(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))/(co s(c + d*x)**4*sec(c + d*x)**2 + 4*cos(c + d*x)**3*sec(c + d*x)**2 + 6*cos( c + d*x)**2*sec(c + d*x)**2 + 4*cos(c + d*x)*sec(c + d*x)**2 + sec(c + d*x )**2),x)*b + int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1))/(cos(c + d*x) **4*sec(c + d*x)**2 + 4*cos(c + d*x)**3*sec(c + d*x)**2 + 6*cos(c + d*x)** 2*sec(c + d*x)**2 + 4*cos(c + d*x)*sec(c + d*x)**2 + sec(c + d*x)**2),x)*a ))/a**4