Integrand size = 35, antiderivative size = 250 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx=-\frac {3 (121 A-21 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{64 \sqrt {2} a^{7/2} d}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{7/2}}-\frac {(19 A-7 B) \sin (c+d x)}{48 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(199 A-43 B) \sin (c+d x)}{192 a^2 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(691 A-103 B) \sin (c+d x)}{192 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \] Output:
-3/128*(121*A-21*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)*sin(d*x+c)/d/cos(d*x+ c)^(1/2)/(a+a*cos(d*x+c))^(7/2)-1/48*(19*A-7*B)*sin(d*x+c)/a/d/cos(d*x+c)^ (1/2)/(a+a*cos(d*x+c))^(5/2)-1/192*(199*A-43*B)*sin(d*x+c)/a^2/d/cos(d*x+c )^(1/2)/(a+a*cos(d*x+c))^(3/2)+1/192*(691*A-103*B)*sin(d*x+c)/a^3/d/cos(d* x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 9.69 (sec) , antiderivative size = 798, normalized size of antiderivative = 3.19 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx =\text {Too large to display} \] Input:
Integrate[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(7 /2)),x]
Output:
-1/24*(B*Cos[c/2 + (d*x)/2]^7*Sec[(c + d*x)/2]^6*Sin[c/2 + (d*x)/2]*(141 - 518*Sin[c/2 + (d*x)/2]^2 + 575*Sin[c/2 + (d*x)/2]^4 - 206*Sin[c/2 + (d*x) /2]^6 - (189*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))]))/(d*(a*(1 + Cos[c + d*x]))^(7/2)*Sqrt[1 - 2*Sin[c/2 + (d*x)/2 ]^2]) + (2*A*Cos[c/2 + (d*x)/2]^7*Sec[(c + d*x)/2]^6*Sin[c/2 + (d*x)/2]*(( 16*Cos[(c + d*x)/2]^8*HypergeometricPFQ[{2, 2, 2, 2, 5/2}, {1, 1, 1, 13/2} , Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^2 )/(3465*(-1 + 2*Sin[c/2 + (d*x)/2]^2)) - (Csc[c/2 + (d*x)/2]^10*(1 - 2*Sin [c/2 + (d*x)/2]^2)^2*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^ 2)]*(105*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]] *Cos[(c + d*x)/2]^6*(2187 - 12908*Sin[c/2 + (d*x)/2]^2 + 27986*Sin[c/2 + ( d*x)/2]^4 - 26380*Sin[c/2 + (d*x)/2]^6 + 8752*Sin[c/2 + (d*x)/2]^8) + Sqrt [Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-229635 + 2120790*Si n[c/2 + (d*x)/2]^2 - 8267707*Sin[c/2 + (d*x)/2]^4 + 17646926*Sin[c/2 + (d* x)/2]^6 - 22251094*Sin[c/2 + (d*x)/2]^8 + 16548816*Sin[c/2 + (d*x)/2]^10 - 6712984*Sin[c/2 + (d*x)/2]^12 + 1144608*Sin[c/2 + (d*x)/2]^14)))/1680))/( d*(a*(1 + Cos[c + d*x]))^(7/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2))
Time = 1.44 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3457, 27, 3042, 3457, 27, 3042, 3457, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {a (13 A-B)-6 a (A-B) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{5/2}}dx}{6 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (13 A-B)-6 a (A-B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{5/2}}dx}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (13 A-B)-6 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {3 a^2 (41 A-5 B)-4 a^2 (19 A-7 B) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {a (19 A-7 B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {3 a^2 (41 A-5 B)-4 a^2 (19 A-7 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {a (19 A-7 B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {3 a^2 (41 A-5 B)-4 a^2 (19 A-7 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {a (19 A-7 B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {a^3 (691 A-103 B)-2 a^3 (199 A-43 B) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a^2 (199 A-43 B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {a (19 A-7 B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {a^3 (691 A-103 B)-2 a^3 (199 A-43 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a^2 (199 A-43 B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {a (19 A-7 B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {a^3 (691 A-103 B)-2 a^3 (199 A-43 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a^2 (199 A-43 B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {a (19 A-7 B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 \int -\frac {9 a^4 (121 A-21 B)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^3 (691 A-103 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {a^2 (199 A-43 B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {a (19 A-7 B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (691 A-103 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-9 a^3 (121 A-21 B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a^2 (199 A-43 B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {a (19 A-7 B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (691 A-103 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-9 a^3 (121 A-21 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a^2 (199 A-43 B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {a (19 A-7 B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {\frac {\frac {\frac {18 a^4 (121 A-21 B) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^3 (691 A-103 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {a^2 (199 A-43 B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {a (19 A-7 B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (691 A-103 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {9 \sqrt {2} a^{5/2} (121 A-21 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a^2 (199 A-43 B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {a (19 A-7 B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{7/2}}\) |
Input:
Int[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(7/2)),x ]
Output:
-1/6*((A - B)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(7/ 2)) + (-1/4*(a*(19*A - 7*B)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos [c + d*x])^(5/2)) + (-1/2*(a^2*(199*A - 43*B)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)) + ((-9*Sqrt[2]*a^(5/2)*(121*A - 21*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*(691*A - 103*B)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]] *Sqrt[a + a*Cos[c + d*x]]))/(4*a^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])
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 = 11.83 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.22
| method | result | size |
| default | \(\frac {\sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\left (1089 \cos \left (d x +c \right )^{4}+4356 \cos \left (d x +c \right )^{3}+6534 \cos \left (d x +c \right )^{2}+4356 \cos \left (d x +c \right )+1089\right ) A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (-189 \cos \left (d x +c \right )^{4}-756 \cos \left (d x +c \right )^{3}-1134 \cos \left (d x +c \right )^{2}-756 \cos \left (d x +c \right )-189\right ) B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right ) \left (691 \cos \left (d x +c \right )^{3}+1874 \cos \left (d x +c \right )^{2}+1599 \cos \left (d x +c \right )+384\right ) \sqrt {2}\, A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-103 \cos \left (d x +c \right )^{2}-266 \cos \left (d x +c \right )-195\right ) \sqrt {2}\, B \right )}{192 d \sqrt {\cos \left (d x +c \right )}\, \left (\cos \left (d x +c \right )+1\right ) \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) a^{4}}\) | \(305\) |
| parts | \(\frac {A \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\sin \left (d x +c \right ) \left (691 \cos \left (d x +c \right )^{3}+1874 \cos \left (d x +c \right )^{2}+1599 \cos \left (d x +c \right )+384\right ) \sqrt {2}+\left (1089 \cos \left (d x +c \right )^{4}+4356 \cos \left (d x +c \right )^{3}+6534 \cos \left (d x +c \right )^{2}+4356 \cos \left (d x +c \right )+1089\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )}{192 d \sqrt {\cos \left (d x +c \right )}\, \left (\cos \left (d x +c \right )+1\right ) \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) a^{4}}-\frac {B \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\, \left (103 \cos \left (d x +c \right )^{2}+266 \cos \left (d x +c \right )+195\right )+\left (189 \cos \left (d x +c \right )^{4}+756 \cos \left (d x +c \right )^{3}+1134 \cos \left (d x +c \right )^{2}+756 \cos \left (d x +c \right )+189\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{192 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\cos \left (d x +c \right )}\, \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) a^{4}}\) | \(376\) |
Input:
int((A+B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(7/2),x,method=_RET URNVERBOSE)
Output:
1/192/d*(a*cos(1/2*d*x+1/2*c)^2)^(1/2)*((1089*cos(d*x+c)^4+4356*cos(d*x+c) ^3+6534*cos(d*x+c)^2+4356*cos(d*x+c)+1089)*A*arcsin(cot(d*x+c)-csc(d*x+c)) *(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+(-189*cos(d*x+c)^4-756*cos(d*x+c)^3-113 4*cos(d*x+c)^2-756*cos(d*x+c)-189)*B*arcsin(cot(d*x+c)-csc(d*x+c))*(cos(d* x+c)/(cos(d*x+c)+1))^(1/2)+sin(d*x+c)*(691*cos(d*x+c)^3+1874*cos(d*x+c)^2+ 1599*cos(d*x+c)+384)*2^(1/2)*A+sin(d*x+c)*cos(d*x+c)*(-103*cos(d*x+c)^2-26 6*cos(d*x+c)-195)*2^(1/2)*B)/cos(d*x+c)^(1/2)/(cos(d*x+c)+1)/(cos(d*x+c)^3 +3*cos(d*x+c)^2+3*cos(d*x+c)+1)/a^4
Time = 0.12 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.19 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx=-\frac {9 \, \sqrt {2} {\left ({\left (121 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (121 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (121 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (121 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (121 \, A - 21 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (691 \, A - 103 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (937 \, A - 133 \, B\right )} \cos \left (d x + c\right )^{2} + 39 \, {\left (41 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 384 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(7/2),x, algo rithm="fricas")
Output:
-1/384*(9*sqrt(2)*((121*A - 21*B)*cos(d*x + c)^5 + 4*(121*A - 21*B)*cos(d* x + c)^4 + 6*(121*A - 21*B)*cos(d*x + c)^3 + 4*(121*A - 21*B)*cos(d*x + c) ^2 + (121*A - 21*B)*cos(d*x + c))*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d* x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a* cos(d*x + c))) - 2*((691*A - 103*B)*cos(d*x + c)^3 + 2*(937*A - 133*B)*cos (d*x + c)^2 + 39*(41*A - 5*B)*cos(d*x + c) + 384*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^4*d*cos(d*x + c)^5 + 4*a^4*d*cos(d* x + c)^4 + 6*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c))
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)**(3/2)/(a+a*cos(d*x+c))**(7/2),x)
Output:
Timed out
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(7/2),x, algo rithm="maxima")
Output:
Timed out
\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(7/2),x, algo rithm="giac")
Output:
integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(7/2)*cos(d*x + c)^(3 /2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \] Input:
int((A + B*cos(c + d*x))/(cos(c + d*x)^(3/2)*(a + a*cos(c + d*x))^(7/2)),x )
Output:
int((A + B*cos(c + d*x))/(cos(c + d*x)^(3/2)*(a + a*cos(c + d*x))^(7/2)), x)
\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{6}+4 \cos \left (d x +c \right )^{5}+6 \cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+\cos \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5}+4 \cos \left (d x +c \right )^{4}+6 \cos \left (d x +c \right )^{3}+4 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )}d x \right ) b \right )}{a^{4}} \] Input:
int((A+B*cos(d*x+c))/cos(d*x+c)^(3/2)/(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)**6 + 4*cos(c + d*x)**5 + 6*cos(c + d*x)**4 + 4*cos(c + d*x)**3 + cos(c + d*x )**2),x)*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)*b))/a**4