Integrand size = 45, antiderivative size = 163 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\frac {(19 A+5 B+3 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(9 A-B-7 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \]
1/32*(19*A+5*B+3*C)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2) /(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/4*(A-B+C)*sin(d*x+c)*cos(d*x+ c)^(1/2)/d/(a+a*cos(d*x+c))^(5/2)-1/16*(9*A-B-7*C)*sin(d*x+c)*cos(d*x+c)^( 1/2)/a/d/(a+a*cos(d*x+c))^(3/2)
Leaf count is larger than twice the leaf count of optimal. \(711\) vs. \(2(163)=326\).
Time = 6.47 (sec) , antiderivative size = 711, normalized size of antiderivative = 4.36 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\frac {3 C \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 ) \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{4 d (a (1+\cos (c+d x)))^{5/2}}+\frac {5 C \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \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 )}}{4 d \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )} (a (1+\cos (c+d x)))^{5/2}}-\frac {C \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \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 )}}{2 d \cos ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a (1+\cos (c+d x)))^{5/2}}-\frac {A \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (11-31 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+18 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )-\frac {19 \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 ^4\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 )}{4 d (a (1+\cos (c+d x)))^{5/2} \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+\frac {B \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )} \left (3-\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-5 \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 ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {c}{2}+\frac {d x}{2}\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 )}{4 d (a (1+\cos (c+d x)))^{5/2}} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)),x]
(3*C*ArcSin[Sin[c/2 + (d*x)/2]/Sqrt[Cos[(c + d*x)/2]^2]]*Cos[c/2 + (d*x)/2 ]^5)/(4*d*(a*(1 + Cos[c + d*x]))^(5/2)) + (5*C*Cos[c/2 + (d*x)/2]^5*Sin[c/ 2 + (d*x)/2]*Sqrt[1 - Sec[(c + d*x)/2]^2*Sin[c/2 + (d*x)/2]^2])/(4*d*Sqrt[ Cos[(c + d*x)/2]^2]*(a*(1 + Cos[c + d*x]))^(5/2)) - (C*Cos[c/2 + (d*x)/2]^ 5*Sin[c/2 + (d*x)/2]^3*Sqrt[1 - Sec[(c + d*x)/2]^2*Sin[c/2 + (d*x)/2]^2])/ (2*d*(Cos[(c + d*x)/2]^2)^(3/2)*(a*(1 + Cos[c + d*x]))^(5/2)) - (A*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*(11 - 31*Sin[c/2 + (d* x)/2]^2 + 18*Sin[c/2 + (d*x)/2]^4 - (19*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^ 2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*Cos[(c + d*x)/2]^4)/Sqrt[-(Sin[c/2 + (d* x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]))/(4*d*(a*(1 + Cos[c + d*x]))^(5/2) *Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (B*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x) /2]^4*Sin[c/2 + (d*x)/2]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*(3 - Sin[c/2 + ( d*x)/2]^2 - 5*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2 ]^2))]]*Cos[(c + d*x)/2]^4*Csc[c/2 + (d*x)/2]^2*Sqrt[-(Sin[c/2 + (d*x)/2]^ 2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]))/(4*d*(a*(1 + Cos[c + d*x]))^(5/2))
Time = 0.78 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3520, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {a (7 A+B-C)-2 a (A-B-3 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (7 A+B-C)-2 a (A-B-3 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (7 A+B-C)-2 a (A-B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (19 A+5 B+3 C)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a (9 A-B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{4} (19 A+5 B+3 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-\frac {a (9 A-B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} (19 A+5 B+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {a (9 A-B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {-\frac {a (19 A+5 B+3 C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{2 d}-\frac {a (9 A-B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {(19 A+5 B+3 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {a (9 A-B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Cos [c + d*x])^(5/2)),x]
-1/4*((A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]) ^(5/2)) + (((19*A + 5*B + 3*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt [Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*Sqrt[a]*d) - (a*(9*A - B - 7*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/ 2)))/(8*a^2)
3.6.19.3.1 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[(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_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x ] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c *(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c ^2 - d^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(484\) vs. \(2(138)=276\).
Time = 14.33 (sec) , antiderivative size = 485, normalized size of antiderivative = 2.98
| method | result | size |
| default | \(-\frac {\left (19 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+5 B \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+3 C \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+38 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+18 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+10 B \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-2 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+6 C \cos \left (d x +c \right ) \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-14 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+19 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+26 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+5 B \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-10 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 C \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-6 C \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{32 a^{3} d \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )^{3}}\) | \(485\) |
| parts | \(\frac {A \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \sqrt {\frac {a}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}}\, \left (-2 \left (\csc ^{3}\left (d x +c \right )\right ) \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (1-\cos \left (d x +c \right )\right )^{3}-11 \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-19 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{32 d \sqrt {-\frac {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}}\, a^{3}}+\frac {B \left (\sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+5 \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-5 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )-10 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-5 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{32 d \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{3}}+\frac {C \left (7 \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-3 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )-6 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-3 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{32 d \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{3}}\) | \(622\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+cos(d*x+c)*a)^(5/2 ),x,method=_RETURNVERBOSE)
-1/32/a^3/d*(19*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^2+5*B*2 ^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^2+3*C*2^(1/2)*arcsin(cot(d *x+c)-csc(d*x+c))*cos(d*x+c)^2+38*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))* cos(d*x+c)+18*A*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+10 *B*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)-2*B*cos(d*x+c)*sin(d*x +c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+6*C*cos(d*x+c)*2^(1/2)*arcsin(cot(d* x+c)-csc(d*x+c))-14*C*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1 /2)+19*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+26*A*(cos(d*x+c)/(1+cos(d*x +c)))^(1/2)*sin(d*x+c)+5*B*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))-10*B*sin( d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3*C*2^(1/2)*arcsin(cot(d*x+c)-csc (d*x+c))-6*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^(1/2 )*(a*(1+cos(d*x+c)))^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c) )^3
Time = 0.30 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.44 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 19 \, A + 5 \, B + 3 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (9 \, A - B - 7 \, C\right )} \cos \left (d x + c\right ) + 13 \, A - 5 \, B - 3 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c) )^(5/2),x, algorithm="fricas")
1/32*(sqrt(2)*((19*A + 5*B + 3*C)*cos(d*x + c)^3 + 3*(19*A + 5*B + 3*C)*co s(d*x + c)^2 + 3*(19*A + 5*B + 3*C)*cos(d*x + c) + 19*A + 5*B + 3*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*((9*A - B - 7*C)*cos (d*x + c) + 13*A - 5*B - 3*C)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))* sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos (d*x + c) + a^3*d)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\cos \left (d x + c\right )}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c) )^(5/2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(5 /2)*sqrt(cos(d*x + c))), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c) )^(5/2),x, algorithm="giac")
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + a*cos (c + d*x))^(5/2)),x)