Integrand size = 35, antiderivative size = 110 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {\sqrt {a} (3 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 d}+\frac {a A \tan (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sqrt {a+a \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{2 d} \]
1/4*(3*A+8*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*a^(1/2)/d +1/4*a*A*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/2*A*sec(d*x+c)*(a+a*cos(d*x +c))^(1/2)*tan(d*x+c)/d
Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.94 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (\sqrt {2} (3 A+8 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2(c+d x)+A \left (\sin \left (\frac {1}{2} (c+d x)\right )+3 \sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{8 d} \]
(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^2*(Sqrt[2]*(3*A + 8*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^2 + A*(Sin[(c + d*x) /2] + 3*Sin[(3*(c + d*x))/2])))/(8*d)
Time = 0.67 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 3523, 27, 3042, 3459, 3042, 3252, 219}
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.
| \(\displaystyle \int \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int \frac {1}{2} \sqrt {\cos (c+d x) a+a} (a A+a (A+4 C) \cos (c+d x)) \sec ^2(c+d x)dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} (a A+a (A+4 C) \cos (c+d x)) \sec ^2(c+d x)dx}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a A+a (A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \frac {\frac {1}{2} a (3 A+8 C) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a^2 A \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} a (3 A+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 A \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {a^2 A \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (3 A+8 C) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {a^{3/2} (3 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a^2 A \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\) |
(A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((a^(3/2)*( 3*A + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + ( a^2*A*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(4*a)
3.1.80.3.1 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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*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_)]]*((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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a *d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* (c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(734\) vs. \(2(94)=188\).
Time = 9.68 (sec) , antiderivative size = 735, normalized size of antiderivative = 6.68
| method | result | size |
| parts | \(\frac {A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (12 a \left (\ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right )+\ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right )\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-12 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-12 \ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a -12 \ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+3 \ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +3 \ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a \right )}{2 \sqrt {a}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\right )^{2} \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\right )^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {C \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\ln \left (\frac {2 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+4 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right )+\ln \left (-\frac {2 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(735\) |
| default | \(\text {Expression too large to display}\) | \(955\) |
1/2*A*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*a*(ln(4/(2*cos (1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2* d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))+ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^ (1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)- 2*a)))*sin(1/2*d*x+1/2*c)^4+(-12*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^ (1/2)-12*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c) +2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a-12*ln(-4/(2*cos(1/ 2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x +1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)*sin(1/2*d*x+1/2*c)^2+10*2^(1/2)*(a*sin(1 /2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+3*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1 /2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2* a))*a+3*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c) -2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)/a^(1/2)/(2*cos(1/ 2*d*x+1/2*c)-2^(1/2))^2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^2/sin(1/2*d*x+1/2*c )/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+C*a^(1/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2 *d*x+1/2*c)^2)^(1/2)*(ln(2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1 /2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))+ln(-2/( 2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin (1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)))/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+ 1/2*c)^2)^(1/2)/d
Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.57 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {{\left ({\left (3 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, A \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{16 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]
1/16*(((3*A + 8*C)*cos(d*x + c)^3 + (3*A + 8*C)*cos(d*x + c)^2)*sqrt(a)*lo g((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt (a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^ 2)) + 4*(3*A*cos(d*x + c) + 2*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d *cos(d*x + c)^3 + d*cos(d*x + c)^2)
\[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 2643 vs. \(2 (94) = 188\).
Time = 3.13 (sec) , antiderivative size = 2643, normalized size of antiderivative = 24.03 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Too large to display} \]
1/16*(3*(log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt( 2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos( 1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/ 2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*si n(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1 /2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(4*d*x + 4*c)^2 + 12*(log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d* x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/ 2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sq rt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*c os(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c) ^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2) *sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 + 3*(log(2*cos(1/2*d*x + 1/ 2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqr t(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2* d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2* sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - lo...
Time = 0.31 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.35 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=-\frac {\sqrt {2} {\left (\sqrt {2} {\left (3 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 8 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) + \frac {4 \, {\left (6 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}\right )} \sqrt {a}}{16 \, d} \]
-1/16*sqrt(2)*(sqrt(2)*(3*A*sgn(cos(1/2*d*x + 1/2*c)) + 8*C*sgn(cos(1/2*d* x + 1/2*c)))*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))) + 4*(6*A*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 5*A*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c))/(2*sin(1/2* d*x + 1/2*c)^2 - 1)^2)*sqrt(a)/d
Timed out. \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^3} \,d x \]