Integrand size = 43, antiderivative size = 144 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a^{3/2} (3 A+2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}-\frac {a^2 (3 A-6 B-8 C) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}-\frac {a (3 A-2 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{d} \]
a^(3/2)*(3*A+2*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d-1/3 *a^2*(3*A-6*B-8*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-1/3*a*(3*A-2*C)*sin (d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+A*(a+a*cos(d*x+c))^(3/2)*tan(d*x+c)/d
Time = 0.54 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.82 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (3 \sqrt {2} (3 A+2 B) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos (c+d x)+2 (3 A+C+2 (3 B+5 C) \cos (c+d x)+C \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d} \]
Integrate[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 2)*Sec[c + d*x]^2,x]
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]*(3*Sqrt[2]*(3* A + 2*B)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x] + 2*(3*A + C + 2*( 3*B + 5*C)*Cos[c + d*x] + C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(6*d)
Time = 0.98 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {3042, 3522, 27, 3042, 3455, 27, 3042, 3460, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (a (3 A+2 B)-a (3 A-2 C) \cos (c+d x)) \sec (c+d x)dx}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{3/2} (a (3 A+2 B)-a (3 A-2 C) \cos (c+d x)) \sec (c+d x)dx}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (3 A+2 B)-a (3 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {2}{3} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (3 a^2 (3 A+2 B)-a^2 (3 A-6 B-8 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \int \sqrt {\cos (c+d x) a+a} \left (3 a^2 (3 A+2 B)-a^2 (3 A-6 B-8 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a^2 (3 A+2 B)-a^2 (3 A-6 B-8 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 a^2 (3 A+2 B) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-\frac {2 a^3 (3 A-6 B-8 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 a^2 (3 A+2 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^3 (3 A-6 B-8 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {1}{3} \left (-\frac {6 a^3 (3 A+2 B) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {2 a^3 (3 A-6 B-8 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {6 a^{5/2} (3 A+2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 a^3 (3 A-6 B-8 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
((-2*a^2*(3*A - 2*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ((6*a^ (5/2)*(3*A + 2*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]] )/d - (2*a^3*(3*A - 6*B - 8*C)*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])) /3)/(2*a) + (A*(a + a*Cos[c + d*x])^(3/2)*Tan[c + d*x])/d
3.4.86.3.1 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[((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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*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] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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 [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]
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[(-(c^2*C - B*c*d + 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 - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(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, B, 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. \(660\) vs. \(2(128)=256\).
Time = 11.34 (sec) , antiderivative size = 661, normalized size of antiderivative = 4.59
| method | result | size |
| parts | \(\frac {A \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 (-6 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 ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \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 )}{\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\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}+\frac {4 C \,a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {B \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 (\sqrt {2}\, \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 ) a +\sqrt {2}\, \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 ) a +4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\right ) \sqrt {2}}{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}\) | \(661\) |
| default | \(\text {Expression too large to display}\) | \(798\) |
int((a+cos(d*x+c)*a)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, method=_RETURNVERBOSE)
A*a^(1/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-6*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)^2+2*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)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))/(2*cos(1/2 *d*x+1/2*c)+2^(1/2))/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+4 /3*C*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)*(cos(1/2*d*x+1/2*c)^2+2)*2^ (1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+1/2*B*a^(1/2)*cos(1/2*d*x+1/2*c)*(a *sin(1/2*d*x+1/2*c)^2)^(1/2)*(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))*a+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))*a+4*a^(1/ 2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2))/sin(1/2*d*x+1/2*c)*2^(1/2)/(a*cos(1/2*d *x+1/2*c)^2)^(1/2)/d
Time = 0.30 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.33 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {3 \, {\left ({\left (3 \, A + 2 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (3 \, A + 2 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (2 \, C a \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, B + 5 \, C\right )} a \cos \left (d x + c\right ) + 3 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^2,x, algorithm="fricas")
1/12*(3*((3*A + 2*B)*a*cos(d*x + c)^2 + (3*A + 2*B)*a*cos(d*x + c))*sqrt(a )*log((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*(2*C*a*cos(d*x + c)^2 + 2*(3*B + 5*C)*a*cos(d*x + c) + 3*A*a)* sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1354 vs. \(2 (128) = 256\).
Time = 0.44 (sec) , antiderivative size = 1354, normalized size of antiderivative = 9.40 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Too large to display} \]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^2,x, algorithm="maxima")
1/12*(4*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c) )*C*sqrt(a) - 3*(2*sqrt(2)*a*cos(7/2*d*x + 7/2*c)*sin(2*d*x + 2*c) + 6*sqr t(2)*a*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) + (2*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 6*sqrt(2)*a*sin(1/2*d*x + 1/2*c) - 3*a*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) + 3*a*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) - 3*a*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) + 3 *a*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 + (2*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 6*sqrt(2)*a*sin(1/2*d*x + 1/2*c) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*c os(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*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) - 3*a*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) + 3*a*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))*sin(2*d*x + 2*c)^2 - 4*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 4...
Time = 0.66 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.37 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {\sqrt {2} {\left (16 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {12 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + 3 \, \sqrt {2} {\left (3 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )\right )} \sqrt {a}}{12 \, d} \]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^2,x, algorithm="giac")
-1/12*sqrt(2)*(16*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 2 4*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 48*C*a*sgn(cos(1/2* d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 12*A*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) + 3*sqrt(2)*(3*A*a*sgn(cos (1/2*d*x + 1/2*c)) + 2*B*a*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))))*sqrt(a) /d
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^2} \,d x \]