Integrand size = 33, antiderivative size = 144 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {a^{5/2} (5 A+2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {a^3 (3 A+14 B) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}-\frac {a^2 (3 A-2 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{d} \]
a^(5/2)*(5*A+2*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+1/3 *a^3*(3*A+14*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-1/3*a^2*(3*A-2*B)*sin( d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+a*A*(a+a*cos(d*x+c))^(3/2)*tan(d*x+c)/d
Time = 0.35 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.83 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (3 \sqrt {2} (5 A+2 B) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos (c+d x)+2 (3 A+B+2 (3 A+8 B) \cos (c+d x)+B \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d} \]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]*(3*Sqrt[2]*( 5*A + 2*B)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x] + 2*(3*A + B + 2 *(3*A + 8*B)*Cos[c + d*x] + B*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(6*d)
Time = 0.92 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3454, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2} (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (a (5 A+2 B)-a (3 A-2 B) \cos (c+d x)) \sec (c+d x)dx+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int (\cos (c+d x) a+a)^{3/2} (a (5 A+2 B)-a (3 A-2 B) \cos (c+d x)) \sec (c+d x)dx+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (5 A+2 B)-a (3 A-2 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {1}{2} \left (\frac {2}{3} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (3 (5 A+2 B) a^2+(3 A+14 B) \cos (c+d x) a^2\right ) \sec (c+d x)dx-\frac {2 a^2 (3 A-2 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \sqrt {\cos (c+d x) a+a} \left (3 (5 A+2 B) a^2+(3 A+14 B) \cos (c+d x) a^2\right ) \sec (c+d x)dx-\frac {2 a^2 (3 A-2 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (5 A+2 B) a^2+(3 A+14 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^2 (3 A-2 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (3 a^2 (5 A+2 B) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {2 a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^2 (3 A-2 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (3 a^2 (5 A+2 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^2 (3 A-2 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {2 a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {6 a^3 (5 A+2 B) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )-\frac {2 a^2 (3 A-2 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {6 a^{5/2} (5 A+2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^2 (3 A-2 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\) |
((-2*a^2*(3*A - 2*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ((6*a^ (5/2)*(5*A + 2*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]] )/d + (2*a^3*(3*A + 14*B)*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
3.1.95.3.1 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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], 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] && GtQ[m, 1/2] && LtQ[n, -1] && 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)*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]
Leaf count of result is larger than twice the leaf count of optimal. \(682\) vs. \(2(128)=256\).
Time = 15.19 (sec) , antiderivative size = 683, normalized size of antiderivative = 4.74
| method | result | size |
| parts | \(\frac {A \,a^{\frac {3}{2}} \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 (-8 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \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 ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -10 \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 ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +6 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+5 \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 +5 \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 \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 {B \,a^{\frac {3}{2}} \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 (-8 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \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 +3 \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 +36 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\right ) \sqrt {2}}{6 \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}\) | \(683\) |
| default | \(\text {Expression too large to display}\) | \(764\) |
A*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-8*2^(1/2)*(a *sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^2-10*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*a-10*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*a+6*2^(1/2)*(a*sin(1/2* d*x+1/2*c)^2)^(1/2)*a^(1/2)+5*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+5*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+1/6*B*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c) ^2)^(1/2)*(-8*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^2+ 3*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+3*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+36*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 = 0.39 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.40 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {3 \, {\left ({\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \]
1/12*(3*((5*A + 2*B)*a^2*cos(d*x + c)^2 + (5*A + 2*B)*a^2*cos(d*x + c))*sq rt(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*B*a^2*cos(d*x + c)^2 + 2*(3*A + 8*B)*a^2*cos(d*x + c) + 3*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos (d*x + c))
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 8114 vs. \(2 (128) = 256\).
Time = 0.65 (sec) , antiderivative size = 8114, normalized size of antiderivative = 56.35 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\text {Too large to display} \]
-1/252*(1449*sqrt(2)*a^2*cos(5/2*d*x + 5/2*c)^3*sin(2*d*x + 2*c) - 1260*sq rt(2)*a^2*sin(1/2*d*x + 1/2*c)^3 - 1449*(sqrt(2)*a^2*cos(2*d*x + 2*c) + sq rt(2)*a^2)*sin(5/2*d*x + 5/2*c)^3 + 21*(25*sqrt(2)*a^2*cos(2*d*x + 2*c)^2* sin(3/2*d*x + 3/2*c) + 25*sqrt(2)*a^2*sin(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2 *c) - 60*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) + 5*(5*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) - 12*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*cos(2*d*x + 2*c) + (25*sqrt (2)*a^2*cos(3/2*d*x + 3/2*c) + 198*sqrt(2)*a^2*cos(1/2*d*x + 1/2*c))*sin(2 *d*x + 2*c))*cos(5/2*d*x + 5/2*c)^2 - 21*(12*sqrt(2)*a^2*sin(1/2*d*x + 1/2 *c) - 25*(sqrt(2)*a^2*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*a^2*sin(1/2*d*x + 1 /2*c)^2)*sin(3/2*d*x + 3/2*c))*cos(2*d*x + 2*c)^2 + 21*(25*sqrt(2)*a^2*cos (2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) + 25*sqrt(2)*a^2*sin(2*d*x + 2*c)^2*s in(3/2*d*x + 3/2*c) + 69*sqrt(2)*a^2*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) - 198*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) + (25*sqrt(2)*a^2*sin(3/2*d*x + 3/ 2*c) - 198*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*cos(2*d*x + 2*c) + 5*(5*sqrt( 2)*a^2*cos(3/2*d*x + 3/2*c) + 12*sqrt(2)*a^2*cos(1/2*d*x + 1/2*c))*sin(2*d *x + 2*c))*sin(5/2*d*x + 5/2*c)^2 - 21*(12*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c ) - 25*(sqrt(2)*a^2*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*a^2*sin(1/2*d*x + 1/2 *c)^2)*sin(3/2*d*x + 3/2*c))*sin(2*d*x + 2*c)^2 - 35*(sqrt(2)*a^2*cos(5/2* d*x + 5/2*c)^2*sin(2*d*x + 2*c) + 2*sqrt(2)*a^2*cos(5/2*d*x + 5/2*c)*cos(1 /2*d*x + 1/2*c)*sin(2*d*x + 2*c) + sqrt(2)*a^2*sin(5/2*d*x + 5/2*c)^2*s...
Time = 0.66 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.45 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=-\frac {\sqrt {2} {\left (16 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 72 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {12 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + 3 \, \sqrt {2} {\left (5 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )\right )} \sqrt {a}}{12 \, d} \]
-1/12*sqrt(2)*(16*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 24*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 72*B*a^2*sgn(co s(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 12*A*a^2*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)*(5*A*a ^2*sgn(cos(1/2*d*x + 1/2*c)) + 2*B*a^2*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. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \]