Integrand size = 37, antiderivative size = 218 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {a^{3/2} (112 A+75 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {a^2 (112 A+75 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (16 A+13 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{32 d \sqrt {a+a \cos (c+d x)}}+\frac {a C \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {C \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{4 d} \]
1/64*a^(3/2)*(112*A+75*C)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2) )/d+1/4*C*cos(d*x+c)^(3/2)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+1/32*a^2*(1 6*A+13*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/64*a^2*(1 12*A+75*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)+1/8*a*C*co s(d*x+c)^(3/2)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d
Time = 0.91 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.59 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2} (112 A+75 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (112 A+95 C+(32 A+62 C) \cos (c+d x)+20 C \cos (2 (c+d x))+4 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{128 d} \]
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(Sqrt[2]*(112*A + 75*C)*Arc Sin[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(112*A + 95*C + (32*A + 62*C)*Cos[c + d*x] + 20*C*Cos[2*(c + d*x)] + 4*C*Cos[3*(c + d*x)])*Sin[ (c + d*x)/2]))/(128*d)
Time = 1.21 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 3525, 27, 3042, 3455, 27, 3042, 3460, 3042, 3249, 3042, 3253, 223}
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.
| \(\displaystyle \int \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3525 |
| \(\displaystyle \frac {\int \frac {1}{2} \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2} (a (8 A+3 C)+3 a C \cos (c+d x))dx}{4 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2} (a (8 A+3 C)+3 a C \cos (c+d x))dx}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (8 A+3 C)+3 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {3}{2} \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a} \left ((16 A+9 C) a^2+(16 A+13 C) \cos (c+d x) a^2\right )dx+\frac {a^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a} \left ((16 A+9 C) a^2+(16 A+13 C) \cos (c+d x) a^2\right )dx+\frac {a^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((16 A+9 C) a^2+(16 A+13 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {a^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (112 A+75 C) \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}dx+\frac {a^3 (16 A+13 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (112 A+75 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^3 (16 A+13 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3249 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (112 A+75 C) \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (16 A+13 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (112 A+75 C) \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (16 A+13 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (112 A+75 C) \left (\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (16 A+13 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {\frac {a^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{d}+\frac {1}{2} \left (\frac {a^3 (16 A+13 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}+\frac {1}{4} a^2 (112 A+75 C) \left (\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )\right )}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
(C*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(4*d) + ((a ^2*C*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/d + ((a^3*( 16*A + 13*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x] ]) + (a^2*(112*A + 75*C)*((Sqrt[a]*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Co s[c + d*x]])))/4)/2)/(8*a)
3.2.80.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[2*n*((b*c + a*d)/(b*( 2*n + 1))) 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}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[-2/f Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E qQ[a^2 - b^2, 0] && EqQ[d, a/b]
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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1 )/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2)) Int[(a + b*Sin[e + f* x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1 )) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
Time = 28.99 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.44
| method | result | size |
| default | \(\frac {a \left (16 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+40 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+32 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+50 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+112 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+75 C \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+112 A \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+75 C \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right )}{64 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) | \(313\) |
| parts | \(\frac {A \left (2 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+7 \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+7 \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) a}{4 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}+\frac {C \left (16 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+40 \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+50 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+75 \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+75 \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) a}{64 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) | \(364\) |
1/64*a/d*(16*C*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+4 0*C*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+32*A*cos(d*x +c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+50*C*cos(d*x+c)*sin(d*x+c )*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+112*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2 )*sin(d*x+c)+75*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+112*A*arcta n(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+75*C*arctan(tan(d*x+c)*(co s(d*x+c)/(1+cos(d*x+c)))^(1/2)))*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^(1/2) /(1+cos(d*x+c))/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
Time = 0.37 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.71 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {{\left (16 \, C a \cos \left (d x + c\right )^{3} + 40 \, C a \cos \left (d x + c\right )^{2} + 2 \, {\left (16 \, A + 25 \, C\right )} a \cos \left (d x + c\right ) + {\left (112 \, A + 75 \, C\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left ({\left (112 \, A + 75 \, C\right )} a \cos \left (d x + c\right ) + {\left (112 \, A + 75 \, C\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{64 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
1/64*((16*C*a*cos(d*x + c)^3 + 40*C*a*cos(d*x + c)^2 + 2*(16*A + 25*C)*a*c os(d*x + c) + (112*A + 75*C)*a)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c) )*sin(d*x + c) - ((112*A + 75*C)*a*cos(d*x + c) + (112*A + 75*C)*a)*sqrt(a )*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c) )))/(d*cos(d*x + c) + d)
Timed out. \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 8041 vs. \(2 (186) = 372\).
Time = 0.93 (sec) , antiderivative size = 8041, normalized size of antiderivative = 36.89 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
1/256*(16*(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((a*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2 *d*x + 2*c) + a*sin(2*d*x + 2*c) - (a*cos(2*d*x + 2*c) - 6*a)*sin(1/2*arct an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d*x + 2*c) , cos(2*d*x + 2*c) + 1)) + (a*sin(2*d*x + 2*c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - a*cos(2*d*x + 2*c) + (a*cos(2*d*x + 2*c) - 6*a )*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 6*a)*sin(1/2*arct an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 7*(a*arctan2((cos( 2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1 /2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d *x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (co s(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos (1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin( 2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2 *d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) + 1) - a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2* c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/ 2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2 *d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), c...
\[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\cos \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \sqrt {\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]