Integrand size = 35, antiderivative size = 227 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {a^{3/2} (88 A+75 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {a^2 (88 A+75 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (88 A+75 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (8 A+9 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a B \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d} \] Output:
1/64*a^(3/2)*(88*A+75*B)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2)) /d+1/64*a^2*(88*A+75*B)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/ 2)+1/96*a^2*(88*A+75*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/ 2)+1/24*a^2*(8*A+9*B)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2) +1/4*a*B*cos(d*x+c)^(5/2)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d
Time = 1.30 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.60 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 \sqrt {2} (88 A+75 B) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (296 A+285 B+2 (88 A+93 B) \cos (c+d x)+4 (8 A+15 B) \cos (2 (c+d x))+12 B \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{384 d} \] Input:
Integrate[Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x ]),x]
Output:
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sqrt[2]*(88*A + 75*B)*Ar cSin[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(296*A + 285*B + 2*( 88*A + 93*B)*Cos[c + d*x] + 4*(8*A + 15*B)*Cos[2*(c + d*x)] + 12*B*Cos[3*( c + d*x)])*Sin[(c + d*x)/2]))/(384*d)
Time = 1.07 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 3455, 27, 3042, 3460, 3042, 3249, 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.
| \(\displaystyle \int \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2} (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {1}{4} \int \frac {1}{2} \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a} (a (8 A+5 B)+a (8 A+9 B) \cos (c+d x))dx+\frac {a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a} (a (8 A+5 B)+a (8 A+9 B) \cos (c+d x))dx+\frac {a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (8 A+5 B)+a (8 A+9 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}dx+\frac {a^2 (8 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 (8 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3249 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3249 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \left (\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{8} \left (\frac {a^2 (8 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \left (\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\right )+\frac {a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\) |
Input:
Int[Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x]),x]
Output:
(a*B*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(4*d) + ((a ^2*(8*A + 9*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d* x]]) + (a*(88*A + 75*B)*((a*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*((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*Cos[ c + d*x]])))/4))/6)/8
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]
Time = 16.38 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00
| method | result | size |
| default | \(\frac {\left (264 A \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+225 B \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\left (64 \cos \left (d x +c \right )^{2}+176 \cos \left (d x +c \right )+264\right ) \sin \left (d x +c \right ) A \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (48 \cos \left (d x +c \right )^{3}+120 \cos \left (d x +c \right )^{2}+150 \cos \left (d x +c \right )+225\right ) \sin \left (d x +c \right ) B \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \sqrt {2}}{192 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\) | \(227\) |
| parts | \(\frac {A \left (33 \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\left (8 \cos \left (d x +c \right )^{2}+22 \cos \left (d x +c \right )+33\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \sqrt {2}}{24 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}+\frac {B \left (75 \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\left (16 \cos \left (d x +c \right )^{3}+40 \cos \left (d x +c \right )^{2}+50 \cos \left (d x +c \right )+75\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \sqrt {2}}{64 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\) | \(288\) |
Input:
int(cos(d*x+c)^(3/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)),x,method=_RET URNVERBOSE)
Output:
1/192/d*(264*A*arctan(tan(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+225*B* arctan(tan(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+(64*cos(d*x+c)^2+176* cos(d*x+c)+264)*sin(d*x+c)*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+(48*cos(d*x +c)^3+120*cos(d*x+c)^2+150*cos(d*x+c)+225)*sin(d*x+c)*B*(cos(d*x+c)/(cos(d *x+c)+1))^(1/2))*cos(d*x+c)^(1/2)*(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/(cos(d*x+ c)+1)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*2^(1/2)
Time = 0.16 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.80 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {{\left (48 \, B a \cos \left (d x + c\right )^{3} + 8 \, {\left (8 \, A + 15 \, B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (88 \, A + 75 \, B\right )} a \cos \left (d x + c\right ) + 3 \, {\left (88 \, A + 75 \, B\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (88 \, A + 75 \, B\right )} a \cos \left (d x + c\right ) + {\left (88 \, A + 75 \, B\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:
integrate(cos(d*x+c)^(3/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)),x, algo rithm="fricas")
Output:
1/192*((48*B*a*cos(d*x + c)^3 + 8*(8*A + 15*B)*a*cos(d*x + c)^2 + 2*(88*A + 75*B)*a*cos(d*x + c) + 3*(88*A + 75*B)*a)*sqrt(a*cos(d*x + c) + a)*sqrt( cos(d*x + c))*sin(d*x + c) + 3*((88*A + 75*B)*a*cos(d*x + c) + (88*A + 75* B)*a)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*s in(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))))/(d*cos(d*x + c) + d)
Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(3/2)*(a+a*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c)),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 8904 vs. \(2 (195) = 390\).
Time = 0.70 (sec) , antiderivative size = 8904, normalized size of antiderivative = 39.22 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^(3/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)),x, algo rithm="maxima")
Output:
1/768*(8*(4*(a*cos(3/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1))*sin( 3*d*x + 3*c) - (a*cos(3*d*x + 3*c) - a)*sin(3/2*arctan2(sin(2/3*arctan2(si n(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos( 3*d*x + 3*c))) + 1)))*(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)) )^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*a rctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(3/4)*sqrt(a) + 6*(cos(2/ 3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d *x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos( 3*d*x + 3*c))) + 1)^(1/4)*((3*a*sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d* x + 3*c))) + 11*a*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))*co s(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/ 3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)) - (3*a*cos(2/3*arctan 2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 5*a*cos(1/3*arctan2(sin(3*d*x + 3 *c), cos(3*d*x + 3*c))) - 8*a)*sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3 *c))) + 1)))*sqrt(a) + 33*(a*arctan2(-(cos(2/3*arctan2(sin(3*d*x + 3*c), c os(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) ^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)*(co s(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos...
\[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(3/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)),x, algo rithm="giac")
Output:
integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(3/2)*cos(d*x + c)^(3/ 2), x)
Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^{3/2}\,\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:
int(cos(c + d*x)^(3/2)*(A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(3/2),x)
Output:
int(cos(c + d*x)^(3/2)*(A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(3/2), x)
\[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) b \right ) \] Input:
int(cos(d*x+c)^(3/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)),x)
Output:
sqrt(a)*a*(int(sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x),x)*a + int(sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**3,x)*b + in t(sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*a + int(sqr t(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*b)