Integrand size = 45, antiderivative size = 201 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {a^{3/2} (8 A+12 B+7 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}-\frac {a^2 (8 A-4 B-5 C) \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a (4 A-C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \] Output:
1/4*a^(3/2)*(8*A+12*B+7*C)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2 ))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d-1/4*a^2*(8*A-4*B-5*C)*sin(d*x+c)/d/ (a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)-1/2*a*(4*A-C)*(a+a*cos(d*x+c))^(1/ 2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(1/ 2)*sin(d*x+c)/d
Time = 0.64 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.63 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (\sqrt {2} (8 A+12 B+7 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 (8 A+C+(4 B+7 C) \cos (c+d x)+C \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \] Input:
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]^(3/2),x]
Output:
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(Sqrt[2] *(8*A + 12*B + 7*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 2*(8*A + C + (4*B + 7*C)*Cos[c + d*x] + C*Cos[2*(c + d*x)])*Sin[(c + d*x)/ 2]))/(8*d)
Time = 1.32 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.04, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 4709, 3042, 3522, 27, 3042, 3455, 27, 3042, 3460, 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.
| \(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^{3/2} (a \cos (c+d x)+a)^{3/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^{3/2} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^{3/2} (a (3 A+B)-a (4 A-C) \cos (c+d x))}{2 \sqrt {\cos (c+d x)}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^{3/2} (a (3 A+B)-a (4 A-C) \cos (c+d x))}{\sqrt {\cos (c+d x)}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (3 A+B)-a (4 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^2 (8 A+4 B+C)-a^2 (8 A-4 B-5 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {a^2 (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^2 (8 A+4 B+C)-a^2 (8 A-4 B-5 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {a^2 (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a^2 (8 A+4 B+C)-a^2 (8 A-4 B-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^2 (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {1}{2} a^2 (8 A+12 B+7 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx-\frac {a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {a^2 (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {1}{2} a^2 (8 A+12 B+7 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {a^2 (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (-\frac {a^2 (8 A+12 B+7 C) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {a^2 (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {a^{5/2} (8 A+12 B+7 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {a^2 (4 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
Input:
Int[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec [c + d*x]^(3/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^(3/2)*Sin [c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (-1/2*(a^2*(4*A - C)*Sqrt[Cos[c + d*x] ]*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/d + ((a^(5/2)*(8*A + 12*B + 7*C)* ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (a^3*(8*A - 4 *B - 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/4 )/a)
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_)]]/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_.) + (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])
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Time = 7.41 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.29
| method | result | size |
| default | \(\frac {a \cos \left (d x +c \right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (2 \cos \left (d x +c \right )+7\right ) C +8 A \sin \left (d x +c \right )+2 B \sin \left (2 d x +2 c \right )+8 \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) A +12 \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) B +7 \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) C \right ) \sec \left (d x +c \right )^{\frac {3}{2}}}{4 d \left (1+\cos \left (d x +c \right )\right )}\) | \(260\) |
| parts | \(\frac {2 A a \cos \left (d x +c \right ) \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right ) \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\sin \left (d x +c \right )\right )}{d \left (1+\cos \left (d x +c \right )\right )}+\frac {B a \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \left (3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}\right )}{d \left (1+\cos \left (d x +c \right )\right )}+\frac {C a \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \left (2 \cos \left (d x +c \right )+7\right )+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \left (7 \cos \left (d x +c \right )^{2}+7 \cos \left (d x +c \right )\right )\right )}{4 d \left (1+\cos \left (d x +c \right )\right )}\) | \(354\) |
Input:
int((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2 ),x,method=_RETURNVERBOSE)
Output:
1/4/d*a*cos(d*x+c)*((1+cos(d*x+c))*a)^(1/2)*(sin(d*x+c)*cos(d*x+c)*(2*cos( d*x+c)+7)*C+8*A*sin(d*x+c)+2*B*sin(2*d*x+2*c)+8*(1+cos(d*x+c))*(cos(d*x+c) /(1+cos(d*x+c)))^(1/2)*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c) )*A+12*(1+cos(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan((cos(d*x+c) /(1+cos(d*x+c)))^(1/2)*tan(d*x+c))*B+7*(1+cos(d*x+c))*(cos(d*x+c)/(1+cos(d *x+c)))^(1/2)*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))*C)*sec( d*x+c)^(3/2)/(1+cos(d*x+c))
Time = 0.23 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.71 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {{\left ({\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right ) + {\left (8 \, A + 12 \, B + 7 \, C\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (2 \, C a \cos \left (d x + c\right )^{2} + {\left (4 \, B + 7 \, C\right )} a \cos \left (d x + c\right ) + 8 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:
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 )^(3/2),x, algorithm="fricas")
Output:
-1/4*(((8*A + 12*B + 7*C)*a*cos(d*x + c) + (8*A + 12*B + 7*C)*a)*sqrt(a)*a rctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (2*C*a*cos(d*x + c)^2 + (4*B + 7*C)*a*cos(d*x + c) + 8*A*a)*sqrt(a*cos(d *x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
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)**(3/2),x)
Output:
Timed out
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
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 )^(3/2),x, algorithm="maxima")
Output:
Timed out
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
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 )^(3/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \] Input:
int((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)
Output:
int((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)
\[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )d x \right ) a \right ) \] Input:
int((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2 ),x)
Output:
sqrt(a)*a*(int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec( c + d*x),x)*a + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x) *sec(c + d*x),x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x),x)*c + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1) *cos(c + d*x)**2*sec(c + d*x),x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d *x) + 1)*cos(c + d*x)**2*sec(c + d*x),x)*c + int(sqrt(sec(c + d*x))*sqrt(c os(c + d*x) + 1)*sec(c + d*x),x)*a)