Integrand size = 43, antiderivative size = 210 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac {2 \left (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 d}+\frac {2 a^2 \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^4 (a+b) d}+\frac {2 (b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac {2 C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b d} \] Output:
2/5*(5*A*b^2-5*B*a*b+5*C*a^2+3*C*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2) )/b^3/d+2/3*(3*B*a^2*b+B*b^3-3*a^3*C-a*b^2*(3*A+C))*InverseJacobiAM(1/2*d* x+1/2*c,2^(1/2))/b^4/d+2*a^2*(A*b^2-a*(B*b-C*a))*EllipticPi(sin(1/2*d*x+1/ 2*c),2*b/(a+b),2^(1/2))/b^4/(a+b)/d+2/3*(B*b-C*a)*cos(d*x+c)^(1/2)*sin(d*x +c)/b^2/d+2/5*C*cos(d*x+c)^(3/2)*sin(d*x+c)/b/d
Time = 3.95 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.30 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\frac {2 b^2 \left (15 A b^2-5 a b B+5 a^2 C+9 b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+2 b^2 (5 b B+4 a C) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )+4 b^2 \sqrt {\cos (c+d x)} (5 b B-5 a C+3 b C \cos (c+d x)) \sin (c+d x)+\frac {6 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a \sqrt {\sin ^2(c+d x)}}}{30 b^4 d} \] Input:
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]
Output:
((2*b^2*(15*A*b^2 - 5*a*b*B + 5*a^2*C + 9*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + 2*b^2*(5*b*B + 4*a*C)*(2*EllipticF[(c + d*x)/2 , 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)) + 4*b^2*Sq rt[Cos[c + d*x]]*(5*b*B - 5*a*C + 3*b*C*Cos[c + d*x])*Sin[c + d*x] + (6*(5 *A*b^2 - 5*a*b*B + 5*a^2*C + 3*b^2*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (- 2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d *x])/(a*Sqrt[Sin[c + d*x]^2]))/(30*b^4*d)
Time = 1.67 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.09, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 3528, 27, 3042, 3528, 27, 3042, 3538, 27, 3042, 3119, 3481, 3042, 3120, 3284}
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.
\(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \frac {2 \int \frac {\sqrt {\cos (c+d x)} \left (5 (b B-a C) \cos ^2(c+d x)+b (5 A+3 C) \cos (c+d x)+3 a C\right )}{2 (a+b \cos (c+d x))}dx}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {\cos (c+d x)} \left (5 (b B-a C) \cos ^2(c+d x)+b (5 A+3 C) \cos (c+d x)+3 a C\right )}{a+b \cos (c+d x)}dx}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (5 (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (5 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a C\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \frac {\frac {2 \int \frac {3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \cos ^2(c+d x)+b (5 b B+4 a C) \cos (c+d x)+5 a (b B-a C)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \cos ^2(c+d x)+b (5 b B+4 a C) \cos (c+d x)+5 a (b B-a C)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (5 b B+4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a (b B-a C)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {5 \left (a b (b B-a C)+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}+\frac {5 \int \frac {a b (b B-a C)+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {5 \int \frac {a b (b B-a C)+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\frac {5 \int \frac {a b (b B-a C)+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\frac {\frac {5 \left (\frac {3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {\left (-3 a^3 C+3 a^2 b B-a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 \left (\frac {3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {\left (-3 a^3 C+3 a^2 b B-a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {5 \left (\frac {3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-3 a^3 C+3 a^2 b B-a b^2 (3 A+C)+b^3 B\right )}{b d}\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {\frac {\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d}+\frac {5 \left (\frac {6 a^2 \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-3 a^3 C+3 a^2 b B-a b^2 (3 A+C)+b^3 B\right )}{b d}\right )}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\) |
Input:
Int[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Co s[c + d*x]),x]
Output:
(2*C*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*b*d) + (((6*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b^2*C)*EllipticE[(c + d*x)/2, 2])/(b*d) + (5*((2*(3*a^2*b*B + b^3*B - 3*a^3*C - a*b^2*(3*A + C))*EllipticF[(c + d*x)/2, 2])/(b*d) + (6*a ^2*(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*( a + b)*d)))/b)/(3*b) + (10*(b*B - a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3 *b*d))/(5*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
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)*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/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(802\) vs. \(2(203)=406\).
Time = 6.09 (sec) , antiderivative size = 803, normalized size of antiderivative = 3.82
Input:
int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x,me thod=_RETURNVERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*a^2*(A*b^2- B*a*b+C*a^2)/b^3/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d *x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)* EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2*(A*a*b^2+A*b^3-B*a^2*b -B*a*b^2-B*b^3+C*a^3+C*a^2*b+C*a*b^2+C*b^3)/b^4*(sin(1/2*d*x+1/2*c)^2)^(1/ 2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+ 1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-4/5*C/b/(-2*sin(1/2* d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x +1/2*c)^6-14*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^ 2*cos(1/2*d*x+1/2*c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^ 2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*(sin(1/2*d*x+1/2*c)^2)^ (1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2 )))+2/b^3*(A*b^2-B*a*b-2*B*b^2+C*a^2+2*C*a*b+3*C*b^2)*(sin(1/2*d*x+1/2*c)^ 2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/ 2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos (1/2*d*x+1/2*c),2^(1/2)))+4/3/b^2*(B*b-C*a-3*C*b)*(2*cos(1/2*d*x+1/2*c)*si n(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+2*(sin(1/2*d*x+ 1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2 *c),2^(1/2))-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/ 2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(...
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c) ),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+ c)),x)
Output:
Timed out
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c) ),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*co s(d*x + c) + a), x)
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c) ),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*co s(d*x + c) + a), x)
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \] Input:
int((cos(c + d*x)^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*co s(c + d*x)),x)
Output:
int((cos(c + d*x)^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*co s(c + d*x)), x)
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right ) b +a}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}}{\cos \left (d x +c \right ) b +a}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right ) b +a}d x \right ) b \] Input:
int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x)
Output:
int((sqrt(cos(c + d*x))*cos(c + d*x))/(cos(c + d*x)*b + a),x)*a + int((sqr t(cos(c + d*x))*cos(c + d*x)**3)/(cos(c + d*x)*b + a),x)*c + int((sqrt(cos (c + d*x))*cos(c + d*x)**2)/(cos(c + d*x)*b + a),x)*b