Integrand size = 41, antiderivative size = 280 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (6 a^2 b B-3 b^3 B-a b^2 (3 A-5 C)-8 a^3 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^3 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (3 A b^2-6 a b B+8 a^2 C+b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 a \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^2 d} \]
2*a*(A*b^2-a*(B*b-C*a))*sin(d*x+c)/b^2/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)+ 2/3*C*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^2/d+2/3*(6*B*a^2*b-3*B*b^3-a*b^2 *(3*A-5*C)-8*a^3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt icE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^3 /(a^2-b^2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2/3*(3*A*b^2-6*B*a*b+8*C*a^2+C *b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d* x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/b^3/d/(a+ b*cos(d*x+c))^(1/2)
Time = 2.76 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.84 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (-\frac {\sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 \left (3 A b^2-3 a b B+2 a^2 C+b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+\left (-6 a^2 b B+3 b^3 B+a b^2 (3 A-5 C)+8 a^3 C\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )\right )}{(a-b) (a+b)}+b \left (\frac {a \left (-3 A b^2+3 a b B-4 a^2 C+b^2 C\right )}{-a^2+b^2}+b C \cos (c+d x)\right ) \sin (c+d x)\right )}{3 b^3 d \sqrt {a+b \cos (c+d x)}} \]
Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Co s[c + d*x])^(3/2),x]
(2*(-((Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(3*A*b^2 - 3*a*b*B + 2*a^2* C + b^2*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + (-6*a^2*b*B + 3*b^3*B + a*b^2*(3*A - 5*C) + 8*a^3*C)*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b )] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])))/((a - b)*(a + b))) + b*((a *(-3*A*b^2 + 3*a*b*B - 4*a^2*C + b^2*C))/(-a^2 + b^2) + b*C*Cos[c + d*x])* Sin[c + d*x]))/(3*b^3*d*Sqrt[a + b*Cos[c + d*x]])
Time = 1.53 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 3510, 27, 3042, 3502, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}
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.
| \(\displaystyle \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {2 \int -\frac {-b \left (a^2-b^2\right ) C \cos ^2(c+d x)-\left (-2 C a^3+2 b B a^2-b^2 (A-C) a-b^3 B\right ) \cos (c+d x)+b \left (A b^2-a (b B-a C)\right )}{2 \sqrt {a+b \cos (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}+\frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {-b \left (a^2-b^2\right ) C \cos ^2(c+d x)-\left (-2 C a^3+2 b B a^2-b^2 (A-C) a-b^3 B\right ) \cos (c+d x)+b \left (A b^2-a (b B-a C)\right )}{\sqrt {a+b \cos (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {-b \left (a^2-b^2\right ) C \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (2 C a^3-2 b B a^2+b^2 (A-C) a+b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+b \left (A b^2-a (b B-a C)\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \int \frac {b^2 \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right )-b \left (-8 C a^3+6 b B a^2-b^2 (3 A-5 C) a-3 b^3 B\right ) \cos (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{3 b}-\frac {2 C \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\int \frac {b^2 \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right )-b \left (-8 C a^3+6 b B a^2-b^2 (3 A-5 C) a-3 b^3 B\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{3 b}-\frac {2 C \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\int \frac {b^2 \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right )-b \left (-8 C a^3+6 b B a^2-b^2 (3 A-5 C) a-3 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}-\frac {2 C \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {-\left (\left (a^2-b^2\right ) \left (8 a^2 C-6 a b B+3 A b^2+b^2 C\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx\right )-\left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \int \sqrt {a+b \cos (c+d x)}dx}{3 b}-\frac {2 C \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {-\left (\left (a^2-b^2\right ) \left (8 a^2 C-6 a b B+3 A b^2+b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}-\frac {2 C \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {-\left (a^2-b^2\right ) \left (8 a^2 C-6 a b B+3 A b^2+b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {\left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}-\frac {2 C \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {-\left (\left (a^2-b^2\right ) \left (8 a^2 C-6 a b B+3 A b^2+b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {\left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}-\frac {2 C \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {-\left (a^2-b^2\right ) \left (8 a^2 C-6 a b B+3 A b^2+b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}-\frac {2 C \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {-\frac {\left (a^2-b^2\right ) \left (8 a^2 C-6 a b B+3 A b^2+b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}-\frac {2 C \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {-\frac {\left (a^2-b^2\right ) \left (8 a^2 C-6 a b B+3 A b^2+b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}-\frac {2 C \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {2 a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {-\frac {2 \left (a^2-b^2\right ) \left (8 a^2 C-6 a b B+3 A b^2+b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}-\frac {2 C \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
(2*a*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(b^2*(a^2 - b^2)*d*Sqrt[a + b*C os[c + d*x]]) - (((-2*(6*a^2*b*B - 3*b^3*B - a*b^2*(3*A - 5*C) - 8*a^3*C)* Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(3*A*b^2 - 6*a*b*B + 8*a^2*C + b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/ (a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]))/(3*b) - (2*(a^2 - b^2)*C*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d))/(b^2*(a^2 - b^2))
3.11.48.3.1 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[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b Int[1/Sqrt[a + b*Sin[e + f*x ]], x], x] + Simp[d/b Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b , c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(1057\) vs. \(2(320)=640\).
Time = 8.45 (sec) , antiderivative size = 1058, normalized size of antiderivative = 3.78
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1058\) |
| parts | \(\text {Expression too large to display}\) | \(1648\) |
int(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x,me thod=_RETURNVERBOSE)
-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a*(A*b^2 -B*a*b+C*a^2)/b^3/sin(1/2*d*x+1/2*c)^2/(2*b*sin(1/2*d*x+1/2*c)^2-a-b)/(a^2 -b^2)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b*si n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/( a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c), (-2*b/(a-b))^(1/2))*a-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x +1/2*c)^2+(a+b)/(a-b))^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^( 1/2)))+2/3/b^3/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2 )*(4*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+3*A*b^2*(sin(1/2*d*x+1/ 2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*Elliptic F(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-6*B*a*b*(sin(1/2*d*x+1/2*c)^2)^(1 /2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2* d*x+1/2*c),(-2*b/(a-b))^(1/2))+3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b )*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2 *b/(a-b))^(1/2))*a*b-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2* d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^ (1/2))*b^2-2*C*b*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*a-2*C*b^2*sin(1/2 *d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+8*a^2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2* b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2* c),(-2*b/(a-b))^(1/2))+C*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*s...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.17 (sec) , antiderivative size = 858, normalized size of antiderivative = 3.06 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2 ),x, algorithm="fricas")
1/9*(6*(4*C*a^3*b^2 - 3*B*a^2*b^3 + (3*A - C)*a*b^4 + (C*a^2*b^3 - C*b^5)* cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*sin(d*x + c) - (sqrt(2)*(16*I*C*a^4 *b - 12*I*B*a^3*b^2 + 2*I*(3*A - 8*C)*a^2*b^3 + 15*I*B*a*b^4 - 3*I*(3*A + C)*b^5)*cos(d*x + c) + sqrt(2)*(16*I*C*a^5 - 12*I*B*a^4*b + 2*I*(3*A - 8*C )*a^3*b^2 + 15*I*B*a^2*b^3 - 3*I*(3*A + C)*a*b^4))*sqrt(b)*weierstrassPInv erse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d* x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) - (sqrt(2)*(-16*I*C*a^4*b + 12*I*B*a ^3*b^2 - 2*I*(3*A - 8*C)*a^2*b^3 - 15*I*B*a*b^4 + 3*I*(3*A + C)*b^5)*cos(d *x + c) + sqrt(2)*(-16*I*C*a^5 + 12*I*B*a^4*b - 2*I*(3*A - 8*C)*a^3*b^2 - 15*I*B*a^2*b^3 + 3*I*(3*A + C)*a*b^4))*sqrt(b)*weierstrassPInverse(4/3*(4* a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I *b*sin(d*x + c) + 2*a)/b) + 3*(sqrt(2)*(-8*I*C*a^3*b^2 + 6*I*B*a^2*b^3 - I *(3*A - 5*C)*a*b^4 - 3*I*B*b^5)*cos(d*x + c) + sqrt(2)*(-8*I*C*a^4*b + 6*I *B*a^3*b^2 - I*(3*A - 5*C)*a^2*b^3 - 3*I*B*a*b^4))*sqrt(b)*weierstrassZeta (4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse (4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b)) + 3*(sqrt(2)*(8*I*C*a^3*b^2 - 6*I*B*a^2 *b^3 + I*(3*A - 5*C)*a*b^4 + 3*I*B*b^5)*cos(d*x + c) + sqrt(2)*(8*I*C*a^4* b - 6*I*B*a^3*b^2 + I*(3*A - 5*C)*a^2*b^3 + 3*I*B*a*b^4))*sqrt(b)*weierstr assZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstra...
Timed out. \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2 ),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)/(b*cos(d*x + c) + a)^(3/2), x)
\[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2 ),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)/(b*cos(d*x + c) + a)^(3/2), x)
Timed out. \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]