Integrand size = 35, antiderivative size = 290 \[ \int (a+b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C (a+b \cos (c+d x))^{5/3} \sin (c+d x)}{8 b d}+\frac {(a+b) (8 b B-3 a C) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {5}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{4 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac {\left (8 A b^2-8 a b B+3 a^2 C+5 b^2 C\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{4 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}} \]
3/8*C*(a+b*cos(d*x+c))^(5/3)*sin(d*x+c)/b/d+1/8*(a+b)*(8*B*b-3*C*a)*Appell F1(1/2,-5/3,1/2,3/2,b*(1-cos(d*x+c))/(a+b),1/2-1/2*cos(d*x+c))*(a+b*cos(d* x+c))^(2/3)*sin(d*x+c)/b^2/d/((a+b*cos(d*x+c))/(a+b))^(2/3)*2^(1/2)/(1+cos (d*x+c))^(1/2)+1/8*(8*A*b^2-8*B*a*b+3*C*a^2+5*C*b^2)*AppellF1(1/2,-2/3,1/2 ,3/2,b*(1-cos(d*x+c))/(a+b),1/2-1/2*cos(d*x+c))*(a+b*cos(d*x+c))^(2/3)*sin (d*x+c)/b^2/d/((a+b*cos(d*x+c))/(a+b))^(2/3)*2^(1/2)/(1+cos(d*x+c))^(1/2)
Time = 3.20 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.02 \[ \int (a+b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 (a+b \cos (c+d x))^{2/3} \csc (c+d x) \left (20 \left (-a^2+b^2\right ) (8 b B-3 a C) \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}}+4 \left (40 A b^2+16 a b B-6 a^2 C+25 b^2 C\right ) \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))-20 b^2 (8 b B+2 a C+5 b C \cos (c+d x)) \sin ^2(c+d x)\right )}{800 b^3 d} \]
(-3*(a + b*Cos[c + d*x])^(2/3)*Csc[c + d*x]*(20*(-a^2 + b^2)*(8*b*B - 3*a* C)*AppellF1[2/3, 1/2, 1/2, 5/3, (a + b*Cos[c + d*x])/(a - b), (a + b*Cos[c + d*x])/(a + b)]*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Cos[c + d*x]))/(a - b))] + 4*(40*A*b^2 + 16*a*b*B - 6*a^2*C + 25*b^2*C)*Ap pellF1[5/3, 1/2, 1/2, 8/3, (a + b*Cos[c + d*x])/(a - b), (a + b*Cos[c + d* x])/(a + b)]*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*(a + b*Cos[c + d*x]) - 20*b^2*(8*b*B + 2*a*C + 5*b*C*Cos [c + d*x])*Sin[c + d*x]^2))/(800*b^3*d)
Time = 0.71 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3502, 27, 3042, 3235, 3042, 3144, 156, 155}
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.
| \(\displaystyle \int (a+b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {3 \int \frac {1}{3} (a+b \cos (c+d x))^{2/3} (b (8 A+5 C)+(8 b B-3 a C) \cos (c+d x))dx}{8 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (a+b \cos (c+d x))^{2/3} (b (8 A+5 C)+(8 b B-3 a C) \cos (c+d x))dx}{8 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3} \left (b (8 A+5 C)+(8 b B-3 a C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{8 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 3235 |
| \(\displaystyle \frac {\frac {\left (3 a^2 C-8 a b B+8 A b^2+5 b^2 C\right ) \int (a+b \cos (c+d x))^{2/3}dx}{b}+\frac {(8 b B-3 a C) \int (a+b \cos (c+d x))^{5/3}dx}{b}}{8 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (3 a^2 C-8 a b B+8 A b^2+5 b^2 C\right ) \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{b}+\frac {(8 b B-3 a C) \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}dx}{b}}{8 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 3144 |
| \(\displaystyle \frac {-\frac {\sin (c+d x) \left (3 a^2 C-8 a b B+8 A b^2+5 b^2 C\right ) \int \frac {(a+b \cos (c+d x))^{2/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-\frac {(8 b B-3 a C) \sin (c+d x) \int \frac {(a+b \cos (c+d x))^{5/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}}{8 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 156 |
| \(\displaystyle \frac {-\frac {\sin (c+d x) \left (3 a^2 C-8 a b B+8 A b^2+5 b^2 C\right ) (a+b \cos (c+d x))^{2/3} \int \frac {\left (\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}\right )^{2/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}-\frac {(a+b) (8 b B-3 a C) \sin (c+d x) (a+b \cos (c+d x))^{2/3} \int \frac {\left (\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}\right )^{5/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}}{8 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 155 |
| \(\displaystyle \frac {\frac {\sqrt {2} \sin (c+d x) \left (3 a^2 C-8 a b B+8 A b^2+5 b^2 C\right ) (a+b \cos (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac {\sqrt {2} (a+b) (8 b B-3 a C) \sin (c+d x) (a+b \cos (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {5}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}}{8 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{5/3}}{8 b d}\) |
(3*C*(a + b*Cos[c + d*x])^(5/3)*Sin[c + d*x])/(8*b*d) + ((Sqrt[2]*(a + b)* (8*b*B - 3*a*C)*AppellF1[1/2, 1/2, -5/3, 3/2, (1 - Cos[c + d*x])/2, (b*(1 - Cos[c + d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(b*d*Sq rt[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(2/3)) + (Sqrt[2]*(8*A *b^2 - 8*a*b*B + 3*a^2*C + 5*b^2*C)*AppellF1[1/2, 1/2, -2/3, 3/2, (1 - Cos [c + d*x])/2, (b*(1 - Cos[c + d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(2/3)*S in[c + d*x])/(b*d*Sqrt[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(2 /3)))/(8*b)
3.4.88.3.1 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[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ (b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p ]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]) Int[(a + b*x)^m*(c + d*x)^n*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]]) Subst[Int[(a + b*x )^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0] && !IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)/b Int[(a + b*Sin[e + f*x])^m, x], x] + Simp[d/b Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, 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]
\[\int \left (a +\cos \left (d x +c \right ) b \right )^{\frac {2}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )d x\]
\[ \int (a+b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]
Timed out. \[ \int (a+b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int (a+b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]
\[ \int (a+b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]
Timed out. \[ \int (a+b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int {\left (a+b\,\cos \left (c+d\,x\right )\right )}^{2/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]