Integrand size = 33, antiderivative size = 154 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C (b \cos (c+d x))^{7/3} \sin (c+d x)}{10 b d}-\frac {3 (10 A+7 C) (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{70 b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 b^2 d \sqrt {\sin ^2(c+d x)}} \]
3/10*C*(b*cos(d*x+c))^(7/3)*sin(d*x+c)/b/d-3/70*(10*A+7*C)*(b*cos(d*x+c))^ (7/3)*hypergeom([1/2, 7/6],[13/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(sin(d*x+c) ^2)^(1/2)-3/10*B*(b*cos(d*x+c))^(10/3)*hypergeom([1/2, 5/3],[8/3],cos(d*x+ c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)
Time = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.74 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 (b \cos (c+d x))^{4/3} \cot (c+d x) \left (-7 C \sin ^2(c+d x)+(10 A+7 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+7 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{70 d} \]
(-3*(b*Cos[c + d*x])^(4/3)*Cot[c + d*x]*(-7*C*Sin[c + d*x]^2 + (10*A + 7*C )*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] + 7*B*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]*Sqrt[Si n[c + d*x]^2]))/(70*d)
Time = 0.48 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 3502, 27, 3042, 3227, 3042, 3122}
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.
\(\displaystyle \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {3 \int \frac {1}{3} (b \cos (c+d x))^{4/3} (b (10 A+7 C)+10 b B \cos (c+d x))dx}{10 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (b \cos (c+d x))^{4/3} (b (10 A+7 C)+10 b B \cos (c+d x))dx}{10 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (b (10 A+7 C)+10 b B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{10 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {b (10 A+7 C) \int (b \cos (c+d x))^{4/3}dx+10 B \int (b \cos (c+d x))^{7/3}dx}{10 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b (10 A+7 C) \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}dx+10 B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3}dx}{10 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {-\frac {3 (10 A+7 C) \sin (c+d x) (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}}{10 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d}\) |
(3*C*(b*Cos[c + d*x])^(7/3)*Sin[c + d*x])/(10*b*d) + ((-3*(10*A + 7*C)*(b* Cos[c + d*x])^(7/3)*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[ c + d*x])/(7*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(10/3)*Hyperg eometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*Sqrt[Sin[c + d*x]^2]))/(10*b)
3.4.46.3.1 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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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]
\[\int \left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )d x\]
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \]
integral((C*b*cos(d*x + c)^3 + B*b*cos(d*x + c)^2 + A*b*cos(d*x + c))*(b*c os(d*x + c))^(1/3), x)
Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \]
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \]
Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int {\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]