Integrand size = 39, antiderivative size = 154 \[ \int \cos (c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C (b \cos (c+d x))^{10/3} \sin (c+d x)}{13 b^2 d}-\frac {3 (13 A+10 C) (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{130 b^2 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{13/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {13}{6},\frac {19}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{13 b^3 d \sqrt {\sin ^2(c+d x)}} \] Output:
3/13*C*(b*cos(d*x+c))^(10/3)*sin(d*x+c)/b^2/d-3/130*(13*A+10*C)*(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)-3/13*B*(b*cos(d*x+c))^(13/3)*hypergeom([1/2, 13/6],[19/6] ,cos(d*x+c)^2)*sin(d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/2)
Time = 0.41 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.76 \[ \int \cos (c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 (b \cos (c+d x))^{7/3} \cot (c+d x) \left (-10 C \sin ^2(c+d x)+(13 A+10 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+10 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {13}{6},\frac {19}{6},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{130 b d} \] Input:
Integrate[Cos[c + d*x]*(b*Cos[c + d*x])^(4/3)*(A + B*Cos[c + d*x] + C*Cos[ c + d*x]^2),x]
Output:
(-3*(b*Cos[c + d*x])^(7/3)*Cot[c + d*x]*(-10*C*Sin[c + d*x]^2 + (13*A + 10 *C)*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] + 10*B*Cos[c + d*x]*Hypergeometric2F1[1/2, 13/6, 19/6, Cos[c + d*x]^2]*Sqr t[Sin[c + d*x]^2]))/(130*b*d)
Time = 0.54 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {2030, 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.
\(\displaystyle \int \cos (c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \cos (c+d x))^{7/3} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{b}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {3 \int \frac {1}{3} (b \cos (c+d x))^{7/3} (b (13 A+10 C)+13 b B \cos (c+d x))dx}{13 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{10/3}}{13 b d}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int (b \cos (c+d x))^{7/3} (b (13 A+10 C)+13 b B \cos (c+d x))dx}{13 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{10/3}}{13 b d}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (b (13 A+10 C)+13 b B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{13 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{10/3}}{13 b d}}{b}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {b (13 A+10 C) \int (b \cos (c+d x))^{7/3}dx+13 B \int (b \cos (c+d x))^{10/3}dx}{13 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{10/3}}{13 b d}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {b (13 A+10 C) \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3}dx+13 B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3}dx}{13 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{10/3}}{13 b d}}{b}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\frac {-\frac {3 (13 A+10 C) \sin (c+d x) (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{13/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {13}{6},\frac {19}{6},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}}{13 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{10/3}}{13 b d}}{b}\) |
Input:
Int[Cos[c + d*x]*(b*Cos[c + d*x])^(4/3)*(A + B*Cos[c + d*x] + C*Cos[c + d* x]^2),x]
Output:
((3*C*(b*Cos[c + d*x])^(10/3)*Sin[c + d*x])/(13*b*d) + ((-3*(13*A + 10*C)* (b*Cos[c + d*x])^(10/3)*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]*S in[c + d*x])/(10*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(13/3)*Hy pergeometric2F1[1/2, 13/6, 19/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*Sqrt[S in[c + d*x]^2]))/(13*b))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
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]
\[\int \cos \left (d x +c \right ) \left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \cos \left (d x +c \right )^{2}\right )d x\]
Input:
int(cos(d*x+c)*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
Output:
int(cos(d*x+c)*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
\[ \int \cos (c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right ) \,d x } \] Input:
integrate(cos(d*x+c)*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2), x, algorithm="fricas")
Output:
integral((C*b*cos(d*x + c)^4 + B*b*cos(d*x + c)^3 + A*b*cos(d*x + c)^2)*(b *cos(d*x + c))^(1/3), x)
Timed out. \[ \int \cos (c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)*(b*cos(d*x+c))**(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2 ),x)
Output:
Timed out
\[ \int \cos (c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right ) \,d x } \] Input:
integrate(cos(d*x+c)*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2), x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(4/3)*c os(d*x + c), x)
\[ \int \cos (c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right ) \,d x } \] Input:
integrate(cos(d*x+c)*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2), x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(4/3)*c os(d*x + c), x)
Timed out. \[ \int \cos (c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int \cos \left (c+d\,x\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \] Input:
int(cos(c + d*x)*(b*cos(c + d*x))^(4/3)*(A + B*cos(c + d*x) + C*cos(c + d* x)^2),x)
Output:
int(cos(c + d*x)*(b*cos(c + d*x))^(4/3)*(A + B*cos(c + d*x) + C*cos(c + d* x)^2), x)
\[ \int \cos (c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=b^{\frac {4}{3}} \left (\left (\int \cos \left (d x +c \right )^{\frac {13}{3}}d x \right ) c +\left (\int \cos \left (d x +c \right )^{\frac {10}{3}}d x \right ) b +\left (\int \cos \left (d x +c \right )^{\frac {7}{3}}d x \right ) a \right ) \] Input:
int(cos(d*x+c)*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
Output:
b**(1/3)*b*(int(cos(c + d*x)**(1/3)*cos(c + d*x)**4,x)*c + int(cos(c + d*x )**(1/3)*cos(c + d*x)**3,x)*b + int(cos(c + d*x)**(1/3)*cos(c + d*x)**2,x) *a)