\(\int \cos (c+d x) (b \cos (c+d x))^{2/3} (A+C \cos ^2(c+d x)) \, dx\) [147]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 31, antiderivative size = 95 \[ \int \cos (c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C (b \cos (c+d x))^{8/3} \sin (c+d x)}{11 b^2 d}-\frac {3 (11 A+8 C) (b \cos (c+d x))^{8/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{88 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/11*C*(b*cos(d*x+c))^(8/3)*sin(d*x+c)/b^2/d-3/88*(11*A+8*C)*(b*cos(d*x+c))^(8/3)*hypergeom([1/2, 4/3],[7/3],c
os(d*x+c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {16, 3093, 2722} \[ \int \cos (c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C \sin (c+d x) (b \cos (c+d x))^{8/3}}{11 b^2 d}-\frac {3 (11 A+8 C) \sin (c+d x) (b \cos (c+d x))^{8/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right )}{88 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[In]

Int[Cos[c + d*x]*(b*Cos[c + d*x])^(2/3)*(A + C*Cos[c + d*x]^2),x]

[Out]

(3*C*(b*Cos[c + d*x])^(8/3)*Sin[c + d*x])/(11*b^2*d) - (3*(11*A + 8*C)*(b*Cos[c + d*x])^(8/3)*Hypergeometric2F
1[1/2, 4/3, 7/3, Cos[c + d*x]^2]*Sin[c + d*x])/(88*b^2*d*Sqrt[Sin[c + d*x]^2])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3093

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos
[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e +
f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (b \cos (c+d x))^{5/3} \left (A+C \cos ^2(c+d x)\right ) \, dx}{b} \\ & = \frac {3 C (b \cos (c+d x))^{8/3} \sin (c+d x)}{11 b^2 d}+\frac {(11 A+8 C) \int (b \cos (c+d x))^{5/3} \, dx}{11 b} \\ & = \frac {3 C (b \cos (c+d x))^{8/3} \sin (c+d x)}{11 b^2 d}-\frac {3 (11 A+8 C) (b \cos (c+d x))^{8/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{88 b^2 d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 (b \cos (c+d x))^{5/3} \cot (c+d x) \left (7 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right )+4 C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{3},\frac {10}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{56 b d} \]

[In]

Integrate[Cos[c + d*x]*(b*Cos[c + d*x])^(2/3)*(A + C*Cos[c + d*x]^2),x]

[Out]

(-3*(b*Cos[c + d*x])^(5/3)*Cot[c + d*x]*(7*A*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2] + 4*C*Cos[c + d*
x]^2*Hypergeometric2F1[1/2, 7/3, 10/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(56*b*d)

Maple [F]

\[\int \cos \left (d x +c \right ) \left (\cos \left (d x +c \right ) b \right )^{\frac {2}{3}} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )d x\]

[In]

int(cos(d*x+c)*(cos(d*x+c)*b)^(2/3)*(A+C*cos(d*x+c)^2),x)

[Out]

int(cos(d*x+c)*(cos(d*x+c)*b)^(2/3)*(A+C*cos(d*x+c)^2),x)

Fricas [F]

\[ \int \cos (c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)*(b*cos(d*x+c))^(2/3)*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^3 + A*cos(d*x + c))*(b*cos(d*x + c))^(2/3), x)

Sympy [F(-1)]

Timed out. \[ \int \cos (c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)*(b*cos(d*x+c))**(2/3)*(A+C*cos(d*x+c)**2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos (c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)*(b*cos(d*x+c))^(2/3)*(A+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(2/3)*cos(d*x + c), x)

Giac [F]

\[ \int \cos (c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)*(b*cos(d*x+c))^(2/3)*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(2/3)*cos(d*x + c), x)

Mupad [F(-1)]

Timed out. \[ \int \cos (c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \cos \left (c+d\,x\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{2/3} \,d x \]

[In]

int(cos(c + d*x)*(A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(2/3),x)

[Out]

int(cos(c + d*x)*(A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(2/3), x)