\(\int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\) [888]

Optimal result

Integrand size = 29, antiderivative size = 119 \[ \int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=-\frac {3 A (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{7 b^2 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 b^3 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3/7*A*(b*cos(d*x+c))^(7/3)*hypergeom([1/2, 7/6],[13/6],cos(d*x+c)^2)*sin(d*x+c)/b^2/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^3/d/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 2827, 2722} \[ \int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=-\frac {3 A \sin (c+d x) (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{7 b^2 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )}{10 b^3 d \sqrt {\sin ^2(c+d x)}} \]

[In]

Int[Cos[c + d*x]*(b*Cos[c + d*x])^(1/3)*(A + B*Cos[c + d*x]),x]

[Out]

(-3*A*(b*Cos[c + d*x])^(7/3)*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*b^2*d*Sqrt[Sin
[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(10/3)*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]*Sin[c + d*x])/(1
0*b^3*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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \, dx}{b} \\ & = \frac {A \int (b \cos (c+d x))^{4/3} \, dx}{b}+\frac {B \int (b \cos (c+d x))^{7/3} \, dx}{b^2} \\ & = -\frac {3 A (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)}{7 b^2 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^3 d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=-\frac {3 (b \cos (c+d x))^{4/3} \cot (c+d x) \left (10 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )+7 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{70 b d} \]

[In]

Integrate[Cos[c + d*x]*(b*Cos[c + d*x])^(1/3)*(A + B*Cos[c + d*x]),x]

[Out]

(-3*(b*Cos[c + d*x])^(4/3)*Cot[c + d*x]*(10*A*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2] + 7*B*Cos[c +
d*x]*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(70*b*d)

Maple [F]

\[\int \cos \left (d x +c \right ) \left (\cos \left (d x +c \right ) b \right )^{\frac {1}{3}} \left (A +B \cos \left (d x +c \right )\right )d x\]

[In]

int(cos(d*x+c)*(cos(d*x+c)*b)^(1/3)*(A+B*cos(d*x+c)),x)

[Out]

int(cos(d*x+c)*(cos(d*x+c)*b)^(1/3)*(A+B*cos(d*x+c)),x)

Fricas [F]

\[ \int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)*(b*cos(d*x+c))^(1/3)*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c)^2 + A*cos(d*x + c))*(b*cos(d*x + c))^(1/3), x)

Sympy [F(-1)]

Timed out. \[ \int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)*(b*cos(d*x+c))**(1/3)*(A+B*cos(d*x+c)),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)*(b*cos(d*x+c))^(1/3)*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(1/3)*cos(d*x + c), x)

Giac [F]

\[ \int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)*(b*cos(d*x+c))^(1/3)*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(1/3)*cos(d*x + c), x)

Mupad [F(-1)]

Timed out. \[ \int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int \cos \left (c+d\,x\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}\,\left (A+B\,\cos \left (c+d\,x\right )\right ) \,d x \]

[In]

int(cos(c + d*x)*(b*cos(c + d*x))^(1/3)*(A + B*cos(c + d*x)),x)

[Out]

int(cos(c + d*x)*(b*cos(c + d*x))^(1/3)*(A + B*cos(c + d*x)), x)