\(\int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [346]

Optimal result

Integrand size = 33, antiderivative size = 154 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C (b \cos (c+d x))^{7/3} \sin (c+d x)}{10 b d}-\frac {3 (10 A+7 C) (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{70 b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

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],co
s(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)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3102, 2827, 2722} \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 (10 A+7 C) \sin (c+d x) (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{70 b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )}{10 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d} \]

[In]

Int[(b*Cos[c + d*x])^(4/3)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(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])/(70*b*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(10/3)*Hyp
ergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]*Sin[c + d*x])/(10*b^2*d*Sqrt[Sin[c + d*x]^2])

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]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {3 C (b \cos (c+d x))^{7/3} \sin (c+d x)}{10 b d}+\frac {3 \int (b \cos (c+d x))^{4/3} \left (\frac {1}{3} b (10 A+7 C)+\frac {10}{3} b B \cos (c+d x)\right ) \, dx}{10 b} \\ & = \frac {3 C (b \cos (c+d x))^{7/3} \sin (c+d x)}{10 b d}+\frac {B \int (b \cos (c+d x))^{7/3} \, dx}{b}+\frac {1}{10} (10 A+7 C) \int (b \cos (c+d x))^{4/3} \, 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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.74 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 (b \cos (c+d x))^{4/3} \cot (c+d x) \left (-7 C \sin ^2(c+d x)+(10 A+7 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+7 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{70 d} \]

[In]

Integrate[(b*Cos[c + d*x])^(4/3)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(-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[
Sin[c + d*x]^2]))/(70*d)

Maple [F]

\[\int \left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )d x\]

[In]

int((cos(d*x+c)*b)^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

int((cos(d*x+c)*b)^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

Fricas [F]

\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \]

[In]

integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate((b*cos(d*x+c))**(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Timed out

Maxima [F]

\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \]

[In]

integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(4/3), x)

Giac [F]

\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \]

[In]

integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(4/3), x)

Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int {\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

[In]

int((b*cos(c + d*x))^(4/3)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

int((b*cos(c + d*x))^(4/3)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)