\(\int (b \cos (c+d x))^{4/3} (A+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [155]

Optimal result

Integrand size = 31, antiderivative size = 89 \[ \int (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {3 C (b \cos (c+d x))^{4/3} \sin (c+d x)}{7 d}-\frac {3 (7 A+4 C) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{28 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/7*C*(b*cos(d*x+c))^(4/3)*sin(d*x+c)/d-3/28*(7*A+4*C)*(b*cos(d*x+c))^(4/3)*hypergeom([1/2, 2/3],[5/3],cos(d*x
+c)^2)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 89, 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} \[ \int (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {3 C \sin (c+d x) (b \cos (c+d x))^{4/3}}{7 d}-\frac {3 (7 A+4 C) \sin (c+d x) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{28 d \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

(3*C*(b*Cos[c + d*x])^(4/3)*Sin[c + d*x])/(7*d) - (3*(7*A + 4*C)*(b*Cos[c + d*x])^(4/3)*Hypergeometric2F1[1/2,
2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(28*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}& = b \int \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx \\ & = \frac {3 C (b \cos (c+d x))^{4/3} \sin (c+d x)}{7 d}+\frac {1}{7} (b (7 A+4 C)) \int \sqrt [3]{b \cos (c+d x)} \, dx \\ & = \frac {3 C (b \cos (c+d x))^{4/3} \sin (c+d x)}{7 d}-\frac {3 (7 A+4 C) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{28 d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=-\frac {3 b \sqrt [3]{b \cos (c+d x)} \cot (c+d x) \left (5 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )+2 C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{20 d} \]

[In]

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

[Out]

(-3*b*(b*Cos[c + d*x])^(1/3)*Cot[c + d*x]*(5*A*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2] + 2*C*Cos[c +
d*x]^2*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(20*d)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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