\(\int (b \cos (c+d x))^{4/3} (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [156]

Optimal result

Integrand size = 33, antiderivative size = 89 \[ \int (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {3 b C \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{4 d}-\frac {3 b (4 A+C) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/4*b*C*(b*cos(d*x+c))^(1/3)*sin(d*x+c)/d-3/4*b*(4*A+C)*(b*cos(d*x+c))^(1/3)*hypergeom([1/6, 1/2],[7/6],cos(d*
x+c)^2)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {16, 3093, 2722} \[ \int (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {3 b C \sin (c+d x) \sqrt [3]{b \cos (c+d x)}}{4 d}-\frac {3 b (4 A+C) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

(3*b*C*(b*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(4*d) - (3*b*(4*A + C)*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/
6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(4*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^2 \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx \\ & = \frac {3 b C \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {1}{4} \left (b^2 (4 A+C)\right ) \int \frac {1}{(b \cos (c+d x))^{2/3}} \, dx \\ & = \frac {3 b C \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{4 d}-\frac {3 b (4 A+C) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {3 b^2 \cot (c+d x) \left (7 A \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )+C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{7 d (b \cos (c+d x))^{2/3}} \]

[In]

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

[Out]

(-3*b^2*Cot[c + d*x]*(7*A*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2] + C*Cos[c + d*x]^2*Hypergeometric2F
1[1/2, 7/6, 13/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(7*d*(b*Cos[c + d*x])^(2/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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