\(\int (b \cos (c+d x))^{2/3} \sec ^3(c+d x) \, dx\) [215]

Optimal result

Integrand size = 21, antiderivative size = 58 \[ \int (b \cos (c+d x))^{2/3} \sec ^3(c+d x) \, dx=\frac {3 b^2 \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {16, 2722} \[ \int (b \cos (c+d x))^{2/3} \sec ^3(c+d x) \, dx=\frac {3 b^2 \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{4/3}} \]

[In]

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

[Out]

(3*b^2*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*Sin[c + d*x])/(4*d*(b*Cos[c + d*x])^(4/3)*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]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09 \[ \int (b \cos (c+d x))^{2/3} \sec ^3(c+d x) \, dx=\frac {3 (b \cos (c+d x))^{2/3} \csc (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) \sec ^2(c+d x) \sqrt {\sin ^2(c+d x)}}{4 d} \]

[In]

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

[Out]

(3*(b*Cos[c + d*x])^(2/3)*Csc[c + d*x]*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*Sec[c + d*x]^2*Sqrt[S
in[c + d*x]^2])/(4*d)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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