\(\int (b \cos (c+d x))^{2/3} (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [150]

Optimal result

Integrand size = 33, antiderivative size = 91 \[ \int (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {3 A b \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}+\frac {3 (2 A-C) (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*A*b*sin(d*x+c)/d/(b*cos(d*x+c))^(1/3)+3/5*(2*A-C)*(b*cos(d*x+c))^(5/3)*hypergeom([1/2, 5/6],[11/6],cos(d*x+c
)^2)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 91, 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, 3091, 2722} \[ \int (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {3 (2 A-C) \sin (c+d x) (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )}{5 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 A b \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}} \]

[In]

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

[Out]

(3*A*b*Sin[c + d*x])/(d*(b*Cos[c + d*x])^(1/3)) + (3*(2*A - C)*(b*Cos[c + d*x])^(5/3)*Hypergeometric2F1[1/2, 5
/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*b*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 3091

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A*Cos[e +
f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, 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))^{4/3}} \, dx \\ & = \frac {3 A b \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}+(-2 A+C) \int (b \cos (c+d x))^{2/3} \, dx \\ & = \frac {3 A b \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}+\frac {3 (2 A-C) (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97 \[ \int (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {3 b \csc (c+d x) \left (-5 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )+C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{5 d \sqrt [3]{b \cos (c+d x)}} \]

[In]

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

[Out]

(-3*b*Csc[c + d*x]*(-5*A*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2] + C*Cos[c + d*x]^2*Hypergeometric2F
1[1/2, 5/6, 11/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(5*d*(b*Cos[c + d*x])^(1/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

Giac [F]

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

[In]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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