\(\int (b \cos (c+d x))^{2/3} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [341]

Optimal result

Integrand size = 39, antiderivative size = 148 \[ \int (b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {3 C (b \cos (c+d x))^{2/3} \sin (c+d x)}{5 d}-\frac {3 (5 A+2 C) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/5*C*(b*cos(d*x+c))^(2/3)*sin(d*x+c)/d-3/10*(5*A+2*C)*(b*cos(d*x+c))^(2/3)*hypergeom([1/3, 1/2],[4/3],cos(d*x
+c)^2)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/2)-3/5*B*(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 = 0.23 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 3102, 2827, 2722} \[ \int (b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=-\frac {3 (5 A+2 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )}{5 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 d} \]

[In]

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

[Out]

(3*C*(b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*d) - (3*(5*A + 2*C)*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[1/3,
1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(10*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(5/3)*Hypergeomet
ric2F1[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 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}& = b \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx \\ & = \frac {3 C (b \cos (c+d x))^{2/3} \sin (c+d x)}{5 d}+\frac {3}{5} \int \frac {\frac {1}{3} b (5 A+2 C)+\frac {5}{3} b B \cos (c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx \\ & = \frac {3 C (b \cos (c+d x))^{2/3} \sin (c+d x)}{5 d}+B \int (b \cos (c+d x))^{2/3} \, dx+\frac {1}{5} (b (5 A+2 C)) \int \frac {1}{\sqrt [3]{b \cos (c+d x)}} \, dx \\ & = \frac {3 C (b \cos (c+d x))^{2/3} \sin (c+d x)}{5 d}-\frac {3 (5 A+2 C) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (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 = 0.38 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.82 \[ \int (b \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {b \left (-3 (5 A+2 C) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}-6 B \cos (c+d x) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+3 C \sin (2 (c+d x))\right )}{10 d \sqrt [3]{b \cos (c+d x)}} \]

[In]

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

[Out]

(b*(-3*(5*A + 2*C)*Cot[c + d*x]*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] - 6*B*Co
s[c + d*x]*Cot[c + d*x]*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] + 3*C*Sin[2*(c
+ d*x)]))/(10*d*(b*Cos[c + d*x])^(1/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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