\(\int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx\) [901]

Optimal result

Integrand size = 23, antiderivative size = 117 \[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=-\frac {3 A \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3*A*(b*cos(d*x+c))^(1/3)*hypergeom([1/6, 1/2],[7/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)-3/4*B*(
b*cos(d*x+c))^(4/3)*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2827, 2722} \[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=-\frac {3 A \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

(-3*A*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*Sqrt[Sin[c +
d*x]^2]) - (3*B*(b*Cos[c + d*x])^(4/3)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(4*b^2*d
*Sqrt[Sin[c + d*x]^2])

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]

Rubi steps \begin{align*} \text {integral}& = A \int \frac {1}{(b \cos (c+d x))^{2/3}} \, dx+\frac {B \int \sqrt [3]{b \cos (c+d x)} \, dx}{b} \\ & = -\frac {3 A \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)}{b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (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)}{4 b^2 d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73 \[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=-\frac {3 \cot (c+d x) \left (4 A \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )+B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{4 d (b \cos (c+d x))^{2/3}} \]

[In]

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

[Out]

(-3*Cot[c + d*x]*(4*A*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2] + B*Cos[c + d*x]*Hypergeometric2F1[1/2,
2/3, 5/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(4*d*(b*Cos[c + d*x])^(2/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((A + B*cos(c + d*x))/(b*cos(c + d*x))**(2/3), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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