\(\int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx\) [903]

Optimal result

Integrand size = 31, antiderivative size = 114 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 A b \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {16, 2827, 2722} \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 A b \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{5/3}}+\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}} \]

[In]

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

[Out]

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

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}& = b^2 \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{8/3}} \, dx \\ & = \left (A b^2\right ) \int \frac {1}{(b \cos (c+d x))^{8/3}} \, dx+(b B) \int \frac {1}{(b \cos (c+d x))^{5/3}} \, dx \\ & = \frac {3 A b \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 b^2 \cot (c+d x) \left (2 A \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right )+5 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{10 d (b \cos (c+d x))^{8/3}} \]

[In]

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

[Out]

(3*b^2*Cot[c + d*x]*(2*A*Hypergeometric2F1[-5/6, 1/2, 1/6, Cos[c + d*x]^2] + 5*B*Cos[c + d*x]*Hypergeometric2F
1[-1/3, 1/2, 2/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(10*d*(b*Cos[c + d*x])^(8/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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