\(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{2/3}} \, dx\) [20]

Optimal result

Integrand size = 29, antiderivative size = 114 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{2/3}} \, dx=-\frac {3 A \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{b d \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3/2*A*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(2/3)/(sin(d*x+c)^2)^(1/2)+3*B*hyp
ergeom([-1/6, 1/2],[5/6],cos(d*x+c)^2)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/b/d/(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 = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {16, 3872, 3857, 2722} \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}-\frac {3 A \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}} \]

[In]

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

[Out]

(-3*A*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2*d*(b*Sec[c + d*x])^(2/3)*Sqrt[Sin[c +
d*x]^2]) + (3*B*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(1/3)*Sin[c + d*x])/(b*d*Sq
rt[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 3857

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sqrt [3]{b \sec (c+d x)} (A+B \sec (c+d x)) \, dx}{b} \\ & = \frac {A \int \sqrt [3]{b \sec (c+d x)} \, dx}{b}+\frac {B \int (b \sec (c+d x))^{4/3} \, dx}{b^2} \\ & = \frac {\left (A \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac {1}{\sqrt [3]{\frac {\cos (c+d x)}{b}}} \, dx}{b}+\frac {\left (B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{4/3}} \, dx}{b^2} \\ & = \frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{b d \sqrt {\sin ^2(c+d x)}}-\frac {3 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{2 b d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.79 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 \csc (c+d x) \left (4 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(c+d x)\right )+B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\sec ^2(c+d x)\right )\right ) \sqrt [3]{b \sec (c+d x)} \sqrt {-\tan ^2(c+d x)}}{4 b d} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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