\(\int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx\) [8]

Optimal result

Integrand size = 29, antiderivative size = 116 \[ \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\frac {3 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{5/3} \sin (c+d x)}{5 b d \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 116, 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} \[ \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\frac {3 A \sin (c+d x) (b \sec (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right )}{5 b d \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.78 \[ \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\frac {3 \csc (c+d x) \left (8 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(c+d x)\right )+5 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{5/3} \sqrt {-\tan ^2(c+d x)}}{40 b d} \]

[In]

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

[Out]

(3*Csc[c + d*x]*(8*A*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 11/6, Sec[c + d*x]^2] + 5*B*Hypergeometric2F1[1/
2, 4/3, 7/3, Sec[c + d*x]^2])*(b*Sec[c + d*x])^(5/3)*Sqrt[-Tan[c + d*x]^2])/(40*b*d)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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