\(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\) [23]

Optimal result

Integrand size = 29, antiderivative size = 114 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d (b \sec (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3/4*A*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(4/3)/(sin(d*x+c)^2)^(1/2)-3*B*hyp
ergeom([1/6, 1/2],[7/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(b*sec(d*x+c))^(1/3)/(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 = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {16, 3872, 3857, 2722} \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 A \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

[In]

Int[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(b*Sec[c + d*x])^(4/3),x]

[Out]

(-3*A*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(4*d*(b*Sec[c + d*x])^(4/3)*Sqrt[Sin[c +
d*x]^2]) - (3*B*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(b*Sec[c + d*x])^(1/3)*Sqr
t[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 \frac {A+B \sec (c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx}{b} \\ & = \frac {A \int \frac {1}{\sqrt [3]{b \sec (c+d x)}} \, dx}{b}+\frac {B \int (b \sec (c+d x))^{2/3} \, 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 \sqrt [3]{\frac {\cos (c+d x)}{b}} \, 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 )^{2/3}} \, dx}{b^2} \\ & = -\frac {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 \sec (c+d x))^{2/3} \sin (c+d x)}{b^2 d \sqrt {\sin ^2(c+d x)}}-\frac {3 A \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{4 b^2 d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.80 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 \csc (c+d x) \left (2 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(c+d x)\right )-B \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sec ^2(c+d x)\right )\right ) \sqrt {-\tan ^2(c+d x)}}{2 b d \sqrt [3]{b \sec (c+d x)}} \]

[In]

Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(b*Sec[c + d*x])^(4/3),x]

[Out]

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

Maple [F]

\[\int \frac {\sec \left (d x +c \right ) \left (A +B \sec \left (d x +c \right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]

[In]

int(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x)

[Out]

int(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x)

Fricas [F]

\[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))**(4/3),x)

[Out]

Integral((A + B*sec(c + d*x))*sec(c + d*x)/(b*sec(c + d*x))**(4/3), x)

Maxima [F]

\[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*sec(d*x + c)/(b*sec(d*x + c))^(4/3), x)

Giac [F]

\[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*sec(d*x + c)/(b*sec(d*x + c))^(4/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\cos \left (c+d\,x\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)*(b/cos(c + d*x))^(4/3)),x)

[Out]

int((A + B/cos(c + d*x))/(cos(c + d*x)*(b/cos(c + d*x))^(4/3)), x)