\(\int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx\) [11]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 31, antiderivative size = 119 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=-\frac {3 A b^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{7 d (b \sec (c+d x))^{7/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b^2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d (b \sec (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3/7*A*b^3*hypergeom([1/2, 7/6],[13/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(7/3)/(sin(d*x+c)^2)^(1/2)-3/
4*b^2*B*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)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {16, 3872, 3857, 2722} \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=-\frac {3 A b^3 \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}}-\frac {3 b^2 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

[In]

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

[Out]

(-3*A*b^3*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*d*(b*Sec[c + d*x])^(7/3)*Sqrt[Sin
[c + d*x]^2]) - (3*b^2*B*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])

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}& = b^2 \int \frac {A+B \sec (c+d x)}{(b \sec (c+d x))^{4/3}} \, dx \\ & = \left (A b^2\right ) \int \frac {1}{(b \sec (c+d x))^{4/3}} \, dx+(b B) \int \frac {1}{\sqrt [3]{b \sec (c+d x)}} \, dx \\ & = \left (A b^2 \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{4/3} \, dx+\left (b B \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 \\ & = -\frac {3 B \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 d \sqrt {\sin ^2(c+d x)}}-\frac {3 A \cos ^3(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{7 d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.74 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=-\frac {3 b \cot (c+d x) \left (A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\sec ^2(c+d x)\right )+4 B \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(c+d x)\right )\right ) \sqrt {-\tan ^2(c+d x)}}{4 d \sqrt [3]{b \sec (c+d x)}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate(cos(d*x+c)^2*(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)*cos(d*x + c)^2, x)

Giac [F]

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

[In]

integrate(cos(d*x+c)^2*(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)*cos(d*x + c)^2, x)

Mupad [F(-1)]

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

[In]

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

[Out]

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