\(\int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [51]

Optimal result

Integrand size = 41, antiderivative size = 150 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 b^3 (A-2 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d (b \sec (c+d x))^{5/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b^2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 b^2 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}} \]

[Out]

-3/5*b^3*(A-2*C)*hypergeom([1/2, 5/6],[11/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(5/3)/(sin(d*x+c)^2)^(1
/2)-3/2*b^2*B*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^2*C*tan(d*x+c)/d/(b*sec(d*x+c))^(2/3)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {16, 4132, 3857, 2722, 4131} \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 b^3 (A-2 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{5/3}}-\frac {3 b^2 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 b^2 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}} \]

[In]

Int[Cos[c + d*x]^2*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(-3*b^3*(A - 2*C)*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*d*(b*Sec[c + d*x])^(5/3)*
Sqrt[Sin[c + d*x]^2]) - (3*b^2*B*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^2*C*Tan[c + d*x])/(d*(b*Sec[c + d*x])^(2/3))

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 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rubi steps \begin{align*} \text {integral}& = b^2 \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{2/3}} \, dx \\ & = b^2 \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{2/3}} \, dx+(b B) \int \sqrt [3]{b \sec (c+d x)} \, dx \\ & = \frac {3 b^2 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}+\left (b^2 (A-2 C)\right ) \int \frac {1}{(b \sec (c+d x))^{2/3}} \, dx+\left (b B \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 \\ & = -\frac {3 b B \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 d \sqrt {\sin ^2(c+d x)}}+\frac {3 b^2 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}+\left (b^2 (A-2 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{2/3} \, dx \\ & = -\frac {3 b B \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 d \sqrt {\sin ^2(c+d x)}}-\frac {3 b (A-2 C) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{5 d \sqrt {\sin ^2(c+d x)}}+\frac {3 b^2 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 \cot (c+d x) \left (2 A \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\sec ^2(c+d x)\right )-4 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(c+d x)\right )-C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{4/3} \sqrt {-\tan ^2(c+d x)}}{4 d} \]

[In]

Integrate[Cos[c + d*x]^2*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(-3*Cot[c + d*x]*(2*A*Cos[c + d*x]^2*Hypergeometric2F1[-1/3, 1/2, 2/3, Sec[c + d*x]^2] - 4*B*Cos[c + d*x]*Hype
rgeometric2F1[1/6, 1/2, 7/6, Sec[c + d*x]^2] - C*Hypergeometric2F1[1/2, 2/3, 5/3, Sec[c + d*x]^2])*(b*Sec[c +
d*x])^(4/3)*Sqrt[-Tan[c + d*x]^2])/(4*d)

Maple [F]

\[\int \cos \left (d x +c \right )^{2} \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]

[In]

int(cos(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

int(cos(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

Fricas [F]

\[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(cos(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b*cos(d*x + c)^2*sec(d*x + c)^3 + B*b*cos(d*x + c)^2*sec(d*x + c)^2 + A*b*cos(d*x + c)^2*sec(d*x +
c))*(b*sec(d*x + c))^(1/3), x)

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**2*(b*sec(d*x+c))**(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(cos(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(4/3)*cos(d*x + c)^2, x)

Giac [F]

\[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(cos(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(4/3)*cos(d*x + c)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

[In]

int(cos(c + d*x)^2*(b/cos(c + d*x))^(4/3)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

int(cos(c + d*x)^2*(b/cos(c + d*x))^(4/3)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)