\(\int \sec (c+d x) (b \sec (c+d x))^{4/3} (A+C \sec ^2(c+d x)) \, dx\) [7]

Optimal result

Integrand size = 31, antiderivative size = 92 \[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 (10 A+7 C) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{40 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b d} \]

[Out]

3/40*(10*A+7*C)*hypergeom([-2/3, 1/2],[1/3],cos(d*x+c)^2)*(b*sec(d*x+c))^(4/3)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/
2)+3/10*C*(b*sec(d*x+c))^(7/3)*tan(d*x+c)/b/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {16, 4131, 3857, 2722} \[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 (10 A+7 C) \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{40 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 b d} \]

[In]

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

[Out]

(3*(10*A + 7*C)*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(4/3)*Sin[c + d*x])/(40*d*S
qrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(7/3)*Tan[c + d*x])/(10*b*d)

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (b \sec (c+d x))^{7/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b} \\ & = \frac {3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b d}+\frac {(10 A+7 C) \int (b \sec (c+d x))^{7/3} \, dx}{10 b} \\ & = \frac {3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b d}+\frac {\left ((10 A+7 C) \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 )^{7/3}} \, dx}{10 b} \\ & = \frac {3 (10 A+7 C) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{40 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.87 \[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \cot (c+d x) (b \sec (c+d x))^{7/3} \left (7 C \tan ^2(c+d x)+(10 A+7 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}\right )}{70 b d} \]

[In]

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

[Out]

(3*Cot[c + d*x]*(b*Sec[c + d*x])^(7/3)*(7*C*Tan[c + d*x]^2 + (10*A + 7*C)*Hypergeometric2F1[1/2, 7/6, 13/6, Se
c[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2]))/(70*b*d)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((C*b*sec(d*x + c)^4 + A*b*sec(d*x + c)^2)*(b*sec(d*x + c))^(1/3), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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