\(\int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} (A+C \sec ^2(c+d x)) \, dx\) [21]

Optimal result

Integrand size = 33, antiderivative size = 146 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {3 b (C (4+3 m)+A (7+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1-3 m),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) (7+3 m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*b*C*sec(d*x+c)^(2+m)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(7+3*m)+3*b*(C*(4+3*m)+A*(7+3*m))*hypergeom([1/2, -1/
6-1/2*m],[5/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^m*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(9*m^2+24*m+7)/(sin(d*x+c)^2
)^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {20, 4131, 3857, 2722} \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b (A (3 m+7)+C (3 m+4)) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-1),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right )}{d (3 m+1) (3 m+7) \sqrt {\sin ^2(c+d x)}}+\frac {3 b C \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+2}(c+d x)}{d (3 m+7)} \]

[In]

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

[Out]

(3*b*C*Sec[c + d*x]^(2 + m)*(b*Sec[c + d*x])^(1/3)*Sin[c + d*x])/(d*(7 + 3*m)) + (3*b*(C*(4 + 3*m) + A*(7 + 3*
m))*Hypergeometric2F1[1/2, (-1 - 3*m)/6, (5 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/3)*Si
n[c + d*x])/(d*(1 + 3*m)*(7 + 3*m)*Sqrt[Sin[c + d*x]^2])

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[b^IntPart[n]*((b*v)^FracPart[n]/(a^IntPart[n]
*(a*v)^FracPart[n])), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

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 {\left (b \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac {4}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{\sqrt [3]{\sec (c+d x)}} \\ & = \frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {\left (b \left (C \left (\frac {4}{3}+m\right )+A \left (\frac {7}{3}+m\right )\right ) \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac {4}{3}+m}(c+d x) \, dx}{\left (\frac {7}{3}+m\right ) \sqrt [3]{\sec (c+d x)}} \\ & = \frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {\left (b \left (C \left (\frac {4}{3}+m\right )+A \left (\frac {7}{3}+m\right )\right ) \cos ^{\frac {1}{3}+m}(c+d x) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)}\right ) \int \cos ^{-\frac {4}{3}-m}(c+d x) \, dx}{\frac {7}{3}+m} \\ & = \frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {3 b (C (4+3 m)+A (7+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1-3 m),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) (7+3 m) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.98 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b \csc (c+d x) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A (10+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {5}{3}+\frac {m}{2},\sec ^2(c+d x)\right )+C (4+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3}+\frac {m}{2},\frac {8}{3}+\frac {m}{2},\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (4+3 m) (10+3 m)} \]

[In]

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

[Out]

(3*b*Csc[c + d*x]*Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/3)*(A*(10 + 3*m)*Hypergeometric2F1[1/2, (4 + 3*m)/6, 5/3
+ m/2, Sec[c + d*x]^2] + C*(4 + 3*m)*Hypergeometric2F1[1/2, 5/3 + m/2, 8/3 + m/2, Sec[c + d*x]^2]*Sec[c + d*x]
^2)*Sqrt[-Tan[c + d*x]^2])/(d*(4 + 3*m)*(10 + 3*m))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

integrate(sec(d*x+c)^m*(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)^3 + A*b*sec(d*x + c))*(b*sec(d*x + c))^(1/3)*sec(d*x + c)^m, x)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate(sec(d*x+c)^m*(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)^m, x)

Giac [F]

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

[In]

integrate(sec(d*x+c)^m*(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)^m, x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \]

[In]

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(4/3)*(1/cos(c + d*x))^m,x)

[Out]

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(4/3)*(1/cos(c + d*x))^m, x)