\(\int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} (A+C \sec ^2(c+d x)) \, dx\) [23]

Optimal result

Integrand size = 33, antiderivative size = 144 \[ \int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 C \sec ^{1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (4+3 m)}-\frac {3 (C+3 C m+A (4+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2-3 m),\frac {1}{6} (8-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (2-3 m) (4+3 m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*C*sec(d*x+c)^(1+m)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(4+3*m)-3*(C+3*C*m+A*(4+3*m))*hypergeom([1/2, 1/3-1/2*m
],[4/3-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(-9*m^2-6*m+8)/(sin(d*x+c)^2)^
(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 144, 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} \[ \int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 C \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m+1}(c+d x)}{d (3 m+4)}-\frac {3 (A (3 m+4)+3 C m+C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^{m-1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2-3 m),\frac {1}{6} (8-3 m),\cos ^2(c+d x)\right )}{d (2-3 m) (3 m+4) \sqrt {\sin ^2(c+d x)}} \]

[In]

Int[Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/3)*(A + C*Sec[c + d*x]^2),x]

[Out]

(3*C*Sec[c + d*x]^(1 + m)*(b*Sec[c + d*x])^(1/3)*Sin[c + d*x])/(d*(4 + 3*m)) - (3*(C + 3*C*m + A*(4 + 3*m))*Hy
pergeometric2F1[1/2, (2 - 3*m)/6, (8 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*(b*Sec[c + d*x])^(1/3)*Si
n[c + d*x])/(d*(2 - 3*m)*(4 + 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 {\sqrt [3]{b \sec (c+d x)} \int \sec ^{\frac {1}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{\sqrt [3]{\sec (c+d x)}} \\ & = \frac {3 C \sec ^{1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (4+3 m)}+\frac {\left (\left (C \left (\frac {1}{3}+m\right )+A \left (\frac {4}{3}+m\right )\right ) \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac {1}{3}+m}(c+d x) \, dx}{\left (\frac {4}{3}+m\right ) \sqrt [3]{\sec (c+d x)}} \\ & = \frac {3 C \sec ^{1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (4+3 m)}+\frac {\left (\left (C \left (\frac {1}{3}+m\right )+A \left (\frac {4}{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 {1}{3}-m}(c+d x) \, dx}{\frac {4}{3}+m} \\ & = \frac {3 C \sec ^{1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (4+3 m)}-\frac {3 (C+3 C m+A (4+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2-3 m),\frac {1}{6} (8-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (2-3 m) (4+3 m) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.99 \[ \int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b \csc (c+d x) \sec ^m(c+d x) \left (A (7+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1+3 m),\frac {1}{6} (7+3 m),\sec ^2(c+d x)\right )+C (1+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (1+3 m) (7+3 m) (b \sec (c+d x))^{2/3}} \]

[In]

Integrate[Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/3)*(A + C*Sec[c + d*x]^2),x]

[Out]

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

Maple [F]

\[\int \sec \left (d x +c \right )^{m} \left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +C \sec \left (d x +c \right )^{2}\right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(sec(d*x+c)**m*(b*sec(d*x+c))**(1/3)*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((b*sec(c + d*x))**(1/3)*(A + C*sec(c + d*x)**2)*sec(c + d*x)**m, x)

Maxima [F]

\[ \int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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