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

Optimal result

Integrand size = 25, antiderivative size = 88 \[ \int \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 b (4 A+C) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 d (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 d} \]

[Out]

-3/8*b*(4*A+C)*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)
+3/4*C*(b*sec(d*x+c))^(1/3)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4131, 3857, 2722} \[ \int \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 C \tan (c+d x) \sqrt [3]{b \sec (c+d x)}}{4 d}-\frac {3 b (4 A+C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}} \]

[In]

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

[Out]

(-3*b*(4*A + C)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*d*(b*Sec[c + d*x])^(2/3)*Sqr
t[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(4*d)

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 {3 C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 d}+\frac {1}{4} (4 A+C) \int \sqrt [3]{b \sec (c+d x)} \, dx \\ & = \frac {3 C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 d}+\frac {1}{4} \left ((4 A+C) \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 (4 A+C) \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)}{8 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \cot (c+d x) \sqrt [3]{b \sec (c+d x)} \left (C \tan ^2(c+d x)+(4 A+C) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}\right )}{4 d} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

integrate((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), x)

Sympy [F]

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

[In]

integrate((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), x)

Maxima [F]

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

[In]

integrate((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), x)

Giac [F]

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

[In]

integrate((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), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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