\(\int \sqrt [3]{c \csc (a+b x)} \, dx\) [33]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 12, antiderivative size = 56 \[ \int \sqrt [3]{c \csc (a+b x)} \, dx=\frac {3 c \cos (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sin ^2(a+b x)\right )}{2 b \sqrt {\cos ^2(a+b x)} (c \csc (a+b x))^{2/3}} \]

[Out]

3/2*c*cos(b*x+a)*hypergeom([1/3, 1/2],[4/3],sin(b*x+a)^2)/b/(c*csc(b*x+a))^(2/3)/(cos(b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3857, 2722} \[ \int \sqrt [3]{c \csc (a+b x)} \, dx=\frac {3 c \cos (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sin ^2(a+b x)\right )}{2 b \sqrt {\cos ^2(a+b x)} (c \csc (a+b x))^{2/3}} \]

[In]

Int[(c*Csc[a + b*x])^(1/3),x]

[Out]

(3*c*Cos[a + b*x]*Hypergeometric2F1[1/3, 1/2, 4/3, Sin[a + b*x]^2])/(2*b*Sqrt[Cos[a + b*x]^2]*(c*Csc[a + b*x])
^(2/3))

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]

Rubi steps \begin{align*} \text {integral}& = \sqrt [3]{c \csc (a+b x)} \sqrt [3]{\frac {\sin (a+b x)}{c}} \int \frac {1}{\sqrt [3]{\frac {\sin (a+b x)}{c}}} \, dx \\ & = \frac {3 \cos (a+b x) \sqrt [3]{c \csc (a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sin ^2(a+b x)\right ) \sin (a+b x)}{2 b \sqrt {\cos ^2(a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.05 \[ \int \sqrt [3]{c \csc (a+b x)} \, dx=-\frac {\cos (a+b x) \sqrt [3]{c \csc (a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\cos ^2(a+b x)\right ) \sin (a+b x)}{b \sqrt [3]{\sin ^2(a+b x)}} \]

[In]

Integrate[(c*Csc[a + b*x])^(1/3),x]

[Out]

-((Cos[a + b*x]*(c*Csc[a + b*x])^(1/3)*Hypergeometric2F1[1/2, 2/3, 3/2, Cos[a + b*x]^2]*Sin[a + b*x])/(b*(Sin[
a + b*x]^2)^(1/3)))

Maple [F]

\[\int \left (c \csc \left (x b +a \right )\right )^{\frac {1}{3}}d x\]

[In]

int((c*csc(b*x+a))^(1/3),x)

[Out]

int((c*csc(b*x+a))^(1/3),x)

Fricas [F]

\[ \int \sqrt [3]{c \csc (a+b x)} \, dx=\int { \left (c \csc \left (b x + a\right )\right )^{\frac {1}{3}} \,d x } \]

[In]

integrate((c*csc(b*x+a))^(1/3),x, algorithm="fricas")

[Out]

integral((c*csc(b*x + a))^(1/3), x)

Sympy [F]

\[ \int \sqrt [3]{c \csc (a+b x)} \, dx=\int \sqrt [3]{c \csc {\left (a + b x \right )}}\, dx \]

[In]

integrate((c*csc(b*x+a))**(1/3),x)

[Out]

Integral((c*csc(a + b*x))**(1/3), x)

Maxima [F]

\[ \int \sqrt [3]{c \csc (a+b x)} \, dx=\int { \left (c \csc \left (b x + a\right )\right )^{\frac {1}{3}} \,d x } \]

[In]

integrate((c*csc(b*x+a))^(1/3),x, algorithm="maxima")

[Out]

integrate((c*csc(b*x + a))^(1/3), x)

Giac [F]

\[ \int \sqrt [3]{c \csc (a+b x)} \, dx=\int { \left (c \csc \left (b x + a\right )\right )^{\frac {1}{3}} \,d x } \]

[In]

integrate((c*csc(b*x+a))^(1/3),x, algorithm="giac")

[Out]

integrate((c*csc(b*x + a))^(1/3), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt [3]{c \csc (a+b x)} \, dx=\int {\left (\frac {c}{\sin \left (a+b\,x\right )}\right )}^{1/3} \,d x \]

[In]

int((c/sin(a + b*x))^(1/3),x)

[Out]

int((c/sin(a + b*x))^(1/3), x)