\(\int \frac {1}{(c \csc (a+b x))^{4/3}} \, dx\) [36]

Optimal result

Integrand size = 12, antiderivative size = 56 \[ \int \frac {1}{(c \csc (a+b x))^{4/3}} \, dx=\frac {3 c \cos (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\sin ^2(a+b x)\right )}{7 b \sqrt {\cos ^2(a+b x)} (c \csc (a+b x))^{7/3}} \]

[Out]

3/7*c*cos(b*x+a)*hypergeom([1/2, 7/6],[13/6],sin(b*x+a)^2)/b/(c*csc(b*x+a))^(7/3)/(cos(b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.04 (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} \[ \int \frac {1}{(c \csc (a+b x))^{4/3}} \, dx=\frac {3 c \cos (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\sin ^2(a+b x)\right )}{7 b \sqrt {\cos ^2(a+b x)} (c \csc (a+b x))^{7/3}} \]

[In]

Int[(c*Csc[a + b*x])^(-4/3),x]

[Out]

(3*c*Cos[a + b*x]*Hypergeometric2F1[1/2, 7/6, 13/6, Sin[a + b*x]^2])/(7*b*Sqrt[Cos[a + b*x]^2]*(c*Csc[a + b*x]
)^(7/3))

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}& = (c \csc (a+b x))^{2/3} \left (\frac {\sin (a+b x)}{c}\right )^{2/3} \int \left (\frac {\sin (a+b x)}{c}\right )^{4/3} \, dx \\ & = \frac {3 \cos (a+b x) (c \csc (a+b x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\sin ^2(a+b x)\right ) \sin ^3(a+b x)}{7 b c^2 \sqrt {\cos ^2(a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.36 \[ \int \frac {1}{(c \csc (a+b x))^{4/3}} \, dx=-\frac {\csc ^2(a+b x) \left (2 \cot (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\cos ^2(a+b x)\right ) \sin ^2(a+b x)^{5/6}+3 \sin (2 (a+b x))\right )}{8 b (c \csc (a+b x))^{4/3}} \]

[In]

Integrate[(c*Csc[a + b*x])^(-4/3),x]

[Out]

-1/8*(Csc[a + b*x]^2*(2*Cot[a + b*x]*Hypergeometric2F1[1/2, 5/6, 3/2, Cos[a + b*x]^2]*(Sin[a + b*x]^2)^(5/6) +
3*Sin[2*(a + b*x)]))/(b*(c*Csc[a + b*x])^(4/3))

Maple [F]

\[\int \frac {1}{\left (c \csc \left (x b +a \right )\right )^{\frac {4}{3}}}d x\]

[In]

int(1/(c*csc(b*x+a))^(4/3),x)

[Out]

int(1/(c*csc(b*x+a))^(4/3),x)

Fricas [F]

\[ \int \frac {1}{(c \csc (a+b x))^{4/3}} \, dx=\int { \frac {1}{\left (c \csc \left (b x + a\right )\right )^{\frac {4}{3}}} \,d x } \]

[In]

integrate(1/(c*csc(b*x+a))^(4/3),x, algorithm="fricas")

[Out]

integral((c*csc(b*x + a))^(2/3)/(c^2*csc(b*x + a)^2), x)

Sympy [F]

\[ \int \frac {1}{(c \csc (a+b x))^{4/3}} \, dx=\int \frac {1}{\left (c \csc {\left (a + b x \right )}\right )^{\frac {4}{3}}}\, dx \]

[In]

integrate(1/(c*csc(b*x+a))**(4/3),x)

[Out]

Integral((c*csc(a + b*x))**(-4/3), x)

Maxima [F]

\[ \int \frac {1}{(c \csc (a+b x))^{4/3}} \, dx=\int { \frac {1}{\left (c \csc \left (b x + a\right )\right )^{\frac {4}{3}}} \,d x } \]

[In]

integrate(1/(c*csc(b*x+a))^(4/3),x, algorithm="maxima")

[Out]

integrate((c*csc(b*x + a))^(-4/3), x)

Giac [F]

\[ \int \frac {1}{(c \csc (a+b x))^{4/3}} \, dx=\int { \frac {1}{\left (c \csc \left (b x + a\right )\right )^{\frac {4}{3}}} \,d x } \]

[In]

integrate(1/(c*csc(b*x+a))^(4/3),x, algorithm="giac")

[Out]

integrate((c*csc(b*x + a))^(-4/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(c \csc (a+b x))^{4/3}} \, dx=\int \frac {1}{{\left (\frac {c}{\sin \left (a+b\,x\right )}\right )}^{4/3}} \,d x \]

[In]

int(1/(c/sin(a + b*x))^(4/3),x)

[Out]

int(1/(c/sin(a + b*x))^(4/3), x)