\(\int \frac {1}{\sqrt [3]{c \sin (a+b x)}} \, dx\) [36]

Optimal result

Integrand size = 12, antiderivative size = 252 \[ \int \frac {1}{\sqrt [3]{c \sin (a+b x)}} \, dx=-\frac {3 \sqrt {3-i \sqrt {3}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt {1-\frac {(c \sin (a+b x))^{2/3}}{c^{2/3}}}}{\sqrt {3-i \sqrt {3}}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right ) \sec (a+b x) \sqrt {1-\frac {(c \sin (a+b x))^{2/3}}{c^{2/3}}} \sqrt {\frac {i+\sqrt {3}}{3 i+\sqrt {3}}+\frac {2 (c \sin (a+b x))^{2/3}}{\left (3-i \sqrt {3}\right ) c^{2/3}}} \sqrt {\frac {i-\sqrt {3}}{3 i-\sqrt {3}}+\frac {2 (c \sin (a+b x))^{2/3}}{\left (3+i \sqrt {3}\right ) c^{2/3}}}}{\sqrt {2} b \sqrt [3]{c}} \]

[Out]

-3/2*EllipticF(2^(1/2)*(1-(c*sin(b*x+a))^(2/3)/c^(2/3))^(1/2)/(3-I*3^(1/2))^(1/2),((3*I+3^(1/2))/(3*I-3^(1/2))
)^(1/2))*sec(b*x+a)*(1-(c*sin(b*x+a))^(2/3)/c^(2/3))^(1/2)*((I-3^(1/2))/(3*I-3^(1/2))+2*(c*sin(b*x+a))^(2/3)/c
^(2/3)/(3+I*3^(1/2)))^(1/2)*(3-I*3^(1/2))^(1/2)*(2*(c*sin(b*x+a))^(2/3)/c^(2/3)/(3-I*3^(1/2))+(3^(1/2)+I)/(3*I
+3^(1/2)))^(1/2)/b/c^(1/3)*2^(1/2)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.23, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2722} \[ \int \frac {1}{\sqrt [3]{c \sin (a+b x)}} \, dx=\frac {3 \cos (a+b x) (c \sin (a+b x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sin ^2(a+b x)\right )}{2 b c \sqrt {\cos ^2(a+b x)}} \]

[In]

Int[(c*Sin[a + b*x])^(-1/3),x]

[Out]

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

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]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.22 \[ \int \frac {1}{\sqrt [3]{c \sin (a+b x)}} \, dx=\frac {3 \sqrt {\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sin ^2(a+b x)\right ) \tan (a+b x)}{2 b \sqrt [3]{c \sin (a+b x)}} \]

[In]

Integrate[(c*Sin[a + b*x])^(-1/3),x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((c*sin(a + b*x))**(-1/3), x)

Maxima [F]

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

[In]

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

[Out]

integrate((c*sin(b*x + a))^(-1/3), x)

Giac [F]

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

[In]

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

[Out]

integrate((c*sin(b*x + a))^(-1/3), x)

Mupad [F(-1)]

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

[In]

int(1/(c*sin(a + b*x))^(1/3),x)

[Out]

int(1/(c*sin(a + b*x))^(1/3), x)