\(\int \frac {1}{(c \sin (a+b x))^{3/2}} \, dx\) [30]

Optimal result

Integrand size = 12, antiderivative size = 73 \[ \int \frac {1}{(c \sin (a+b x))^{3/2}} \, dx=-\frac {2 \cos (a+b x)}{b c \sqrt {c \sin (a+b x)}}-\frac {2 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{b c^2 \sqrt {\sin (a+b x)}} \]

[Out]

-2*cos(b*x+a)/b/c/(c*sin(b*x+a))^(1/2)+2*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*x)*Ellipti
cE(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*(c*sin(b*x+a))^(1/2)/b/c^2/sin(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2716, 2721, 2719} \[ \int \frac {1}{(c \sin (a+b x))^{3/2}} \, dx=-\frac {2 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{b c^2 \sqrt {\sin (a+b x)}}-\frac {2 \cos (a+b x)}{b c \sqrt {c \sin (a+b x)}} \]

[In]

Int[(c*Sin[a + b*x])^(-3/2),x]

[Out]

(-2*Cos[a + b*x])/(b*c*Sqrt[c*Sin[a + b*x]]) - (2*EllipticE[(a - Pi/2 + b*x)/2, 2]*Sqrt[c*Sin[a + b*x]])/(b*c^
2*Sqrt[Sin[a + b*x]])

Rule 2716

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
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cos (a+b x)}{b c \sqrt {c \sin (a+b x)}}-\frac {\int \sqrt {c \sin (a+b x)} \, dx}{c^2} \\ & = -\frac {2 \cos (a+b x)}{b c \sqrt {c \sin (a+b x)}}-\frac {\sqrt {c \sin (a+b x)} \int \sqrt {\sin (a+b x)} \, dx}{c^2 \sqrt {\sin (a+b x)}} \\ & = -\frac {2 \cos (a+b x)}{b c \sqrt {c \sin (a+b x)}}-\frac {2 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{b c^2 \sqrt {\sin (a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(c \sin (a+b x))^{3/2}} \, dx=-\frac {2 \left (\cos (a+b x)-E\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right ) \sqrt {\sin (a+b x)}\right )}{b c \sqrt {c \sin (a+b x)}} \]

[In]

Integrate[(c*Sin[a + b*x])^(-3/2),x]

[Out]

(-2*(Cos[a + b*x] - EllipticE[(-2*a + Pi - 2*b*x)/4, 2]*Sqrt[Sin[a + b*x]]))/(b*c*Sqrt[c*Sin[a + b*x]])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.93

method	result	size
default	\(\frac {2 \sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sqrt {\sin }\left (b x +a \right )\right ) E\left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sqrt {\sin }\left (b x +a \right )\right ) F\left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\cos ^{2}\left (b x +a \right )\right )}{c \cos \left (b x +a \right ) \sqrt {c \sin \left (b x +a \right )}\, b}\)	\(141\)

[In]

int(1/(c*sin(b*x+a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/c*(2*(-sin(b*x+a)+1)^(1/2)*(2*sin(b*x+a)+2)^(1/2)*sin(b*x+a)^(1/2)*EllipticE((-sin(b*x+a)+1)^(1/2),1/2*2^(1/
2))-(-sin(b*x+a)+1)^(1/2)*(2*sin(b*x+a)+2)^(1/2)*sin(b*x+a)^(1/2)*EllipticF((-sin(b*x+a)+1)^(1/2),1/2*2^(1/2))
-2*cos(b*x+a)^2)/cos(b*x+a)/(c*sin(b*x+a))^(1/2)/b

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.48 \[ \int \frac {1}{(c \sin (a+b x))^{3/2}} \, dx=\frac {-i \, \sqrt {2} \sqrt {-i \, c} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + i \, \sqrt {2} \sqrt {i \, c} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) - 2 \, \sqrt {c \sin \left (b x + a\right )} \cos \left (b x + a\right )}{b c^{2} \sin \left (b x + a\right )} \]

[In]

integrate(1/(c*sin(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

(-I*sqrt(2)*sqrt(-I*c)*sin(b*x + a)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x +
a))) + I*sqrt(2)*sqrt(I*c)*sin(b*x + a)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(b*x + a) - I*sin(
b*x + a))) - 2*sqrt(c*sin(b*x + a))*cos(b*x + a))/(b*c^2*sin(b*x + a))

Sympy [F]

\[ \int \frac {1}{(c \sin (a+b x))^{3/2}} \, dx=\int \frac {1}{\left (c \sin {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(c*sin(b*x+a))**(3/2),x)

[Out]

Integral((c*sin(a + b*x))**(-3/2), x)

Maxima [F]

\[ \int \frac {1}{(c \sin (a+b x))^{3/2}} \, dx=\int { \frac {1}{\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(c*sin(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a))^(-3/2), x)

Giac [F]

\[ \int \frac {1}{(c \sin (a+b x))^{3/2}} \, dx=\int { \frac {1}{\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(c*sin(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a))^(-3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(c \sin (a+b x))^{3/2}} \, dx=\int \frac {1}{{\left (c\,\sin \left (a+b\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(1/(c*sin(a + b*x))^(3/2),x)

[Out]

int(1/(c*sin(a + b*x))^(3/2), x)