\(\int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx\) [32]

Optimal result

Integrand size = 12, antiderivative size = 105 \[ \int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx=-\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}}-\frac {6 \cos (a+b x)}{5 b c^3 \sqrt {c \sin (a+b x)}}-\frac {6 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b c^4 \sqrt {\sin (a+b x)}} \]

[Out]

-2/5*cos(b*x+a)/b/c/(c*sin(b*x+a))^(5/2)-6/5*cos(b*x+a)/b/c^3/(c*sin(b*x+a))^(1/2)+6/5*(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)*EllipticE(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*(c*sin(b*x+a))^(1/2)/b/c^4
/sin(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2716, 2721, 2719} \[ \int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx=-\frac {6 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b c^4 \sqrt {\sin (a+b x)}}-\frac {6 \cos (a+b x)}{5 b c^3 \sqrt {c \sin (a+b x)}}-\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}} \]

[In]

Int[(c*Sin[a + b*x])^(-7/2),x]

[Out]

(-2*Cos[a + b*x])/(5*b*c*(c*Sin[a + b*x])^(5/2)) - (6*Cos[a + b*x])/(5*b*c^3*Sqrt[c*Sin[a + b*x]]) - (6*Ellipt
icE[(a - Pi/2 + b*x)/2, 2]*Sqrt[c*Sin[a + b*x]])/(5*b*c^4*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)}{5 b c (c \sin (a+b x))^{5/2}}+\frac {3 \int \frac {1}{(c \sin (a+b x))^{3/2}} \, dx}{5 c^2} \\ & = -\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}}-\frac {6 \cos (a+b x)}{5 b c^3 \sqrt {c \sin (a+b x)}}-\frac {3 \int \sqrt {c \sin (a+b x)} \, dx}{5 c^4} \\ & = -\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}}-\frac {6 \cos (a+b x)}{5 b c^3 \sqrt {c \sin (a+b x)}}-\frac {\left (3 \sqrt {c \sin (a+b x)}\right ) \int \sqrt {\sin (a+b x)} \, dx}{5 c^4 \sqrt {\sin (a+b x)}} \\ & = -\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}}-\frac {6 \cos (a+b x)}{5 b c^3 \sqrt {c \sin (a+b x)}}-\frac {6 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b c^4 \sqrt {\sin (a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx=-\frac {2 \left (\cot (a+b x)-3 E\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right ) \sin ^{\frac {3}{2}}(a+b x)+\frac {3}{2} \sin (2 (a+b x))\right )}{5 b c^2 (c \sin (a+b x))^{3/2}} \]

[In]

Integrate[(c*Sin[a + b*x])^(-7/2),x]

[Out]

(-2*(Cot[a + b*x] - 3*EllipticE[(-2*a + Pi - 2*b*x)/4, 2]*Sin[a + b*x]^(3/2) + (3*Sin[2*(a + b*x)])/2))/(5*b*c
^2*(c*Sin[a + b*x])^(3/2))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.60

method	result	size
default	\(\frac {6 \sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (b x +a \right )\right ) E\left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (b x +a \right )\right ) F\left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+6 \left (\sin ^{5}\left (b x +a \right )\right )-4 \left (\sin ^{3}\left (b x +a \right )\right )-2 \sin \left (b x +a \right )}{5 c^{3} \sin \left (b x +a \right )^{3} \cos \left (b x +a \right ) \sqrt {c \sin \left (b x +a \right )}\, b}\)	\(168\)

[In]

int(1/(c*sin(b*x+a))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/5/c^3*(6*(-sin(b*x+a)+1)^(1/2)*(2*sin(b*x+a)+2)^(1/2)*sin(b*x+a)^(7/2)*EllipticE((-sin(b*x+a)+1)^(1/2),1/2*2
^(1/2))-3*(-sin(b*x+a)+1)^(1/2)*(2*sin(b*x+a)+2)^(1/2)*sin(b*x+a)^(7/2)*EllipticF((-sin(b*x+a)+1)^(1/2),1/2*2^
(1/2))+6*sin(b*x+a)^5-4*sin(b*x+a)^3-2*sin(b*x+a))/sin(b*x+a)^3/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 = 0.11 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.62 \[ \int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx=-\frac {3 \, {\left (i \, \sqrt {2} \cos \left (b x + a\right )^{2} - i \, \sqrt {2}\right )} \sqrt {-i \, c} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 \, {\left (-i \, \sqrt {2} \cos \left (b x + a\right )^{2} + i \, \sqrt {2}\right )} \sqrt {i \, c} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + 2 \, {\left (3 \, \cos \left (b x + a\right )^{3} - 4 \, \cos \left (b x + a\right )\right )} \sqrt {c \sin \left (b x + a\right )}}{5 \, {\left (b c^{4} \cos \left (b x + a\right )^{2} - b c^{4}\right )} \sin \left (b x + a\right )} \]

[In]

integrate(1/(c*sin(b*x+a))^(7/2),x, algorithm="fricas")

[Out]

-1/5*(3*(I*sqrt(2)*cos(b*x + a)^2 - I*sqrt(2))*sqrt(-I*c)*sin(b*x + a)*weierstrassZeta(4, 0, weierstrassPInver
se(4, 0, cos(b*x + a) + I*sin(b*x + a))) + 3*(-I*sqrt(2)*cos(b*x + a)^2 + I*sqrt(2))*sqrt(I*c)*sin(b*x + a)*we
ierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a))) + 2*(3*cos(b*x + a)^3 - 4*cos(b*
x + a))*sqrt(c*sin(b*x + a)))/((b*c^4*cos(b*x + a)^2 - b*c^4)*sin(b*x + a))

Sympy [F]

\[ \int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx=\int \frac {1}{\left (c \sin {\left (a + b x \right )}\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate(1/(c*sin(b*x+a))**(7/2),x)

[Out]

Integral((c*sin(a + b*x))**(-7/2), x)

Maxima [F]

\[ \int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx=\int { \frac {1}{\left (c \sin \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate(1/(c*sin(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a))^(-7/2), x)

Giac [F]

\[ \int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx=\int { \frac {1}{\left (c \sin \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate(1/(c*sin(b*x+a))^(7/2),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a))^(-7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx=\int \frac {1}{{\left (c\,\sin \left (a+b\,x\right )\right )}^{7/2}} \,d x \]

[In]

int(1/(c*sin(a + b*x))^(7/2),x)

[Out]

int(1/(c*sin(a + b*x))^(7/2), x)