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

Optimal result

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

-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*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)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx=-\frac {2 \left (\cot (a+b x)-3 E\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right ) \sin ^{\frac {3}{2}}(a+b x)+\frac {3}{2} \sin (2 (a+b x))\right )}{5 b c^2 (c \sin (a+b x))^{3/2}} \] Input:

Integrate[(c*Sin[a + b*x])^(-7/2),x]
 

Output:

(-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))
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3116, 3042, 3116, 3042, 3121, 3042, 3119}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(c \sin (a+b x))^{7/2}}dx\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {3 \int \frac {1}{(c \sin (a+b x))^{3/2}}dx}{5 c^2}-\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int \frac {1}{(c \sin (a+b x))^{3/2}}dx}{5 c^2}-\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {3 \left (-\frac {\int \sqrt {c \sin (a+b x)}dx}{c^2}-\frac {2 \cos (a+b x)}{b c \sqrt {c \sin (a+b x)}}\right )}{5 c^2}-\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (-\frac {\int \sqrt {c \sin (a+b x)}dx}{c^2}-\frac {2 \cos (a+b x)}{b c \sqrt {c \sin (a+b x)}}\right )}{5 c^2}-\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {3 \left (-\frac {\sqrt {c \sin (a+b x)} \int \sqrt {\sin (a+b x)}dx}{c^2 \sqrt {\sin (a+b x)}}-\frac {2 \cos (a+b x)}{b c \sqrt {c \sin (a+b x)}}\right )}{5 c^2}-\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (-\frac {\sqrt {c \sin (a+b x)} \int \sqrt {\sin (a+b x)}dx}{c^2 \sqrt {\sin (a+b x)}}-\frac {2 \cos (a+b x)}{b c \sqrt {c \sin (a+b x)}}\right )}{5 c^2}-\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {3 \left (-\frac {2 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{b c^2 \sqrt {\sin (a+b x)}}-\frac {2 \cos (a+b x)}{b c \sqrt {c \sin (a+b x)}}\right )}{5 c^2}-\frac {2 \cos (a+b x)}{5 b c (c \sin (a+b x))^{5/2}}\)

Input:

Int[(c*Sin[a + b*x])^(-7/2),x]
 

Output:

(-2*Cos[a + b*x])/(5*b*c*(c*Sin[a + b*x])^(5/2)) + (3*((-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]])))/(5*c^2)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3116

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

rule 3119

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 3121

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

Maple [A] (verified)

Time = 3.07 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.60

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

Input:

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

Output:

1/5/c^3*(6*(1-sin(b*x+a))^(1/2)*(2+2*sin(b*x+a))^(1/2)*sin(b*x+a)^(7/2)*El 
lipticE((1-sin(b*x+a))^(1/2),1/2*2^(1/2))-3*(1-sin(b*x+a))^(1/2)*(2+2*sin( 
b*x+a))^(1/2)*sin(b*x+a)^(7/2)*EllipticF((1-sin(b*x+a))^(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*si 
n(b*x+a))^(1/2)/b
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.48 \[ \int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx=-\frac {2 \, {\left (3 \, {\left (i \, \cos \left (b x + a\right )^{2} - i\right )} \sqrt {-\frac {1}{2} i \, c} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 \, {\left (-i \, \cos \left (b x + a\right )^{2} + i\right )} \sqrt {\frac {1}{2} i \, c} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + {\left (3 \, \cos \left (b x + a\right )^{3} - 4 \, \cos \left (b x + a\right )\right )} \sqrt {c \sin \left (b x + a\right )}\right )}}{5 \, {\left (b c^{4} \cos \left (b x + a\right )^{2} - b c^{4}\right )} \sin \left (b x + a\right )} \] Input:

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

Output:

-2/5*(3*(I*cos(b*x + a)^2 - I)*sqrt(-1/2*I*c)*sin(b*x + a)*weierstrassZeta 
(4, 0, weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a))) + 3*(-I*c 
os(b*x + a)^2 + I)*sqrt(1/2*I*c)*sin(b*x + a)*weierstrassZeta(4, 0, weiers 
trassPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a))) + (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]

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

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

Output:

Integral((c*sin(a + b*x))**(-7/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(c \sin (a+b x))^{7/2}} \, dx=\frac {\sqrt {c}\, \left (\int \frac {\sqrt {\sin \left (b x +a \right )}}{\sin \left (b x +a \right )^{4}}d x \right )}{c^{4}} \] Input:

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

Output:

(sqrt(c)*int(sqrt(sin(a + b*x))/sin(a + b*x)**4,x))/c**4