Integrand size = 10, antiderivative size = 70 \[ \int \frac {1}{\sin ^{\frac {7}{2}}(a+b x)} \, dx=-\frac {6 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right )}{5 b}-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(a+b x)}-\frac {6 \cos (a+b x)}{5 b \sqrt {\sin (a+b x)}} \] Output:
6/5*EllipticE(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))/b-2/5*cos(b*x+a)/b/sin(b* x+a)^(5/2)-6/5*cos(b*x+a)/b/sin(b*x+a)^(1/2)
Time = 0.35 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\sin ^{\frac {7}{2}}(a+b x)} \, dx=\frac {2 \left (3 E\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right )-\frac {\cos (a+b x) \left (1+3 \sin ^2(a+b x)\right )}{\sin ^{\frac {5}{2}}(a+b x)}\right )}{5 b} \] Input:
Integrate[Sin[a + b*x]^(-7/2),x]
Output:
(2*(3*EllipticE[(-2*a + Pi - 2*b*x)/4, 2] - (Cos[a + b*x]*(1 + 3*Sin[a + b *x]^2))/Sin[a + b*x]^(5/2)))/(5*b)
Time = 0.52 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3116, 3042, 3116, 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.
\(\displaystyle \int \frac {1}{\sin ^{\frac {7}{2}}(a+b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x)^{7/2}}dx\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {3}{5} \int \frac {1}{\sin ^{\frac {3}{2}}(a+b x)}dx-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{5} \int \frac {1}{\sin (a+b x)^{3/2}}dx-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {3}{5} \left (-\int \sqrt {\sin (a+b x)}dx-\frac {2 \cos (a+b x)}{b \sqrt {\sin (a+b x)}}\right )-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{5} \left (-\int \sqrt {\sin (a+b x)}dx-\frac {2 \cos (a+b x)}{b \sqrt {\sin (a+b x)}}\right )-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {3}{5} \left (-\frac {2 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{b}-\frac {2 \cos (a+b x)}{b \sqrt {\sin (a+b x)}}\right )-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(a+b x)}\) |
Input:
Int[Sin[a + b*x]^(-7/2),x]
Output:
(3*((-2*EllipticE[(a - Pi/2 + b*x)/2, 2])/b - (2*Cos[a + b*x])/(b*Sqrt[Sin [a + b*x]])))/5 - (2*Cos[a + b*x])/(5*b*Sin[a + b*x]^(5/2))
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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(159\) vs. \(2(60)=120\).
Time = 2.26 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.29
method | result | size |
default | \(\frac {6 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \sin \left (b x +a \right )^{2} \operatorname {EllipticE}\left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \sin \left (b x +a \right )^{2} \operatorname {EllipticF}\left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+6 \sin \left (b x +a \right )^{4}-4 \sin \left (b x +a \right )^{2}-2}{5 \sin \left (b x +a \right )^{\frac {5}{2}} \cos \left (b x +a \right ) b}\) | \(160\) |
Input:
int(1/sin(b*x+a)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/5/sin(b*x+a)^(5/2)*(6*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin (b*x+a))^(1/2)*sin(b*x+a)^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))^(1/2)*sin(b*x+a )^2*EllipticF((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))+6*sin(b*x+a)^4-4*sin(b*x+a )^2-2)/cos(b*x+a)/b
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.04 \[ \int \frac {1}{\sin ^{\frac {7}{2}}(a+b x)} \, dx=-\frac {2 \, {\left (3 \, \sqrt {-\frac {1}{2} i} {\left (i \, \cos \left (b x + a\right )^{2} - i\right )} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 \, \sqrt {\frac {1}{2} i} {\left (-i \, \cos \left (b x + a\right )^{2} + i\right )} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + {\left (3 \, \cos \left (b x + a\right )^{3} - 4 \, \cos \left (b x + a\right )\right )} \sqrt {\sin \left (b x + a\right )}\right )}}{5 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \] Input:
integrate(1/sin(b*x+a)^(7/2),x, algorithm="fricas")
Output:
-2/5*(3*sqrt(-1/2*I)*(I*cos(b*x + a)^2 - I)*sin(b*x + a)*weierstrassZeta(4 , 0, weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a))) + 3*sqrt(1/ 2*I)*(-I*cos(b*x + a)^2 + I)*sin(b*x + a)*weierstrassZeta(4, 0, weierstras sPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a))) + (3*cos(b*x + a)^3 - 4*co s(b*x + a))*sqrt(sin(b*x + a)))/((b*cos(b*x + a)^2 - b)*sin(b*x + a))
\[ \int \frac {1}{\sin ^{\frac {7}{2}}(a+b x)} \, dx=\int \frac {1}{\sin ^{\frac {7}{2}}{\left (a + b x \right )}}\, dx \] Input:
integrate(1/sin(b*x+a)**(7/2),x)
Output:
Integral(sin(a + b*x)**(-7/2), x)
\[ \int \frac {1}{\sin ^{\frac {7}{2}}(a+b x)} \, dx=\int { \frac {1}{\sin \left (b x + a\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate(1/sin(b*x+a)^(7/2),x, algorithm="maxima")
Output:
integrate(sin(b*x + a)^(-7/2), x)
\[ \int \frac {1}{\sin ^{\frac {7}{2}}(a+b x)} \, dx=\int { \frac {1}{\sin \left (b x + a\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate(1/sin(b*x+a)^(7/2),x, algorithm="giac")
Output:
integrate(sin(b*x + a)^(-7/2), x)
Time = 26.34 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.60 \[ \int \frac {1}{\sin ^{\frac {7}{2}}(a+b x)} \, dx=-\frac {\cos \left (a+b\,x\right )\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {3}{2};\ {\cos \left (a+b\,x\right )}^2\right )}{b\,{\sin \left (a+b\,x\right )}^{5/2}} \] Input:
int(1/sin(a + b*x)^(7/2),x)
Output:
-(cos(a + b*x)*(sin(a + b*x)^2)^(5/4)*hypergeom([1/2, 9/4], 3/2, cos(a + b *x)^2))/(b*sin(a + b*x)^(5/2))
\[ \int \frac {1}{\sin ^{\frac {7}{2}}(a+b x)} \, dx=\int \frac {\sqrt {\sin \left (b x +a \right )}}{\sin \left (b x +a \right )^{4}}d x \] Input:
int(1/sin(b*x+a)^(7/2),x)
Output:
int(sqrt(sin(a + b*x))/sin(a + b*x)**4,x)