Integrand size = 20, antiderivative size = 53 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {2 \sin (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}} \] Output:
-1/3*cos(b*x+a)/b/sin(2*b*x+2*a)^(3/2)+2/3*sin(b*x+a)/b/sin(2*b*x+2*a)^(1/ 2)
Time = 0.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {\left (-\frac {1}{12} \cot (a+b x) \csc (a+b x)+\frac {1}{4} \sec (a+b x)\right ) \sqrt {\sin (2 (a+b x))}}{b} \] Input:
Integrate[Cos[a + b*x]/Sin[2*a + 2*b*x]^(5/2),x]
Output:
((-1/12*(Cot[a + b*x]*Csc[a + b*x]) + Sec[a + b*x]/4)*Sqrt[Sin[2*(a + b*x) ]])/b
Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4791, 3042, 4780}
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.
| \(\displaystyle \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{5/2}}dx\) |
| \(\Big \downarrow \) 4791 |
| \(\displaystyle \frac {2}{3} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{3} \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^{3/2}}dx-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
| \(\Big \downarrow \) 4780 |
| \(\displaystyle \frac {2 \sin (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
Input:
Int[Cos[a + b*x]/Sin[2*a + 2*b*x]^(5/2),x]
Output:
-1/3*Cos[a + b*x]/(b*Sin[2*a + 2*b*x]^(3/2)) + (2*Sin[a + b*x])/(3*b*Sqrt[ Sin[2*a + 2*b*x]])
Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^( p_), x_Symbol] :> Simp[(e*Sin[a + b*x])^m*((g*Sin[c + d*x])^(p + 1)/(b*g*m) ), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b , 2] && !IntegerQ[p] && EqQ[m + 2*p + 2, 0]
Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[Cos[a + b*x]*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + Simp [(2*p + 3)/(2*g*(p + 1)) Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p + 1), x], x ] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !Int egerQ[p] && LtQ[p, -1] && IntegerQ[2*p]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 71.53 (sec) , antiderivative size = 194, normalized size of antiderivative = 3.66
| method | result | size |
| default | \(-\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}{\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1}}\, \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1\right ) \left (2 \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+1\right )}{24 b \tan \left (\frac {a}{2}+\frac {b x}{2}\right ) \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}}\) | \(194\) |
Input:
int(cos(b*x+a)/sin(2*b*x+2*a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/24/b*(-tan(1/2*a+1/2*b*x)/(tan(1/2*a+1/2*b*x)^2-1))^(1/2)*(tan(1/2*a+1/ 2*b*x)^2-1)/tan(1/2*a+1/2*b*x)*(2*(tan(1/2*a+1/2*b*x)+1)^(1/2)*(-2*tan(1/2 *a+1/2*b*x)+2)^(1/2)*(-tan(1/2*a+1/2*b*x))^(1/2)*EllipticF((tan(1/2*a+1/2* b*x)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*b*x)-tan(1/2*a+1/2*b*x)^4+1)/(tan (1/2*a+1/2*b*x)*(tan(1/2*a+1/2*b*x)^2-1))^(1/2)/(tan(1/2*a+1/2*b*x)^3-tan( 1/2*a+1/2*b*x))^(1/2)
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.40 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {4 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 3\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 4 \, \cos \left (b x + a\right )}{12 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \] Input:
integrate(cos(b*x+a)/sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")
Output:
1/12*(4*cos(b*x + a)^3 + sqrt(2)*(4*cos(b*x + a)^2 - 3)*sqrt(cos(b*x + a)* sin(b*x + a)) - 4*cos(b*x + a))/(b*cos(b*x + a)^3 - b*cos(b*x + a))
Timed out. \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \] Input:
integrate(cos(b*x+a)/sin(2*b*x+2*a)**(5/2),x)
Output:
Timed out
\[ \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(cos(b*x+a)/sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")
Output:
integrate(cos(b*x + a)/sin(2*b*x + 2*a)^(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 7875 vs. \(2 (45) = 90\).
Time = 46.02 (sec) , antiderivative size = 7875, normalized size of antiderivative = 148.58 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)/sin(2*b*x+2*a)^(5/2),x, algorithm="giac")
Output:
1/48*sqrt(2)*sqrt(-tan(1/2*b*x)^4*tan(1/2*a)^3 - tan(1/2*b*x)^3*tan(1/2*a) ^4 + tan(1/2*b*x)^4*tan(1/2*a) + 6*tan(1/2*b*x)^3*tan(1/2*a)^2 + 6*tan(1/2 *b*x)^2*tan(1/2*a)^3 + tan(1/2*b*x)*tan(1/2*a)^4 - tan(1/2*b*x)^3 - 6*tan( 1/2*b*x)^2*tan(1/2*a) - 6*tan(1/2*b*x)*tan(1/2*a)^2 - tan(1/2*a)^3 + tan(1 /2*b*x) + tan(1/2*a))*(((((((sqrt(2)*tan(1/2*a)^57 + 18*sqrt(2)*tan(1/2*a) ^55 + 132*sqrt(2)*tan(1/2*a)^53 + 374*sqrt(2)*tan(1/2*a)^51 - 1375*sqrt(2) *tan(1/2*a)^49 - 19620*sqrt(2)*tan(1/2*a)^47 - 108560*sqrt(2)*tan(1/2*a)^4 5 - 399740*sqrt(2)*tan(1/2*a)^43 - 1096755*sqrt(2)*tan(1/2*a)^41 - 2340250 *sqrt(2)*tan(1/2*a)^39 - 3941740*sqrt(2)*tan(1/2*a)^37 - 5204670*sqrt(2)*t an(1/2*a)^35 - 5163155*sqrt(2)*tan(1/2*a)^33 - 3268760*sqrt(2)*tan(1/2*a)^ 31 + 3268760*sqrt(2)*tan(1/2*a)^27 + 5163155*sqrt(2)*tan(1/2*a)^25 + 52046 70*sqrt(2)*tan(1/2*a)^23 + 3941740*sqrt(2)*tan(1/2*a)^21 + 2340250*sqrt(2) *tan(1/2*a)^19 + 1096755*sqrt(2)*tan(1/2*a)^17 + 399740*sqrt(2)*tan(1/2*a) ^15 + 108560*sqrt(2)*tan(1/2*a)^13 + 19620*sqrt(2)*tan(1/2*a)^11 + 1375*sq rt(2)*tan(1/2*a)^9 - 374*sqrt(2)*tan(1/2*a)^7 - 132*sqrt(2)*tan(1/2*a)^5 - 18*sqrt(2)*tan(1/2*a)^3 - sqrt(2)*tan(1/2*a))*tan(1/2*b*x)/(tan(1/2*a)^51 + 23*tan(1/2*a)^49 + 252*tan(1/2*a)^47 + 1748*tan(1/2*a)^45 + 8602*tan(1/ 2*a)^43 + 31878*tan(1/2*a)^41 + 92092*tan(1/2*a)^39 + 211508*tan(1/2*a)^37 + 389367*tan(1/2*a)^35 + 572033*tan(1/2*a)^33 + 653752*tan(1/2*a)^31 + 53 4888*tan(1/2*a)^29 + 208012*tan(1/2*a)^27 - 208012*tan(1/2*a)^25 - 5348...
Time = 22.77 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.96 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {2\,\sqrt {\sin \left (2\,a+2\,b\,x\right )}\,\left (3\,\cos \left (a+b\,x\right )-6\,\cos \left (3\,a+3\,b\,x\right )+4\,\cos \left (5\,a+5\,b\,x\right )-\cos \left (7\,a+7\,b\,x\right )\right )}{3\,b\,\left (4\,\cos \left (2\,a+2\,b\,x\right )+4\,\cos \left (4\,a+4\,b\,x\right )-4\,\cos \left (6\,a+6\,b\,x\right )+\cos \left (8\,a+8\,b\,x\right )-5\right )} \] Input:
int(cos(a + b*x)/sin(2*a + 2*b*x)^(5/2),x)
Output:
-(2*sin(2*a + 2*b*x)^(1/2)*(3*cos(a + b*x) - 6*cos(3*a + 3*b*x) + 4*cos(5* a + 5*b*x) - cos(7*a + 7*b*x)))/(3*b*(4*cos(2*a + 2*b*x) + 4*cos(4*a + 4*b *x) - 4*cos(6*a + 6*b*x) + cos(8*a + 8*b*x) - 5))
\[ \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int \frac {\sqrt {\sin \left (2 b x +2 a \right )}\, \cos \left (b x +a \right )}{\sin \left (2 b x +2 a \right )^{3}}d x \] Input:
int(cos(b*x+a)/sin(2*b*x+2*a)^(5/2),x)
Output:
int((sqrt(sin(2*a + 2*b*x))*cos(a + b*x))/sin(2*a + 2*b*x)**3,x)