Integrand size = 22, antiderivative size = 77 \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=-\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{10 b}-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {3 \cos (2 a+2 b x)}{10 b \sqrt {\sin (2 a+2 b x)}} \]
3/10*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+ b*x),2^(1/2))/b-1/5*cos(b*x+a)^2/b/sin(2*b*x+2*a)^(5/2)-3/10*cos(2*b*x+2*a )/b/sin(2*b*x+2*a)^(1/2)
Time = 1.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\frac {-12 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\frac {2 (1-6 \cos (2 (a+b x))+3 \cos (4 (a+b x))) \cot (a+b x)}{\sin ^{\frac {3}{2}}(2 (a+b x))}}{40 b} \]
(-12*EllipticE[a - Pi/4 + b*x, 2] + (2*(1 - 6*Cos[2*(a + b*x)] + 3*Cos[4*( a + b*x)])*Cot[a + b*x])/Sin[2*(a + b*x)]^(3/2))/(40*b)
Time = 0.34 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4783, 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.
\(\displaystyle \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (a+b x)^2}{\sin (2 a+2 b x)^{7/2}}dx\) |
\(\Big \downarrow \) 4783 |
\(\displaystyle \frac {3}{10} \int \frac {1}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{10} \int \frac {1}{\sin (2 a+2 b x)^{3/2}}dx-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {3}{10} \left (-\int \sqrt {\sin (2 a+2 b x)}dx-\frac {\cos (2 a+2 b x)}{b \sqrt {\sin (2 a+2 b x)}}\right )-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{10} \left (-\int \sqrt {\sin (2 a+2 b x)}dx-\frac {\cos (2 a+2 b x)}{b \sqrt {\sin (2 a+2 b x)}}\right )-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {3}{10} \left (-\frac {E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b}-\frac {\cos (2 a+2 b x)}{b \sqrt {\sin (2 a+2 b x)}}\right )-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\) |
(3*(-(EllipticE[a - Pi/4 + b*x, 2]/b) - Cos[2*a + 2*b*x]/(b*Sqrt[Sin[2*a + 2*b*x]])))/10 - Cos[a + b*x]^2/(5*b*Sin[2*a + 2*b*x]^(5/2))
3.2.76.3.1 Defintions of rubi rules used
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]
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p _), x_Symbol] :> Simp[(e*Cos[a + b*x])^m*((g*Sin[c + d*x])^(p + 1)/(2*b*g*( p + 1))), x] + Simp[e^2*((m + 2*p + 2)/(4*g^2*(p + 1))) Int[(e*Cos[a + b* x])^(m - 2)*(g*Sin[c + d*x])^(p + 2), x], x] /; FreeQ[{a, b, c, d, e, g}, x ] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p] && GtQ[m, 1] && LtQ[ p, -1] && NeQ[m + 2*p + 2, 0] && (LtQ[p, -2] || EqQ[m, 2]) && IntegersQ[2*m , 2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(226\) vs. \(2(92)=184\).
Time = 145.20 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.95
method | result | size |
default | \(\frac {\sqrt {2}\, \left (-\frac {8 \sqrt {2}}{5 \sin \left (2 x b +2 a \right )^{\frac {5}{2}}}+\frac {4 \sqrt {2}\, \left (6 \sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \sin \left (2 x b +2 a \right )^{2} \operatorname {EllipticE}\left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \sin \left (2 x b +2 a \right )^{2} \operatorname {EllipticF}\left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )+6 \sin \left (2 x b +2 a \right )^{4}-4 \sin \left (2 x b +2 a \right )^{2}-2\right )}{5 \sin \left (2 x b +2 a \right )^{\frac {5}{2}} \cos \left (2 x b +2 a \right )}\right )}{32 b}\) | \(227\) |
1/32*2^(1/2)*(-8/5*2^(1/2)/sin(2*b*x+2*a)^(5/2)+4/5*2^(1/2)/sin(2*b*x+2*a) ^(5/2)*(6*(sin(2*b*x+2*a)+1)^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x +2*a))^(1/2)*sin(2*b*x+2*a)^2*EllipticE((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1/ 2))-3*(sin(2*b*x+2*a)+1)^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a ))^(1/2)*sin(2*b*x+2*a)^2*EllipticF((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1/2))+ 6*sin(2*b*x+2*a)^4-4*sin(2*b*x+2*a)^2-2)/cos(2*b*x+2*a))/b
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 266, normalized size of antiderivative = 3.45 \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=-\frac {6 \, \sqrt {2 i} {\left (i \, \cos \left (b x + a\right )^{3} - i \, \cos \left (b x + a\right )\right )} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 \, \sqrt {-2 i} {\left (-i \, \cos \left (b x + a\right )^{3} + i \, \cos \left (b x + a\right )\right )} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 \, \sqrt {2 i} {\left (-i \, \cos \left (b x + a\right )^{3} + i \, \cos \left (b x + a\right )\right )} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 \, \sqrt {-2 i} {\left (i \, \cos \left (b x + a\right )^{3} - i \, \cos \left (b x + a\right )\right )} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + \sqrt {2} {\left (12 \, \cos \left (b x + a\right )^{4} - 18 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{40 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )} \]
-1/40*(6*sqrt(2*I)*(I*cos(b*x + a)^3 - I*cos(b*x + a))*elliptic_e(arcsin(c os(b*x + a) + I*sin(b*x + a)), -1)*sin(b*x + a) + 6*sqrt(-2*I)*(-I*cos(b*x + a)^3 + I*cos(b*x + a))*elliptic_e(arcsin(cos(b*x + a) - I*sin(b*x + a)) , -1)*sin(b*x + a) + 6*sqrt(2*I)*(-I*cos(b*x + a)^3 + I*cos(b*x + a))*elli ptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1)*sin(b*x + a) + 6*sqrt(-2 *I)*(I*cos(b*x + a)^3 - I*cos(b*x + a))*elliptic_f(arcsin(cos(b*x + a) - I *sin(b*x + a)), -1)*sin(b*x + a) + sqrt(2)*(12*cos(b*x + a)^4 - 18*cos(b*x + a)^2 + 5)*sqrt(cos(b*x + a)*sin(b*x + a)))/((b*cos(b*x + a)^3 - b*cos(b *x + a))*sin(b*x + a))
Timed out. \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\int \frac {{\cos \left (a+b\,x\right )}^2}{{\sin \left (2\,a+2\,b\,x\right )}^{7/2}} \,d x \]