Integrand size = 22, antiderivative size = 55 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=-\frac {\cos ^3(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{5 b \sqrt {\sin (2 a+2 b x)}} \]
Time = 0.56 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.64 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=-\frac {\csc (a+b x) \left (4+\csc ^2(a+b x)\right ) \sqrt {\sin (2 (a+b x))}}{40 b} \]
Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 4783, 3042, 4779}
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.
\(\displaystyle \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (a+b x)^3}{\sin (2 a+2 b x)^{7/2}}dx\) |
\(\Big \downarrow \) 4783 |
\(\displaystyle \frac {1}{5} \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx-\frac {\cos ^3(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{3/2}}dx-\frac {\cos ^3(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 4779 |
\(\displaystyle -\frac {\cos ^3(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{5 b \sqrt {\sin (2 a+2 b x)}}\) |
3.2.82.3.1 Defintions of rubi rules used
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)/(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_)]*(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]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 230.85 (sec) , antiderivative size = 482, normalized size of antiderivative = 8.76
method | result | size |
default | \(\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1}}\, \left (16 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticE}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-8 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}+\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}+8 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}+\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-8 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\right )}{160 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3} \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, b}\) | \(482\) |
1/160*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(1/2)/tan(1/2*a+1/2*x *b)^3*(16*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*(tan(1/2*a+1 /2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/ 2)*EllipticE((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b)^ 2-8*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*(tan(1/2*a+1/2*x*b )+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/2)*Ell ipticF((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b)^2-(tan (1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b)^6+(tan( 1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b)^4+8*(tan (1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^4+(tan(1/2* a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b)^2-8*(tan(1/2 *a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^2-(tan(1/2*a+1/ 2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2))/(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/ 2*x*b))^(1/2)/b
Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 4 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )}{40 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \]
-1/40*(sqrt(2)*(4*cos(b*x + a)^2 - 5)*sqrt(cos(b*x + a)*sin(b*x + a)) + 4* (cos(b*x + a)^2 - 1)*sin(b*x + a))/((b*cos(b*x + a)^2 - b)*sin(b*x + a))
Timed out. \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
Time = 23.66 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.69 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left (-{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}{5\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^3} \]