Integrand size = 22, antiderivative size = 81 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=-\frac {\cos ^3(a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {4 \sin (a+b x)}{21 b \sqrt {\sin (2 a+2 b x)}} \]
-1/7*cos(b*x+a)^3/b/sin(2*b*x+2*a)^(7/2)-2/21*cos(b*x+a)/b/sin(2*b*x+2*a)^ (3/2)+4/21*sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)
Time = 0.80 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=\frac {(5-12 \cos (2 (a+b x))+4 \cos (4 (a+b x))) \csc ^4(a+b x) \sec (a+b x) \sqrt {\sin (2 (a+b x))}}{336 b} \]
((5 - 12*Cos[2*(a + b*x)] + 4*Cos[4*(a + b*x)])*Csc[a + b*x]^4*Sec[a + b*x ]*Sqrt[Sin[2*(a + b*x)]])/(336*b)
Time = 0.39 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06, 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, 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.
\(\displaystyle \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (a+b x)^3}{\sin (2 a+2 b x)^{9/2}}dx\) |
\(\Big \downarrow \) 4783 |
\(\displaystyle \frac {2}{7} \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)}dx-\frac {\cos ^3(a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{7} \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{5/2}}dx-\frac {\cos ^3(a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 4791 |
\(\displaystyle \frac {2}{7} \left (\frac {2}{3} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\cos ^3(a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{7} \left (\frac {2}{3} \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^{3/2}}dx-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\cos ^3(a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 4780 |
\(\displaystyle \frac {2}{7} \left (\frac {2 \sin (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\cos ^3(a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\) |
(2*(-1/3*Cos[a + b*x]/(b*Sin[2*a + 2*b*x]^(3/2)) + (2*Sin[a + b*x])/(3*b*S qrt[Sin[2*a + 2*b*x]])))/7 - Cos[a + b*x]^3/(7*b*Sin[2*a + 2*b*x]^(7/2))
3.2.83.3.1 Defintions of rubi rules used
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_)]*(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]
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]
Timed out.
\[\int \frac {\cos \left (x b +a \right )^{3}}{\sin \left (2 x b +2 a \right )^{\frac {9}{2}}}d x\]
Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.28 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=\frac {32 \, \cos \left (b x + a\right )^{5} - 64 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 56 \, \cos \left (b x + a\right )^{2} + 21\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )}{336 \, {\left (b \cos \left (b x + a\right )^{5} - 2 \, b \cos \left (b x + a\right )^{3} + b \cos \left (b x + a\right )\right )}} \]
1/336*(32*cos(b*x + a)^5 - 64*cos(b*x + a)^3 + sqrt(2)*(32*cos(b*x + a)^4 - 56*cos(b*x + a)^2 + 21)*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*cos(b*x + a ))/(b*cos(b*x + a)^5 - 2*b*cos(b*x + a)^3 + b*cos(b*x + a))
Timed out. \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
Time = 23.99 (sec) , antiderivative size = 302, normalized size of antiderivative = 3.73 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=-\frac {5\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{84\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,3{}\mathrm {i}}{14\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^3}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{7\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^4}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {5{}\mathrm {i}}{84\,b}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,4{}\mathrm {i}}{21\,b}\right )\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )} \]
(exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i) /2)^(1/2)*3i)/(14*b*(exp(a*2i + b*x*2i)*1i - 1i)^3) - (5*exp(a*1i + b*x*1i )*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/(84*b*( exp(a*2i + b*x*2i)*1i - 1i)^2) - (exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i )*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/(7*b*(exp(a*2i + b*x*2i)*1i - 1i)^4) - (exp(a*1i + b*x*1i)*(5i/(84*b) - (exp(a*2i + b*x*2i)*4i)/(21*b))* ((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/((exp(a*2 i + b*x*2i) + 1)*(exp(a*2i + b*x*2i)*1i - 1i))