Integrand size = 22, antiderivative size = 82 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{4 b}+\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{4 b}-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \] Output:
1/4*arcsin(cos(b*x+a)-sin(b*x+a))/b+1/4*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x +2*a)^(1/2))/b-cos(b*x+a)/b/sin(2*b*x+2*a)^(1/2)
Time = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )-2 \csc (a+b x) \sqrt {\sin (2 (a+b x))}}{4 b} \] Input:
Integrate[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(3/2),x]
Output:
(ArcSin[Cos[a + b*x] - Sin[a + b*x]] + Log[Cos[a + b*x] + Sin[a + b*x] + S qrt[Sin[2*(a + b*x)]]] - 2*Csc[a + b*x]*Sqrt[Sin[2*(a + b*x)]])/(4*b)
Time = 0.41 (sec) , antiderivative size = 87, 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, 4781, 3042, 4795, 3042, 4794}
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.
\(\displaystyle \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (a+b x)^3}{\sin (2 a+2 b x)^{3/2}}dx\) |
\(\Big \downarrow \) 4781 |
\(\displaystyle -\frac {1}{4} \int \sec (a+b x) \sqrt {\sin (2 a+2 b x)}dx-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{4} \int \frac {\sqrt {\sin (2 a+2 b x)}}{\cos (a+b x)}dx-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}\) |
\(\Big \downarrow \) 4795 |
\(\displaystyle -\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}\) |
\(\Big \downarrow \) 4794 |
\(\displaystyle \frac {1}{2} \left (\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}+\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}\right )-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}\) |
Input:
Int[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(3/2),x]
Output:
(ArcSin[Cos[a + b*x] - Sin[a + b*x]]/(2*b) + Log[Cos[a + b*x] + Sin[a + b* x] + Sqrt[Sin[2*a + 2*b*x]]]/(2*b))/2 - Cos[a + b*x]/(b*Sqrt[Sin[2*a + 2*b *x]])
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p _), x_Symbol] :> Simp[e^2*(e*Cos[a + b*x])^(m - 2)*((g*Sin[c + d*x])^(p + 1 )/(2*b*g*(p + 1))), x] + Simp[e^4*((m + p - 1)/(4*g^2*(p + 1))) Int[(e*Co s[a + b*x])^(m - 4)*(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, 2] && LtQ[p, -1] && (GtQ[m, 3] || EqQ[p, -3/2]) && IntegersQ[2*m, 2*p]
Int[sin[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Sim p[-ArcSin[Cos[a + b*x] - Sin[a + b*x]]/d, x] - Simp[Log[Cos[a + b*x] + Sin[ a + b*x] + Sqrt[Sin[c + d*x]]]/d, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/cos[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[2*g Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; FreeQ[{ a, b, c, d, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p] && IntegerQ[2*p]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 8.30 (sec) , antiderivative size = 69948460, normalized size of antiderivative = 853030.00
Input:
int(cos(b*x+a)^3/sin(2*b*x+2*a)^(3/2),x,method=_RETURNVERBOSE)
Output:
result too large to display
Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (74) = 148\).
Time = 0.09 (sec) , antiderivative size = 295, normalized size of antiderivative = 3.60 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {2 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) \sin \left (b x + a\right ) - 2 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) \sin \left (b x + a\right ) + \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 8 \, \sin \left (b x + a\right )}{16 \, b \sin \left (b x + a\right )} \] Input:
integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")
Output:
-1/16*(2*arctan(-(sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*(cos(b*x + a) - sin(b*x + a)) + cos(b*x + a)*sin(b*x + a))/(cos(b*x + a)^2 + 2*cos(b*x + a )*sin(b*x + a) - 1))*sin(b*x + a) - 2*arctan(-(2*sqrt(2)*sqrt(cos(b*x + a) *sin(b*x + a)) - cos(b*x + a) - sin(b*x + a))/(cos(b*x + a) - sin(b*x + a) ))*sin(b*x + a) + log(-32*cos(b*x + a)^4 + 4*sqrt(2)*(4*cos(b*x + a)^3 - ( 4*cos(b*x + a)^2 + 1)*sin(b*x + a) - 5*cos(b*x + a))*sqrt(cos(b*x + a)*sin (b*x + a)) + 32*cos(b*x + a)^2 + 16*cos(b*x + a)*sin(b*x + a) + 1)*sin(b*x + a) + 8*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)) + 8*sin(b*x + a))/(b*sin (b*x + a))
Timed out. \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \] Input:
integrate(cos(b*x+a)**3/sin(2*b*x+2*a)**(3/2),x)
Output:
Timed out
\[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(3/2),x, algorithm="maxima")
Output:
integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(3/2), x)
\[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(3/2),x, algorithm="giac")
Output:
integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(3/2), x)
Timed out. \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int \frac {{\cos \left (a+b\,x\right )}^3}{{\sin \left (2\,a+2\,b\,x\right )}^{3/2}} \,d x \] Input:
int(cos(a + b*x)^3/sin(2*a + 2*b*x)^(3/2),x)
Output:
int(cos(a + b*x)^3/sin(2*a + 2*b*x)^(3/2), x)
\[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int \frac {\sqrt {\sin \left (2 b x +2 a \right )}\, \cos \left (b x +a \right )^{3}}{\sin \left (2 b x +2 a \right )^{2}}d x \] Input:
int(cos(b*x+a)^3/sin(2*b*x+2*a)^(3/2),x)
Output:
int((sqrt(sin(2*a + 2*b*x))*cos(a + b*x)**3)/sin(2*a + 2*b*x)**2,x)