Integrand size = 22, antiderivative size = 136 \[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {7 \arcsin (\cos (a+b x)-\sin (a+b x))}{64 b}+\frac {7 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{64 b}-\frac {7 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{32 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{48 b}+\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{12 b} \] Output:
-7/64*arcsin(cos(b*x+a)-sin(b*x+a))/b+7/64*ln(cos(b*x+a)+sin(b*x+a)+sin(2* b*x+2*a)^(1/2))/b-7/32*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b+7/48*sin(b*x+a)*s in(2*b*x+2*a)^(3/2)/b+1/12*cos(b*x+a)*sin(2*b*x+2*a)^(5/2)/b
Time = 0.35 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.73 \[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {-7 \arcsin (\cos (a+b x)-\sin (a+b x))+7 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )-\frac {2}{3} (10 \cos (a+b x)+9 \cos (3 (a+b x))+2 \cos (5 (a+b x))) \sqrt {\sin (2 (a+b x))}}{64 b} \] Input:
Integrate[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^(3/2),x]
Output:
(-7*ArcSin[Cos[a + b*x] - Sin[a + b*x]] + 7*Log[Cos[a + b*x] + Sin[a + b*x ] + Sqrt[Sin[2*(a + b*x)]]] - (2*(10*Cos[a + b*x] + 9*Cos[3*(a + b*x)] + 2 *Cos[5*(a + b*x)])*Sqrt[Sin[2*(a + b*x)]])/3)/(64*b)
Time = 0.55 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4785, 3042, 4789, 3042, 4790, 3042, 4793}
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.
\(\displaystyle \int \sin ^{\frac {3}{2}}(2 a+2 b x) \cos ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (2 a+2 b x)^{3/2} \cos (a+b x)^3dx\) |
\(\Big \downarrow \) 4785 |
\(\displaystyle \frac {7}{12} \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {7}{12} \int \cos (a+b x) \sin (2 a+2 b x)^{3/2}dx+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}\) |
\(\Big \downarrow \) 4789 |
\(\displaystyle \frac {7}{12} \left (\frac {3}{4} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {7}{12} \left (\frac {3}{4} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}\) |
\(\Big \downarrow \) 4790 |
\(\displaystyle \frac {7}{12} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {7}{12} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}\) |
\(\Big \downarrow \) 4793 |
\(\displaystyle \frac {7}{12} \left (\frac {3}{4} \left (\frac {1}{2} \left (\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}\right )-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}\) |
Input:
Int[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^(3/2),x]
Output:
(Cos[a + b*x]*Sin[2*a + 2*b*x]^(5/2))/(12*b) + (7*((3*((-1/2*ArcSin[Cos[a + b*x] - Sin[a + b*x]]/b + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]/(2*b))/2 - (Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(2*b)))/4 + (S in[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(4*b)))/12
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*(m + 2*p))), x] + Simp[e^2*((m + p - 1)/(m + 2*p)) Int[(e*Cos[a + b*x])^(m - 2)*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, e, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p] && GtQ[m, 1] && NeQ [m + 2*p, 0] && IntegersQ[2*m, 2*p]
Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[2*Sin[a + b*x]*((g*Sin[c + d*x])^p/(d*(2*p + 1))), x] + Simp[2*p*( g/(2*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] && !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]
Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[-2*Cos[a + b*x]*((g*Sin[c + d*x])^p/(d*(2*p + 1))), x] + Simp[2*p* (g/(2*p + 1)) Int[Cos[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] && !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]
Int[cos[(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]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 202.21 (sec) , antiderivative size = 375719658, normalized size of antiderivative = 2762644.54
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. 291 vs. \(2 (118) = 236\).
Time = 0.10 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.14 \[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {8 \, \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{5} - 4 \, \cos \left (b x + a\right )^{3} - 7 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 42 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 42 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 21 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{768 \, b} \] Input:
integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")
Output:
-1/768*(8*sqrt(2)*(32*cos(b*x + a)^5 - 4*cos(b*x + a)^3 - 7*cos(b*x + a))* sqrt(cos(b*x + a)*sin(b*x + a)) - 42*arctan(-(sqrt(2)*sqrt(cos(b*x + a)*si n(b*x + a))*(cos(b*x + a) - sin(b*x + a)) + cos(b*x + a)*sin(b*x + a))/(co s(b*x + a)^2 + 2*cos(b*x + a)*sin(b*x + a) - 1)) + 42*arctan(-(2*sqrt(2)*s qrt(cos(b*x + a)*sin(b*x + a)) - cos(b*x + a) - sin(b*x + a))/(cos(b*x + a ) - sin(b*x + a))) + 21*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))/ b
Timed out. \[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:
integrate(cos(b*x+a)**3*sin(2*b*x+2*a)**(3/2),x)
Output:
Timed out
\[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \,d x } \] 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)
\[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \,d x } \] 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. \[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int {\cos \left (a+b\,x\right )}^3\,{\sin \left (2\,a+2\,b\,x\right )}^{3/2} \,d x \] 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)
\[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \cos \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )d x \] 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),x)