\(\int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\) [162]

Optimal result

Integrand size = 20, antiderivative size = 110 \[ \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{16 b}+\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{16 b}-\frac {3 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b} \]

[Out]

-3/16*arcsin(cos(b*x+a)-sin(b*x+a))/b+3/16*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b+1/4*sin(b*x+a)*sin
(2*b*x+2*a)^(3/2)/b-3/8*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4386, 4387, 4390} \[ \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{16 b}+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}-\frac {3 \sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{8 b}+\frac {3 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{16 b} \]

[In]

Int[Cos[a + b*x]*Sin[2*a + 2*b*x]^(3/2),x]

[Out]

(-3*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(16*b) + (3*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]
)/(16*b) - (3*Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(8*b) + (Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(4*b)

Rule 4386

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] + Dist[2*p*(g/(2*p + 1)), Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; Fre
eQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]

Rule 4387

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] + Dist[2*p*(g/(2*p + 1)), Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; Fr
eeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]

Rule 4390

Int[cos[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}+\frac {3}{4} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx \\ & = -\frac {3 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}+\frac {3}{8} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = -\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{16 b}+\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{16 b}-\frac {3 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.78 \[ \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {3 \left (-\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )-2 (2 \cos (a+b x)+\cos (3 (a+b x))) \sqrt {\sin (2 (a+b x))}}{16 b} \]

[In]

Integrate[Cos[a + b*x]*Sin[2*a + 2*b*x]^(3/2),x]

[Out]

(3*(-ArcSin[Cos[a + b*x] - Sin[a + b*x]] + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]]) - 2*(2*C
os[a + b*x] + Cos[3*(a + b*x)])*Sqrt[Sin[2*(a + b*x)]])/(16*b)

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 16.62 (sec) , antiderivative size = 85939054, normalized size of antiderivative = 781264.13

method	result	size
default	\(\text {Expression too large to display}\)	\(85939054\)

[In]

int(cos(b*x+a)*sin(2*b*x+2*a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (96) = 192\).

Time = 0.26 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.55 \[ \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {8 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 6 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 6 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 3 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{64 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")

[Out]

-1/64*(8*sqrt(2)*(4*cos(b*x + a)^3 - cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) - 6*arctan(-(sqrt(2)*sqrt(c
os(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)) + 6*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))) + 3*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

Sympy [F(-1)]

Timed out. \[ \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\text {Timed out} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)*sin(2*b*x + 2*a)^(3/2), x)

Giac [F]

\[ \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^(3/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)*sin(2*b*x + 2*a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int \cos \left (a+b\,x\right )\,{\sin \left (2\,a+2\,b\,x\right )}^{3/2} \,d x \]

[In]

int(cos(a + b*x)*sin(2*a + 2*b*x)^(3/2),x)

[Out]

int(cos(a + b*x)*sin(2*a + 2*b*x)^(3/2), x)