\(\int \cos ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx\) [178]

Optimal result

Integrand size = 22, antiderivative size = 110 \[ \int \cos ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{32 b}-\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{32 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}+\frac {\cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{8 b} \]

[Out]

-5/32*arcsin(cos(b*x+a)-sin(b*x+a))/b-5/32*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b+1/8*cos(b*x+a)*sin
(2*b*x+2*a)^(3/2)/b+5/16*sin(b*x+a)*sin(2*b*x+2*a)^(1/2)/b

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4382, 4386, 4391} \[ \int \cos ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{32 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}+\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{8 b}-\frac {5 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{32 b} \]

[In]

Int[Cos[a + b*x]^3*Sqrt[Sin[2*a + 2*b*x]],x]

[Out]

(-5*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(32*b) - (5*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]
)/(32*b) + (5*Sin[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(16*b) + (Cos[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(8*b)

Rule 4382

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] + Dist[e^2*((m + p - 1)/(m + 2*p)), Int[(e*Co
s[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]

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 4391

Int[sin[(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 {\cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{8 b}+\frac {5}{8} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx \\ & = \frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}+\frac {\cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{8 b}+\frac {5}{16} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = -\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{32 b}-\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{32 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}+\frac {\cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{8 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int \cos ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {-5 \left (\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )+2 \sqrt {\sin (2 (a+b x))} (6 \sin (a+b x)+\sin (3 (a+b x)))}{32 b} \]

[In]

Integrate[Cos[a + b*x]^3*Sqrt[Sin[2*a + 2*b*x]],x]

[Out]

(-5*(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*Sqrt
[Sin[2*(a + b*x)]]*(6*Sin[a + b*x] + Sin[3*(a + b*x)]))/(32*b)

Maple [B] (warning: unable to verify)

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

Time = 18.15 (sec) , antiderivative size = 103475646, normalized size of antiderivative = 940687.69

method	result	size
default	\(\text {Expression too large to display}\)	\(103475646\)

[In]

int(cos(b*x+a)^3*sin(2*b*x+2*a)^(1/2),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. 280 vs. \(2 (96) = 192\).

Time = 0.28 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.55 \[ \int \cos ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {8 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right ) + 10 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 10 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 5 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{128 \, b} \]

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="fricas")

[Out]

1/128*(8*sqrt(2)*(4*cos(b*x + a)^2 + 5)*sqrt(cos(b*x + a)*sin(b*x + a))*sin(b*x + a) + 10*arctan(-(sqrt(2)*sqr
t(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*co
s(b*x + a)*sin(b*x + a) - 1)) - 10*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))) + 5*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. \[ \int \cos ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(b*x+a)**3*sin(2*b*x+2*a)**(1/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \cos ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:The choice was done assuming 0=[0]ext_reduce Error: Bad Argument TypeDone

Mupad [F(-1)]

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

[In]

int(cos(a + b*x)^3*sin(2*a + 2*b*x)^(1/2),x)

[Out]

int(cos(a + b*x)^3*sin(2*a + 2*b*x)^(1/2), x)