\(\int \frac {\cos ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\) [179]

Optimal result

Integrand size = 22, antiderivative size = 84 \[ \int \frac {\cos ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{8 b}+\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{8 b}+\frac {\cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4382, 4390} \[ \int \frac {\cos ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{8 b}+\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{4 b}+\frac {3 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{8 b} \]

[In]

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

[Out]

(-3*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(8*b) + (3*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]])
/(8*b) + (Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(4*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 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 {\cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}+\frac {3}{4} \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))}{8 b}+\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{8 b}+\frac {\cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\frac {-3 \arcsin (\cos (a+b x)-\sin (a+b x))+3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )+\csc (a+b x) \sin ^{\frac {3}{2}}(2 (a+b x))}{8 b} \]

[In]

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

[Out]

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

Maple [B] (warning: unable to verify)

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

Time = 36.61 (sec) , antiderivative size = 213968312, normalized size of antiderivative = 2547241.81

method	result	size
default	\(\text {Expression too large to display}\)	\(213968312\)

[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. 268 vs. \(2 (74) = 148\).

Time = 0.26 (sec) , antiderivative size = 268, normalized size of antiderivative = 3.19 \[ \int \frac {\cos ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\frac {8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \cos \left (b x + a\right ) + 6 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 6 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - 3 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{32 \, b} \]

[In]

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

[Out]

1/32*(8*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*cos(b*x + a) + 6*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)) - 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. \[ \int \frac {\cos ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)