\(\int \frac {\cos ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) [175]

Optimal result

Integrand size = 22, antiderivative size = 48 \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {\operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{6 b}-\frac {\cos ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

[Out]

-1/6*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))/b-1/3*cos(b*x+a)^2/b/s
in(2*b*x+2*a)^(3/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 48, 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 = {4380, 2720} \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {\operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{6 b}-\frac {\cos ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

[In]

Int[Cos[a + b*x]^2/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

EllipticF[a - Pi/4 + b*x, 2]/(6*b) - Cos[a + b*x]^2/(3*b*Sin[2*a + 2*b*x]^(3/2))

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 4380

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(e*Cos[a + b*
x])^m*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + Dist[e^2*((m + 2*p + 2)/(4*g^2*(p + 1))), Int[(e*Cos[a
+ b*x])^(m - 2)*(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, 1] && LtQ[p, -1] && NeQ[m + 2*p + 2, 0] && (LtQ[p, -2] || EqQ[m, 2]) && Inte
gersQ[2*m, 2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {1}{6} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = \frac {\operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{6 b}-\frac {\cos ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.89 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.29 \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\left (\csc ^2(a+b x)-2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sqrt {\sin (2 (a+b x))}}{12 b} \]

[In]

Integrate[Cos[a + b*x]^2/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

-1/12*((Csc[a + b*x]^2 - 2*Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*Sqrt[Sin[2*
(a + b*x)]])/b

Maple [A] (verified)

Time = 27.51 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.56

method	result	size
default	\(\frac {\sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sin \left (2 x b +2 a \right )-2 \cos \left (2 x b +2 a \right )^{2}-2 \cos \left (2 x b +2 a \right )}{12 \sin \left (2 x b +2 a \right )^{\frac {3}{2}} \cos \left (2 x b +2 a \right ) b}\)	\(123\)

[In]

int(cos(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/12/sin(2*b*x+2*a)^(3/2)/cos(2*b*x+2*a)*((sin(2*b*x+2*a)+1)^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a
))^(1/2)*EllipticF((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1/2))*sin(2*b*x+2*a)-2*cos(2*b*x+2*a)^2-2*cos(2*b*x+2*a))/b

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.15 \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {2 i} {\left (\cos \left (b x + a\right )^{2} - 1\right )} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + \sqrt {-2 i} {\left (\cos \left (b x + a\right )^{2} - 1\right )} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{12 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \]

[In]

integrate(cos(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(sqrt(2*I)*(cos(b*x + a)^2 - 1)*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1) + sqrt(-2*I)*(cos(
b*x + a)^2 - 1)*elliptic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)
))/(b*cos(b*x + a)^2 - b)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(b*x+a)**2/sin(2*b*x+2*a)**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(cos(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^2/sin(2*b*x + 2*a)^(5/2), x)

Giac [F]

\[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(cos(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^2/sin(2*b*x + 2*a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int \frac {{\cos \left (a+b\,x\right )}^2}{{\sin \left (2\,a+2\,b\,x\right )}^{5/2}} \,d x \]

[In]

int(cos(a + b*x)^2/sin(2*a + 2*b*x)^(5/2),x)

[Out]

int(cos(a + b*x)^2/sin(2*a + 2*b*x)^(5/2), x)