\(\int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) [87]

Optimal result

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

[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*sin(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 = {4381, 2720} \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {\sin ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {\operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{6 b} \]

[In]

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

[Out]

EllipticF[a - Pi/4 + b*x, 2]/(6*b) + Sin[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 4381

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-(e*Sin[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*Sin
[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] && Eq
Q[d/b, 2] &&  !IntegerQ[p] && GtQ[m, 1] && LtQ[p, -1] && NeQ[m + 2*p + 2, 0] && (LtQ[p, -2] || EqQ[m, 2]) && I
ntegersQ[2*m, 2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {\sin ^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 {\sin ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.73 \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {\sec ^2(a+b x) \sqrt {\sin (2 (a+b x))}-\frac {\sqrt {2} \operatorname {EllipticF}\left (\arcsin (\cos (a+b x)-\sin (a+b x)),\frac {1}{2}\right ) (\cos (a+b x)+\sin (a+b x))}{\sqrt {1+\sin (2 (a+b x))}}}{12 b} \]

[In]

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

[Out]

(Sec[a + b*x]^2*Sqrt[Sin[2*(a + b*x)]] - (Sqrt[2]*EllipticF[ArcSin[Cos[a + b*x] - Sin[a + b*x]], 1/2]*(Cos[a +
b*x] + Sin[a + b*x]))/Sqrt[1 + Sin[2*(a + b*x)]])/(12*b)

Maple [A] (verified)

Time = 28.15 (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(sin(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 = 94, normalized size of antiderivative = 1.96 \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {2 i} \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + \sqrt {-2 i} \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{12 \, b \cos \left (b x + a\right )^{2}} \]

[In]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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