\(\int \frac {\csc ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\) [120]

Optimal result

Integrand size = 22, antiderivative size = 55 \[ \int \frac {\csc ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {4 \csc (a+b x) \sqrt {\sin (2 a+2 b x)}}{5 b}-\frac {\csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)}}{5 b} \]

[Out]

-4/5*csc(b*x+a)*sin(2*b*x+2*a)^(1/2)/b-1/5*csc(b*x+a)^3*sin(2*b*x+2*a)^(1/2)/b

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 55, 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 = {4385, 4377} \[ \int \frac {\csc ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {\sqrt {\sin (2 a+2 b x)} \csc ^3(a+b x)}{5 b}-\frac {4 \sqrt {\sin (2 a+2 b x)} \csc (a+b x)}{5 b} \]

[In]

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

[Out]

(-4*Csc[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(5*b) - (Csc[a + b*x]^3*Sqrt[Sin[2*a + 2*b*x]])/(5*b)

Rule 4377

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)/(b*g*m)), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && Eq
Q[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 4385

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*(m + p + 1))), x] + Dist[(m + 2*p + 2)/(e^2*(m + p + 1)), Int[(e*Sin[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] && LtQ[m, -1] && NeQ[m + 2*p + 2, 0] && NeQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {\csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)}}{5 b}+\frac {4}{5} \int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = -\frac {4 \csc (a+b x) \sqrt {\sin (2 a+2 b x)}}{5 b}-\frac {\csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)}}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.64 \[ \int \frac {\csc ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {\csc (a+b x) \left (4+\csc ^2(a+b x)\right ) \sqrt {\sin (2 (a+b x))}}{5 b} \]

[In]

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

[Out]

-1/5*(Csc[a + b*x]*(4 + Csc[a + b*x]^2)*Sqrt[Sin[2*(a + b*x)]])/b

Maple [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 13.76 (sec) , antiderivative size = 482, normalized size of antiderivative = 8.76

method	result	size
default	\(\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1}}\, \left (16 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticE}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-8 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}+\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}+8 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}+\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-8 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\right )}{20 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3} \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, b}\)	\(482\)

[In]

int(csc(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/20*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(1/2)/tan(1/2*a+1/2*x*b)^3*(16*(tan(1/2*a+1/2*x*b)*(tan(1/
2*a+1/2*x*b)^2-1))^(1/2)*(tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1
/2)*EllipticE((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b)^2-8*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+
1/2*x*b)^2-1))^(1/2)*(tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/2)*
EllipticF((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b)^2-(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*
b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b)^6+(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b)^4+8
*(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^4+(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^
2-1))^(1/2)*tan(1/2*a+1/2*x*b)^2-8*(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^2-(tan(1
/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2))/(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)/b

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.38 \[ \int \frac {\csc ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {\sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 4 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )}{5 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \]

[In]

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

[Out]

-1/5*(sqrt(2)*(4*cos(b*x + a)^2 - 5)*sqrt(cos(b*x + a)*sin(b*x + a)) + 4*(cos(b*x + a)^2 - 1)*sin(b*x + a))/((
b*cos(b*x + a)^2 - b)*sin(b*x + a))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 24.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.69 \[ \int \frac {\csc ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {8\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left (-{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}{5\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^3} \]

[In]

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

[Out]

-(8*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2)*(exp(a*4i + b*x*4i)*1i
- exp(a*2i + b*x*2i)*3i + 1i))/(5*b*(exp(a*2i + b*x*2i) - 1)^3)