\(\int \csc ^2(2 a+2 b x) \sin ^3(a+b x) \, dx\) [29]

Optimal result

Integrand size = 20, antiderivative size = 13 \[ \int \csc ^2(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\sec (a+b x)}{4 b} \]

[Out]

1/4*sec(b*x+a)/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4373, 2686, 8} \[ \int \csc ^2(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\sec (a+b x)}{4 b} \]

[In]

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

[Out]

Sec[a + b*x]/(4*b)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 4373

Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/f^p, Int[Cos[a
+ b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b, c, d, f, n}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \sec (a+b x) \tan (a+b x) \, dx \\ & = \frac {\text {Subst}(\int 1 \, dx,x,\sec (a+b x))}{4 b} \\ & = \frac {\sec (a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \csc ^2(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\sec (a+b x)}{4 b} \]

[In]

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

[Out]

Sec[a + b*x]/(4*b)

Maple [A] (verified)

Time = 1.72 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {1}{4 \cos \left (x b +a \right ) b}\)	\(14\)
risch	\(\frac {{\mathrm e}^{i \left (x b +a \right )}}{2 b \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}\)	\(28\)

[In]

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

[Out]

1/4/cos(b*x+a)/b

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \csc ^2(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {1}{4 \, b \cos \left (b x + a\right )} \]

[In]

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

[Out]

1/4/(b*cos(b*x + a))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (11) = 22\).

Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 6.38 \[ \int \csc ^2(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\cos \left (2 \, b x + 2 \, a\right ) \cos \left (b x + a\right ) + \sin \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) + \cos \left (b x + a\right )}{2 \, {\left (b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \]

[In]

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

[Out]

1/2*(cos(2*b*x + 2*a)*cos(b*x + a) + sin(2*b*x + 2*a)*sin(b*x + a) + cos(b*x + a))/(b*cos(2*b*x + 2*a)^2 + b*s
in(2*b*x + 2*a)^2 + 2*b*cos(2*b*x + 2*a) + b)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 28 vs. \(2 (11) = 22\).

Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 2.15 \[ \int \csc ^2(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {1}{2 \, b {\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}} \]

[In]

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

[Out]

1/2/(b*((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \csc ^2(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {1}{4\,b\,\cos \left (a+b\,x\right )} \]

[In]

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

[Out]

1/(4*b*cos(a + b*x))