3.29 \(\int \csc ^2(2 a+2 b x) \sin ^3(a+b x) \, dx\)

Optimal. Leaf size=13 \[ \frac {\sec (a+b x)}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4288, 2606, 8} \[ \frac {\sec (a+b x)}{4 b} \]

Antiderivative was successfully verified.

[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 2606

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 4288

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*} \int \csc ^2(2 a+2 b x) \sin ^3(a+b x) \, dx &=\frac {1}{4} \int \sec (a+b x) \tan (a+b x) \, dx\\ &=\frac {\operatorname {Subst}(\int 1 \, dx,x,\sec (a+b x))}{4 b}\\ &=\frac {\sec (a+b x)}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \[ \frac {\sec (a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sec[a + b*x]/(4*b)

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fricas [A]  time = 0.44, size = 13, normalized size = 1.00 \[ \frac {1}{4 \, b \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [B]  time = 3.87, size = 319, normalized size = 24.54 \[ -\frac {6 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{11} - \tan \left (\frac {1}{2} \, a\right )^{12} - 2 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{9} + 12 \, \tan \left (\frac {1}{2} \, a\right )^{10} - 36 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{7} + 27 \, \tan \left (\frac {1}{2} \, a\right )^{8} - 36 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{5} - 2 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{3} - 27 \, \tan \left (\frac {1}{2} \, a\right )^{4} + 6 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right ) - 12 \, \tan \left (\frac {1}{2} \, a\right )^{2} + 1}{2 \, {\left (\tan \left (\frac {1}{2} \, b x + 2 \, a\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{6} - 15 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{4} + 12 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{5} - \tan \left (\frac {1}{2} \, a\right )^{6} + 15 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} - 40 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{3} + 15 \, \tan \left (\frac {1}{2} \, a\right )^{4} - \tan \left (\frac {1}{2} \, b x + 2 \, a\right )^{2} + 12 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right ) - 15 \, \tan \left (\frac {1}{2} \, a\right )^{2} + 1\right )} {\left (\tan \left (\frac {1}{2} \, a\right )^{6} - 15 \, \tan \left (\frac {1}{2} \, a\right )^{4} + 15 \, \tan \left (\frac {1}{2} \, a\right )^{2} - 1\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*tan(1/2*b*x + 2*a)*tan(1/2*a)^11 - tan(1/2*a)^12 - 2*tan(1/2*b*x + 2*a)*tan(1/2*a)^9 + 12*tan(1/2*a)^1
0 - 36*tan(1/2*b*x + 2*a)*tan(1/2*a)^7 + 27*tan(1/2*a)^8 - 36*tan(1/2*b*x + 2*a)*tan(1/2*a)^5 - 2*tan(1/2*b*x
+ 2*a)*tan(1/2*a)^3 - 27*tan(1/2*a)^4 + 6*tan(1/2*b*x + 2*a)*tan(1/2*a) - 12*tan(1/2*a)^2 + 1)/((tan(1/2*b*x +
2*a)^2*tan(1/2*a)^6 - 15*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^4 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^5 - tan(1/2*a)^
6 + 15*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^2 - 40*tan(1/2*b*x + 2*a)*tan(1/2*a)^3 + 15*tan(1/2*a)^4 - tan(1/2*b*x
+ 2*a)^2 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a) - 15*tan(1/2*a)^2 + 1)*(tan(1/2*a)^6 - 15*tan(1/2*a)^4 + 15*tan(1/
2*a)^2 - 1)*b)

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maple [A]  time = 0.68, size = 14, normalized size = 1.08 \[ \frac {1}{4 b \cos \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.33, size = 83, normalized size = 6.38 \[ \frac {\cos \left (2 \, b x + 2 \, a\right ) \cos \left (b x + a\right ) + \sin \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) + \cos \left (b x + a\right )}{2 \, {\left (b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.09, size = 13, normalized size = 1.00 \[ \frac {1}{4\,b\,\cos \left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(4*b*cos(a + b*x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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