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

Optimal. Leaf size=28 \[ \frac {\tanh ^{-1}(\sin (a+b x))}{2 b}-\frac {\sin (a+b x)}{2 b} \]

[Out]

1/2*arctanh(sin(b*x+a))/b-1/2*sin(b*x+a)/b

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4288, 2592, 321, 206} \[ \frac {\tanh ^{-1}(\sin (a+b x))}{2 b}-\frac {\sin (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[a + b*x]]/(2*b) - Sin[a + b*x]/(2*b)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

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 a+2 b x) \sin ^3(a+b x) \, dx &=\frac {1}{2} \int \sin (a+b x) \tan (a+b x) \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (a+b x)\right )}{2 b}\\ &=-\frac {\sin (a+b x)}{2 b}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (a+b x)\right )}{2 b}\\ &=\frac {\tanh ^{-1}(\sin (a+b x))}{2 b}-\frac {\sin (a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.96 \[ \frac {1}{2} \left (\frac {\tanh ^{-1}(\sin (a+b x))}{b}-\frac {\sin (a+b x)}{b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTanh[Sin[a + b*x]]/b - Sin[a + b*x]/b)/2

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fricas [A]  time = 0.49, size = 36, normalized size = 1.29 \[ \frac {\log \left (\sin \left (b x + a\right ) + 1\right ) - \log \left (-\sin \left (b x + a\right ) + 1\right ) - 2 \, \sin \left (b x + a\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(log(sin(b*x + a) + 1) - log(-sin(b*x + a) + 1) - 2*sin(b*x + a))/b

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giac [B]  time = 0.99, size = 618, normalized size = 22.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((tan(1/2*a)^15 + 3*tan(1/2*a)^14 + 3*tan(1/2*a)^13 + 17*tan(1/2*a)^12 - 3*tan(1/2*a)^11 + 39*tan(1/2*a)^1
0 - 25*tan(1/2*a)^9 + 45*tan(1/2*a)^8 - 45*tan(1/2*a)^7 + 25*tan(1/2*a)^6 - 39*tan(1/2*a)^5 + 3*tan(1/2*a)^4 -
17*tan(1/2*a)^3 - 3*tan(1/2*a)^2 - 3*tan(1/2*a) - 1)*log(abs(tan(1/2*b*x + 2*a)*tan(1/2*a)^3 + 3*tan(1/2*b*x
+ 2*a)*tan(1/2*a)^2 - tan(1/2*a)^3 - 3*tan(1/2*b*x + 2*a)*tan(1/2*a) + 3*tan(1/2*a)^2 - tan(1/2*b*x + 2*a) + 3
*tan(1/2*a) - 1))/(tan(1/2*a)^3 + 3*tan(1/2*a)^2 - 3*tan(1/2*a) - 1) - (tan(1/2*a)^15 - 3*tan(1/2*a)^14 + 3*ta
n(1/2*a)^13 - 17*tan(1/2*a)^12 - 3*tan(1/2*a)^11 - 39*tan(1/2*a)^10 - 25*tan(1/2*a)^9 - 45*tan(1/2*a)^8 - 45*t
an(1/2*a)^7 - 25*tan(1/2*a)^6 - 39*tan(1/2*a)^5 - 3*tan(1/2*a)^4 - 17*tan(1/2*a)^3 + 3*tan(1/2*a)^2 - 3*tan(1/
2*a) + 1)*log(abs(tan(1/2*b*x + 2*a)*tan(1/2*a)^3 - 3*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 + tan(1/2*a)^3 - 3*tan(1
/2*b*x + 2*a)*tan(1/2*a) + 3*tan(1/2*a)^2 + tan(1/2*b*x + 2*a) - 3*tan(1/2*a) - 1))/(tan(1/2*a)^3 - 3*tan(1/2*
a)^2 - 3*tan(1/2*a) + 1) + 2*(tan(1/2*b*x + 2*a)*tan(1/2*a)^12 - 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^10 + 6*tan(1
/2*a)^11 - 27*tan(1/2*b*x + 2*a)*tan(1/2*a)^8 - 2*tan(1/2*a)^9 - 36*tan(1/2*a)^7 + 27*tan(1/2*b*x + 2*a)*tan(1
/2*a)^4 - 36*tan(1/2*a)^5 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - 2*tan(1/2*a)^3 - tan(1/2*b*x + 2*a) + 6*tan(1
/2*a))/(tan(1/2*b*x + 2*a)^2 + 1))/b

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maple [A]  time = 0.52, size = 32, normalized size = 1.14 \[ -\frac {\sin \left (b x +a \right )}{2 b}+\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sin(b*x+a)/b+1/2/b*ln(sec(b*x+a)+tan(b*x+a))

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maxima [B]  time = 0.47, size = 124, normalized size = 4.43 \[ -\frac {\log \left (\frac {\cos \left (b x + 2 \, a\right )^{2} + \cos \relax (a)^{2} - 2 \, \cos \relax (a) \sin \left (b x + 2 \, a\right ) + \sin \left (b x + 2 \, a\right )^{2} + 2 \, \cos \left (b x + 2 \, a\right ) \sin \relax (a) + \sin \relax (a)^{2}}{\cos \left (b x + 2 \, a\right )^{2} + \cos \relax (a)^{2} + 2 \, \cos \relax (a) \sin \left (b x + 2 \, a\right ) + \sin \left (b x + 2 \, a\right )^{2} - 2 \, \cos \left (b x + 2 \, a\right ) \sin \relax (a) + \sin \relax (a)^{2}}\right ) + 2 \, \sin \left (b x + a\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.12, size = 23, normalized size = 0.82 \[ -\frac {\frac {\sin \left (a+b\,x\right )}{2}-\frac {\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sin(a + b*x)/2 - atanh(sin(a + b*x))/2)/b

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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)*sin(b*x+a)**3,x)

[Out]

Timed out

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