\(\int \csc ^3(a+b x) \csc (2 a+2 b x) \, dx\) [69]

Optimal result

Integrand size = 18, antiderivative size = 43 \[ \int \csc ^3(a+b x) \csc (2 a+2 b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{2 b}-\frac {\csc (a+b x)}{2 b}-\frac {\csc ^3(a+b x)}{6 b} \]

[Out]

1/2*arctanh(sin(b*x+a))/b-1/2*csc(b*x+a)/b-1/6*csc(b*x+a)^3/b

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4373, 2701, 308, 213} \[ \int \csc ^3(a+b x) \csc (2 a+2 b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{2 b}-\frac {\csc ^3(a+b x)}{6 b}-\frac {\csc (a+b x)}{2 b} \]

[In]

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

[Out]

ArcTanh[Sin[a + b*x]]/(2*b) - Csc[a + b*x]/(2*b) - Csc[a + b*x]^3/(6*b)

Rule 213

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

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2701

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

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}{2} \int \csc ^4(a+b x) \sec (a+b x) \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{2 b} \\ & = -\frac {\text {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (a+b x)\right )}{2 b} \\ & = -\frac {\csc (a+b x)}{2 b}-\frac {\csc ^3(a+b x)}{6 b}-\frac {\text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{2 b} \\ & = \frac {\text {arctanh}(\sin (a+b x))}{2 b}-\frac {\csc (a+b x)}{2 b}-\frac {\csc ^3(a+b x)}{6 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.72 \[ \int \csc ^3(a+b x) \csc (2 a+2 b x) \, dx=-\frac {\csc ^3(a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\sin ^2(a+b x)\right )}{6 b} \]

[In]

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

[Out]

-1/6*(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Sin[a + b*x]^2])/b

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {-\frac {1}{3 \sin \left (x b +a \right )^{3}}-\frac {1}{\sin \left (x b +a \right )}+\ln \left (\sec \left (x b +a \right )+\tan \left (x b +a \right )\right )}{2 b}\)	\(41\)
risch	\(-\frac {i \left (3 \,{\mathrm e}^{5 i \left (x b +a \right )}-10 \,{\mathrm e}^{3 i \left (x b +a \right )}+3 \,{\mathrm e}^{i \left (x b +a \right )}\right )}{3 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (x b +a \right )}-i\right )}{2 b}+\frac {\ln \left (i+{\mathrm e}^{i \left (x b +a \right )}\right )}{2 b}\)	\(91\)

[In]

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

[Out]

1/2/b*(-1/3/sin(b*x+a)^3-1/sin(b*x+a)+ln(sec(b*x+a)+tan(b*x+a)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (37) = 74\).

Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.19 \[ \int \csc ^3(a+b x) \csc (2 a+2 b x) \, dx=\frac {3 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 3 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 6 \, \cos \left (b x + a\right )^{2} + 8}{12 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \]

[In]

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

[Out]

1/12*(3*(cos(b*x + a)^2 - 1)*log(sin(b*x + a) + 1)*sin(b*x + a) - 3*(cos(b*x + a)^2 - 1)*log(-sin(b*x + a) + 1
)*sin(b*x + a) - 6*cos(b*x + a)^2 + 8)/((b*cos(b*x + a)^2 - b)*sin(b*x + a))

Sympy [F]

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

[In]

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

[Out]

Integral(csc(a + b*x)**3*csc(2*a + 2*b*x), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 834 vs. \(2 (37) = 74\).

Time = 0.33 (sec) , antiderivative size = 834, normalized size of antiderivative = 19.40 \[ \int \csc ^3(a+b x) \csc (2 a+2 b x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/12*(4*(3*sin(5*b*x + 5*a) - 10*sin(3*b*x + 3*a) + 3*sin(b*x + a))*cos(6*b*x + 6*a) + 36*(sin(4*b*x + 4*a) -
sin(2*b*x + 2*a))*cos(5*b*x + 5*a) + 12*(10*sin(3*b*x + 3*a) - 3*sin(b*x + a))*cos(4*b*x + 4*a) + 3*(2*(3*cos(
4*b*x + 4*a) - 3*cos(2*b*x + 2*a) + 1)*cos(6*b*x + 6*a) - cos(6*b*x + 6*a)^2 + 6*(3*cos(2*b*x + 2*a) - 1)*cos(
4*b*x + 4*a) - 9*cos(4*b*x + 4*a)^2 - 9*cos(2*b*x + 2*a)^2 + 6*(sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*sin(6*b*x
+ 6*a) - sin(6*b*x + 6*a)^2 - 9*sin(4*b*x + 4*a)^2 + 18*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) - 9*sin(2*b*x + 2*a
)^2 + 6*cos(2*b*x + 2*a) - 1)*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)) - 4*(3*cos(5*b*x + 5*a) - 10*cos(3*b*x + 3*a) + 3*cos(b*x + a))*sin(6*b
*x + 6*a) - 12*(3*cos(4*b*x + 4*a) - 3*cos(2*b*x + 2*a) + 1)*sin(5*b*x + 5*a) - 12*(10*cos(3*b*x + 3*a) - 3*co
s(b*x + a))*sin(4*b*x + 4*a) - 40*(3*cos(2*b*x + 2*a) - 1)*sin(3*b*x + 3*a) + 120*cos(3*b*x + 3*a)*sin(2*b*x +
2*a) - 36*cos(b*x + a)*sin(2*b*x + 2*a) + 36*cos(2*b*x + 2*a)*sin(b*x + a) - 12*sin(b*x + a))/(b*cos(6*b*x +
6*a)^2 + 9*b*cos(4*b*x + 4*a)^2 + 9*b*cos(2*b*x + 2*a)^2 + b*sin(6*b*x + 6*a)^2 + 9*b*sin(4*b*x + 4*a)^2 - 18*
b*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 9*b*sin(2*b*x + 2*a)^2 - 2*(3*b*cos(4*b*x + 4*a) - 3*b*cos(2*b*x + 2*a)
+ b)*cos(6*b*x + 6*a) - 6*(3*b*cos(2*b*x + 2*a) - b)*cos(4*b*x + 4*a) - 6*b*cos(2*b*x + 2*a) - 6*(b*sin(4*b*x
+ 4*a) - b*sin(2*b*x + 2*a))*sin(6*b*x + 6*a) + b)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.21 \[ \int \csc ^3(a+b x) \csc (2 a+2 b x) \, dx=-\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{12 \, b} \]

[In]

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

[Out]

-1/12*(2*(3*sin(b*x + a)^2 + 1)/sin(b*x + a)^3 - 3*log(sin(b*x + a) + 1) + 3*log(-sin(b*x + a) + 1))/b

Mupad [B] (verification not implemented)

Time = 19.91 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int \csc ^3(a+b x) \csc (2 a+2 b x) \, dx=\frac {\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{2\,b}-\frac {\frac {{\sin \left (a+b\,x\right )}^2}{2}+\frac {1}{6}}{b\,{\sin \left (a+b\,x\right )}^3} \]

[In]

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

[Out]

atanh(sin(a + b*x))/(2*b) - (sin(a + b*x)^2/2 + 1/6)/(b*sin(a + b*x)^3)