\(\int \cos (a+b x) \csc ^2(2 a+2 b x) \, dx\) [137]

Optimal result

Integrand size = 18, antiderivative size = 28 \[ \int \cos (a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{4 b}-\frac {\csc (a+b x)}{4 b} \]

[Out]

1/4*arctanh(sin(b*x+a))/b-1/4*csc(b*x+a)/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4372, 2701, 327, 213} \[ \int \cos (a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{4 b}-\frac {\csc (a+b x)}{4 b} \]

[In]

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

[Out]

ArcTanh[Sin[a + b*x]]/(4*b) - Csc[a + b*x]/(4*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 327

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 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 4372

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

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

Mathematica [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \cos (a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\csc (a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\sin ^2(a+b x)\right )}{4 b} \]

[In]

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

[Out]

-1/4*(Csc[a + b*x]*Hypergeometric2F1[-1/2, 1, 1/2, Sin[a + b*x]^2])/b

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11

method	result	size
default	\(\frac {-\frac {1}{\sin \left (x b +a \right )}+\ln \left (\sec \left (x b +a \right )+\tan \left (x b +a \right )\right )}{4 b}\)	\(31\)
risch	\(-\frac {i {\mathrm e}^{i \left (x b +a \right )}}{2 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (x b +a \right )}-i\right )}{4 b}+\frac {\ln \left (i+{\mathrm e}^{i \left (x b +a \right )}\right )}{4 b}\)	\(66\)

[In]

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

[Out]

1/4/b*(-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. 50 vs. \(2 (24) = 48\).

Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.79 \[ \int \cos (a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {\log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 2}{8 \, b \sin \left (b x + a\right )} \]

[In]

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

[Out]

1/8*(log(sin(b*x + a) + 1)*sin(b*x + a) - log(-sin(b*x + a) + 1)*sin(b*x + a) - 2)/(b*sin(b*x + a))

Sympy [F(-1)]

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

[In]

integrate(cos(b*x+a)/sin(2*b*x+2*a)**2,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (24) = 48\).

Time = 0.31 (sec) , antiderivative size = 233, normalized size of antiderivative = 8.32 \[ \int \cos (a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\frac {\cos \left (b x + 2 \, a\right )^{2} + \cos \left (a\right )^{2} - 2 \, \cos \left (a\right ) \sin \left (b x + 2 \, a\right ) + \sin \left (b x + 2 \, a\right )^{2} + 2 \, \cos \left (b x + 2 \, a\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}}{\cos \left (b x + 2 \, a\right )^{2} + \cos \left (a\right )^{2} + 2 \, \cos \left (a\right ) \sin \left (b x + 2 \, a\right ) + \sin \left (b x + 2 \, a\right )^{2} - 2 \, \cos \left (b x + 2 \, a\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}}\right ) + 4 \, \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right ) - 4 \, \cos \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) + 4 \, \sin \left (b x + a\right )}{8 \, {\left (b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \]

[In]

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

[Out]

-1/8*((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*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*cos(b*x + a)*sin(2*b*x +
2*a) - 4*cos(2*b*x + 2*a)*sin(b*x + a) + 4*sin(b*x + a))/(b*cos(2*b*x + 2*a)^2 + b*sin(2*b*x + 2*a)^2 - 2*b*c
os(2*b*x + 2*a) + b)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \cos (a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\frac {2}{\sin \left (b x + a\right )} - \log \left (\sin \left (b x + a\right ) + 1\right ) + \log \left (-\sin \left (b x + a\right ) + 1\right )}{8 \, b} \]

[In]

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

[Out]

-1/8*(2/sin(b*x + a) - log(sin(b*x + a) + 1) + log(-sin(b*x + a) + 1))/b

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \cos (a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{4\,b}-\frac {1}{4\,b\,\sin \left (a+b\,x\right )} \]

[In]

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

[Out]

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