\(\int \cos (a+b x) \csc ^3(2 a+2 b x) \, dx\) [138]

Optimal result

Integrand size = 18, antiderivative size = 49 \[ \int \cos (a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {3 \text {arctanh}(\cos (a+b x))}{16 b}+\frac {3 \sec (a+b x)}{16 b}-\frac {\csc ^2(a+b x) \sec (a+b x)}{16 b} \]

[Out]

-3/16*arctanh(cos(b*x+a))/b+3/16*sec(b*x+a)/b-1/16*csc(b*x+a)^2*sec(b*x+a)/b

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {4372, 2702, 294, 327, 213} \[ \int \cos (a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {3 \text {arctanh}(\cos (a+b x))}{16 b}+\frac {3 \sec (a+b x)}{16 b}-\frac {\csc ^2(a+b x) \sec (a+b x)}{16 b} \]

[In]

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

[Out]

(-3*ArcTanh[Cos[a + b*x]])/(16*b) + (3*Sec[a + b*x])/(16*b) - (Csc[a + b*x]^2*Sec[a + b*x])/(16*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 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(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}{8} \int \csc ^3(a+b x) \sec ^2(a+b x) \, dx \\ & = \frac {\text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (a+b x)\right )}{8 b} \\ & = -\frac {\csc ^2(a+b x) \sec (a+b x)}{16 b}+\frac {3 \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{16 b} \\ & = \frac {3 \sec (a+b x)}{16 b}-\frac {\csc ^2(a+b x) \sec (a+b x)}{16 b}+\frac {3 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{16 b} \\ & = -\frac {3 \text {arctanh}(\cos (a+b x))}{16 b}+\frac {3 \sec (a+b x)}{16 b}-\frac {\csc ^2(a+b x) \sec (a+b x)}{16 b} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(143\) vs. \(2(49)=98\).

Time = 0.52 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.92 \[ \int \cos (a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {\csc ^4(a+b x) \left (2-6 \cos (2 (a+b x))+2 \cos (3 (a+b x))+3 \cos (3 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-3 \cos (3 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )+\cos (a+b x) \left (-2-3 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+3 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )}{16 b \left (\csc ^2\left (\frac {1}{2} (a+b x)\right )-\sec ^2\left (\frac {1}{2} (a+b x)\right )\right )} \]

[In]

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

[Out]

(Csc[a + b*x]^4*(2 - 6*Cos[2*(a + b*x)] + 2*Cos[3*(a + b*x)] + 3*Cos[3*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 3*Co
s[3*(a + b*x)]*Log[Sin[(a + b*x)/2]] + Cos[a + b*x]*(-2 - 3*Log[Cos[(a + b*x)/2]] + 3*Log[Sin[(a + b*x)/2]])))
/(16*b*(Csc[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2))

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {-\frac {1}{2 \sin \left (x b +a \right )^{2} \cos \left (x b +a \right )}+\frac {3}{2 \cos \left (x b +a \right )}+\frac {3 \ln \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}{2}}{8 b}\)	\(53\)
risch	\(\frac {3 \,{\mathrm e}^{5 i \left (x b +a \right )}-2 \,{\mathrm e}^{3 i \left (x b +a \right )}+3 \,{\mathrm e}^{i \left (x b +a \right )}}{8 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{16 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{16 b}\)	\(101\)

[In]

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

[Out]

1/8/b*(-1/2/sin(b*x+a)^2/cos(b*x+a)+3/2/cos(b*x+a)+3/2*ln(csc(b*x+a)-cot(b*x+a)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (43) = 86\).

Time = 0.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.96 \[ \int \cos (a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {6 \, \cos \left (b x + a\right )^{2} - 3 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 4}{32 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \]

[In]

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

[Out]

1/32*(6*cos(b*x + a)^2 - 3*(cos(b*x + a)^3 - cos(b*x + a))*log(1/2*cos(b*x + a) + 1/2) + 3*(cos(b*x + a)^3 - c
os(b*x + a))*log(-1/2*cos(b*x + a) + 1/2) - 4)/(b*cos(b*x + a)^3 - b*cos(b*x + a))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 974 vs. \(2 (43) = 86\).

Time = 0.21 (sec) , antiderivative size = 974, normalized size of antiderivative = 19.88 \[ \int \cos (a+b x) \csc ^3(2 a+2 b x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/32*(4*(3*cos(5*b*x + 5*a) - 2*cos(3*b*x + 3*a) + 3*cos(b*x + a))*cos(6*b*x + 6*a) - 12*(cos(4*b*x + 4*a) + c
os(2*b*x + 2*a) - 1)*cos(5*b*x + 5*a) + 4*(2*cos(3*b*x + 3*a) - 3*cos(b*x + a))*cos(4*b*x + 4*a) + 8*(cos(2*b*
x + 2*a) - 1)*cos(3*b*x + 3*a) - 12*cos(2*b*x + 2*a)*cos(b*x + a) + 3*(2*(cos(4*b*x + 4*a) + cos(2*b*x + 2*a)
- 1)*cos(6*b*x + 6*a) - cos(6*b*x + 6*a)^2 - 2*(cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a) - cos(4*b*x + 4*a)^2 -
cos(2*b*x + 2*a)^2 + 2*(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 - sin(4*b*x
+ 4*a)^2 - 2*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) - sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) - 1)*log(cos(b*x)^2
+ 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(a) + sin(a)^2) - 3*(2*(cos(4*b*x + 4*a) + cos(2*b
*x + 2*a) - 1)*cos(6*b*x + 6*a) - cos(6*b*x + 6*a)^2 - 2*(cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a) - cos(4*b*x +
4*a)^2 - cos(2*b*x + 2*a)^2 + 2*(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 -
sin(4*b*x + 4*a)^2 - 2*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) - sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) - 1)*log(c
os(b*x)^2 - 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(a) + sin(a)^2) + 4*(3*sin(5*b*x + 5*a)
- 2*sin(3*b*x + 3*a) + 3*sin(b*x + a))*sin(6*b*x + 6*a) - 12*(sin(4*b*x + 4*a) + sin(2*b*x + 2*a))*sin(5*b*x +
5*a) + 4*(2*sin(3*b*x + 3*a) - 3*sin(b*x + a))*sin(4*b*x + 4*a) + 8*sin(3*b*x + 3*a)*sin(2*b*x + 2*a) - 12*si
n(2*b*x + 2*a)*sin(b*x + a) + 12*cos(b*x + a))/(b*cos(6*b*x + 6*a)^2 + b*cos(4*b*x + 4*a)^2 + b*cos(2*b*x + 2*
a)^2 + b*sin(6*b*x + 6*a)^2 + b*sin(4*b*x + 4*a)^2 + 2*b*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + b*sin(2*b*x + 2*a
)^2 - 2*(b*cos(4*b*x + 4*a) + b*cos(2*b*x + 2*a) - b)*cos(6*b*x + 6*a) + 2*(b*cos(2*b*x + 2*a) - b)*cos(4*b*x
+ 4*a) - 2*b*cos(2*b*x + 2*a) - 2*(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 = 0.41 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.29 \[ \int \cos (a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {\frac {2 \, {\left (3 \, \cos \left (b x + a\right )^{2} - 2\right )}}{\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )} - 3 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 3 \, \log \left (-\cos \left (b x + a\right ) + 1\right )}{32 \, b} \]

[In]

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

[Out]

1/32*(2*(3*cos(b*x + a)^2 - 2)/(cos(b*x + a)^3 - cos(b*x + a)) - 3*log(cos(b*x + a) + 1) + 3*log(-cos(b*x + a)
+ 1))/b

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \cos (a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {3\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{16\,b}-\frac {\frac {3\,{\cos \left (a+b\,x\right )}^2}{16}-\frac {1}{8}}{b\,\left (\cos \left (a+b\,x\right )-{\cos \left (a+b\,x\right )}^3\right )} \]

[In]

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

[Out]

- (3*atanh(cos(a + b*x)))/(16*b) - ((3*cos(a + b*x)^2)/16 - 1/8)/(b*(cos(a + b*x) - cos(a + b*x)^3))