\(\int \cot ^2(a+b x) \csc ^3(a+b x) \, dx\) [178]

Optimal result

Integrand size = 17, antiderivative size = 55 \[ \int \cot ^2(a+b x) \csc ^3(a+b x) \, dx=\frac {\text {arctanh}(\cos (a+b x))}{8 b}+\frac {\cot (a+b x) \csc (a+b x)}{8 b}-\frac {\cot (a+b x) \csc ^3(a+b x)}{4 b} \]

[Out]

1/8*arctanh(cos(b*x+a))/b+1/8*cot(b*x+a)*csc(b*x+a)/b-1/4*cot(b*x+a)*csc(b*x+a)^3/b

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2691, 3853, 3855} \[ \int \cot ^2(a+b x) \csc ^3(a+b x) \, dx=\frac {\text {arctanh}(\cos (a+b x))}{8 b}-\frac {\cot (a+b x) \csc ^3(a+b x)}{4 b}+\frac {\cot (a+b x) \csc (a+b x)}{8 b} \]

[In]

Int[Cot[a + b*x]^2*Csc[a + b*x]^3,x]

[Out]

ArcTanh[Cos[a + b*x]]/(8*b) + (Cot[a + b*x]*Csc[a + b*x])/(8*b) - (Cot[a + b*x]*Csc[a + b*x]^3)/(4*b)

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cot (a+b x) \csc ^3(a+b x)}{4 b}-\frac {1}{4} \int \csc ^3(a+b x) \, dx \\ & = \frac {\cot (a+b x) \csc (a+b x)}{8 b}-\frac {\cot (a+b x) \csc ^3(a+b x)}{4 b}-\frac {1}{8} \int \csc (a+b x) \, dx \\ & = \frac {\text {arctanh}(\cos (a+b x))}{8 b}+\frac {\cot (a+b x) \csc (a+b x)}{8 b}-\frac {\cot (a+b x) \csc ^3(a+b x)}{4 b} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(113\) vs. \(2(55)=110\).

Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 2.05 \[ \int \cot ^2(a+b x) \csc ^3(a+b x) \, dx=\frac {\csc ^2\left (\frac {1}{2} (a+b x)\right )}{32 b}-\frac {\csc ^4\left (\frac {1}{2} (a+b x)\right )}{64 b}+\frac {\log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{8 b}-\frac {\log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{8 b}-\frac {\sec ^2\left (\frac {1}{2} (a+b x)\right )}{32 b}+\frac {\sec ^4\left (\frac {1}{2} (a+b x)\right )}{64 b} \]

[In]

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

[Out]

Csc[(a + b*x)/2]^2/(32*b) - Csc[(a + b*x)/2]^4/(64*b) + Log[Cos[(a + b*x)/2]]/(8*b) - Log[Sin[(a + b*x)/2]]/(8
*b) - Sec[(a + b*x)/2]^2/(32*b) + Sec[(a + b*x)/2]^4/(64*b)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93

method	result	size
norman	\(\frac {-\frac {1}{64 b}+\frac {\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )}{64 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}\)	\(51\)
parallelrisch	\(\frac {\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )-8 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}{64 b \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}\)	\(53\)
derivativedivides	\(\frac {-\frac {\cos ^{3}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}-\frac {\cos ^{3}\left (b x +a \right )}{8 \sin \left (b x +a \right )^{2}}-\frac {\cos \left (b x +a \right )}{8}-\frac {\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{8}}{b}\)	\(68\)
default	\(\frac {-\frac {\cos ^{3}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}-\frac {\cos ^{3}\left (b x +a \right )}{8 \sin \left (b x +a \right )^{2}}-\frac {\cos \left (b x +a \right )}{8}-\frac {\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{8}}{b}\)	\(68\)
risch	\(-\frac {{\mathrm e}^{7 i \left (b x +a \right )}+7 \,{\mathrm e}^{5 i \left (b x +a \right )}+7 \,{\mathrm e}^{3 i \left (b x +a \right )}+{\mathrm e}^{i \left (b x +a \right )}}{4 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{8 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{8 b}\)	\(95\)

[In]

int(cos(b*x+a)^2/sin(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

(-1/64/b+1/64/b*tan(1/2*b*x+1/2*a)^8)/tan(1/2*b*x+1/2*a)^4-1/8/b*ln(tan(1/2*b*x+1/2*a))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (49) = 98\).

Time = 0.51 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.02 \[ \int \cot ^2(a+b x) \csc ^3(a+b x) \, dx=-\frac {2 \, \cos \left (b x + a\right )^{3} - {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 2 \, \cos \left (b x + a\right )}{16 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \]

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^5,x, algorithm="fricas")

[Out]

-1/16*(2*cos(b*x + a)^3 - (cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(1/2*cos(b*x + a) + 1/2) + (cos(b*x + a)^
4 - 2*cos(b*x + a)^2 + 1)*log(-1/2*cos(b*x + a) + 1/2) + 2*cos(b*x + a))/(b*cos(b*x + a)^4 - 2*b*cos(b*x + a)^
2 + b)

Sympy [A] (verification not implemented)

Time = 0.93 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.05 \[ \int \cot ^2(a+b x) \csc ^3(a+b x) \, dx=\begin {cases} - \frac {\log {\left (\tan {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )}}{8 b} + \frac {\tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{64 b} - \frac {1}{64 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{2}{\left (a \right )}}{\sin ^{5}{\left (a \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)**2/sin(b*x+a)**5,x)

[Out]

Piecewise((-log(tan(a/2 + b*x/2))/(8*b) + tan(a/2 + b*x/2)**4/(64*b) - 1/(64*b*tan(a/2 + b*x/2)**4), Ne(b, 0))
, (x*cos(a)**2/sin(a)**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.18 \[ \int \cot ^2(a+b x) \csc ^3(a+b x) \, dx=-\frac {\frac {2 \, {\left (\cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )\right )}}{\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1} - \log \left (\cos \left (b x + a\right ) + 1\right ) + \log \left (\cos \left (b x + a\right ) - 1\right )}{16 \, b} \]

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/16*(2*(cos(b*x + a)^3 + cos(b*x + a))/(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1) - log(cos(b*x + a) + 1) + log
(cos(b*x + a) - 1))/b

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.78 \[ \int \cot ^2(a+b x) \csc ^3(a+b x) \, dx=\frac {\frac {{\left (\frac {2 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 4 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{64 \, b} \]

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^5,x, algorithm="giac")

[Out]

1/64*((2*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 - 1)*(cos(b*x + a) + 1)^2/(cos(b*x + a) - 1)^2 + (cos(b*x +
a) - 1)^2/(cos(b*x + a) + 1)^2 - 4*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \cot ^2(a+b x) \csc ^3(a+b x) \, dx=\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4}{64\,b}-\frac {1}{64\,b\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4}-\frac {\ln \left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{8\,b} \]

[In]

int(cos(a + b*x)^2/sin(a + b*x)^5,x)

[Out]

tan(a/2 + (b*x)/2)^4/(64*b) - 1/(64*b*tan(a/2 + (b*x)/2)^4) - log(tan(a/2 + (b*x)/2))/(8*b)