3.2.50 \(\int \cot ^2(a+b x) \csc (a+b x) \, dx\) [150]

Optimal. Leaf size=34 \[ \frac {\tanh ^{-1}(\cos (a+b x))}{2 b}-\frac {\cot (a+b x) \csc (a+b x)}{2 b} \]

[Out]

1/2*arctanh(cos(b*x+a))/b-1/2*cot(b*x+a)*csc(b*x+a)/b

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Rubi [A]
time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2691, 3855} \begin {gather*} \frac {\tanh ^{-1}(\cos (a+b x))}{2 b}-\frac {\cot (a+b x) \csc (a+b x)}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[a + b*x]^2*Csc[a + b*x],x]

[Out]

ArcTanh[Cos[a + b*x]]/(2*b) - (Cot[a + b*x]*Csc[a + b*x])/(2*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 3855

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

Rubi steps

\begin {align*} \int \cot ^2(a+b x) \csc (a+b x) \, dx &=-\frac {\cot (a+b x) \csc (a+b x)}{2 b}-\frac {1}{2} \int \csc (a+b x) \, dx\\ &=\frac {\tanh ^{-1}(\cos (a+b x))}{2 b}-\frac {\cot (a+b x) \csc (a+b x)}{2 b}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(75\) vs. \(2(34)=68\).
time = 0.02, size = 75, normalized size = 2.21 \begin {gather*} -\frac {\csc ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}+\frac {\log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}-\frac {\log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}+\frac {\sec ^2\left (\frac {1}{2} (a+b x)\right )}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*Csc[(a + b*x)/2]^2/b + Log[Cos[(a + b*x)/2]]/(2*b) - Log[Sin[(a + b*x)/2]]/(2*b) + Sec[(a + b*x)/2]^2/(8*
b)

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Maple [A]
time = 0.03, size = 50, normalized size = 1.47

method	result	size
derivativedivides	\(\frac {-\frac {\cos ^{3}\left (b x +a \right )}{2 \sin \left (b x +a \right )^{2}}-\frac {\cos \left (b x +a \right )}{2}-\frac {\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\)	\(50\)
default	\(\frac {-\frac {\cos ^{3}\left (b x +a \right )}{2 \sin \left (b x +a \right )^{2}}-\frac {\cos \left (b x +a \right )}{2}-\frac {\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\)	\(50\)
norman	\(\frac {-\frac {1}{8 b}+\frac {\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2 b}\)	\(51\)
risch	\(\frac {{\mathrm e}^{3 i \left (b x +a \right )}+{\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{2 b}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{2 b}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 46, normalized size = 1.35 \begin {gather*} \frac {\frac {2 \, \cos \left (b x + a\right )}{\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 )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (30) = 60\).
time = 0.37, size = 72, normalized size = 2.12 \begin {gather*} \frac {{\left (\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 )^{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 )}{4 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (27) = 54\).
time = 0.53, size = 58, normalized size = 1.71 \begin {gather*} \begin {cases} - \frac {\log {\left (\tan {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )}}{2 b} + \frac {\tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{8 b} - \frac {1}{8 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{2}{\left (a \right )}}{\sin ^{3}{\left (a \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-log(tan(a/2 + b*x/2))/(2*b) + tan(a/2 + b*x/2)**2/(8*b) - 1/(8*b*tan(a/2 + b*x/2)**2), Ne(b, 0)),
(x*cos(a)**2/sin(a)**3, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (30) = 60\).
time = 3.17, size = 93, normalized size = 2.74 \begin {gather*} \frac {\frac {{\left (\frac {2 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}}{\cos \left (b x + a\right ) - 1} - \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 2 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.45, size = 48, normalized size = 1.41 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{8\,b}-\frac {1}{8\,b\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}-\frac {\ln \left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{2\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2/sin(a + b*x)^3,x)

[Out]

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

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