3.2.48 \(\int \cos (a+b x) \cot ^3(a+b x) \, dx\) [148]

Optimal. Leaf size=49 \[ \frac {3 \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac {3 \cos (a+b x)}{2 b}-\frac {\cos (a+b x) \cot ^2(a+b x)}{2 b} \]

[Out]

3/2*arctanh(cos(b*x+a))/b-3/2*cos(b*x+a)/b-1/2*cos(b*x+a)*cot(b*x+a)^2/b

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Rubi [A]
time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2672, 294, 327, 212} \begin {gather*} -\frac {3 \cos (a+b x)}{2 b}-\frac {\cos (a+b x) \cot ^2(a+b x)}{2 b}+\frac {3 \tanh ^{-1}(\cos (a+b x))}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Cot[a + b*x]^3,x]

[Out]

(3*ArcTanh[Cos[a + b*x]])/(2*b) - (3*Cos[a + b*x])/(2*b) - (Cos[a + b*x]*Cot[a + b*x]^2)/(2*b)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rubi steps

\begin {align*} \int \cos (a+b x) \cot ^3(a+b x) \, dx &=-\frac {\text {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^2} \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {\cos (a+b x) \cot ^2(a+b x)}{2 b}+\frac {3 \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\cos (a+b x)\right )}{2 b}\\ &=-\frac {3 \cos (a+b x)}{2 b}-\frac {\cos (a+b x) \cot ^2(a+b x)}{2 b}+\frac {3 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (a+b x)\right )}{2 b}\\ &=\frac {3 \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac {3 \cos (a+b x)}{2 b}-\frac {\cos (a+b x) \cot ^2(a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 86, normalized size = 1.76 \begin {gather*} -\frac {\cos (a+b x)}{b}-\frac {\csc ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}+\frac {3 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}-\frac {3 \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[Cos[a + b*x]*Cot[a + b*x]^3,x]

[Out]

-(Cos[a + b*x]/b) - Csc[(a + b*x)/2]^2/(8*b) + (3*Log[Cos[(a + b*x)/2]])/(2*b) - (3*Log[Sin[(a + b*x)/2]])/(2*
b) + Sec[(a + b*x)/2]^2/(8*b)

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Maple [A]
time = 0.04, size = 60, normalized size = 1.22

method	result	size
derivativedivides	\(\frac {-\frac {\cos ^{5}\left (b x +a \right )}{2 \sin \left (b x +a \right )^{2}}-\frac {\left (\cos ^{3}\left (b x +a \right )\right )}{2}-\frac {3 \cos \left (b x +a \right )}{2}-\frac {3 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\)	\(60\)
default	\(\frac {-\frac {\cos ^{5}\left (b x +a \right )}{2 \sin \left (b x +a \right )^{2}}-\frac {\left (\cos ^{3}\left (b x +a \right )\right )}{2}-\frac {3 \cos \left (b x +a \right )}{2}-\frac {3 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\)	\(60\)
norman	\(\frac {-\frac {1}{8 b}+\frac {\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}-\frac {9 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}-\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2 b}\)	\(82\)
risch	\(-\frac {{\mathrm e}^{i \left (b x +a \right )}}{2 b}-\frac {{\mathrm e}^{-i \left (b x +a \right )}}{2 b}+\frac {{\mathrm e}^{3 i \left (b x +a \right )}+{\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{2 b}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^4/sin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/b*(-1/2*cos(b*x+a)^5/sin(b*x+a)^2-1/2*cos(b*x+a)^3-3/2*cos(b*x+a)-3/2*ln(csc(b*x+a)-cot(b*x+a)))

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Maxima [A]
time = 0.29, size = 56, normalized size = 1.14 \begin {gather*} \frac {\frac {2 \, \cos \left (b x + a\right )}{\cos \left (b x + a\right )^{2} - 1} - 4 \, \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 )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^4/sin(b*x+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.41, size = 83, normalized size = 1.69 \begin {gather*} -\frac {4 \, \cos \left (b x + a\right )^{3} - 3 \, {\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 ) + 3 \, {\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 ) - 6 \, \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)^4/sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/4*(4*cos(b*x + a)^3 - 3*(cos(b*x + a)^2 - 1)*log(1/2*cos(b*x + a) + 1/2) + 3*(cos(b*x + a)^2 - 1)*log(-1/2*
cos(b*x + a) + 1/2) - 6*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. 241 vs. \(2 (42) = 84\).
time = 0.94, size = 241, normalized size = 4.92 \begin {gather*} \begin {cases} - \frac {12 \log {\left (\tan {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )} \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{8 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 8 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} - \frac {12 \log {\left (\tan {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )} \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{8 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 8 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} + \frac {\tan ^{6}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{8 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 8 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} - \frac {18 \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{8 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 8 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} - \frac {1}{8 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 8 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{4}{\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)**4/sin(b*x+a)**3,x)

[Out]

Piecewise((-12*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) -
12*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(8*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + tan(a/2 +
b*x/2)**6/(8*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 18*tan(a/2 + b*x/2)**2/(8*b*tan(a/2 + b*x/2)**
4 + 8*b*tan(a/2 + b*x/2)**2) - 1/(8*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2), Ne(b, 0)), (x*cos(a)**4/
sin(a)**3, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (43) = 86\).
time = 4.22, size = 140, normalized size = 2.86 \begin {gather*} -\frac {\frac {\frac {14 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {3 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1}{\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}}} + \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 6 \, \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)^4/sin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/8*((14*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 3*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 - 1)/((cos(b*x +
a) - 1)/(cos(b*x + a) + 1) - (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2) + (cos(b*x + a) - 1)/(cos(b*x + a) +
1) + 6*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b

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Mupad [B]
time = 0.54, size = 77, normalized size = 1.57 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{8\,b}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{2\,b}-\frac {\frac {17\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{8}+\frac {1}{8}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^4/sin(a + b*x)^3,x)

[Out]

tan(a/2 + (b*x)/2)^2/(8*b) - (3*log(tan(a/2 + (b*x)/2)))/(2*b) - ((17*tan(a/2 + (b*x)/2)^2)/8 + 1/8)/(b*(tan(a
/2 + (b*x)/2)^2 + tan(a/2 + (b*x)/2)^4))

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