3.2.35 \(\int \cos ^3(a+b x) \cot ^2(a+b x) \, dx\) [135]

Optimal. Leaf size=38 \[ -\frac {\csc (a+b x)}{b}-\frac {2 \sin (a+b x)}{b}+\frac {\sin ^3(a+b x)}{3 b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2670, 276} \begin {gather*} \frac {\sin ^3(a+b x)}{3 b}-\frac {2 \sin (a+b x)}{b}-\frac {\csc (a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Cot[a + b*x]^2,x]

[Out]

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

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2670

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 38, normalized size = 1.00 \begin {gather*} -\frac {\csc (a+b x)}{b}-\frac {2 \sin (a+b x)}{b}+\frac {\sin ^3(a+b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.05, size = 52, normalized size = 1.37

method	result	size
derivativedivides	\(\frac {-\frac {\cos ^{6}\left (b x +a \right )}{\sin \left (b x +a \right )}-\left (\frac {8}{3}+\cos ^{4}\left (b x +a \right )+\frac {4 \left (\cos ^{2}\left (b x +a \right )\right )}{3}\right ) \sin \left (b x +a \right )}{b}\)	\(52\)
default	\(\frac {-\frac {\cos ^{6}\left (b x +a \right )}{\sin \left (b x +a \right )}-\left (\frac {8}{3}+\cos ^{4}\left (b x +a \right )+\frac {4 \left (\cos ^{2}\left (b x +a \right )\right )}{3}\right ) \sin \left (b x +a \right )}{b}\)	\(52\)
risch	\(\frac {7 i {\mathrm e}^{i \left (b x +a \right )}}{8 b}-\frac {7 i {\mathrm e}^{-i \left (b x +a \right )}}{8 b}-\frac {2 i {\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {\sin \left (3 b x +3 a \right )}{12 b}\)	\(74\)
norman	\(\frac {-\frac {1}{2 b}-\frac {6 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {25 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {6 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )}{2 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{3} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-cos(b*x+a)^6/sin(b*x+a)-(8/3+cos(b*x+a)^4+4/3*cos(b*x+a)^2)*sin(b*x+a))

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Maxima [A]
time = 0.28, size = 32, normalized size = 0.84 \begin {gather*} \frac {\sin \left (b x + a\right )^{3} - \frac {3}{\sin \left (b x + a\right )} - 6 \, \sin \left (b x + a\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sin(b*x + a)^3 - 3/sin(b*x + a) - 6*sin(b*x + a))/b

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Fricas [A]
time = 0.36, size = 33, normalized size = 0.87 \begin {gather*} \frac {\cos \left (b x + a\right )^{4} + 4 \, \cos \left (b x + a\right )^{2} - 8}{3 \, b \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(cos(b*x + a)^4 + 4*cos(b*x + a)^2 - 8)/(b*sin(b*x + a))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (29) = 58\).
time = 0.55, size = 61, normalized size = 1.61 \begin {gather*} \begin {cases} - \frac {8 \sin ^{3}{\left (a + b x \right )}}{3 b} - \frac {4 \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} - \frac {\cos ^{4}{\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{5}{\left (a \right )}}{\sin ^{2}{\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)**5/sin(b*x+a)**2,x)

[Out]

Piecewise((-8*sin(a + b*x)**3/(3*b) - 4*sin(a + b*x)*cos(a + b*x)**2/b - cos(a + b*x)**4/(b*sin(a + b*x)), Ne(
b, 0)), (x*cos(a)**5/sin(a)**2, True))

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Giac [A]
time = 3.49, size = 32, normalized size = 0.84 \begin {gather*} \frac {\sin \left (b x + a\right )^{3} - \frac {3}{\sin \left (b x + a\right )} - 6 \, \sin \left (b x + a\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sin(b*x + a)^3 - 3/sin(b*x + a) - 6*sin(b*x + a))/b

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Mupad [B]
time = 0.45, size = 35, normalized size = 0.92 \begin {gather*} -\frac {-{\sin \left (a+b\,x\right )}^4+6\,{\sin \left (a+b\,x\right )}^2+3}{3\,b\,\sin \left (a+b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^5/sin(a + b*x)^2,x)

[Out]

-(6*sin(a + b*x)^2 - sin(a + b*x)^4 + 3)/(3*b*sin(a + b*x))

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