3.108 \(\int (c+d x) \cot ^2(a+b x) \, dx\)

Optimal. Leaf size=41 \[ \frac{d \log (\sin (a+b x))}{b^2}-\frac{(c+d x) \cot (a+b x)}{b}-c x-\frac{d x^2}{2} \]

[Out]

-(c*x) - (d*x^2)/2 - ((c + d*x)*Cot[a + b*x])/b + (d*Log[Sin[a + b*x]])/b^2

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Rubi [A]  time = 0.0261395, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3720, 3475} \[ \frac{d \log (\sin (a+b x))}{b^2}-\frac{(c+d x) \cot (a+b x)}{b}-c x-\frac{d x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Cot[a + b*x]^2,x]

[Out]

-(c*x) - (d*x^2)/2 - ((c + d*x)*Cot[a + b*x])/b + (d*Log[Sin[a + b*x]])/b^2

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rule 3475

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

Rubi steps

\begin{align*} \int (c+d x) \cot ^2(a+b x) \, dx &=-\frac{(c+d x) \cot (a+b x)}{b}+\frac{d \int \cot (a+b x) \, dx}{b}-\int (c+d x) \, dx\\ &=-c x-\frac{d x^2}{2}-\frac{(c+d x) \cot (a+b x)}{b}+\frac{d \log (\sin (a+b x))}{b^2}\\ \end{align*}

Mathematica [C]  time = 0.450996, size = 82, normalized size = 2. \[ -\frac{c \cot (a+b x) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-\tan ^2(a+b x)\right )}{b}+\frac{d \log (\sin (a+b x))}{b^2}+\frac{d x \csc (a) \sin (b x) \csc (a+b x)}{b}-\frac{d x \csc (a) (b x \sin (a)+2 \cos (a))}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Cot[a + b*x]^2,x]

[Out]

-((c*Cot[a + b*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[a + b*x]^2])/b) + (d*Log[Sin[a + b*x]])/b^2 - (d*x*Csc[
a]*(2*Cos[a] + b*x*Sin[a]))/(2*b) + (d*x*Csc[a]*Csc[a + b*x]*Sin[b*x])/b

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Maple [A]  time = 0.041, size = 49, normalized size = 1.2 \begin{align*} -{\frac{d{x}^{2}}{2}}-cx-{\frac{d\cot \left ( bx+a \right ) x}{b}}+{\frac{d\ln \left ( \sin \left ( bx+a \right ) \right ) }{{b}^{2}}}-{\frac{c\cot \left ( bx+a \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*cot(b*x+a)^2,x)

[Out]

-1/2*d*x^2-c*x-1/b*d*cot(b*x+a)*x+d*ln(sin(b*x+a))/b^2-1/b*c*cot(b*x+a)

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Maxima [B]  time = 1.68273, size = 394, normalized size = 9.61 \begin{align*} -\frac{2 \,{\left (b x + a + \frac{1}{\tan \left (b x + a\right )}\right )} c - \frac{2 \,{\left (b x + a + \frac{1}{\tan \left (b x + a\right )}\right )} a d}{b} + \frac{{\left ({\left (b x + a\right )}^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} +{\left (b x + a\right )}^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \,{\left (b x + a\right )}^{2} \cos \left (2 \, b x + 2 \, a\right ) +{\left (b x + a\right )}^{2} -{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) -{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) + 4 \,{\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d}{{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} b}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*cot(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/2*(2*(b*x + a + 1/tan(b*x + a))*c - 2*(b*x + a + 1/tan(b*x + a))*a*d/b + ((b*x + a)^2*cos(2*b*x + 2*a)^2 +
(b*x + a)^2*sin(2*b*x + 2*a)^2 - 2*(b*x + a)^2*cos(2*b*x + 2*a) + (b*x + a)^2 - (cos(2*b*x + 2*a)^2 + sin(2*b*
x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - (cos(2*b*x +
2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1
) + 4*(b*x + a)*sin(2*b*x + 2*a))*d/((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*b))/b

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Fricas [B]  time = 0.490468, size = 242, normalized size = 5.9 \begin{align*} -\frac{2 \, b d x - d \log \left (-\frac{1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac{1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) + 2 \, b c + 2 \,{\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (b^{2} d x^{2} + 2 \, b^{2} c x\right )} \sin \left (2 \, b x + 2 \, a\right )}{2 \, b^{2} \sin \left (2 \, b x + 2 \, a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*cot(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/2*(2*b*d*x - d*log(-1/2*cos(2*b*x + 2*a) + 1/2)*sin(2*b*x + 2*a) + 2*b*c + 2*(b*d*x + b*c)*cos(2*b*x + 2*a)
+ (b^2*d*x^2 + 2*b^2*c*x)*sin(2*b*x + 2*a))/(b^2*sin(2*b*x + 2*a))

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Sympy [A]  time = 1.29077, size = 100, normalized size = 2.44 \begin{align*} \begin{cases} \tilde{\infty } \left (c x + \frac{d x^{2}}{2}\right ) & \text{for}\: \left (a = 0 \vee a = - b x\right ) \wedge \left (a = - b x \vee b = 0\right ) \\\left (c x + \frac{d x^{2}}{2}\right ) \cot ^{2}{\left (a \right )} & \text{for}\: b = 0 \\- c x - \frac{d x^{2}}{2} - \frac{c}{b \tan{\left (a + b x \right )}} - \frac{d x}{b \tan{\left (a + b x \right )}} - \frac{d \log{\left (\tan ^{2}{\left (a + b x \right )} + 1 \right )}}{2 b^{2}} + \frac{d \log{\left (\tan{\left (a + b x \right )} \right )}}{b^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*cot(b*x+a)**2,x)

[Out]

Piecewise((zoo*(c*x + d*x**2/2), (Eq(a, 0) | Eq(a, -b*x)) & (Eq(b, 0) | Eq(a, -b*x))), ((c*x + d*x**2/2)*cot(a
)**2, Eq(b, 0)), (-c*x - d*x**2/2 - c/(b*tan(a + b*x)) - d*x/(b*tan(a + b*x)) - d*log(tan(a + b*x)**2 + 1)/(2*
b**2) + d*log(tan(a + b*x))/b**2, True))

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Giac [B]  time = 1.60698, size = 1856, normalized size = 45.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*cot(b*x+a)^2,x, algorithm="giac")

[Out]

-1/2*(b^2*d*x^2*tan(1/2*b*x)^2*tan(1/2*a) + b^2*d*x^2*tan(1/2*b*x)*tan(1/2*a)^2 + 2*b^2*c*x*tan(1/2*b*x)^2*tan
(1/2*a) + 2*b^2*c*x*tan(1/2*b*x)*tan(1/2*a)^2 - b*d*x*tan(1/2*b*x)^2*tan(1/2*a)^2 - b^2*d*x^2*tan(1/2*b*x) - b
^2*d*x^2*tan(1/2*a) - b*c*tan(1/2*b*x)^2*tan(1/2*a)^2 - 2*b^2*c*x*tan(1/2*b*x) + b*d*x*tan(1/2*b*x)^2 - 2*b^2*
c*x*tan(1/2*a) + 4*b*d*x*tan(1/2*b*x)*tan(1/2*a) - d*log(16*(tan(1/2*a)^4 + 2*tan(1/2*a)^2 + 1)/(tan(1/2*b*x)^
8*tan(1/2*a)^2 + 2*tan(1/2*b*x)^7*tan(1/2*a)^3 + tan(1/2*b*x)^6*tan(1/2*a)^4 - 2*tan(1/2*b*x)^7*tan(1/2*a) - 2
*tan(1/2*b*x)^6*tan(1/2*a)^2 + 2*tan(1/2*b*x)^5*tan(1/2*a)^3 + 2*tan(1/2*b*x)^4*tan(1/2*a)^4 + tan(1/2*b*x)^6
- 2*tan(1/2*b*x)^5*tan(1/2*a) - 6*tan(1/2*b*x)^4*tan(1/2*a)^2 - 2*tan(1/2*b*x)^3*tan(1/2*a)^3 + tan(1/2*b*x)^2
*tan(1/2*a)^4 + 2*tan(1/2*b*x)^4 + 2*tan(1/2*b*x)^3*tan(1/2*a) - 2*tan(1/2*b*x)^2*tan(1/2*a)^2 - 2*tan(1/2*b*x
)*tan(1/2*a)^3 + tan(1/2*b*x)^2 + 2*tan(1/2*b*x)*tan(1/2*a) + tan(1/2*a)^2))*tan(1/2*b*x)^2*tan(1/2*a) + b*d*x
*tan(1/2*a)^2 - d*log(16*(tan(1/2*a)^4 + 2*tan(1/2*a)^2 + 1)/(tan(1/2*b*x)^8*tan(1/2*a)^2 + 2*tan(1/2*b*x)^7*t
an(1/2*a)^3 + tan(1/2*b*x)^6*tan(1/2*a)^4 - 2*tan(1/2*b*x)^7*tan(1/2*a) - 2*tan(1/2*b*x)^6*tan(1/2*a)^2 + 2*ta
n(1/2*b*x)^5*tan(1/2*a)^3 + 2*tan(1/2*b*x)^4*tan(1/2*a)^4 + tan(1/2*b*x)^6 - 2*tan(1/2*b*x)^5*tan(1/2*a) - 6*t
an(1/2*b*x)^4*tan(1/2*a)^2 - 2*tan(1/2*b*x)^3*tan(1/2*a)^3 + tan(1/2*b*x)^2*tan(1/2*a)^4 + 2*tan(1/2*b*x)^4 +
2*tan(1/2*b*x)^3*tan(1/2*a) - 2*tan(1/2*b*x)^2*tan(1/2*a)^2 - 2*tan(1/2*b*x)*tan(1/2*a)^3 + tan(1/2*b*x)^2 + 2
*tan(1/2*b*x)*tan(1/2*a) + tan(1/2*a)^2))*tan(1/2*b*x)*tan(1/2*a)^2 + b*c*tan(1/2*b*x)^2 + 4*b*c*tan(1/2*b*x)*
tan(1/2*a) + b*c*tan(1/2*a)^2 - b*d*x + d*log(16*(tan(1/2*a)^4 + 2*tan(1/2*a)^2 + 1)/(tan(1/2*b*x)^8*tan(1/2*a
)^2 + 2*tan(1/2*b*x)^7*tan(1/2*a)^3 + tan(1/2*b*x)^6*tan(1/2*a)^4 - 2*tan(1/2*b*x)^7*tan(1/2*a) - 2*tan(1/2*b*
x)^6*tan(1/2*a)^2 + 2*tan(1/2*b*x)^5*tan(1/2*a)^3 + 2*tan(1/2*b*x)^4*tan(1/2*a)^4 + tan(1/2*b*x)^6 - 2*tan(1/2
*b*x)^5*tan(1/2*a) - 6*tan(1/2*b*x)^4*tan(1/2*a)^2 - 2*tan(1/2*b*x)^3*tan(1/2*a)^3 + tan(1/2*b*x)^2*tan(1/2*a)
^4 + 2*tan(1/2*b*x)^4 + 2*tan(1/2*b*x)^3*tan(1/2*a) - 2*tan(1/2*b*x)^2*tan(1/2*a)^2 - 2*tan(1/2*b*x)*tan(1/2*a
)^3 + tan(1/2*b*x)^2 + 2*tan(1/2*b*x)*tan(1/2*a) + tan(1/2*a)^2))*tan(1/2*b*x) + d*log(16*(tan(1/2*a)^4 + 2*ta
n(1/2*a)^2 + 1)/(tan(1/2*b*x)^8*tan(1/2*a)^2 + 2*tan(1/2*b*x)^7*tan(1/2*a)^3 + tan(1/2*b*x)^6*tan(1/2*a)^4 - 2
*tan(1/2*b*x)^7*tan(1/2*a) - 2*tan(1/2*b*x)^6*tan(1/2*a)^2 + 2*tan(1/2*b*x)^5*tan(1/2*a)^3 + 2*tan(1/2*b*x)^4*
tan(1/2*a)^4 + tan(1/2*b*x)^6 - 2*tan(1/2*b*x)^5*tan(1/2*a) - 6*tan(1/2*b*x)^4*tan(1/2*a)^2 - 2*tan(1/2*b*x)^3
*tan(1/2*a)^3 + tan(1/2*b*x)^2*tan(1/2*a)^4 + 2*tan(1/2*b*x)^4 + 2*tan(1/2*b*x)^3*tan(1/2*a) - 2*tan(1/2*b*x)^
2*tan(1/2*a)^2 - 2*tan(1/2*b*x)*tan(1/2*a)^3 + tan(1/2*b*x)^2 + 2*tan(1/2*b*x)*tan(1/2*a) + tan(1/2*a)^2))*tan
(1/2*a) - b*c)/(b^2*tan(1/2*b*x)^2*tan(1/2*a) + b^2*tan(1/2*b*x)*tan(1/2*a)^2 - b^2*tan(1/2*b*x) - b^2*tan(1/2
*a))