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

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

[Out]

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

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Rubi [A]  time = 0.0645253, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3542, 3531, 3475} \[ -x \left (a^2-b^2\right )-\frac{a^2 \cot (c+d x)}{d}+\frac{2 a b \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*(a + b*Tan[c + d*x])^2,x]

[Out]

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

Rule 3542

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

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 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 \cot ^2(c+d x) (a+b \tan (c+d x))^2 \, dx &=-\frac{a^2 \cot (c+d x)}{d}+\int \cot (c+d x) \left (2 a b-\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=-\left (a^2-b^2\right ) x-\frac{a^2 \cot (c+d x)}{d}+(2 a b) \int \cot (c+d x) \, dx\\ &=-\left (a^2-b^2\right ) x-\frac{a^2 \cot (c+d x)}{d}+\frac{2 a b \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [C]  time = 0.151921, size = 82, normalized size = 2. \[ \frac{-2 a^2 \cot (c+d x)+i \left ((a-i b)^2 (-\log (\tan (c+d x)+i))+(a+i b)^2 \log (-\tan (c+d x)+i)-4 i a b \log (\tan (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*(a + b*Tan[c + d*x])^2,x]

[Out]

(-2*a^2*Cot[c + d*x] + I*((a + I*b)^2*Log[I - Tan[c + d*x]] - (4*I)*a*b*Log[Tan[c + d*x]] - (a - I*b)^2*Log[I
+ Tan[c + d*x]]))/(2*d)

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Maple [A]  time = 0.037, size = 58, normalized size = 1.4 \begin{align*} -{a}^{2}x+{b}^{2}x-{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}+2\,{\frac{ab\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}c}{d}}+{\frac{{b}^{2}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.63351, size = 78, normalized size = 1.9 \begin{align*} -\frac{a b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, a b \log \left (\tan \left (d x + c\right )\right ) +{\left (a^{2} - b^{2}\right )}{\left (d x + c\right )} + \frac{a^{2}}{\tan \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.7548, size = 157, normalized size = 3.83 \begin{align*} -\frac{{\left (a^{2} - b^{2}\right )} d x \tan \left (d x + c\right ) - a b \log \left (\frac{\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) + a^{2}}{d \tan \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.55002, size = 85, normalized size = 2.07 \begin{align*} \begin{cases} \tilde{\infty } a^{2} x & \text{for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (a + b \tan{\left (c \right )}\right )^{2} \cot ^{2}{\left (c \right )} & \text{for}\: d = 0 \\- a^{2} x - \frac{a^{2}}{d \tan{\left (c + d x \right )}} - \frac{a b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac{2 a b \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + b^{2} x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.4756, size = 132, normalized size = 3.22 \begin{align*} -\frac{4 \, a b \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 4 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \,{\left (a^{2} - b^{2}\right )}{\left (d x + c\right )} + \frac{4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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