3.431 \(\int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=58 \[ -\frac{\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 \cot ^2(c+d x)}{2 d}-\frac{2 a b \cot (c+d x)}{d}-2 a b x \]

[Out]

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

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Rubi [A]  time = 0.100025, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3542, 3529, 3531, 3475} \[ -\frac{\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 \cot ^2(c+d x)}{2 d}-\frac{2 a b \cot (c+d x)}{d}-2 a b x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^2,x]

[Out]

-2*a*b*x - (2*a*b*Cot[c + d*x])/d - (a^2*Cot[c + d*x]^2)/(2*d) - ((a^2 - b^2)*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 3529

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

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 ^3(c+d x) (a+b \tan (c+d x))^2 \, dx &=-\frac{a^2 \cot ^2(c+d x)}{2 d}+\int \cot ^2(c+d x) \left (2 a b-\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=-\frac{2 a b \cot (c+d x)}{d}-\frac{a^2 \cot ^2(c+d x)}{2 d}+\int \cot (c+d x) \left (-a^2+b^2-2 a b \tan (c+d x)\right ) \, dx\\ &=-2 a b x-\frac{2 a b \cot (c+d x)}{d}-\frac{a^2 \cot ^2(c+d x)}{2 d}+\left (-a^2+b^2\right ) \int \cot (c+d x) \, dx\\ &=-2 a b x-\frac{2 a b \cot (c+d x)}{d}-\frac{a^2 \cot ^2(c+d x)}{2 d}-\frac{\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [C]  time = 0.26111, size = 92, normalized size = 1.59 \[ \frac{-a^2 \cot ^2(c+d x)-4 a b \cot (c+d x)+(a-i b)^2 \log (\tan (c+d x)+i)+(a+i b)^2 \log (-\tan (c+d x)+i)-2 (a-b) (a+b) \log (\tan (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^2,x]

[Out]

(-4*a*b*Cot[c + d*x] - a^2*Cot[c + d*x]^2 + (a + I*b)^2*Log[I - Tan[c + d*x]] - 2*(a - b)*(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 = 0.052, size = 73, normalized size = 1.3 \begin{align*}{\frac{{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-2\,abx-2\,{\frac{ab\cot \left ( dx+c \right ) }{d}}-2\,{\frac{abc}{d}}-{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.64902, size = 105, normalized size = 1.81 \begin{align*} -\frac{4 \,{\left (d x + c\right )} a b -{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac{4 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.79176, size = 212, normalized size = 3.66 \begin{align*} -\frac{{\left (a^{2} - b^{2}\right )} \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 )^{2} + 4 \, a b \tan \left (d x + c\right ) +{\left (4 \, a b d x + a^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2}}{2 \, d \tan \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.92323, size = 131, normalized size = 2.26 \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 ^{3}{\left (c \right )} & \text{for}\: d = 0 \\\frac{a^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac{a^{2} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} - \frac{a^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - 2 a b x - \frac{2 a b}{d \tan{\left (c + d x \right )}} - \frac{b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{b^{2} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(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)**3, E
q(d, 0)), (a**2*log(tan(c + d*x)**2 + 1)/(2*d) - a**2*log(tan(c + d*x))/d - a**2/(2*d*tan(c + d*x)**2) - 2*a*b
*x - 2*a*b/(d*tan(c + d*x)) - b**2*log(tan(c + d*x)**2 + 1)/(2*d) + b**2*log(tan(c + d*x))/d, True))

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Giac [B]  time = 1.52623, size = 208, normalized size = 3.59 \begin{align*} -\frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 16 \,{\left (d x + c\right )} a b - 8 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \,{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) + 8 \,{\left (a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, 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}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 + 16*(d*x + c)*a*b - 8*a*b*tan(1/2*d*x + 1/2*c) - 8*(a^2 - b^2)*log(tan(1/2*d
*x + 1/2*c)^2 + 1) + 8*(a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c))) - (12*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*ta
n(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^2)/d