3.432 \(\int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=78 \[ \frac{\left (a^2-b^2\right ) \cot (c+d x)}{d}+x \left (a^2-b^2\right )-\frac{a^2 \cot ^3(c+d x)}{3 d}-\frac{a b \cot ^2(c+d x)}{d}-\frac{2 a b \log (\sin (c+d x))}{d} \]

[Out]

(a^2 - b^2)*x + ((a^2 - b^2)*Cot[c + d*x])/d - (a*b*Cot[c + d*x]^2)/d - (a^2*Cot[c + d*x]^3)/(3*d) - (2*a*b*Lo
g[Sin[c + d*x]])/d

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Rubi [A]  time = 0.131527, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3542, 3529, 3531, 3475} \[ \frac{\left (a^2-b^2\right ) \cot (c+d x)}{d}+x \left (a^2-b^2\right )-\frac{a^2 \cot ^3(c+d x)}{3 d}-\frac{a b \cot ^2(c+d x)}{d}-\frac{2 a b \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^2,x]

[Out]

(a^2 - b^2)*x + ((a^2 - b^2)*Cot[c + d*x])/d - (a*b*Cot[c + d*x]^2)/d - (a^2*Cot[c + d*x]^3)/(3*d) - (2*a*b*Lo
g[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 ^4(c+d x) (a+b \tan (c+d x))^2 \, dx &=-\frac{a^2 \cot ^3(c+d x)}{3 d}+\int \cot ^3(c+d x) \left (2 a b-\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=-\frac{a b \cot ^2(c+d x)}{d}-\frac{a^2 \cot ^3(c+d x)}{3 d}+\int \cot ^2(c+d x) \left (-a^2+b^2-2 a b \tan (c+d x)\right ) \, dx\\ &=\frac{\left (a^2-b^2\right ) \cot (c+d x)}{d}-\frac{a b \cot ^2(c+d x)}{d}-\frac{a^2 \cot ^3(c+d x)}{3 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{\left (a^2-b^2\right ) \cot (c+d x)}{d}-\frac{a b \cot ^2(c+d x)}{d}-\frac{a^2 \cot ^3(c+d x)}{3 d}-(2 a b) \int \cot (c+d x) \, dx\\ &=\left (a^2-b^2\right ) x+\frac{\left (a^2-b^2\right ) \cot (c+d x)}{d}-\frac{a b \cot ^2(c+d x)}{d}-\frac{a^2 \cot ^3(c+d x)}{3 d}-\frac{2 a b \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [C]  time = 0.68775, size = 103, normalized size = 1.32 \[ -\frac{a^2 \cot ^3(c+d x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\tan ^2(c+d x)\right )}{3 d}-\frac{a b \left (\cot ^2(c+d x)+2 \log (\tan (c+d x))+2 \log (\cos (c+d x))\right )}{d}-\frac{b^2 \cot (c+d x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\tan ^2(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^2,x]

[Out]

-(a^2*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2])/(3*d) - (b^2*Cot[c + d*x]*Hypergeometr
ic2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2])/d - (a*b*(Cot[c + d*x]^2 + 2*Log[Cos[c + d*x]] + 2*Log[Tan[c + d*x]]))/d

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Maple [A]  time = 0.044, size = 102, normalized size = 1.3 \begin{align*} -{b}^{2}x-{\frac{{b}^{2}\cot \left ( dx+c \right ) }{d}}-{\frac{{b}^{2}c}{d}}-{\frac{ab \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{d}}-2\,{\frac{ab\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}+{a}^{2}x+{\frac{{a}^{2}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+b*tan(d*x+c))^2,x)

[Out]

-b^2*x-1/d*b^2*cot(d*x+c)-1/d*b^2*c-a*b*cot(d*x+c)^2/d-2*a*b*ln(sin(d*x+c))/d-1/3*a^2*cot(d*x+c)^3/d+a^2*cot(d
*x+c)/d+a^2*x+1/d*a^2*c

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Maxima [A]  time = 1.60489, size = 123, normalized size = 1.58 \begin{align*} \frac{3 \, a b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, a b \log \left (\tan \left (d x + c\right )\right ) + 3 \,{\left (a^{2} - b^{2}\right )}{\left (d x + c\right )} - \frac{3 \, a b \tan \left (d x + c\right ) - 3 \,{\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/3*(3*a*b*log(tan(d*x + c)^2 + 1) - 6*a*b*log(tan(d*x + c)) + 3*(a^2 - b^2)*(d*x + c) - (3*a*b*tan(d*x + c) -
3*(a^2 - b^2)*tan(d*x + c)^2 + a^2)/tan(d*x + c)^3)/d

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Fricas [A]  time = 1.76239, size = 257, normalized size = 3.29 \begin{align*} -\frac{3 \, 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 )^{3} - 3 \,{\left ({\left (a^{2} - b^{2}\right )} d x - a b\right )} \tan \left (d x + c\right )^{3} + 3 \, a b \tan \left (d x + c\right ) - 3 \,{\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2}}{3 \, d \tan \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(3*a*b*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^3 - 3*((a^2 - b^2)*d*x - a*b)*tan(d*x + c)^3
+ 3*a*b*tan(d*x + c) - 3*(a^2 - b^2)*tan(d*x + c)^2 + a^2)/(d*tan(d*x + c)^3)

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Sympy [A]  time = 4.90077, size = 126, normalized size = 1.62 \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 ^{4}{\left (c \right )} & \text{for}\: d = 0 \\a^{2} x + \frac{a^{2}}{d \tan{\left (c + d x \right )}} - \frac{a^{2}}{3 d \tan ^{3}{\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} - \frac{a b}{d \tan ^{2}{\left (c + d x \right )}} - b^{2} x - \frac{b^{2}}{d \tan{\left (c + d x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(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)**4, E
q(d, 0)), (a**2*x + a**2/(d*tan(c + d*x)) - a**2/(3*d*tan(c + d*x)**3) + a*b*log(tan(c + d*x)**2 + 1)/d - 2*a*
b*log(tan(c + d*x))/d - a*b/(d*tan(c + d*x)**2) - b**2*x - b**2/(d*tan(c + d*x)), True))

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Giac [B]  time = 1.53491, size = 258, normalized size = 3.31 \begin{align*} \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 48 \, a b \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 48 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \,{\left (a^{2} - b^{2}\right )}{\left (d x + c\right )} + \frac{88 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, 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} - 6 \, 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 )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 - 6*a*b*tan(1/2*d*x + 1/2*c)^2 + 48*a*b*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 48*
a*b*log(abs(tan(1/2*d*x + 1/2*c))) - 15*a^2*tan(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2*d*x + 1/2*c) + 24*(a^2 - b^2
)*(d*x + c) + (88*a*b*tan(1/2*d*x + 1/2*c)^3 + 15*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 1/2*c)^2 -
6*a*b*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^3)/d