3.433 \(\int \cot ^5(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=98 \[ \frac{\left (a^2-b^2\right ) \cot ^2(c+d x)}{2 d}+\frac{\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 \cot ^4(c+d x)}{4 d}-\frac{2 a b \cot ^3(c+d x)}{3 d}+\frac{2 a b \cot (c+d x)}{d}+2 a b x \]

[Out]

2*a*b*x + (2*a*b*Cot[c + d*x])/d + ((a^2 - b^2)*Cot[c + d*x]^2)/(2*d) - (2*a*b*Cot[c + d*x]^3)/(3*d) - (a^2*Co
t[c + d*x]^4)/(4*d) + ((a^2 - b^2)*Log[Sin[c + d*x]])/d

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Rubi [A]  time = 0.156156, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3542, 3529, 3531, 3475} \[ \frac{\left (a^2-b^2\right ) \cot ^2(c+d x)}{2 d}+\frac{\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 \cot ^4(c+d x)}{4 d}-\frac{2 a b \cot ^3(c+d x)}{3 d}+\frac{2 a b \cot (c+d x)}{d}+2 a b x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]

[Out]

2*a*b*x + (2*a*b*Cot[c + d*x])/d + ((a^2 - b^2)*Cot[c + d*x]^2)/(2*d) - (2*a*b*Cot[c + d*x]^3)/(3*d) - (a^2*Co
t[c + d*x]^4)/(4*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 ^5(c+d x) (a+b \tan (c+d x))^2 \, dx &=-\frac{a^2 \cot ^4(c+d x)}{4 d}+\int \cot ^4(c+d x) \left (2 a b-\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=-\frac{2 a b \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot ^4(c+d x)}{4 d}+\int \cot ^3(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 ^2(c+d x)}{2 d}-\frac{2 a b \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot ^4(c+d x)}{4 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{\left (a^2-b^2\right ) \cot ^2(c+d x)}{2 d}-\frac{2 a b \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot ^4(c+d x)}{4 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{\left (a^2-b^2\right ) \cot ^2(c+d x)}{2 d}-\frac{2 a b \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot ^4(c+d x)}{4 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{\left (a^2-b^2\right ) \cot ^2(c+d x)}{2 d}-\frac{2 a b \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot ^4(c+d x)}{4 d}+\frac{\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [C]  time = 0.548745, size = 122, normalized size = 1.24 \[ \frac{a^2 \left (-\cot ^4(c+d x)+2 \cot ^2(c+d x)+4 \log (\tan (c+d x))+4 \log (\cos (c+d x))\right )}{4 d}-\frac{2 a b \cot ^3(c+d x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\tan ^2(c+d x)\right )}{3 d}-\frac{b^2 \left (\cot ^2(c+d x)+2 \log (\tan (c+d x))+2 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]

[Out]

(-2*a*b*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2])/(3*d) - (b^2*(Cot[c + d*x]^2 + 2*Log
[Cos[c + d*x]] + 2*Log[Tan[c + d*x]]))/(2*d) + (a^2*(2*Cot[c + d*x]^2 - Cot[c + d*x]^4 + 4*Log[Cos[c + d*x]] +
4*Log[Tan[c + d*x]]))/(4*d)

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Maple [A]  time = 0.046, size = 120, normalized size = 1.2 \begin{align*} -{\frac{{b}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{2\,ab \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+2\,{\frac{ab\cot \left ( dx+c \right ) }{d}}+2\,abx+2\,{\frac{abc}{d}}-{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,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)^5*(a+b*tan(d*x+c))^2,x)

[Out]

-1/2/d*b^2*cot(d*x+c)^2-1/d*b^2*ln(sin(d*x+c))-2/3*a*b*cot(d*x+c)^3/d+2*a*b*cot(d*x+c)/d+2*a*b*x+2/d*a*b*c-1/4
*a^2*cot(d*x+c)^4/d+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 = 1.60176, size = 150, normalized size = 1.53 \begin{align*} \frac{24 \,{\left (d x + c\right )} a b - 6 \,{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \,{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac{24 \, a b \tan \left (d x + c\right )^{3} - 8 \, a b \tan \left (d x + c\right ) + 6 \,{\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} - 3 \, a^{2}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(24*(d*x + c)*a*b - 6*(a^2 - b^2)*log(tan(d*x + c)^2 + 1) + 12*(a^2 - b^2)*log(tan(d*x + c)) + (24*a*b*ta
n(d*x + c)^3 - 8*a*b*tan(d*x + c) + 6*(a^2 - b^2)*tan(d*x + c)^2 - 3*a^2)/tan(d*x + c)^4)/d

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Fricas [A]  time = 1.7871, size = 308, normalized size = 3.14 \begin{align*} \frac{6 \,{\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 )^{4} + 24 \, a b \tan \left (d x + c\right )^{3} + 3 \,{\left (8 \, a b d x + 3 \, a^{2} - 2 \, b^{2}\right )} \tan \left (d x + c\right )^{4} - 8 \, a b \tan \left (d x + c\right ) + 6 \,{\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} - 3 \, a^{2}}{12 \, d \tan \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(6*(a^2 - b^2)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^4 + 24*a*b*tan(d*x + c)^3 + 3*(8*a*b
*d*x + 3*a^2 - 2*b^2)*tan(d*x + c)^4 - 8*a*b*tan(d*x + c) + 6*(a^2 - b^2)*tan(d*x + c)^2 - 3*a^2)/(d*tan(d*x +
c)^4)

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Sympy [A]  time = 9.33855, size = 178, normalized size = 1.82 \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 ^{5}{\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 )}} - \frac{a^{2}}{4 d \tan ^{4}{\left (c + d x \right )}} + 2 a b x + \frac{2 a b}{d \tan{\left (c + d x \right )}} - \frac{2 a b}{3 d \tan ^{3}{\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} - \frac{b^{2}}{2 d \tan ^{2}{\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)**5*(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)**5, 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) - a**2
/(4*d*tan(c + d*x)**4) + 2*a*b*x + 2*a*b/(d*tan(c + d*x)) - 2*a*b/(3*d*tan(c + d*x)**3) + b**2*log(tan(c + d*x
)**2 + 1)/(2*d) - b**2*log(tan(c + d*x))/d - b**2/(2*d*tan(c + d*x)**2), True))

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Giac [B]  time = 1.57947, size = 335, normalized size = 3.42 \begin{align*} -\frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 16 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 384 \,{\left (d x + c\right )} a b + 240 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 192 \,{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 192 \,{\left (a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{400 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 400 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 240 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 16 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 - 16*a*b*tan(1/2*d*x + 1/2*c)^3 - 36*a^2*tan(1/2*d*x + 1/2*c)^2 + 24*b^2*
tan(1/2*d*x + 1/2*c)^2 - 384*(d*x + c)*a*b + 240*a*b*tan(1/2*d*x + 1/2*c) + 192*(a^2 - b^2)*log(tan(1/2*d*x +
1/2*c)^2 + 1) - 192*(a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + (400*a^2*tan(1/2*d*x + 1/2*c)^4 - 400*b^2*tan
(1/2*d*x + 1/2*c)^4 - 240*a*b*tan(1/2*d*x + 1/2*c)^3 - 36*a^2*tan(1/2*d*x + 1/2*c)^2 + 24*b^2*tan(1/2*d*x + 1/
2*c)^2 + 16*a*b*tan(1/2*d*x + 1/2*c) + 3*a^2)/tan(1/2*d*x + 1/2*c)^4)/d