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

Optimal. Leaf size=83 \[ -\frac{a \left (a^2-3 b^2\right ) \log (\sin (c+d x))}{d}-b x \left (3 a^2-b^2\right )-\frac{5 a^2 b \cot (c+d x)}{2 d}-\frac{a^2 \cot ^2(c+d x) (a+b \tan (c+d x))}{2 d} \]

[Out]

-(b*(3*a^2 - b^2)*x) - (5*a^2*b*Cot[c + d*x])/(2*d) - (a*(a^2 - 3*b^2)*Log[Sin[c + d*x]])/d - (a^2*Cot[c + d*x
]^2*(a + b*Tan[c + d*x]))/(2*d)

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Rubi [A]  time = 0.141154, antiderivative size = 83, 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 = {3565, 3628, 3531, 3475} \[ -\frac{a \left (a^2-3 b^2\right ) \log (\sin (c+d x))}{d}-b x \left (3 a^2-b^2\right )-\frac{5 a^2 b \cot (c+d x)}{2 d}-\frac{a^2 \cot ^2(c+d x) (a+b \tan (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(3*a^2 - b^2)*x) - (5*a^2*b*Cot[c + d*x])/(2*d) - (a*(a^2 - 3*b^2)*Log[Sin[c + d*x]])/d - (a^2*Cot[c + d*x
]^2*(a + b*Tan[c + d*x]))/(2*d)

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3628

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*(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*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 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 ^3(c+d x) (a+b \tan (c+d x))^3 \, dx &=-\frac{a^2 \cot ^2(c+d x) (a+b \tan (c+d x))}{2 d}+\frac{1}{2} \int \cot ^2(c+d x) \left (5 a^2 b-2 a \left (a^2-3 b^2\right ) \tan (c+d x)-b \left (a^2-2 b^2\right ) \tan ^2(c+d x)\right ) \, dx\\ &=-\frac{5 a^2 b \cot (c+d x)}{2 d}-\frac{a^2 \cot ^2(c+d x) (a+b \tan (c+d x))}{2 d}+\frac{1}{2} \int \cot (c+d x) \left (-2 a \left (a^2-3 b^2\right )-2 b \left (3 a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=-b \left (3 a^2-b^2\right ) x-\frac{5 a^2 b \cot (c+d x)}{2 d}-\frac{a^2 \cot ^2(c+d x) (a+b \tan (c+d x))}{2 d}-\left (a \left (a^2-3 b^2\right )\right ) \int \cot (c+d x) \, dx\\ &=-b \left (3 a^2-b^2\right ) x-\frac{5 a^2 b \cot (c+d x)}{2 d}-\frac{a \left (a^2-3 b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 \cot ^2(c+d x) (a+b \tan (c+d x))}{2 d}\\ \end{align*}

Mathematica [C]  time = 0.329958, size = 96, normalized size = 1.16 \[ \frac{-2 a \left (a^2-3 b^2\right ) \log (\tan (c+d x))-6 a^2 b \cot (c+d x)+a^3 \left (-\cot ^2(c+d x)\right )+(a+i b)^3 \log (-\tan (c+d x)+i)+(a-i b)^3 \log (\tan (c+d x)+i)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a^2*b*Cot[c + d*x] - a^3*Cot[c + d*x]^2 + (a + I*b)^3*Log[I - Tan[c + d*x]] - 2*a*(a^2 - 3*b^2)*Log[Tan[c
+ d*x]] + (a - I*b)^3*Log[I + Tan[c + d*x]])/(2*d)

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Maple [A]  time = 0.058, size = 94, normalized size = 1.1 \begin{align*}{b}^{3}x+{\frac{{b}^{3}c}{d}}+3\,{\frac{a{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-3\,x{a}^{2}b-3\,{\frac{b{a}^{2}\cot \left ( dx+c \right ) }{d}}-3\,{\frac{{a}^{2}bc}{d}}-{\frac{{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{a}^{3}\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))^3,x)

[Out]

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

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Maxima [A]  time = 1.66, size = 124, normalized size = 1.49 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )}{\left (d x + c\right )} -{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac{6 \, a^{2} b \tan \left (d x + c\right ) + a^{3}}{\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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.77601, size = 236, normalized size = 2.84 \begin{align*} -\frac{6 \, a^{2} b \tan \left (d x + c\right ) +{\left (a^{3} - 3 \, a 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} + a^{3} +{\left (a^{3} + 2 \,{\left (3 \, a^{2} b - b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{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))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.7936, size = 146, normalized size = 1.76 \begin{align*} \begin{cases} \tilde{\infty } a^{3} 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 )^{3} \cot ^{3}{\left (c \right )} & \text{for}\: d = 0 \\\frac{a^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac{a^{3} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} - \frac{a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 a^{2} b x - \frac{3 a^{2} b}{d \tan{\left (c + d x \right )}} - \frac{3 a b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{3 a b^{2} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + b^{3} x & \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))**3,x)

[Out]

Piecewise((zoo*a**3*x, (Eq(c, 0) | Eq(c, -d*x)) & (Eq(d, 0) | Eq(c, -d*x))), (x*(a + b*tan(c))**3*cot(c)**3, E
q(d, 0)), (a**3*log(tan(c + d*x)**2 + 1)/(2*d) - a**3*log(tan(c + d*x))/d - a**3/(2*d*tan(c + d*x)**2) - 3*a**
2*b*x - 3*a**2*b/(d*tan(c + d*x)) - 3*a*b**2*log(tan(c + d*x)**2 + 1)/(2*d) + 3*a*b**2*log(tan(c + d*x))/d + b
**3*x, True))

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Giac [B]  time = 1.99117, size = 231, normalized size = 2.78 \begin{align*} -\frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 \,{\left (3 \, a^{2} b - b^{3}\right )}{\left (d x + c\right )} - 8 \,{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) + 8 \,{\left (a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{12 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 36 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{3}}{\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))^3,x, algorithm="giac")

[Out]

-1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a^2*b*tan(1/2*d*x + 1/2*c) + 8*(3*a^2*b - b^3)*(d*x + c) - 8*(a^3 - 3*a*
b^2)*log(tan(1/2*d*x + 1/2*c)^2 + 1) + 8*(a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) - (12*a^3*tan(1/2*d*x
+ 1/2*c)^2 - 36*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 12*a^2*b*tan(1/2*d*x + 1/2*c) - a^3)/tan(1/2*d*x + 1/2*c)^2)/d