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

Optimal. Leaf size=117 \[ -\frac {4 a^3 b \cot ^2(c+d x)}{3 d}+\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{3 d}-\frac {4 a b \left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}+x \left (a^4-6 a^2 b^2+b^4\right ) \]

[Out]

(a^4-6*a^2*b^2+b^4)*x+1/3*a^2*(3*a^2-17*b^2)*cot(d*x+c)/d-4/3*a^3*b*cot(d*x+c)^2/d-4*a*b*(a^2-b^2)*ln(sin(d*x+
c))/d-1/3*a^2*cot(d*x+c)^3*(a+b*tan(d*x+c))^2/d

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Rubi [A]  time = 0.30, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3565, 3635, 3628, 3531, 3475} \[ \frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{3 d}-\frac {4 a b \left (a^2-b^2\right ) \log (\sin (c+d x))}{d}+x \left (-6 a^2 b^2+a^4+b^4\right )-\frac {4 a^3 b \cot ^2(c+d x)}{3 d}-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4 - 6*a^2*b^2 + b^4)*x + (a^2*(3*a^2 - 17*b^2)*Cot[c + d*x])/(3*d) - (4*a^3*b*Cot[c + d*x]^2)/(3*d) - (4*a*
b*(a^2 - b^2)*Log[Sin[c + d*x]])/d - (a^2*Cot[c + d*x]^3*(a + b*Tan[c + d*x])^2)/(3*d)

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 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 3635

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

Rubi steps

\begin {align*} \int \cot ^4(c+d x) (a+b \tan (c+d x))^4 \, dx &=-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}+\frac {1}{3} \int \cot ^3(c+d x) (a+b \tan (c+d x)) \left (8 a^2 b-3 a \left (a^2-3 b^2\right ) \tan (c+d x)-b \left (a^2-3 b^2\right ) \tan ^2(c+d x)\right ) \, dx\\ &=-\frac {4 a^3 b \cot ^2(c+d x)}{3 d}-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}+\frac {1}{3} \int \cot ^2(c+d x) \left (-a^2 \left (3 a^2-17 b^2\right )-12 a b \left (a^2-b^2\right ) \tan (c+d x)-b^2 \left (a^2-3 b^2\right ) \tan ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{3 d}-\frac {4 a^3 b \cot ^2(c+d x)}{3 d}-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}+\frac {1}{3} \int \cot (c+d x) \left (-12 a b \left (a^2-b^2\right )+3 \left (a^4-6 a^2 b^2+b^4\right ) \tan (c+d x)\right ) \, dx\\ &=\left (a^4-6 a^2 b^2+b^4\right ) x+\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{3 d}-\frac {4 a^3 b \cot ^2(c+d x)}{3 d}-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}-\left (4 a b \left (a^2-b^2\right )\right ) \int \cot (c+d x) \, dx\\ &=\left (a^4-6 a^2 b^2+b^4\right ) x+\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{3 d}-\frac {4 a^3 b \cot ^2(c+d x)}{3 d}-\frac {4 a b \left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\\ \end {align*}

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Mathematica [C]  time = 1.27, size = 125, normalized size = 1.07 \[ -\frac {2 a^4 \cot ^3(c+d x)+12 a^3 b \cot ^2(c+d x)-6 a^2 \left (a^2-6 b^2\right ) \cot (c+d x)+24 a b \left (a^2-b^2\right ) \log (\tan (c+d x))+3 i (a+i b)^4 \log (-\tan (c+d x)+i)-3 i (a-i b)^4 \log (\tan (c+d x)+i)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(-6*a^2*(a^2 - 6*b^2)*Cot[c + d*x] + 12*a^3*b*Cot[c + d*x]^2 + 2*a^4*Cot[c + d*x]^3 + (3*I)*(a + I*b)^4*L
og[I - Tan[c + d*x]] + 24*a*b*(a^2 - b^2)*Log[Tan[c + d*x]] - (3*I)*(a - I*b)^4*Log[I + Tan[c + d*x]])/d

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fricas [A]  time = 0.47, size = 131, normalized size = 1.12 \[ -\frac {6 \, a^{3} b \tan \left (d x + c\right ) + 6 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} + a^{4} + 3 \, {\left (2 \, a^{3} b - {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} d x\right )} \tan \left (d x + c\right )^{3} - 3 \, {\left (a^{4} - 6 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{3 \, d \tan \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 7.09, size = 246, normalized size = 2.10 \[ \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )} + 96 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 96 \, {\left (a^{3} b - a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {176 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 176 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{4}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^4*tan(1/2*d*x + 1/2*c)^3 - 12*a^3*b*tan(1/2*d*x + 1/2*c)^2 - 15*a^4*tan(1/2*d*x + 1/2*c) + 72*a^2*b^2*
tan(1/2*d*x + 1/2*c) + 24*(a^4 - 6*a^2*b^2 + b^4)*(d*x + c) + 96*(a^3*b - a*b^3)*log(tan(1/2*d*x + 1/2*c)^2 +
1) - 96*(a^3*b - a*b^3)*log(abs(tan(1/2*d*x + 1/2*c))) + (176*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 176*a*b^3*tan(1/2
*d*x + 1/2*c)^3 + 15*a^4*tan(1/2*d*x + 1/2*c)^2 - 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 12*a^3*b*tan(1/2*d*x + 1
/2*c) - a^4)/tan(1/2*d*x + 1/2*c)^3)/d

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maple [A]  time = 0.33, size = 144, normalized size = 1.23 \[ -\frac {a^{4} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{4} \cot \left (d x +c \right )}{d}+a^{4} x +\frac {a^{4} c}{d}-\frac {2 a^{3} b \left (\cot ^{2}\left (d x +c \right )\right )}{d}-\frac {4 a^{3} b \ln \left (\sin \left (d x +c \right )\right )}{d}-6 a^{2} b^{2} x -\frac {6 \cot \left (d x +c \right ) a^{2} b^{2}}{d}-\frac {6 a^{2} b^{2} c}{d}+\frac {4 a \,b^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}+b^{4} x +\frac {b^{4} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^4*cot(d*x+c)^3/d+a^4*cot(d*x+c)/d+a^4*x+1/d*a^4*c-2*a^3*b*cot(d*x+c)^2/d-4*a^3*b*ln(sin(d*x+c))/d-6*a^2
*b^2*x-6/d*cot(d*x+c)*a^2*b^2-6/d*a^2*b^2*c+4/d*a*b^3*ln(sin(d*x+c))+b^4*x+1/d*b^4*c

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maxima [A]  time = 0.50, size = 122, normalized size = 1.04 \[ \frac {3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )} + 6 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 12 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {6 \, a^{3} b \tan \left (d x + c\right ) + a^{4} - 3 \, {\left (a^{4} - 6 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.98, size = 127, normalized size = 1.09 \[ -\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (\frac {a^4}{3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^4-6\,a^2\,b^2\right )+2\,a^3\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {4\,a\,b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^2-b^2\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,{\left (a-b\,1{}\mathrm {i}\right )}^4\,1{}\mathrm {i}}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^4\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c + d*x) + 1i)*(a - b*1i)^4*1i)/(2*d) - (cot(c + d*x)^3*(a^4/3 - tan(c + d*x)^2*(a^4 - 6*a^2*b^2) + 2
*a^3*b*tan(c + d*x)))/d - (log(tan(c + d*x) - 1i)*(a*1i - b)^4*1i)/(2*d) - (4*a*b*log(tan(c + d*x))*(a^2 - b^2
))/d

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sympy [A]  time = 4.43, size = 187, normalized size = 1.60 \[ \begin {cases} \tilde {\infty } a^{4} x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (a + b \tan {\relax (c )}\right )^{4} \cot ^{4}{\relax (c )} & \text {for}\: d = 0 \\a^{4} x + \frac {a^{4}}{d \tan {\left (c + d x \right )}} - \frac {a^{4}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {2 a^{3} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - \frac {4 a^{3} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {2 a^{3} b}{d \tan ^{2}{\left (c + d x \right )}} - 6 a^{2} b^{2} x - \frac {6 a^{2} b^{2}}{d \tan {\left (c + d x \right )}} - \frac {2 a b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {4 a b^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + b^{4} x & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(a+b*tan(d*x+c))**4,x)

[Out]

Piecewise((zoo*a**4*x, (Eq(c, 0) | Eq(c, -d*x)) & (Eq(d, 0) | Eq(c, -d*x))), (x*(a + b*tan(c))**4*cot(c)**4, E
q(d, 0)), (a**4*x + a**4/(d*tan(c + d*x)) - a**4/(3*d*tan(c + d*x)**3) + 2*a**3*b*log(tan(c + d*x)**2 + 1)/d -
4*a**3*b*log(tan(c + d*x))/d - 2*a**3*b/(d*tan(c + d*x)**2) - 6*a**2*b**2*x - 6*a**2*b**2/(d*tan(c + d*x)) -
2*a*b**3*log(tan(c + d*x)**2 + 1)/d + 4*a*b**3*log(tan(c + d*x))/d + b**4*x, True))

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