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

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

[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)

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Rubi [A]  time = 0.297712, antiderivative size = 117, normalized size of antiderivative = 1., 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 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 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]

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 ^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*}

Mathematica [C]  time = 1.27623, size = 125, normalized size = 1.07 \[ -\frac{-6 a^2 \left (a^2-6 b^2\right ) \cot (c+d x)+24 a b \left (a^2-b^2\right ) \log (\tan (c+d x))+12 a^3 b \cot ^2(c+d x)+2 a^4 \cot ^3(c+d x)+3 i (a+i b)^4 \log (-\tan (c+d x)+i)-3 i (a-i b)^4 \log (\tan (c+d x)+i)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-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*Log[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]])/(6*d)

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Maple [A]  time = 0.056, size = 144, normalized size = 1.2 \begin{align*}{b}^{4}x+{\frac{{b}^{4}c}{d}}+4\,{\frac{{b}^{3}a\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-6\,{a}^{2}{b}^{2}x-6\,{\frac{{a}^{2}{b}^{2}\cot \left ( dx+c \right ) }{d}}-6\,{\frac{{a}^{2}{b}^{2}c}{d}}-2\,{\frac{b{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{d}}-4\,{\frac{b{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{a}^{4}\cot \left ( dx+c \right ) }{d}}+{a}^{4}x+{\frac{{a}^{4}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))^4,x)

[Out]

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

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Maxima [A]  time = 1.6247, size = 165, normalized size = 1.41 \begin{align*} \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} \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))^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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Fricas [A]  time = 1.98553, size = 305, normalized size = 2.61 \begin{align*} -\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}} \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))^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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**4,x)

[Out]

Timed out

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Giac [B]  time = 2.84492, size = 332, normalized size = 2.84 \begin{align*} \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} \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))^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