3.451 \(\int \cot ^3(c+d x) (a+b \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=99 \[ -\frac{a^2 \left (a^2-6 b^2\right ) \log (\sin (c+d x))}{d}-4 a b x \left (a^2-b^2\right )-\frac{3 a^3 b \cot (c+d x)}{d}-\frac{a^2 \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}-\frac{b^4 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.199208, antiderivative size = 99, 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 = {3565, 3635, 3624, 3475} \[ -\frac{a^2 \left (a^2-6 b^2\right ) \log (\sin (c+d x))}{d}-4 a b x \left (a^2-b^2\right )-\frac{3 a^3 b \cot (c+d x)}{d}-\frac{a^2 \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}-\frac{b^4 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^4,x]

[Out]

-4*a*b*(a^2 - b^2)*x - (3*a^3*b*Cot[c + d*x])/d - (b^4*Log[Cos[c + d*x]])/d - (a^2*(a^2 - 6*b^2)*Log[Sin[c + d
*x]])/d - (a^2*Cot[c + d*x]^2*(a + b*Tan[c + d*x])^2)/(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 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 3624

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/tan[(e_.) + (f_.)*(x_)], x_Symbol
] :> Simp[B*x, x] + (Dist[A, Int[1/Tan[e + f*x], x], x] + Dist[C, Int[Tan[e + f*x], x], x]) /; FreeQ[{e, f, A,
B, C}, x] && NeQ[A, C]

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

Mathematica [C]  time = 0.304061, size = 90, normalized size = 0.91 \[ -\frac{8 a^3 b \cot (c+d x)+a^4 \cot ^2(c+d x)-(a-i b)^4 \log (-\cot (c+d x)+i)-(a+i b)^4 \log (\cot (c+d x)+i)-2 b^4 \log (\tan (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^4,x]

[Out]

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

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Maple [A]  time = 0.066, size = 115, normalized size = 1.2 \begin{align*} -{\frac{{b}^{4}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+4\,{b}^{3}ax+4\,{\frac{a{b}^{3}c}{d}}+6\,{\frac{{a}^{2}{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-4\,x{a}^{3}b-4\,{\frac{b{a}^{3}\cot \left ( dx+c \right ) }{d}}-4\,{\frac{{a}^{3}bc}{d}}-{\frac{{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{a}^{4}\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))^4,x)

[Out]

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

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Maxima [A]  time = 1.62397, size = 134, normalized size = 1.35 \begin{align*} -\frac{8 \,{\left (a^{3} b - a b^{3}\right )}{\left (d x + c\right )} -{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (a^{4} - 6 \, a^{2} b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac{8 \, a^{3} b \tan \left (d x + c\right ) + a^{4}}{\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))^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.0346, size = 305, normalized size = 3.08 \begin{align*} -\frac{b^{4} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + 8 \, a^{3} b \tan \left (d x + c\right ) + a^{4} +{\left (a^{4} - 6 \, 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 )^{2} +{\left (a^{4} + 8 \,{\left (a^{3} b - a 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))^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 9.26062, size = 170, normalized size = 1.72 \begin{align*} \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{\left (c \right )}\right )^{4} \cot ^{3}{\left (c \right )} & \text{for}\: d = 0 \\\frac{a^{4} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac{a^{4} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} - \frac{a^{4}}{2 d \tan ^{2}{\left (c + d x \right )}} - 4 a^{3} b x - \frac{4 a^{3} b}{d \tan{\left (c + d x \right )}} - \frac{3 a^{2} b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac{6 a^{2} b^{2} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + 4 a b^{3} x + \frac{b^{4} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \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))**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)**3, E
q(d, 0)), (a**4*log(tan(c + d*x)**2 + 1)/(2*d) - a**4*log(tan(c + d*x))/d - a**4/(2*d*tan(c + d*x)**2) - 4*a**
3*b*x - 4*a**3*b/(d*tan(c + d*x)) - 3*a**2*b**2*log(tan(c + d*x)**2 + 1)/d + 6*a**2*b**2*log(tan(c + d*x))/d +
4*a*b**3*x + b**4*log(tan(c + d*x)**2 + 1)/(2*d), True))

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Giac [A]  time = 2.59846, size = 178, normalized size = 1.8 \begin{align*} -\frac{8 \,{\left (a^{3} b - a b^{3}\right )}{\left (d x + c\right )} -{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (a^{4} - 6 \, a^{2} b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac{3 \, a^{4} \tan \left (d x + c\right )^{2} - 18 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} - 8 \, a^{3} b \tan \left (d x + c\right ) - a^{4}}{\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))^4,x, algorithm="giac")

[Out]

-1/2*(8*(a^3*b - a*b^3)*(d*x + c) - (a^4 - 6*a^2*b^2 + b^4)*log(tan(d*x + c)^2 + 1) + 2*(a^4 - 6*a^2*b^2)*log(
abs(tan(d*x + c))) - (3*a^4*tan(d*x + c)^2 - 18*a^2*b^2*tan(d*x + c)^2 - 8*a^3*b*tan(d*x + c) - a^4)/tan(d*x +
c)^2)/d