3.21 \(\int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=127 \[ -\frac{a \left (a^2 B-3 a b C-3 b^2 B\right ) \log (\sin (c+d x))}{d}-x \left (3 a^2 b B+a^3 C-3 a b^2 C-b^3 B\right )-\frac{a^2 (a C+2 b B) \cot (c+d x)}{d}-\frac{a B \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}-\frac{b^3 C \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.354905, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3632, 3605, 3635, 3624, 3475} \[ -\frac{a \left (a^2 B-3 a b C-3 b^2 B\right ) \log (\sin (c+d x))}{d}-x \left (3 a^2 b B+a^3 C-3 a b^2 C-b^3 B\right )-\frac{a^2 (a C+2 b B) \cot (c+d x)}{d}-\frac{a B \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}-\frac{b^3 C \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

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

Rule 3632

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

Rule 3605

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

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

Mathematica [C]  time = 0.447553, size = 126, normalized size = 0.99 \[ \frac{-2 a \left (a^2 B-3 a b C-3 b^2 B\right ) \log (\tan (c+d x))-2 a^2 (a C+3 b B) \cot (c+d x)+a^3 (-B) \cot ^2(c+d x)+(a+i b)^3 (B+i C) \log (-\tan (c+d x)+i)+(a-i b)^3 (B-i C) \log (\tan (c+d x)+i)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

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

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Maple [A]  time = 0.137, size = 186, normalized size = 1.5 \begin{align*} Bx{b}^{3}+{\frac{B{b}^{3}c}{d}}-{\frac{C{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{Ba{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+3\,Ca{b}^{2}x+3\,{\frac{Ca{b}^{2}c}{d}}-3\,B{a}^{2}bx-3\,{\frac{B\cot \left ( dx+c \right ){a}^{2}b}{d}}-3\,{\frac{B{a}^{2}bc}{d}}+3\,{\frac{C{a}^{2}b\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{B{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{B{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-Cx{a}^{3}-{\frac{C\cot \left ( dx+c \right ){a}^{3}}{d}}-{\frac{C{a}^{3}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))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 1.78442, size = 192, normalized size = 1.51 \begin{align*} -\frac{2 \,{\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )}{\left (d x + c\right )} -{\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac{B a^{3} + 2 \,{\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{\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)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.76333, size = 383, normalized size = 3.02 \begin{align*} -\frac{C b^{3} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + B a^{3} +{\left (B a^{3} - 3 \, C a^{2} b - 3 \, B 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} +{\left (B a^{3} + 2 \,{\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{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)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 78.7063, size = 253, normalized size = 1.99 \begin{align*} \begin{cases} \text{NaN} & \text{for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan{\left (c \right )}\right )^{3} \left (B \tan{\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{4}{\left (c \right )} & \text{for}\: d = 0 \\\text{NaN} & \text{for}\: c = - d x \\\frac{B a^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac{B a^{3} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} - \frac{B a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 B a^{2} b x - \frac{3 B a^{2} b}{d \tan{\left (c + d x \right )}} - \frac{3 B a b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{3 B a b^{2} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + B b^{3} x - C a^{3} x - \frac{C a^{3}}{d \tan{\left (c + d x \right )}} - \frac{3 C a^{2} b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{3 C a^{2} b \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + 3 C a b^{2} x + \frac{C b^{3} \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)**4*(a+b*tan(d*x+c))**3*(B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

Piecewise((nan, Eq(c, 0) & Eq(d, 0)), (x*(a + b*tan(c))**3*(B*tan(c) + C*tan(c)**2)*cot(c)**4, Eq(d, 0)), (nan
, Eq(c, -d*x)), (B*a**3*log(tan(c + d*x)**2 + 1)/(2*d) - B*a**3*log(tan(c + d*x))/d - B*a**3/(2*d*tan(c + d*x)
**2) - 3*B*a**2*b*x - 3*B*a**2*b/(d*tan(c + d*x)) - 3*B*a*b**2*log(tan(c + d*x)**2 + 1)/(2*d) + 3*B*a*b**2*log
(tan(c + d*x))/d + B*b**3*x - C*a**3*x - C*a**3/(d*tan(c + d*x)) - 3*C*a**2*b*log(tan(c + d*x)**2 + 1)/(2*d) +
3*C*a**2*b*log(tan(c + d*x))/d + 3*C*a*b**2*x + C*b**3*log(tan(c + d*x)**2 + 1)/(2*d), True))

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Giac [A]  time = 2.51342, size = 261, normalized size = 2.06 \begin{align*} -\frac{2 \,{\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )}{\left (d x + c\right )} -{\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac{3 \, B a^{3} \tan \left (d x + c\right )^{2} - 9 \, C a^{2} b \tan \left (d x + c\right )^{2} - 9 \, B a b^{2} \tan \left (d x + c\right )^{2} - 2 \, C a^{3} \tan \left (d x + c\right ) - 6 \, B a^{2} b \tan \left (d x + c\right ) - B 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)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

-1/2*(2*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*(d*x + c) - (B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*log(tan(d*
x + c)^2 + 1) + 2*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2)*log(abs(tan(d*x + c))) - (3*B*a^3*tan(d*x + c)^2 - 9*C*a^2*b
*tan(d*x + c)^2 - 9*B*a*b^2*tan(d*x + c)^2 - 2*C*a^3*tan(d*x + c) - 6*B*a^2*b*tan(d*x + c) - B*a^3)/tan(d*x +
c)^2)/d