3.252 \(\int \cot ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=119 \[ -x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )+\frac{a^2 (a B+3 A b) \log (\sin (c+d x))}{d}+\frac{b^2 (a A+b B) \tan (c+d x)}{d}-\frac{b^2 (3 a B+A b) \log (\cos (c+d x))}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.262912, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3605, 3637, 3624, 3475} \[ -x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )+\frac{a^2 (a B+3 A b) \log (\sin (c+d x))}{d}+\frac{b^2 (a A+b B) \tan (c+d x)}{d}-\frac{b^2 (3 a B+A b) \log (\cos (c+d x))}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3637

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*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, 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 ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^2}{d}+\int \cot (c+d x) (a+b \tan (c+d x)) \left (a (3 A b+a B)-\left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+b (a A+b B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{b^2 (a A+b B) \tan (c+d x)}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^2}{d}-\int \cot (c+d x) \left (-a^2 (3 A b+a B)+\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)-b^2 (A b+3 a B) \tan ^2(c+d x)\right ) \, dx\\ &=-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x+\frac{b^2 (a A+b B) \tan (c+d x)}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^2}{d}+\left (a^2 (3 A b+a B)\right ) \int \cot (c+d x) \, dx+\left (b^2 (A b+3 a B)\right ) \int \tan (c+d x) \, dx\\ &=-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x-\frac{b^2 (A b+3 a B) \log (\cos (c+d x))}{d}+\frac{a^2 (3 A b+a B) \log (\sin (c+d x))}{d}+\frac{b^2 (a A+b B) \tan (c+d x)}{d}-\frac{a A \cot (c+d x) (a+b \tan (c+d x))^2}{d}\\ \end{align*}

Mathematica [C]  time = 0.506833, size = 113, normalized size = 0.95 \[ \frac{2 a^2 (a B+3 A b) \log (\tan (c+d x))-2 a^3 A \cot (c+d x)+i (a+i b)^3 (A+i B) \log (-\tan (c+d x)+i)+(b+i a)^3 (A-i B) \log (\tan (c+d x)+i)+2 b^3 B \tan (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.067, size = 168, normalized size = 1.4 \begin{align*} -A{a}^{3}x+3\,Aa{b}^{2}x+3\,B{a}^{2}bx-B{b}^{3}x-{\frac{A\cot \left ( dx+c \right ){a}^{3}}{d}}+3\,{\frac{A{a}^{2}b\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{A{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-{\frac{A{a}^{3}c}{d}}+3\,{\frac{Aa{b}^{2}c}{d}}+{\frac{B{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{B{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-3\,{\frac{Ba{b}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{B{a}^{2}bc}{d}}-{\frac{B{b}^{3}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.47403, size = 169, normalized size = 1.42 \begin{align*} \frac{2 \, B b^{3} \tan \left (d x + c\right ) - \frac{2 \, A a^{3}}{\tan \left (d x + c\right )} - 2 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )}{\left (d x + c\right )} -{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left (\tan \left (d x + c\right )\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.16643, size = 347, normalized size = 2.92 \begin{align*} \frac{2 \, B b^{3} \tan \left (d x + c\right )^{2} - 2 \, A a^{3} - 2 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} d x \tan \left (d x + c\right ) +{\left (B a^{3} + 3 \, A a^{2} b\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 ) -{\left (3 \, B a b^{2} + A b^{3}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.89423, size = 223, normalized size = 1.87 \begin{align*} \begin{cases} \tilde{\infty } A 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 ) \left (a + b \tan{\left (c \right )}\right )^{3} \cot ^{2}{\left (c \right )} & \text{for}\: d = 0 \\- A a^{3} x - \frac{A a^{3}}{d \tan{\left (c + d x \right )}} - \frac{3 A a^{2} b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{3 A a^{2} b \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + 3 A a b^{2} x + \frac{A b^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \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} + 3 B a^{2} b x + \frac{3 B a b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - B b^{3} x + \frac{B b^{3} \tan{\left (c + d x \right )}}{d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(a+b*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.90259, size = 205, normalized size = 1.72 \begin{align*} \frac{2 \, B b^{3} \tan \left (d x + c\right ) - 2 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )}{\left (d x + c\right )} -{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac{2 \,{\left (B a^{3} \tan \left (d x + c\right ) + 3 \, A a^{2} b \tan \left (d x + c\right ) + A a^{3}\right )}}{\tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

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