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

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

[Out]

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

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Rubi [A]  time = 0.289626, antiderivative size = 127, 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, 3635, 3624, 3475} \[ -\frac{a \left (a^2 A-3 a b B-3 A b^2\right ) \log (\sin (c+d x))}{d}-x \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right )-\frac{a^2 (a B+2 A b) \cot (c+d x)}{d}-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}-\frac{b^3 B \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*x) - (a^2*(2*A*b + a*B)*Cot[c + d*x])/d - (b^3*B*Log[Cos[c + d*x]])/
d - (a*(a^2*A - 3*A*b^2 - 3*a*b*B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c + d*x]^2*(a + b*Tan[c + d*x])^2)/(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 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))^3 (A+B \tan (c+d x)) \, dx &=-\frac{a A \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 A b+a B)-2 \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+2 b^2 B \tan ^2(c+d x)\right ) \, dx\\ &=-\frac{a^2 (2 A b+a B) \cot (c+d x)}{d}-\frac{a A \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 A-3 A b^2-3 a b B\right )-2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)+2 b^3 B \tan ^2(c+d x)\right ) \, dx\\ &=-\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x-\frac{a^2 (2 A b+a B) \cot (c+d x)}{d}-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}+\left (b^3 B\right ) \int \tan (c+d x) \, dx-\left (a \left (a^2 A-3 A b^2-3 a b B\right )\right ) \int \cot (c+d x) \, dx\\ &=-\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x-\frac{a^2 (2 A b+a B) \cot (c+d x)}{d}-\frac{b^3 B \log (\cos (c+d x))}{d}-\frac{a \left (a^2 A-3 A b^2-3 a b B\right ) \log (\sin (c+d x))}{d}-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}\\ \end{align*}

Mathematica [C]  time = 0.425828, size = 126, normalized size = 0.99 \[ \frac{-2 a \left (a^2 A-3 a b B-3 A b^2\right ) \log (\tan (c+d x))-2 a^2 (a B+3 A b) \cot (c+d x)+a^3 (-A) \cot ^2(c+d x)+(a+i b)^3 (A+i B) \log (-\tan (c+d x)+i)+(a-i b)^3 (A-i B) \log (\tan (c+d x)+i)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.092, size = 186, normalized size = 1.5 \begin{align*} A{b}^{3}x+{\frac{A{b}^{3}c}{d}}-{\frac{B{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{Aa{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+3\,Ba{b}^{2}x+3\,{\frac{Ba{b}^{2}c}{d}}-3\,Ax{a}^{2}b-3\,{\frac{A\cot \left ( dx+c \right ){a}^{2}b}{d}}-3\,{\frac{A{a}^{2}bc}{d}}+3\,{\frac{B{a}^{2}b\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{A{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{A{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-B{a}^{3}x-{\frac{B\cot \left ( dx+c \right ){a}^{3}}{d}}-{\frac{B{a}^{3}c}{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))^3*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.47555, size = 192, normalized size = 1.51 \begin{align*} -\frac{2 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )}{\left (d x + c\right )} -{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac{A a^{3} + 2 \,{\left (B a^{3} + 3 \, A 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)^3*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.12356, size = 383, normalized size = 3.02 \begin{align*} -\frac{B b^{3} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + A a^{3} +{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A 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 (A a^{3} + 2 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (B a^{3} + 3 \, A 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)^3*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 7.83668, size = 262, normalized size = 2.06 \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 ^{3}{\left (c \right )} & \text{for}\: d = 0 \\\frac{A a^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac{A a^{3} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} - \frac{A a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 A a^{2} b x - \frac{3 A a^{2} b}{d \tan{\left (c + d x \right )}} - \frac{3 A a b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{3 A a b^{2} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + A b^{3} x - B a^{3} x - \frac{B a^{3}}{d \tan{\left (c + d x \right )}} - \frac{3 B a^{2} b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{3 B a^{2} b \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + 3 B a b^{2} x + \frac{B 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)**3*(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)**3, Eq(d, 0)), (A*a**3*log(tan(c + d*x)**2 + 1)/(2*d) - A*a**3*log(tan(c + d*x))/d - A*a**3/(2*d*t
an(c + d*x)**2) - 3*A*a**2*b*x - 3*A*a**2*b/(d*tan(c + d*x)) - 3*A*a*b**2*log(tan(c + d*x)**2 + 1)/(2*d) + 3*A
*a*b**2*log(tan(c + d*x))/d + A*b**3*x - B*a**3*x - B*a**3/(d*tan(c + d*x)) - 3*B*a**2*b*log(tan(c + d*x)**2 +
1)/(2*d) + 3*B*a**2*b*log(tan(c + d*x))/d + 3*B*a*b**2*x + B*b**3*log(tan(c + d*x)**2 + 1)/(2*d), True))

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Giac [A]  time = 1.97813, size = 261, normalized size = 2.06 \begin{align*} -\frac{2 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )}{\left (d x + c\right )} -{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac{3 \, A a^{3} \tan \left (d x + c\right )^{2} - 9 \, B a^{2} b \tan \left (d x + c\right )^{2} - 9 \, A a b^{2} \tan \left (d x + c\right )^{2} - 2 \, B a^{3} \tan \left (d x + c\right ) - 6 \, A a^{2} b \tan \left (d x + c\right ) - A 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)^3*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

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