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

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 127, normalized size of antiderivative = 1.00, 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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 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 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]

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*}

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Mathematica [C]  time = 0.44, size = 126, normalized size = 0.99 \[ \frac {a^3 (-A) \cot ^2(c+d x)-2 a \left (a^2 A-3 a b B-3 A b^2\right ) \log (\tan (c+d x))-2 a^2 (a B+3 A b) \cot (c+d x)+(a+i b)^3 (A+i B) \log (-\tan (c+d x)+i)+(a-i b)^3 (A-i B) \log (\tan (c+d x)+i)}{2 d} \]

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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fricas [A]  time = 0.58, size = 162, normalized size = 1.28 \[ -\frac {B b^{3} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + A a^{3} + {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + {\left (A a^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \]

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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giac [A]  time = 3.04, size = 193, normalized size = 1.52 \[ -\frac {2 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} {\left (d x + c\right )} - {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {3 \, A a^{3} \tan \left (d x + c\right )^{2} - 9 \, B a^{2} b \tan \left (d x + c\right )^{2} - 9 \, A a b^{2} \tan \left (d x + c\right )^{2} - 2 \, B a^{3} \tan \left (d x + c\right ) - 6 \, A a^{2} b \tan \left (d x + c\right ) - A a^{3}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

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

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maple [A]  time = 0.50, size = 186, normalized size = 1.46 \[ -\frac {A \,a^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{3} A \ln \left (\sin \left (d x +c \right )\right )}{d}-a^{3} B x -\frac {B \cot \left (d x +c \right ) a^{3}}{d}-\frac {a^{3} B c}{d}-3 A x \,a^{2} b -\frac {3 A \cot \left (d x +c \right ) a^{2} b}{d}-\frac {3 A \,a^{2} b c}{d}+\frac {3 a^{2} b B \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {3 A a \,b^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}+3 B x a \,b^{2}+\frac {3 B a \,b^{2} c}{d}+A x \,b^{3}+\frac {A \,b^{3} c}{d}-\frac {b^{3} B \ln \left (\cos \left (d x +c \right )\right )}{d} \]

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]

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

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maxima [A]  time = 0.94, size = 142, normalized size = 1.12 \[ -\frac {2 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} {\left (d x + c\right )} - {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {A a^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

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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mupad [B]  time = 6.39, size = 135, normalized size = 1.06 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-A\,a^3+3\,B\,a^2\,b+3\,A\,a\,b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^3+3\,A\,b\,a^2\right )+\frac {A\,a^3}{2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.47, size = 262, normalized size = 2.06 \[ \begin {cases} \tilde {\infty } A a^{3} x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (A + B \tan {\relax (c )}\right ) \left (a + b \tan {\relax (c )}\right )^{3} \cot ^{3}{\relax (c )} & \text {for}\: d = 0 \\\frac {A a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {A a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {A a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 A a^{2} b x - \frac {3 A a^{2} b}{d \tan {\left (c + d x \right )}} - \frac {3 A a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 A a b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + A b^{3} x - B a^{3} x - \frac {B a^{3}}{d \tan {\left (c + d x \right )}} - \frac {3 B a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 B a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 B a b^{2} x + \frac {B b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text {otherwise} \end {cases} \]

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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