3.255 \(\int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=191 \[ \frac{a \left (2 a^2 A-6 a b B-5 A b^2\right ) \cot ^2(c+d x)}{4 d}+\frac{\left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \cot (c+d x)}{d}+\frac{\left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \log (\sin (c+d x))}{d}+x \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right )-\frac{a^2 (2 a B+3 A b) \cot ^3(c+d x)}{6 d}-\frac{a A \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d} \]

[Out]

(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*x + ((3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Cot[c + d*x])/d + (a*(2*a
^2*A - 5*A*b^2 - 6*a*b*B)*Cot[c + d*x]^2)/(4*d) - (a^2*(3*A*b + 2*a*B)*Cot[c + d*x]^3)/(6*d) + ((a^3*A - 3*a*A
*b^2 - 3*a^2*b*B + b^3*B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c + d*x]^4*(a + b*Tan[c + d*x])^2)/(4*d)

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Rubi [A]  time = 0.452533, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3605, 3635, 3628, 3529, 3531, 3475} \[ \frac{a \left (2 a^2 A-6 a b B-5 A b^2\right ) \cot ^2(c+d x)}{4 d}+\frac{\left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \cot (c+d x)}{d}+\frac{\left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \log (\sin (c+d x))}{d}+x \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right )-\frac{a^2 (2 a B+3 A b) \cot ^3(c+d x)}{6 d}-\frac{a A \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(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 + ((3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Cot[c + d*x])/d + (a*(2*a
^2*A - 5*A*b^2 - 6*a*b*B)*Cot[c + d*x]^2)/(4*d) - (a^2*(3*A*b + 2*a*B)*Cot[c + d*x]^3)/(6*d) + ((a^3*A - 3*a*A
*b^2 - 3*a^2*b*B + b^3*B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c + d*x]^4*(a + b*Tan[c + d*x])^2)/(4*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 3628

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

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

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 ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac{1}{4} \int \cot ^4(c+d x) (a+b \tan (c+d x)) \left (2 a (3 A b+2 a B)-4 \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-2 b (a A-2 b B) \tan ^2(c+d x)\right ) \, dx\\ &=-\frac{a^2 (3 A b+2 a B) \cot ^3(c+d x)}{6 d}-\frac{a A \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac{1}{4} \int \cot ^3(c+d x) \left (-2 a \left (2 a^2 A-5 A b^2-6 a b B\right )-4 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-2 b^2 (a A-2 b B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (2 a^2 A-5 A b^2-6 a b B\right ) \cot ^2(c+d x)}{4 d}-\frac{a^2 (3 A b+2 a B) \cot ^3(c+d x)}{6 d}-\frac{a A \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac{1}{4} \int \cot ^2(c+d x) \left (-4 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+4 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)\right ) \, dx\\ &=\frac{\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot (c+d x)}{d}+\frac{a \left (2 a^2 A-5 A b^2-6 a b B\right ) \cot ^2(c+d x)}{4 d}-\frac{a^2 (3 A b+2 a B) \cot ^3(c+d x)}{6 d}-\frac{a A \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac{1}{4} \int \cot (c+d x) \left (4 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+4 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (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{\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot (c+d x)}{d}+\frac{a \left (2 a^2 A-5 A b^2-6 a b B\right ) \cot ^2(c+d x)}{4 d}-\frac{a^2 (3 A b+2 a B) \cot ^3(c+d x)}{6 d}-\frac{a A \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}+\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\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{\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot (c+d x)}{d}+\frac{a \left (2 a^2 A-5 A b^2-6 a b B\right ) \cot ^2(c+d x)}{4 d}-\frac{a^2 (3 A b+2 a B) \cot ^3(c+d x)}{6 d}+\frac{\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \log (\sin (c+d x))}{d}-\frac{a A \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\\ \end{align*}

Mathematica [C]  time = 0.703405, size = 199, normalized size = 1.04 \[ \frac{6 a \left (a^2 A-3 a b B-3 A b^2\right ) \cot ^2(c+d x)+12 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \cot (c+d x)+12 \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \log (\tan (c+d x))-4 a^2 (a B+3 A b) \cot ^3(c+d x)-3 a^3 A \cot ^4(c+d x)-6 (a+i b)^3 (A+i B) \log (-\tan (c+d x)+i)-6 (a-i b)^3 (A-i B) \log (\tan (c+d x)+i)}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(12*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Cot[c + d*x] + 6*a*(a^2*A - 3*A*b^2 - 3*a*b*B)*Cot[c + d*x]^2 - 4*
a^2*(3*A*b + a*B)*Cot[c + d*x]^3 - 3*a^3*A*Cot[c + d*x]^4 - 6*(a + I*b)^3*(A + I*B)*Log[I - Tan[c + d*x]] + 12
*(a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*Log[Tan[c + d*x]] - 6*(a - I*b)^3*(A - I*B)*Log[I + Tan[c + d*x]])/(1
2*d)

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Maple [A]  time = 0.093, size = 302, normalized size = 1.6 \begin{align*} -A{b}^{3}x-{\frac{A\cot \left ( dx+c \right ){b}^{3}}{d}}-{\frac{A{b}^{3}c}{d}}+{\frac{B{b}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{3\,Aa{b}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-3\,{\frac{Aa{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-3\,Ba{b}^{2}x-3\,{\frac{B\cot \left ( dx+c \right ) a{b}^{2}}{d}}-3\,{\frac{Ba{b}^{2}c}{d}}-{\frac{A{a}^{2}b \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{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}}-{\frac{3\,B{a}^{2}b \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,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 ) ^{4}}{4\,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}}-{\frac{B{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{B\cot \left ( dx+c \right ){a}^{3}}{d}}+B{a}^{3}x+{\frac{B{a}^{3}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

-A*b^3*x-1/d*A*cot(d*x+c)*b^3-1/d*A*b^3*c+1/d*B*b^3*ln(sin(d*x+c))-3/2/d*A*a*b^2*cot(d*x+c)^2-3/d*A*a*b^2*ln(s
in(d*x+c))-3*B*a*b^2*x-3/d*B*cot(d*x+c)*a*b^2-3/d*B*a*b^2*c-1/d*A*a^2*b*cot(d*x+c)^3+3*A*x*a^2*b+3/d*A*cot(d*x
+c)*a^2*b+3/d*A*a^2*b*c-3/2/d*B*a^2*b*cot(d*x+c)^2-3/d*B*a^2*b*ln(sin(d*x+c))-1/4/d*A*a^3*cot(d*x+c)^4+1/2/d*A
*a^3*cot(d*x+c)^2+a^3*A*ln(sin(d*x+c))/d-1/3/d*B*a^3*cot(d*x+c)^3+1/d*B*cot(d*x+c)*a^3+B*a^3*x+1/d*B*a^3*c

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Maxima [A]  time = 1.4885, size = 290, normalized size = 1.52 \begin{align*} \frac{12 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )}{\left (d x + c\right )} - 6 \,{\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 ) + 12 \,{\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 )\right ) - \frac{3 \, A a^{3} - 12 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{3} - 6 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 4 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*(d*x + c) - 6*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*log(tan
(d*x + c)^2 + 1) + 12*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*log(tan(d*x + c)) - (3*A*a^3 - 12*(B*a^3 + 3*A*a
^2*b - 3*B*a*b^2 - A*b^3)*tan(d*x + c)^3 - 6*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2)*tan(d*x + c)^2 + 4*(B*a^3 + 3*A*a
^2*b)*tan(d*x + c))/tan(d*x + c)^4)/d

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Fricas [A]  time = 2.00943, size = 518, normalized size = 2.71 \begin{align*} \frac{6 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\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 )^{4} + 3 \,{\left (3 \, A a^{3} - 6 \, B a^{2} b - 6 \, A a b^{2} + 4 \,{\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 )^{4} - 3 \, A a^{3} + 12 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{3} + 6 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} - 4 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{12 \, d \tan \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(6*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^4 + 3*(3
*A*a^3 - 6*B*a^2*b - 6*A*a*b^2 + 4*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*d*x)*tan(d*x + c)^4 - 3*A*a^3 + 12*
(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*tan(d*x + c)^3 + 6*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2)*tan(d*x + c)^2 - 4*
(B*a^3 + 3*A*a^2*b)*tan(d*x + c))/(d*tan(d*x + c)^4)

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Sympy [A]  time = 41.5826, size = 400, normalized size = 2.09 \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 ^{5}{\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 )}} - \frac{A a^{3}}{4 d \tan ^{4}{\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{A a^{2} b}{d \tan ^{3}{\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} - \frac{3 A a b^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - A b^{3} x - \frac{A b^{3}}{d \tan{\left (c + d x \right )}} + B a^{3} x + \frac{B a^{3}}{d \tan{\left (c + d x \right )}} - \frac{B a^{3}}{3 d \tan ^{3}{\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} - \frac{3 B a^{2} b}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 B a b^{2} x - \frac{3 B a b^{2}}{d \tan{\left (c + d x \right )}} - \frac{B b^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{B b^{3} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(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)**5, 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*
tan(c + d*x)**2) - A*a**3/(4*d*tan(c + d*x)**4) + 3*A*a**2*b*x + 3*A*a**2*b/(d*tan(c + d*x)) - A*a**2*b/(d*tan
(c + d*x)**3) + 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 - 3*A*a*b**2/(2*d*t
an(c + d*x)**2) - A*b**3*x - A*b**3/(d*tan(c + d*x)) + B*a**3*x + B*a**3/(d*tan(c + d*x)) - B*a**3/(3*d*tan(c
+ d*x)**3) + 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**2*b/(2*d*tan(
c + d*x)**2) - 3*B*a*b**2*x - 3*B*a*b**2/(d*tan(c + d*x)) - B*b**3*log(tan(c + d*x)**2 + 1)/(2*d) + B*b**3*log
(tan(c + d*x))/d, True))

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Giac [B]  time = 2.10846, size = 713, normalized size = 3.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 8*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 24*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 -
36*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 72*B*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 120*
B*a^3*tan(1/2*d*x + 1/2*c) + 360*A*a^2*b*tan(1/2*d*x + 1/2*c) - 288*B*a*b^2*tan(1/2*d*x + 1/2*c) - 96*A*b^3*ta
n(1/2*d*x + 1/2*c) - 192*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*(d*x + c) + 192*(A*a^3 - 3*B*a^2*b - 3*A*a*b^
2 + B*b^3)*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 192*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x +
1/2*c))) + (400*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 1200*B*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 1200*A*a*b^2*tan(1/2*d*x
+ 1/2*c)^4 + 400*B*b^3*tan(1/2*d*x + 1/2*c)^4 - 120*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 360*A*a^2*b*tan(1/2*d*x +
1/2*c)^3 + 288*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 96*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 36*A*a^3*tan(1/2*d*x + 1/2*c
)^2 + 72*B*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 8*B*a^3*tan(1/2*d*x + 1/2*c) + 2
4*A*a^2*b*tan(1/2*d*x + 1/2*c) + 3*A*a^3)/tan(1/2*d*x + 1/2*c)^4)/d