3.247 \(\int \cot ^5(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=151 \[ \frac {\left (a^2 A-2 a b B-A b^2\right ) \cot ^2(c+d x)}{2 d}+\frac {\left (a^2 A-2 a b B-A b^2\right ) \log (\sin (c+d x))}{d}+x \left (a^2 B+2 a A b-b^2 B\right )-\frac {a^2 A \cot ^4(c+d x)}{4 d}-\frac {\left (b^2 B-a (a B+2 A b)\right ) \cot (c+d x)}{d}-\frac {a (a B+2 A b) \cot ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3604, 3628, 3529, 3531, 3475} \[ \frac {\left (a^2 A-2 a b B-A b^2\right ) \cot ^2(c+d x)}{2 d}+\frac {\left (a^2 A-2 a b B-A b^2\right ) \log (\sin (c+d x))}{d}+x \left (a^2 B+2 a A b-b^2 B\right )-\frac {a^2 A \cot ^4(c+d x)}{4 d}-\frac {\left (b^2 B-a (a B+2 A b)\right ) \cot (c+d x)}{d}-\frac {a (a B+2 A b) \cot ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

(2*a*A*b + a^2*B - b^2*B)*x - ((b^2*B - a*(2*A*b + a*B))*Cot[c + d*x])/d + ((a^2*A - A*b^2 - 2*a*b*B)*Cot[c +
d*x]^2)/(2*d) - (a*(2*A*b + a*B)*Cot[c + d*x]^3)/(3*d) - (a^2*A*Cot[c + d*x]^4)/(4*d) + ((a^2*A - A*b^2 - 2*a*
b*B)*Log[Sin[c + d*x]])/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 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 3604

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((B*c - A*d)*(b*c - a*d)^2*(c + d*Tan[e + f*x])^(n + 1))/(f*d^2*(n +
1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[B*(b*c - a*d)^2 + A*d*(a^2
*c - b^2*c + 2*a*b*d) + d*(B*(a^2*c - b^2*c + 2*a*b*d) + A*(2*a*b*c - a^2*d + b^2*d))*Tan[e + f*x] + b^2*B*(c^
2 + d^2)*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] && 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]

Rubi steps

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

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Mathematica [C]  time = 2.92, size = 180, normalized size = 1.19 \[ \frac {6 \left (a^2 A-2 a b B-A b^2\right ) \cot ^2(c+d x)+12 \left (a^2 B+2 a A b-b^2 B\right ) \cot (c+d x)-6 \left (\left (-2 a^2 A+4 a b B+2 A b^2\right ) \log (\tan (c+d x))+(a-i b)^2 (A-i B) \log (\tan (c+d x)+i)+(a+i b)^2 (A+i B) \log (-\tan (c+d x)+i)\right )-3 a^2 A \cot ^4(c+d x)-4 a (a B+2 A b) \cot ^3(c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

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

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fricas [A]  time = 0.52, size = 191, normalized size = 1.26 \[ \frac {6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{4} + 3 \, {\left (3 \, A a^{2} - 4 \, B a b - 2 \, A b^{2} + 4 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} d x\right )} \tan \left (d x + c\right )^{4} + 12 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \tan \left (d x + c\right )^{3} - 3 \, A a^{2} + 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} - 4 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{12 \, d \tan \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 2.88, size = 435, normalized size = 2.88 \[ -\frac {3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 48 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 240 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 192 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} {\left (d x + c\right )} + 192 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 192 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {400 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 800 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 400 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 240 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 96 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 48 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 16 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*A*a^2*tan(1/2*d*x + 1/2*c)^4 - 8*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 16*A*a*b*tan(1/2*d*x + 1/2*c)^3 - 36
*A*a^2*tan(1/2*d*x + 1/2*c)^2 + 48*B*a*b*tan(1/2*d*x + 1/2*c)^2 + 24*A*b^2*tan(1/2*d*x + 1/2*c)^2 + 120*B*a^2*
tan(1/2*d*x + 1/2*c) + 240*A*a*b*tan(1/2*d*x + 1/2*c) - 96*B*b^2*tan(1/2*d*x + 1/2*c) - 192*(B*a^2 + 2*A*a*b -
B*b^2)*(d*x + c) + 192*(A*a^2 - 2*B*a*b - A*b^2)*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 192*(A*a^2 - 2*B*a*b - A*b
^2)*log(abs(tan(1/2*d*x + 1/2*c))) + (400*A*a^2*tan(1/2*d*x + 1/2*c)^4 - 800*B*a*b*tan(1/2*d*x + 1/2*c)^4 - 40
0*A*b^2*tan(1/2*d*x + 1/2*c)^4 - 120*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 240*A*a*b*tan(1/2*d*x + 1/2*c)^3 + 96*B*b^
2*tan(1/2*d*x + 1/2*c)^3 - 36*A*a^2*tan(1/2*d*x + 1/2*c)^2 + 48*B*a*b*tan(1/2*d*x + 1/2*c)^2 + 24*A*b^2*tan(1/
2*d*x + 1/2*c)^2 + 8*B*a^2*tan(1/2*d*x + 1/2*c) + 16*A*a*b*tan(1/2*d*x + 1/2*c) + 3*A*a^2)/tan(1/2*d*x + 1/2*c
)^4)/d

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maple [A]  time = 0.42, size = 238, normalized size = 1.58 \[ -\frac {a^{2} A \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a^{2} A \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a^{2} A \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a^{2} B \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {B \cot \left (d x +c \right ) a^{2}}{d}+a^{2} B x +\frac {B \,a^{2} c}{d}-\frac {2 A a b \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {2 A \cot \left (d x +c \right ) a b}{d}+2 A x a b +\frac {2 A a b c}{d}-\frac {B a b \left (\cot ^{2}\left (d x +c \right )\right )}{d}-\frac {2 B a b \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {A \,b^{2} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {A \,b^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}-B x \,b^{2}-\frac {B \cot \left (d x +c \right ) b^{2}}{d}-\frac {B \,b^{2} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.79, size = 175, normalized size = 1.16 \[ \frac {12 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} {\left (d x + c\right )} - 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {12 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \tan \left (d x + c\right )^{3} - 3 \, A a^{2} + 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} - 4 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.34, size = 182, normalized size = 1.21 \[ -\frac {{\mathrm {cot}\left (c+d\,x\right )}^4\,\left (\frac {A\,a^2}{4}+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {A\,a^2}{2}+B\,a\,b+\frac {A\,b^2}{2}\right )-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (B\,a^2+2\,A\,a\,b-B\,b^2\right )+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^2}{3}+\frac {2\,A\,b\,a}{3}\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-A\,a^2+2\,B\,a\,b+A\,b^2\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^5*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^2,x)

[Out]

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

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sympy [A]  time = 4.66, size = 313, normalized size = 2.07 \[ \begin {cases} \tilde {\infty } A a^{2} x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (A + B \tan {\relax (c )}\right ) \left (a + b \tan {\relax (c )}\right )^{2} \cot ^{5}{\relax (c )} & \text {for}\: d = 0 \\- \frac {A a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {A a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {A a^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {A a^{2}}{4 d \tan ^{4}{\left (c + d x \right )}} + 2 A a b x + \frac {2 A a b}{d \tan {\left (c + d x \right )}} - \frac {2 A a b}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {A b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {A b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {A b^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} + B a^{2} x + \frac {B a^{2}}{d \tan {\left (c + d x \right )}} - \frac {B a^{2}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {B a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - \frac {2 B a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B a b}{d \tan ^{2}{\left (c + d x \right )}} - B b^{2} x - \frac {B b^{2}}{d \tan {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+b*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

Piecewise((zoo*A*a**2*x, (Eq(c, 0) | Eq(c, -d*x)) & (Eq(d, 0) | Eq(c, -d*x))), (x*(A + B*tan(c))*(a + b*tan(c)
)**2*cot(c)**5, Eq(d, 0)), (-A*a**2*log(tan(c + d*x)**2 + 1)/(2*d) + A*a**2*log(tan(c + d*x))/d + A*a**2/(2*d*
tan(c + d*x)**2) - A*a**2/(4*d*tan(c + d*x)**4) + 2*A*a*b*x + 2*A*a*b/(d*tan(c + d*x)) - 2*A*a*b/(3*d*tan(c +
d*x)**3) + A*b**2*log(tan(c + d*x)**2 + 1)/(2*d) - A*b**2*log(tan(c + d*x))/d - A*b**2/(2*d*tan(c + d*x)**2) +
B*a**2*x + B*a**2/(d*tan(c + d*x)) - B*a**2/(3*d*tan(c + d*x)**3) + B*a*b*log(tan(c + d*x)**2 + 1)/d - 2*B*a*
b*log(tan(c + d*x))/d - B*a*b/(d*tan(c + d*x)**2) - B*b**2*x - B*b**2/(d*tan(c + d*x)), True))

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