3.262 \(\int \cot ^3(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=186 \[ \frac {b^2 \left (a^2 B+3 a A b+b^2 B\right ) \tan (c+d x)}{d}-\frac {a^2 \left (a^2 A-4 a b B-6 A b^2\right ) \log (\sin (c+d x))}{d}-x \left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right )-\frac {b^3 (4 a B+A b) \log (\cos (c+d x))}{d}-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{2 d}-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d} \]

[Out]

-(4*A*a^3*b-4*A*a*b^3+B*a^4-6*B*a^2*b^2+B*b^4)*x-b^3*(A*b+4*B*a)*ln(cos(d*x+c))/d-a^2*(A*a^2-6*A*b^2-4*B*a*b)*
ln(sin(d*x+c))/d+b^2*(3*A*a*b+B*a^2+B*b^2)*tan(d*x+c)/d-1/2*a*(5*A*b+2*B*a)*cot(d*x+c)*(a+b*tan(d*x+c))^2/d-1/
2*a*A*cot(d*x+c)^2*(a+b*tan(d*x+c))^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.51, antiderivative size = 186, 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 = {3605, 3645, 3637, 3624, 3475} \[ \frac {b^2 \left (a^2 B+3 a A b+b^2 B\right ) \tan (c+d x)}{d}-\frac {a^2 \left (a^2 A-4 a b B-6 A b^2\right ) \log (\sin (c+d x))}{d}-x \left (4 a^3 A b-6 a^2 b^2 B+a^4 B-4 a A b^3+b^4 B\right )-\frac {b^3 (4 a B+A b) \log (\cos (c+d x))}{d}-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{2 d}-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

-((4*a^3*A*b - 4*a*A*b^3 + a^4*B - 6*a^2*b^2*B + b^4*B)*x) - (b^3*(A*b + 4*a*B)*Log[Cos[c + d*x]])/d - (a^2*(a
^2*A - 6*A*b^2 - 4*a*b*B)*Log[Sin[c + d*x]])/d + (b^2*(3*a*A*b + a^2*B + b^2*B)*Tan[c + d*x])/d - (a*(5*A*b +
2*a*B)*Cot[c + d*x]*(a + b*Tan[c + d*x])^2)/(2*d) - (a*A*Cot[c + d*x]^2*(a + b*Tan[c + d*x])^3)/(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 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 3645

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*d^2 + c*(c*C - B*d))*(a + b*T
an[e + f*x])^m*(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)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*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[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.69, size = 140, normalized size = 0.75 \[ \frac {a^4 (-A) \cot ^2(c+d x)-2 a^3 (a B+4 A b) \cot (c+d x)-2 a^2 \left (a^2 A-4 a b B-6 A b^2\right ) \log (\tan (c+d x))+(a+i b)^4 (A+i B) \log (-\tan (c+d x)+i)+(a-i b)^4 (A-i B) \log (\tan (c+d x)+i)+2 b^4 B \tan (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 199, normalized size = 1.07 \[ \frac {2 \, B b^{4} \tan \left (d x + c\right )^{3} - A a^{4} - {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} - {\left (4 \, B a b^{3} + A b^{4}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} - {\left (A a^{4} + 2 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} d x\right )} \tan \left (d x + c\right )^{2} - 2 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 5.16, size = 224, normalized size = 1.20 \[ \frac {2 \, B b^{4} \tan \left (d x + c\right ) - 2 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} {\left (d x + c\right )} + {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + \frac {3 \, A a^{4} \tan \left (d x + c\right )^{2} - 12 \, B a^{3} b \tan \left (d x + c\right )^{2} - 18 \, A a^{2} b^{2} \tan \left (d x + c\right )^{2} - 2 \, B a^{4} \tan \left (d x + c\right ) - 8 \, A a^{3} b \tan \left (d x + c\right ) - A a^{4}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*B*b^4*tan(d*x + c) - 2*(B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3 + B*b^4)*(d*x + c) + (A*a^4 - 4*B*a
^3*b - 6*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*log(tan(d*x + c)^2 + 1) - 2*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2)*log(abs(
tan(d*x + c))) + (3*A*a^4*tan(d*x + c)^2 - 12*B*a^3*b*tan(d*x + c)^2 - 18*A*a^2*b^2*tan(d*x + c)^2 - 2*B*a^4*t
an(d*x + c) - 8*A*a^3*b*tan(d*x + c) - A*a^4)/tan(d*x + c)^2)/d

________________________________________________________________________________________

maple [A]  time = 0.48, size = 244, normalized size = 1.31 \[ -\frac {A \,a^{4} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{4} A \ln \left (\sin \left (d x +c \right )\right )}{d}-a^{4} B x -\frac {B \cot \left (d x +c \right ) a^{4}}{d}-\frac {a^{4} B c}{d}-4 A \,a^{3} b x -\frac {4 A \cot \left (d x +c \right ) a^{3} b}{d}-\frac {4 A \,a^{3} b c}{d}+\frac {4 B \,a^{3} b \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {6 A \,a^{2} b^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}+6 B \,a^{2} b^{2} x +\frac {6 B \,a^{2} b^{2} c}{d}+4 A a \,b^{3} x +\frac {4 A a \,b^{3} c}{d}-\frac {4 B a \,b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}-\frac {A \,b^{4} \ln \left (\cos \left (d x +c \right )\right )}{d}-B \,b^{4} x +\frac {B \,b^{4} \tan \left (d x +c \right )}{d}-\frac {B \,b^{4} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)

[Out]

-1/2/d*A*a^4*cot(d*x+c)^2-a^4*A*ln(sin(d*x+c))/d-a^4*B*x-1/d*B*cot(d*x+c)*a^4-1/d*a^4*B*c-4*A*a^3*b*x-4/d*A*co
t(d*x+c)*a^3*b-4/d*A*a^3*b*c+4/d*B*a^3*b*ln(sin(d*x+c))+6/d*A*a^2*b^2*ln(sin(d*x+c))+6*B*a^2*b^2*x+6/d*B*a^2*b
^2*c+4*A*a*b^3*x+4/d*A*a*b^3*c-4/d*B*a*b^3*ln(cos(d*x+c))-1/d*A*b^4*ln(cos(d*x+c))-B*b^4*x+1/d*B*b^4*tan(d*x+c
)-1/d*B*b^4*c

________________________________________________________________________________________

maxima [A]  time = 0.94, size = 173, normalized size = 0.93 \[ \frac {2 \, B b^{4} \tan \left (d x + c\right ) - 2 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} {\left (d x + c\right )} + {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {A a^{4} + 2 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 6.45, size = 149, normalized size = 0.80 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-A\,a^4+4\,B\,a^3\,b+6\,A\,a^2\,b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^4+4\,A\,b\,a^3\right )+\frac {A\,a^4}{2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^4}{2\,d}+\frac {B\,b^4\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^4}{2\,d} \]

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))^4,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 4.61, size = 309, normalized size = 1.66 \[ \begin {cases} \tilde {\infty } A a^{4} x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (A + B \tan {\relax (c )}\right ) \left (a + b \tan {\relax (c )}\right )^{4} \cot ^{3}{\relax (c )} & \text {for}\: d = 0 \\\frac {A a^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {A a^{4} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {A a^{4}}{2 d \tan ^{2}{\left (c + d x \right )}} - 4 A a^{3} b x - \frac {4 A a^{3} b}{d \tan {\left (c + d x \right )}} - \frac {3 A a^{2} b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {6 A a^{2} b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 4 A a b^{3} x + \frac {A b^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - B a^{4} x - \frac {B a^{4}}{d \tan {\left (c + d x \right )}} - \frac {2 B a^{3} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {4 B a^{3} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 6 B a^{2} b^{2} x + \frac {2 B a b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - B b^{4} x + \frac {B b^{4} \tan {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(a+b*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

Piecewise((zoo*A*a**4*x, (Eq(c, 0) | Eq(c, -d*x)) & (Eq(d, 0) | Eq(c, -d*x))), (x*(A + B*tan(c))*(a + b*tan(c)
)**4*cot(c)**3, Eq(d, 0)), (A*a**4*log(tan(c + d*x)**2 + 1)/(2*d) - A*a**4*log(tan(c + d*x))/d - A*a**4/(2*d*t
an(c + d*x)**2) - 4*A*a**3*b*x - 4*A*a**3*b/(d*tan(c + d*x)) - 3*A*a**2*b**2*log(tan(c + d*x)**2 + 1)/d + 6*A*
a**2*b**2*log(tan(c + d*x))/d + 4*A*a*b**3*x + A*b**4*log(tan(c + d*x)**2 + 1)/(2*d) - B*a**4*x - B*a**4/(d*ta
n(c + d*x)) - 2*B*a**3*b*log(tan(c + d*x)**2 + 1)/d + 4*B*a**3*b*log(tan(c + d*x))/d + 6*B*a**2*b**2*x + 2*B*a
*b**3*log(tan(c + d*x)**2 + 1)/d - B*b**4*x + B*b**4*tan(c + d*x)/d, True))

________________________________________________________________________________________