\(\int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\) [21]

Optimal result

Integrand size = 40, antiderivative size = 127 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\left (\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) x\right )-\frac {a^2 (2 b B+a C) \cot (c+d x)}{d}-\frac {b^3 C \log (\cos (c+d x))}{d}-\frac {a \left (a^2 B-3 b^2 B-3 a b C\right ) \log (\sin (c+d x))}{d}-\frac {a B \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3713, 3686, 3716, 3705, 3556} \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {a \left (a^2 B-3 a b C-3 b^2 B\right ) \log (\sin (c+d x))}{d}-\frac {a^2 (a C+2 b B) \cot (c+d x)}{d}-x \left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right )-\frac {a B \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}-\frac {b^3 C \log (\cos (c+d x))}{d} \]

[In]

Int[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

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

Rule 3556

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

Rule 3686

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 3705

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 3713

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

Rule 3716

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*} \text {integral}& = \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 (B+C \tan (c+d x)) \, dx \\ & = -\frac {a B \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 b B+a C)-2 \left (a^2 B-b^2 B-2 a b C\right ) \tan (c+d x)+2 b^2 C \tan ^2(c+d x)\right ) \, dx \\ & = -\frac {a^2 (2 b B+a C) \cot (c+d x)}{d}-\frac {a B \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 B-3 b^2 B-3 a b C\right )-2 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \tan (c+d x)+2 b^3 C \tan ^2(c+d x)\right ) \, dx \\ & = -\left (\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) x\right )-\frac {a^2 (2 b B+a C) \cot (c+d x)}{d}-\frac {a B \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}+\left (b^3 C\right ) \int \tan (c+d x) \, dx-\left (a \left (a^2 B-3 b^2 B-3 a b C\right )\right ) \int \cot (c+d x) \, dx \\ & = -\left (\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) x\right )-\frac {a^2 (2 b B+a C) \cot (c+d x)}{d}-\frac {b^3 C \log (\cos (c+d x))}{d}-\frac {a \left (a^2 B-3 b^2 B-3 a b C\right ) \log (\sin (c+d x))}{d}-\frac {a B \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.49 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.99 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {-2 a^2 (3 b B+a C) \cot (c+d x)-a^3 B \cot ^2(c+d x)+(a+i b)^3 (B+i C) \log (i-\tan (c+d x))-2 a \left (a^2 B-3 b^2 B-3 a b C\right ) \log (\tan (c+d x))+(a-i b)^3 (B-i C) \log (i+\tan (c+d x))}{2 d} \]

[In]

Integrate[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(\frac {\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+\left (-2 B \,a^{3}+6 B a \,b^{2}+6 C \,a^{2} b \right ) \ln \left (\tan \left (d x +c \right )\right )-B \,a^{3} \cot \left (d x +c \right )^{2}+\left (-6 B \,a^{2} b -2 C \,a^{3}\right ) \cot \left (d x +c \right )-6 d \left (B \,a^{2} b -\frac {1}{3} B \,b^{3}+\frac {1}{3} C \,a^{3}-C a \,b^{2}\right ) x}{2 d}\)	\(136\)
derivativedivides	\(\frac {-\frac {B \,a^{3}}{2 \tan \left (d x +c \right )^{2}}-\frac {a^{2} \left (3 B b +C a \right )}{\tan \left (d x +c \right )}-a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right ) \ln \left (\tan \left (d x +c \right )\right )+\frac {\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(140\)
default	\(\frac {-\frac {B \,a^{3}}{2 \tan \left (d x +c \right )^{2}}-\frac {a^{2} \left (3 B b +C a \right )}{\tan \left (d x +c \right )}-a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right ) \ln \left (\tan \left (d x +c \right )\right )+\frac {\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(140\)
norman	\(\frac {\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) x \tan \left (d x +c \right )^{3}-\frac {B \,a^{3} \tan \left (d x +c \right )}{2 d}-\frac {a^{2} \left (3 B b +C a \right ) \tan \left (d x +c \right )^{2}}{d}}{\tan \left (d x +c \right )^{3}}+\frac {\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}-\frac {a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(158\)
risch	\(i B \,a^{3} x -3 i C \,a^{2} b x -3 i B a \,b^{2} x +\frac {2 i C \,b^{3} c}{d}-3 B \,a^{2} b x +B \,b^{3} x -C \,a^{3} x +3 C a \,b^{2} x -\frac {2 i a^{2} \left (3 B b \,{\mathrm e}^{2 i \left (d x +c \right )}+C a \,{\mathrm e}^{2 i \left (d x +c \right )}+i B a \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B b -C a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+i C \,b^{3} x +\frac {2 i B \,a^{3} c}{d}-\frac {6 i C \,a^{2} b c}{d}-\frac {6 i B a \,b^{2} c}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) C \,b^{3}}{d}-\frac {B \,a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B \,b^{2}}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C b}{d}\)	\(267\)

[In]

int(cot(d*x+c)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/2*((B*a^3-3*B*a*b^2-3*C*a^2*b+C*b^3)*ln(sec(d*x+c)^2)+(-2*B*a^3+6*B*a*b^2+6*C*a^2*b)*ln(tan(d*x+c))-B*a^3*co
t(d*x+c)^2+(-6*B*a^2*b-2*C*a^3)*cot(d*x+c)-6*d*(B*a^2*b-1/3*B*b^3+1/3*C*a^3-C*a*b^2)*x)/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.28 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {C b^{3} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + B a^{3} + {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + {\left (B a^{3} + 2 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \]

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (121) = 242\).

Time = 2.33 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.99 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{3} \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{4}{\left (c \right )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\\frac {B a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 B a^{2} b x - \frac {3 B a^{2} b}{d \tan {\left (c + d x \right )}} - \frac {3 B a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 B a b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + B b^{3} x - C a^{3} x - \frac {C a^{3}}{d \tan {\left (c + d x \right )}} - \frac {3 C a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 C a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 C a b^{2} x + \frac {C b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text {otherwise} \end {cases} \]

[In]

integrate(cot(d*x+c)**4*(a+b*tan(d*x+c))**3*(B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.12 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} {\left (d x + c\right )} - {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {B a^{3} + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.28 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.52 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} {\left (d x + c\right )} - {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {3 \, B a^{3} \tan \left (d x + c\right )^{2} - 9 \, C a^{2} b \tan \left (d x + c\right )^{2} - 9 \, B a b^{2} \tan \left (d x + c\right )^{2} - 2 \, C a^{3} \tan \left (d x + c\right ) - 6 \, B a^{2} b \tan \left (d x + c\right ) - B a^{3}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.77 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B\,a^3+3\,C\,a^2\,b+3\,B\,a\,b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (C\,a^3+3\,B\,b\,a^2\right )+\frac {B\,a^3}{2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \]

[In]

int(cot(c + d*x)^4*(B*tan(c + d*x) + C*tan(c + d*x)^2)*(a + b*tan(c + d*x))^3,x)

[Out]

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