\(\int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \, dx\) [442]

Optimal result

Integrand size = 21, antiderivative size = 104 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=a \left (a^2-3 b^2\right ) x+\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}-\frac {7 a^2 b \cot ^2(c+d x)}{6 d}-\frac {b \left (3 a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d} \] Output:

a*(a^2-3*b^2)*x+a*(a^2-3*b^2)*cot(d*x+c)/d-7/6*a^2*b*cot(d*x+c)^2/d-b*(3*a 
^2-b^2)*ln(sin(d*x+c))/d-1/3*a^2*cot(d*x+c)^3*(a+b*tan(d*x+c))/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.77 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.15 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {6 a \left (a^2-3 b^2\right ) \cot (c+d x)-9 a^2 b \cot ^2(c+d x)-2 a^3 \cot ^3(c+d x)+3 (i a-b)^3 \log (i-\tan (c+d x))+6 b \left (-3 a^2+b^2\right ) \log (\tan (c+d x))-3 (i a+b)^3 \log (i+\tan (c+d x))}{6 d} \] Input:

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

Output:

(6*a*(a^2 - 3*b^2)*Cot[c + d*x] - 9*a^2*b*Cot[c + d*x]^2 - 2*a^3*Cot[c + d 
*x]^3 + 3*(I*a - b)^3*Log[I - Tan[c + d*x]] + 6*b*(-3*a^2 + b^2)*Log[Tan[c 
 + d*x]] - 3*(I*a + b)^3*Log[I + Tan[c + d*x]])/(6*d)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4048, 3042, 4111, 27, 3042, 4012, 3042, 4014, 3042, 25, 3956}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^3}{\tan (c+d x)^4}dx\)

\(\Big \downarrow \) 4048

\(\displaystyle \frac {1}{3} \int \cot ^3(c+d x) \left (7 b a^2-3 \left (a^2-3 b^2\right ) \tan (c+d x) a-b \left (2 a^2-3 b^2\right ) \tan ^2(c+d x)\right )dx-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {7 b a^2-3 \left (a^2-3 b^2\right ) \tan (c+d x) a-b \left (2 a^2-3 b^2\right ) \tan (c+d x)^2}{\tan (c+d x)^3}dx-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d}\)

\(\Big \downarrow \) 4111

\(\displaystyle \frac {1}{3} \left (\int -3 \cot ^2(c+d x) \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right )dx-\frac {7 a^2 b \cot ^2(c+d x)}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-3 \int \cot ^2(c+d x) \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right )dx-\frac {7 a^2 b \cot ^2(c+d x)}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-3 \int \frac {a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)}{\tan (c+d x)^2}dx-\frac {7 a^2 b \cot ^2(c+d x)}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {1}{3} \left (-3 \left (\int \cot (c+d x) \left (b \left (3 a^2-b^2\right )-a \left (a^2-3 b^2\right ) \tan (c+d x)\right )dx-\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}\right )-\frac {7 a^2 b \cot ^2(c+d x)}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-3 \left (\int \frac {b \left (3 a^2-b^2\right )-a \left (a^2-3 b^2\right ) \tan (c+d x)}{\tan (c+d x)}dx-\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}\right )-\frac {7 a^2 b \cot ^2(c+d x)}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d}\)

\(\Big \downarrow \) 4014

\(\displaystyle \frac {1}{3} \left (-3 \left (b \left (3 a^2-b^2\right ) \int \cot (c+d x)dx-\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}-a x \left (a^2-3 b^2\right )\right )-\frac {7 a^2 b \cot ^2(c+d x)}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-3 \left (b \left (3 a^2-b^2\right ) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}-a x \left (a^2-3 b^2\right )\right )-\frac {7 a^2 b \cot ^2(c+d x)}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (-3 \left (-b \left (3 a^2-b^2\right ) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}-a x \left (a^2-3 b^2\right )\right )-\frac {7 a^2 b \cot ^2(c+d x)}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {1}{3} \left (-3 \left (-\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{d}+\frac {b \left (3 a^2-b^2\right ) \log (-\sin (c+d x))}{d}-a x \left (a^2-3 b^2\right )\right )-\frac {7 a^2 b \cot ^2(c+d x)}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))}{3 d}\)

Input:

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

Output:

((-7*a^2*b*Cot[c + d*x]^2)/(2*d) - 3*(-(a*(a^2 - 3*b^2)*x) - (a*(a^2 - 3*b 
^2)*Cot[c + d*x])/d + (b*(3*a^2 - b^2)*Log[-Sin[c + d*x]])/d))/3 - (a^2*Co 
t[c + d*x]^3*(a + b*Tan[c + d*x]))/(3*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956

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

rule 4012

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] + Simp[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 4014

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] + Simp[(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] && N 
eQ[a*c + b*d, 0]
 

rule 4048

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m 
 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 
/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + 
f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c 
*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) 
*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( 
n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 2] && LtQ 
[n, -1] && IntegerQ[2*m]
 

rule 4111

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 
] + Simp[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 
]
 

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {-\frac {a^{3}}{3 \tan \left (d x +c \right )^{3}}+\frac {a \left (a^{2}-3 b^{2}\right )}{\tan \left (d x +c \right )}-\frac {3 a^{2} b}{2 \tan \left (d x +c \right )^{2}}-b \left (3 a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )+\frac {\left (3 a^{2} b -b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (a^{3}-3 a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(116\)
default	\(\frac {-\frac {a^{3}}{3 \tan \left (d x +c \right )^{3}}+\frac {a \left (a^{2}-3 b^{2}\right )}{\tan \left (d x +c \right )}-\frac {3 a^{2} b}{2 \tan \left (d x +c \right )^{2}}-b \left (3 a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )+\frac {\left (3 a^{2} b -b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (a^{3}-3 a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(116\)
parallelrisch	\(\frac {-9 a^{2} b \left (2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec \left (d x +c \right )^{2}\right )\right )+3 b^{3} \left (2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec \left (d x +c \right )^{2}\right )\right )-2 \cot \left (d x +c \right )^{3} a^{3}-9 \cot \left (d x +c \right )^{2} a^{2} b +6 a^{3} d x -18 a \,b^{2} d x +6 a^{3} \cot \left (d x +c \right )-18 \cot \left (d x +c \right ) a \,b^{2}}{6 d}\)	\(125\)
norman	\(\frac {\frac {a \left (a^{2}-3 b^{2}\right ) \tan \left (d x +c \right )^{2}}{d}+a \left (a^{2}-3 b^{2}\right ) x \tan \left (d x +c \right )^{3}-\frac {a^{3}}{3 d}-\frac {3 a^{2} b \tan \left (d x +c \right )}{2 d}}{\tan \left (d x +c \right )^{3}}-\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(129\)
risch	\(3 i a^{2} b x -i b^{3} x +a^{3} x -3 a \,b^{2} x +\frac {6 i b \,a^{2} c}{d}-\frac {2 i b^{3} c}{d}+\frac {2 i a \left (6 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-9 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-9 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+18 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+9 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a^{2}-9 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(205\)

Input:

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

Output:

1/d*(-1/3*a^3/tan(d*x+c)^3+a*(a^2-3*b^2)/tan(d*x+c)-3/2*a^2*b/tan(d*x+c)^2 
-b*(3*a^2-b^2)*ln(tan(d*x+c))+1/2*(3*a^2*b-b^3)*ln(1+tan(d*x+c)^2)+(a^3-3* 
a*b^2)*arctan(tan(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.21 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {3 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} + 9 \, a^{2} b \tan \left (d x + c\right ) + 3 \, {\left (3 \, a^{2} b - 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} d x\right )} \tan \left (d x + c\right )^{3} + 2 \, a^{3} - 6 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.07 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.68 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\begin {cases} \tilde {\infty } a^{3} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{3} \cot ^{4}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } a^{3} x & \text {for}\: c = - d x \\a^{3} x + \frac {a^{3}}{d \tan {\left (c + d x \right )}} - \frac {a^{3}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {3 a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {3 a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {3 a^{2} b}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 a b^{2} x - \frac {3 a b^{2}}{d \tan {\left (c + d x \right )}} - \frac {b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.12 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {6 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )} + 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {9 \, a^{2} b \tan \left (d x + c\right ) + 2 \, a^{3} - 6 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{3}}}{6 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.19 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {{\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{d} + \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} - \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {9 \, a^{2} b \tan \left (d x + c\right ) + 2 \, a^{3} - 6 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.19 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (3\,a^2\,b-b^3\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^3}{2\,d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (3\,a\,b^2-a^3\right )+\frac {a^3}{3}+\frac {3\,a^2\,b\,\mathrm {tan}\left (c+d\,x\right )}{2}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.14 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-36 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-4 \cos \left (d x +c \right ) a^{3}+36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} a^{2} b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} b^{3}-36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a^{2} b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} b^{3}+12 \sin \left (d x +c \right )^{3} a^{3} d x +9 \sin \left (d x +c \right )^{3} a^{2} b -36 \sin \left (d x +c \right )^{3} a \,b^{2} d x -18 \sin \left (d x +c \right ) a^{2} b}{12 \sin \left (d x +c \right )^{3} d} \] Input:

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

Output:

(16*cos(c + d*x)*sin(c + d*x)**2*a**3 - 36*cos(c + d*x)*sin(c + d*x)**2*a* 
b**2 - 4*cos(c + d*x)*a**3 + 36*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)* 
*3*a**2*b - 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3*b**3 - 36*log( 
tan((c + d*x)/2))*sin(c + d*x)**3*a**2*b + 12*log(tan((c + d*x)/2))*sin(c 
+ d*x)**3*b**3 + 12*sin(c + d*x)**3*a**3*d*x + 9*sin(c + d*x)**3*a**2*b - 
36*sin(c + d*x)**3*a*b**2*d*x - 18*sin(c + d*x)*a**2*b)/(12*sin(c + d*x)** 
3*d)