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

3.5.42.1 Optimal result
3.5.42.2 Mathematica [C] (verified)
3.5.42.3 Rubi [A] (verified)
3.5.42.4 Maple [A] (verified)
3.5.42.5 Fricas [A] (verification not implemented)
3.5.42.6 Sympy [A] (verification not implemented)
3.5.42.7 Maxima [A] (verification not implemented)
3.5.42.8 Giac [B] (verification not implemented)
3.5.42.9 Mupad [B] (verification not implemented)

3.5.42.1 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
 
3.5.42.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.24 (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)
 
3.5.42.3 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)
 

3.5.42.3.1 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 
]
 
3.5.42.4 Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.12

method result size
derivativedivides \(\frac {\frac {\left (3 a^{2} b -b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (a^{3}-3 a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {a^{3}}{3 \tan \left (d x +c \right )^{3}}+\frac {a \left (a^{2}-3 b^{2}\right )}{\tan \left (d x +c \right )}-\frac {3 a^{2} b}{2 \tan \left (d x +c \right )^{2}}-b \left (3 a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\) \(116\)
default \(\frac {\frac {\left (3 a^{2} b -b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (a^{3}-3 a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {a^{3}}{3 \tan \left (d x +c \right )^{3}}+\frac {a \left (a^{2}-3 b^{2}\right )}{\tan \left (d x +c \right )}-\frac {3 a^{2} b}{2 \tan \left (d x +c \right )^{2}}-b \left (3 a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\) \(116\)
parallelrisch \(\frac {-2 \left (\cot ^{3}\left (d x +c \right )\right ) a^{3}-9 \left (\cot ^{2}\left (d x +c \right )\right ) a^{2} b +6 a^{3} d x -18 a \,b^{2} d x +6 a^{3} \cot \left (d x +c \right )-18 \cot \left (d x +c \right ) a \,b^{2}-18 a^{2} b \ln \left (\tan \left (d x +c \right )\right )+6 \ln \left (\tan \left (d x +c \right )\right ) b^{3}+9 \ln \left (\sec ^{2}\left (d x +c \right )\right ) a^{2} b -3 \ln \left (\sec ^{2}\left (d x +c \right )\right ) b^{3}}{6 d}\) \(126\)
norman \(\frac {\frac {a \left (a^{2}-3 b^{2}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{d}+a \left (a^{2}-3 b^{2}\right ) x \left (\tan ^{3}\left (d x +c \right )\right )-\frac {a^{3}}{3 d}-\frac {3 a^{2} b \tan \left (d x +c \right )}{2 d}}{\tan \left (d x +c \right )^{3}}-\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\) \(129\)
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*(1/2*(3*a^2*b-b^3)*ln(1+tan(d*x+c)^2)+(a^3-3*a*b^2)*arctan(tan(d*x+c)) 
-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)))
 
3.5.42.5 Fricas [A] (verification not implemented)

Time = 0.25 (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)
 
3.5.42.6 Sympy [A] (verification not implemented)

Time = 1.05 (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))
 
3.5.42.7 Maxima [A] (verification not implemented)

Time = 0.33 (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
 
3.5.42.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (100) = 200\).

Time = 1.34 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.27 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )} + 24 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 24 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {132 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

input
integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^3,x, algorithm="giac")
 
output
1/24*(a^3*tan(1/2*d*x + 1/2*c)^3 - 9*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 15*a^3 
*tan(1/2*d*x + 1/2*c) + 36*a*b^2*tan(1/2*d*x + 1/2*c) + 24*(a^3 - 3*a*b^2) 
*(d*x + c) + 24*(3*a^2*b - b^3)*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 24*(3*a^ 
2*b - b^3)*log(abs(tan(1/2*d*x + 1/2*c))) + (132*a^2*b*tan(1/2*d*x + 1/2*c 
)^3 - 44*b^3*tan(1/2*d*x + 1/2*c)^3 + 15*a^3*tan(1/2*d*x + 1/2*c)^2 - 36*a 
*b^2*tan(1/2*d*x + 1/2*c)^2 - 9*a^2*b*tan(1/2*d*x + 1/2*c) - a^3)/tan(1/2* 
d*x + 1/2*c)^3)/d
 
3.5.42.9 Mupad [B] (verification not implemented)

Time = 5.50 (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