\(\int \cot ^3(c+d x) (a+b \tan (c+d x))^2 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\) [13]

Optimal result

Integrand size = 40, antiderivative size = 72 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\left (\left (a^2 B-b^2 B-2 a b C\right ) x\right )-\frac {a^2 B \cot (c+d x)}{d}-\frac {b^2 C \log (\cos (c+d x))}{d}+\frac {a (2 b B+a C) \log (\sin (c+d x))}{d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.39 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {-2 a^2 B \cot (c+d x)+i (a+i b)^2 (B+i C) \log (i-\tan (c+d x))+2 a (2 b B+a C) \log (\tan (c+d x))-(a-i b)^2 (i B+C) \log (i+\tan (c+d x))}{2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {3042, 4115, 3042, 4087, 3042, 4107, 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.

\(\displaystyle \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan (c+d x)^2\right )}{\tan (c+d x)^3}dx\)

\(\Big \downarrow \) 4115

\(\displaystyle \int \cot ^2(c+d x) (a+b \tan (c+d x))^2 (B+C \tan (c+d x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^2 (B+C \tan (c+d x))}{\tan (c+d x)^2}dx\)

\(\Big \downarrow \) 4087

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4107

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

\(\displaystyle -x \left (a^2 B-2 a b C-b^2 B\right )-\frac {a^2 B \cot (c+d x)}{d}+\frac {a (a C+2 b B) \log (-\sin (c+d x))}{d}-\frac {b^2 C \log (\cos (c+d x))}{d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4087

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] + Simp[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 4107

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2 
)/tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[B*x, x] + (Simp[A   Int[1/Tan[ 
e + f*x], x], x] + Simp[C   Int[Tan[e + f*x], x], x]) /; FreeQ[{e, f, A, B, 
 C}, x] && NeQ[A, C]
 

rule 4115

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] :> Simp[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]
 

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {B \,b^{2} \left (d x +c \right )-C \,b^{2} \ln \left (\cos \left (d x +c \right )\right )+2 B a b \ln \left (\sin \left (d x +c \right )\right )+2 C a b \left (d x +c \right )+B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+C \,a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}\)	\(84\)
default	\(\frac {B \,b^{2} \left (d x +c \right )-C \,b^{2} \ln \left (\cos \left (d x +c \right )\right )+2 B a b \ln \left (\sin \left (d x +c \right )\right )+2 C a b \left (d x +c \right )+B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+C \,a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}\)	\(84\)
parallelrisch	\(\frac {\left (-2 B a b -C \,a^{2}+C \,b^{2}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+\left (4 B a b +2 C \,a^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-2 B \cot \left (d x +c \right ) a^{2}-2 d x \left (B \,a^{2}-B \,b^{2}-2 C a b \right )}{2 d}\)	\(87\)
norman	\(\frac {\left (-B \,a^{2}+B \,b^{2}+2 C a b \right ) x \tan \left (d x +c \right )^{2}-\frac {B \,a^{2} \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}+\frac {a \left (2 B b +C a \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (2 B a b +C \,a^{2}-C \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(107\)
risch	\(-B \,a^{2} x +B \,b^{2} x +2 C a b x +\frac {2 i C \,b^{2} c}{d}-\frac {2 i C \,a^{2} c}{d}-i C \,a^{2} x +i C \,b^{2} x -\frac {4 i B a b c}{d}-\frac {2 i B \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-2 i B a b x +\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) C \,b^{2}}{d}\)	\(160\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.56 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {C b^{2} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) + 2 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} d x \tan \left (d x + c\right ) + 2 \, B a^{2} - {\left (C a^{2} + 2 \, B a b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )} \] Input:

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (66) = 132\).

Time = 1.07 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.19 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{2} \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{3}{\left (c \right )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\- B a^{2} x - \frac {B a^{2}}{d \tan {\left (c + d x \right )}} - \frac {B a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {2 B a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + B b^{2} x - \frac {C a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 2 C a b x + \frac {C b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.29 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} {\left (d x + c\right )} + {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (C a^{2} + 2 \, B a b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {2 \, B a^{2}}{\tan \left (d x + c\right )}}{2 \, d} \] Input:

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

Output:

-1/2*(2*(B*a^2 - 2*C*a*b - B*b^2)*(d*x + c) + (C*a^2 + 2*B*a*b - C*b^2)*lo 
g(tan(d*x + c)^2 + 1) - 2*(C*a^2 + 2*B*a*b)*log(tan(d*x + c)) + 2*B*a^2/ta 
n(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.88 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.40 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {{\left (B a^{2} - 2 \, C a b - B b^{2}\right )} {\left (d x + c\right )}}{d} - \frac {{\left (C a^{2} + 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} + \frac {{\left (C a^{2} + 2 \, B a b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {B a^{2}}{d \tan \left (d x + c\right )} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.39 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (C\,a^2+2\,B\,b\,a\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (C+B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {B\,a^2\,\mathrm {cot}\left (c+d\,x\right )}{d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 232, normalized size of antiderivative = 3.22 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {-\cos \left (d x +c \right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right ) a^{2} c -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right ) a \,b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right ) b^{2} c -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b^{2} c -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b^{2} c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a^{2} c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a \,b^{2}-\sin \left (d x +c \right ) a^{2} b d x +2 \sin \left (d x +c \right ) a b c d x +\sin \left (d x +c \right ) b^{3} d x}{\sin \left (d x +c \right ) d} \] Input:

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

Output:

( - cos(c + d*x)*a**2*b - log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**2*c 
 - 2*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a*b**2 + log(tan((c + d*x)/ 
2)**2 + 1)*sin(c + d*x)*b**2*c - log(tan((c + d*x)/2) - 1)*sin(c + d*x)*b* 
*2*c - log(tan((c + d*x)/2) + 1)*sin(c + d*x)*b**2*c + log(tan((c + d*x)/2 
))*sin(c + d*x)*a**2*c + 2*log(tan((c + d*x)/2))*sin(c + d*x)*a*b**2 - sin 
(c + d*x)*a**2*b*d*x + 2*sin(c + d*x)*a*b*c*d*x + sin(c + d*x)*b**3*d*x)/( 
sin(c + d*x)*d)