\(\int \tan ^3(c+d x) (a+b \tan (c+d x))^2 \, dx\) [425]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=2 a b x+\frac {\left (a^2-b^2\right ) \log (\cos (c+d x))}{d}-\frac {2 a b \tan (c+d x)}{d}+\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 d}+\frac {2 a b \tan ^3(c+d x)}{3 d}+\frac {b^2 \tan ^4(c+d x)}{4 d} \] Output:

2*a*b*x+(a^2-b^2)*ln(cos(d*x+c))/d-2*a*b*tan(d*x+c)/d+1/2*(a^2-b^2)*tan(d* 
x+c)^2/d+2/3*a*b*tan(d*x+c)^3/d+1/4*b^2*tan(d*x+c)^4/d
 

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.93 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {24 a b \arctan (\tan (c+d x))+12 \left (a^2-b^2\right ) \log (\cos (c+d x))+6 \left (a^2-2 b^2\right ) \sec ^2(c+d x)+3 b^2 \sec ^4(c+d x)-24 a b \tan (c+d x)+8 a b \tan ^3(c+d x)}{12 d} \] Input:

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

Output:

(24*a*b*ArcTan[Tan[c + d*x]] + 12*(a^2 - b^2)*Log[Cos[c + d*x]] + 6*(a^2 - 
 2*b^2)*Sec[c + d*x]^2 + 3*b^2*Sec[c + d*x]^4 - 24*a*b*Tan[c + d*x] + 8*a* 
b*Tan[c + d*x]^3)/(12*d)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 4026, 3042, 4011, 3042, 4011, 3042, 4008, 3042, 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.

\(\displaystyle \int \tan ^3(c+d x) (a+b \tan (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^3 (a+b \tan (c+d x))^2dx\)

\(\Big \downarrow \) 4026

\(\displaystyle \int \tan ^3(c+d x) \left (a^2+2 b \tan (c+d x) a-b^2\right )dx+\frac {b^2 \tan ^4(c+d x)}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^3 \left (a^2+2 b \tan (c+d x) a-b^2\right )dx+\frac {b^2 \tan ^4(c+d x)}{4 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \int \tan ^2(c+d x) \left (\left (a^2-b^2\right ) \tan (c+d x)-2 a b\right )dx+\frac {2 a b \tan ^3(c+d x)}{3 d}+\frac {b^2 \tan ^4(c+d x)}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^2 \left (\left (a^2-b^2\right ) \tan (c+d x)-2 a b\right )dx+\frac {2 a b \tan ^3(c+d x)}{3 d}+\frac {b^2 \tan ^4(c+d x)}{4 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \int \tan (c+d x) \left (-a^2-2 b \tan (c+d x) a+b^2\right )dx+\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 d}+\frac {2 a b \tan ^3(c+d x)}{3 d}+\frac {b^2 \tan ^4(c+d x)}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x) \left (-a^2-2 b \tan (c+d x) a+b^2\right )dx+\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 d}+\frac {2 a b \tan ^3(c+d x)}{3 d}+\frac {b^2 \tan ^4(c+d x)}{4 d}\)

\(\Big \downarrow \) 4008

\(\displaystyle -\left (a^2-b^2\right ) \int \tan (c+d x)dx+\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 d}+\frac {2 a b \tan ^3(c+d x)}{3 d}-\frac {2 a b \tan (c+d x)}{d}+2 a b x+\frac {b^2 \tan ^4(c+d x)}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\left (a^2-b^2\right ) \int \tan (c+d x)dx+\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 d}+\frac {2 a b \tan ^3(c+d x)}{3 d}-\frac {2 a b \tan (c+d x)}{d}+2 a b x+\frac {b^2 \tan ^4(c+d x)}{4 d}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 d}+\frac {\left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {2 a b \tan ^3(c+d x)}{3 d}-\frac {2 a b \tan (c+d x)}{d}+2 a b x+\frac {b^2 \tan ^4(c+d x)}{4 d}\)

Input:

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

Output:

2*a*b*x + ((a^2 - b^2)*Log[Cos[c + d*x]])/d - (2*a*b*Tan[c + d*x])/d + ((a 
^2 - b^2)*Tan[c + d*x]^2)/(2*d) + (2*a*b*Tan[c + d*x]^3)/(3*d) + (b^2*Tan[ 
c + d*x]^4)/(4*d)
 

Defintions of rubi rules used

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 4008

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) 
*(x_)]), x_Symbol] :> Simp[(a*c - b*d)*x, x] + (Simp[b*d*(Tan[e + f*x]/f), 
x] + Simp[(b*c + a*d)   Int[Tan[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, 
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
 

rule 4011

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]
 

rule 4026

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

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00

method	result	size
norman	\(2 a b x +\frac {b^{2} \tan \left (d x +c \right )^{4}}{4 d}+\frac {\left (a^{2}-b^{2}\right ) \tan \left (d x +c \right )^{2}}{2 d}-\frac {2 a b \tan \left (d x +c \right )}{d}+\frac {2 a b \tan \left (d x +c \right )^{3}}{3 d}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(98\)
derivativedivides	\(\frac {\frac {\tan \left (d x +c \right )^{4} b^{2}}{4}+\frac {2 \tan \left (d x +c \right )^{3} a b}{3}+\frac {\tan \left (d x +c \right )^{2} a^{2}}{2}-\frac {\tan \left (d x +c \right )^{2} b^{2}}{2}-2 \tan \left (d x +c \right ) a b +\frac {\left (-a^{2}+b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+2 a b \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(100\)
default	\(\frac {\frac {\tan \left (d x +c \right )^{4} b^{2}}{4}+\frac {2 \tan \left (d x +c \right )^{3} a b}{3}+\frac {\tan \left (d x +c \right )^{2} a^{2}}{2}-\frac {\tan \left (d x +c \right )^{2} b^{2}}{2}-2 \tan \left (d x +c \right ) a b +\frac {\left (-a^{2}+b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+2 a b \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(100\)
parallelrisch	\(-\frac {-3 \tan \left (d x +c \right )^{4} b^{2}-8 \tan \left (d x +c \right )^{3} a b -24 a b d x -6 \tan \left (d x +c \right )^{2} a^{2}+6 \tan \left (d x +c \right )^{2} b^{2}+6 \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a^{2}-6 \ln \left (1+\tan \left (d x +c \right )^{2}\right ) b^{2}+24 \tan \left (d x +c \right ) a b}{12 d}\)	\(106\)
parts	\(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {b^{2} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {2 a b \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(107\)
risch	\(2 a b x -i a^{2} x +i b^{2} x -\frac {2 i a^{2} c}{d}+\frac {2 i b^{2} c}{d}+\frac {-8 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+2 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-4 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-16 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-4 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-\frac {40 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}}{3}+2 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-\frac {16 i a b}{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b^{2}}{d}\)	\(230\)

Input:

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

Output:

2*a*b*x+1/4*b^2*tan(d*x+c)^4/d+1/2*(a^2-b^2)*tan(d*x+c)^2/d-2*a*b*tan(d*x+ 
c)/d+2/3*a*b*tan(d*x+c)^3/d-1/2*(a^2-b^2)/d*ln(1+tan(d*x+c)^2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.92 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {3 \, b^{2} \tan \left (d x + c\right )^{4} + 8 \, a b \tan \left (d x + c\right )^{3} + 24 \, a b d x - 24 \, a b \tan \left (d x + c\right ) + 6 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} + 6 \, {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{12 \, d} \] Input:

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

Output:

1/12*(3*b^2*tan(d*x + c)^4 + 8*a*b*tan(d*x + c)^3 + 24*a*b*d*x - 24*a*b*ta 
n(d*x + c) + 6*(a^2 - b^2)*tan(d*x + c)^2 + 6*(a^2 - b^2)*log(1/(tan(d*x + 
 c)^2 + 1)))/d
 

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.37 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\begin {cases} - \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + 2 a b x + \frac {2 a b \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 a b \tan {\left (c + d x \right )}}{d} + \frac {b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right )^{2} \tan ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-a**2*log(tan(c + d*x)**2 + 1)/(2*d) + a**2*tan(c + d*x)**2/(2* 
d) + 2*a*b*x + 2*a*b*tan(c + d*x)**3/(3*d) - 2*a*b*tan(c + d*x)/d + b**2*l 
og(tan(c + d*x)**2 + 1)/(2*d) + b**2*tan(c + d*x)**4/(4*d) - b**2*tan(c + 
d*x)**2/(2*d), Ne(d, 0)), (x*(a + b*tan(c))**2*tan(c)**3, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.93 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {3 \, b^{2} \tan \left (d x + c\right )^{4} + 8 \, a b \tan \left (d x + c\right )^{3} + 24 \, {\left (d x + c\right )} a b - 24 \, a b \tan \left (d x + c\right ) + 6 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} - 6 \, {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{12 \, d} \] Input:

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

Output:

1/12*(3*b^2*tan(d*x + c)^4 + 8*a*b*tan(d*x + c)^3 + 24*(d*x + c)*a*b - 24* 
a*b*tan(d*x + c) + 6*(a^2 - b^2)*tan(d*x + c)^2 - 6*(a^2 - b^2)*log(tan(d* 
x + c)^2 + 1))/d
                                                                                    
                                                                                    
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.22 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {2 \, {\left (d x + c\right )} a b}{d} - \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} + \frac {3 \, b^{2} d^{3} \tan \left (d x + c\right )^{4} + 8 \, a b d^{3} \tan \left (d x + c\right )^{3} + 6 \, a^{2} d^{3} \tan \left (d x + c\right )^{2} - 6 \, b^{2} d^{3} \tan \left (d x + c\right )^{2} - 24 \, a b d^{3} \tan \left (d x + c\right )}{12 \, d^{4}} \] Input:

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

Output:

2*(d*x + c)*a*b/d - 1/2*(a^2 - b^2)*log(tan(d*x + c)^2 + 1)/d + 1/12*(3*b^ 
2*d^3*tan(d*x + c)^4 + 8*a*b*d^3*tan(d*x + c)^3 + 6*a^2*d^3*tan(d*x + c)^2 
 - 6*b^2*d^3*tan(d*x + c)^2 - 24*a*b*d^3*tan(d*x + c))/d^4
 

Mupad [B] (verification not implemented)

Time = 1.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.92 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )-\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}-2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+\frac {2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+2\,a\,b\,d\,x}{d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.07 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {-6 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a^{2}+6 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) b^{2}+3 \tan \left (d x +c \right )^{4} b^{2}+8 \tan \left (d x +c \right )^{3} a b +6 \tan \left (d x +c \right )^{2} a^{2}-6 \tan \left (d x +c \right )^{2} b^{2}-24 \tan \left (d x +c \right ) a b +24 a b d x}{12 d} \] Input:

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

Output:

( - 6*log(tan(c + d*x)**2 + 1)*a**2 + 6*log(tan(c + d*x)**2 + 1)*b**2 + 3* 
tan(c + d*x)**4*b**2 + 8*tan(c + d*x)**3*a*b + 6*tan(c + d*x)**2*a**2 - 6* 
tan(c + d*x)**2*b**2 - 24*tan(c + d*x)*a*b + 24*a*b*d*x)/(12*d)