\(\int (a+b \tan (c+d x))^4 \, dx\) [448]

Optimal result

Integrand size = 12, antiderivative size = 103 \[ \int (a+b \tan (c+d x))^4 \, dx=\left (a^4-6 a^2 b^2+b^4\right ) x-\frac {4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}+\frac {a b (a+b \tan (c+d x))^2}{d}+\frac {b (a+b \tan (c+d x))^3}{3 d} \] Output:

(a^4-6*a^2*b^2+b^4)*x-4*a*b*(a^2-b^2)*ln(cos(d*x+c))/d+b^2*(3*a^2-b^2)*tan 
(d*x+c)/d+a*b*(a+b*tan(d*x+c))^2/d+1/3*b*(a+b*tan(d*x+c))^3/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02 \[ \int (a+b \tan (c+d x))^4 \, dx=\frac {-3 i (a+i b)^4 \log (i-\tan (c+d x))+3 i (a-i b)^4 \log (i+\tan (c+d x))-6 b^2 \left (-6 a^2+b^2\right ) \tan (c+d x)+12 a b^3 \tan ^2(c+d x)+2 b^4 \tan ^3(c+d x)}{6 d} \] Input:

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

Output:

((-3*I)*(a + I*b)^4*Log[I - Tan[c + d*x]] + (3*I)*(a - I*b)^4*Log[I + Tan[ 
c + d*x]] - 6*b^2*(-6*a^2 + b^2)*Tan[c + d*x] + 12*a*b^3*Tan[c + d*x]^2 + 
2*b^4*Tan[c + d*x]^3)/(6*d)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3963, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3963

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4008

\(\displaystyle 4 a b \left (a^2-b^2\right ) \int \tan (c+d x)dx+\frac {b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}+x \left (a^4-6 a^2 b^2+b^4\right )+\frac {b (a+b \tan (c+d x))^3}{3 d}+\frac {a b (a+b \tan (c+d x))^2}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle 4 a b \left (a^2-b^2\right ) \int \tan (c+d x)dx+\frac {b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}+x \left (a^4-6 a^2 b^2+b^4\right )+\frac {b (a+b \tan (c+d x))^3}{3 d}+\frac {a b (a+b \tan (c+d x))^2}{d}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}-\frac {4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+x \left (a^4-6 a^2 b^2+b^4\right )+\frac {b (a+b \tan (c+d x))^3}{3 d}+\frac {a b (a+b \tan (c+d x))^2}{d}\)

Input:

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

Output:

(a^4 - 6*a^2*b^2 + b^4)*x - (4*a*b*(a^2 - b^2)*Log[Cos[c + d*x]])/d + (b^2 
*(3*a^2 - b^2)*Tan[c + d*x])/d + (a*b*(a + b*Tan[c + d*x])^2)/d + (b*(a + 
b*Tan[c + d*x])^3)/(3*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 3963

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

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]
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00

method	result	size
norman	\(\left (a^{4}-6 b^{2} a^{2}+b^{4}\right ) x +\frac {b^{2} \left (6 a^{2}-b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{4} \tan \left (d x +c \right )^{3}}{3 d}+\frac {2 a \,b^{3} \tan \left (d x +c \right )^{2}}{d}+\frac {2 a b \left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\)	\(103\)
derivativedivides	\(\frac {\frac {b^{4} \tan \left (d x +c \right )^{3}}{3}+2 a \,b^{3} \tan \left (d x +c \right )^{2}+6 b^{2} a^{2} \tan \left (d x +c \right )-b^{4} \tan \left (d x +c \right )+\frac {\left (4 a^{3} b -4 a \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (a^{4}-6 b^{2} a^{2}+b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(107\)
default	\(\frac {\frac {b^{4} \tan \left (d x +c \right )^{3}}{3}+2 a \,b^{3} \tan \left (d x +c \right )^{2}+6 b^{2} a^{2} \tan \left (d x +c \right )-b^{4} \tan \left (d x +c \right )+\frac {\left (4 a^{3} b -4 a \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (a^{4}-6 b^{2} a^{2}+b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(107\)
parallelrisch	\(\frac {b^{4} \tan \left (d x +c \right )^{3}+3 a^{4} x d -18 a^{2} b^{2} d x +3 b^{4} d x +6 a \,b^{3} \tan \left (d x +c \right )^{2}+6 \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a^{3} b -6 \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a \,b^{3}+18 b^{2} a^{2} \tan \left (d x +c \right )-3 b^{4} \tan \left (d x +c \right )}{3 d}\)	\(116\)
parts	\(a^{4} x +\frac {b^{4} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {2 a \,b^{3} \tan \left (d x +c \right )^{2}}{d}-\frac {2 a \,b^{3} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}+\frac {2 a^{3} b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}+\frac {6 b^{2} a^{2} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(124\)
risch	\(4 i a^{3} b x -4 i a \,b^{3} x +a^{4} x -6 a^{2} b^{2} x +b^{4} x +\frac {8 i a^{3} b c}{d}-\frac {8 i a \,b^{3} c}{d}-\frac {4 i b^{2} \left (-9 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-18 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-9 a^{2}+2 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {4 a^{3} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(218\)

Input:

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

Output:

(a^4-6*a^2*b^2+b^4)*x+b^2*(6*a^2-b^2)/d*tan(d*x+c)+1/3*b^4*tan(d*x+c)^3/d+ 
2*a*b^3*tan(d*x+c)^2/d+2*a*b*(a^2-b^2)/d*ln(1+tan(d*x+c)^2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int (a+b \tan (c+d x))^4 \, dx=\frac {b^{4} \tan \left (d x + c\right )^{3} + 6 \, a b^{3} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} d x - 6 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 3 \, {\left (6 \, a^{2} b^{2} - b^{4}\right )} \tan \left (d x + c\right )}{3 \, d} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.27 \[ \int (a+b \tan (c+d x))^4 \, dx=\begin {cases} a^{4} x + \frac {2 a^{3} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - 6 a^{2} b^{2} x + \frac {6 a^{2} b^{2} \tan {\left (c + d x \right )}}{d} - \frac {2 a b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {2 a b^{3} \tan ^{2}{\left (c + d x \right )}}{d} + b^{4} x + \frac {b^{4} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{4} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right )^{4} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a**4*x + 2*a**3*b*log(tan(c + d*x)**2 + 1)/d - 6*a**2*b**2*x + 
6*a**2*b**2*tan(c + d*x)/d - 2*a*b**3*log(tan(c + d*x)**2 + 1)/d + 2*a*b** 
3*tan(c + d*x)**2/d + b**4*x + b**4*tan(c + d*x)**3/(3*d) - b**4*tan(c + d 
*x)/d, Ne(d, 0)), (x*(a + b*tan(c))**4, True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.10 \[ \int (a+b \tan (c+d x))^4 \, dx=a^{4} x - \frac {6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} b^{2}}{d} + \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} b^{4}}{3 \, d} - \frac {2 \, a b^{3} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{d} + \frac {4 \, a^{3} b \log \left (\sec \left (d x + c\right )\right )}{d} \] Input:

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

Output:

a^4*x - 6*(d*x + c - tan(d*x + c))*a^2*b^2/d + 1/3*(tan(d*x + c)^3 + 3*d*x 
 + 3*c - 3*tan(d*x + c))*b^4/d - 2*a*b^3*(1/(sin(d*x + c)^2 - 1) - log(sin 
(d*x + c)^2 - 1))/d + 4*a^3*b*log(sec(d*x + c))/d
                                                                                    
                                                                                    
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.18 \[ \int (a+b \tan (c+d x))^4 \, dx=\frac {{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )}}{d} + \frac {2 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{d} + \frac {b^{4} d^{2} \tan \left (d x + c\right )^{3} + 6 \, a b^{3} d^{2} \tan \left (d x + c\right )^{2} + 18 \, a^{2} b^{2} d^{2} \tan \left (d x + c\right ) - 3 \, b^{4} d^{2} \tan \left (d x + c\right )}{3 \, d^{3}} \] Input:

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

Output:

(a^4 - 6*a^2*b^2 + b^4)*(d*x + c)/d + 2*(a^3*b - a*b^3)*log(tan(d*x + c)^2 
 + 1)/d + 1/3*(b^4*d^2*tan(d*x + c)^3 + 6*a*b^3*d^2*tan(d*x + c)^2 + 18*a^ 
2*b^2*d^2*tan(d*x + c) - 3*b^4*d^2*tan(d*x + c))/d^3
 

Mupad [B] (verification not implemented)

Time = 1.02 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.62 \[ \int (a+b \tan (c+d x))^4 \, dx=\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (b^4-6\,a^2\,b^2\right )}{d}-\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (2\,a\,b^3-2\,a^3\,b\right )}{d}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-a^2+2\,a\,b+b^2\right )\,\left (a^2+2\,a\,b-b^2\right )}{a^4-6\,a^2\,b^2+b^4}\right )\,\left (-a^2+2\,a\,b+b^2\right )\,\left (a^2+2\,a\,b-b^2\right )}{d}+\frac {2\,a\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.12 \[ \int (a+b \tan (c+d x))^4 \, dx=\frac {6 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a^{3} b -6 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a \,b^{3}+\tan \left (d x +c \right )^{3} b^{4}+6 \tan \left (d x +c \right )^{2} a \,b^{3}+18 \tan \left (d x +c \right ) a^{2} b^{2}-3 \tan \left (d x +c \right ) b^{4}+3 a^{4} d x -18 a^{2} b^{2} d x +3 b^{4} d x}{3 d} \] Input:

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

Output:

(6*log(tan(c + d*x)**2 + 1)*a**3*b - 6*log(tan(c + d*x)**2 + 1)*a*b**3 + t 
an(c + d*x)**3*b**4 + 6*tan(c + d*x)**2*a*b**3 + 18*tan(c + d*x)*a**2*b**2 
 - 3*tan(c + d*x)*b**4 + 3*a**4*d*x - 18*a**2*b**2*d*x + 3*b**4*d*x)/(3*d)