\(\int (c+d x)^3 \tan (a+b x) \, dx\) [210]

Optimal result

Integrand size = 14, antiderivative size = 132 \[ \int (c+d x)^3 \tan (a+b x) \, dx=\frac {i (c+d x)^4}{4 d}-\frac {(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i d^3 \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4} \] Output:

1/4*I*(d*x+c)^4/d-(d*x+c)^3*ln(1+exp(2*I*(b*x+a)))/b+3/2*I*d*(d*x+c)^2*pol 
ylog(2,-exp(2*I*(b*x+a)))/b^2-3/2*d^2*(d*x+c)*polylog(3,-exp(2*I*(b*x+a))) 
/b^3-3/4*I*d^3*polylog(4,-exp(2*I*(b*x+a)))/b^4
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.95 \[ \int (c+d x)^3 \tan (a+b x) \, dx=\frac {1}{4} i \left (\frac {(c+d x)^4}{d}+\frac {4 i (c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 d \left (2 b^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )+d \left (2 i b (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )-d \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )\right )\right )}{b^4}\right ) \] Input:

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

Output:

(I/4)*((c + d*x)^4/d + ((4*I)*(c + d*x)^3*Log[1 + E^((2*I)*(a + b*x))])/b 
+ (3*d*(2*b^2*(c + d*x)^2*PolyLog[2, -E^((2*I)*(a + b*x))] + d*((2*I)*b*(c 
 + d*x)*PolyLog[3, -E^((2*I)*(a + b*x))] - d*PolyLog[4, -E^((2*I)*(a + b*x 
))])))/b^4)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4202, 2620, 3011, 7163, 2720, 7143}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4202

\(\displaystyle \frac {i (c+d x)^4}{4 d}-2 i \int \frac {e^{2 i (a+b x)} (c+d x)^3}{1+e^{2 i (a+b x)}}dx\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {i (c+d x)^4}{4 d}-2 i \left (\frac {3 i d \int (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {i (c+d x)^4}{4 d}-2 i \left (\frac {3 i d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \int (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {i (c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\)

\(\Big \downarrow \) 7163

\(\displaystyle \frac {i (c+d x)^4}{4 d}-2 i \left (\frac {3 i d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \left (\frac {i d \int \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i (c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {i (c+d x)^4}{4 d}-2 i \left (\frac {3 i d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \left (\frac {d \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i (c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {i (c+d x)^4}{4 d}-2 i \left (\frac {3 i d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \left (\frac {d \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i (c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )\)

Input:

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

Output:

((I/4)*(c + d*x)^4)/d - (2*I)*(((-1/2*I)*(c + d*x)^3*Log[1 + E^((2*I)*(a + 
 b*x))])/b + (((3*I)/2)*d*(((I/2)*(c + d*x)^2*PolyLog[2, -E^((2*I)*(a + b* 
x))])/b - (I*d*(((-1/2*I)*(c + d*x)*PolyLog[3, -E^((2*I)*(a + b*x))])/b + 
(d*PolyLog[4, -E^((2*I)*(a + b*x))])/(4*b^2)))/b))/b)
 

Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 431 vs. \(2 (113 ) = 226\).

Time = 0.37 (sec) , antiderivative size = 432, normalized size of antiderivative = 3.27

method	result	size
risch	\(-\frac {3 d \,c^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x}{b}-\frac {3 c \,d^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {3 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{2 b^{2}}+\frac {3 i d \,c^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}+\frac {3 i c \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {3 i d^{3} \operatorname {polylog}\left (4, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{4 b^{4}}-\frac {3 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{2 b^{3}}-\frac {3 c \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}-\frac {d^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{3}}{b}+\frac {6 i d \,c^{2} a x}{b}-\frac {6 i c \,d^{2} a^{2} x}{b^{2}}-\frac {c^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}+\frac {3 i d^{3} a^{4}}{2 b^{4}}+i d^{2} c \,x^{3}+\frac {3 i d \,c^{2} x^{2}}{2}-\frac {2 d^{3} a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-i c^{3} x -\frac {i c^{4}}{4 d}+\frac {6 c \,d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 c^{2} d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 i d \,c^{2} a^{2}}{b^{2}}+\frac {2 i d^{3} a^{3} x}{b^{3}}-\frac {4 i c \,d^{2} a^{3}}{b^{3}}+\frac {i d^{3} x^{4}}{4}+\frac {2 c^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}\)	\(432\)

Input:

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

Output:

6/b^3*c*d^2*a^2*ln(exp(I*(b*x+a)))+3*I/b^2*c*d^2*polylog(2,-exp(2*I*(b*x+a 
)))*x-6*I/b^2*c*d^2*a^2*x+6*I/b*d*c^2*a*x-I*c^3*x-1/4*I/d*c^4-3/b*d*c^2*ln 
(exp(2*I*(b*x+a))+1)*x-3/b*c*d^2*ln(exp(2*I*(b*x+a))+1)*x^2-4*I/b^3*c*d^2* 
a^3+3*I/b^2*d*c^2*a^2+3/2*I/b^2*d^3*polylog(2,-exp(2*I*(b*x+a)))*x^2+3/2*I 
/b^2*d*c^2*polylog(2,-exp(2*I*(b*x+a)))+2*I/b^3*d^3*a^3*x+I*d^2*c*x^3-3/2/ 
b^3*d^3*polylog(3,-exp(2*I*(b*x+a)))*x-3/2/b^3*c*d^2*polylog(3,-exp(2*I*(b 
*x+a)))+3/2*I/b^4*d^3*a^4-1/b*d^3*ln(exp(2*I*(b*x+a))+1)*x^3+3/2*I*d*c^2*x 
^2+1/4*I*d^3*x^4-1/b*c^3*ln(exp(2*I*(b*x+a))+1)-2/b^4*d^3*a^3*ln(exp(I*(b* 
x+a)))-6/b^2*c^2*d*a*ln(exp(I*(b*x+a)))+2/b*c^3*ln(exp(I*(b*x+a)))-3/4*I*d 
^3*polylog(4,-exp(2*I*(b*x+a)))/b^4
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 432 vs. \(2 (109) = 218\).

Time = 0.09 (sec) , antiderivative size = 432, normalized size of antiderivative = 3.27 \[ \int (c+d x)^3 \tan (a+b x) \, dx=\frac {3 i \, d^{3} {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 3 i \, d^{3} {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 \, {\left (i \, b^{2} d^{3} x^{2} + 2 i \, b^{2} c d^{2} x + i \, b^{2} c^{2} d\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 6 \, {\left (-i \, b^{2} d^{3} x^{2} - 2 i \, b^{2} c d^{2} x - i \, b^{2} c^{2} d\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 4 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 4 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 \, {\left (b d^{3} x + b c d^{2}\right )} {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 \, {\left (b d^{3} x + b c d^{2}\right )} {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{8 \, b^{4}} \] Input:

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

Output:

1/8*(3*I*d^3*polylog(4, (tan(b*x + a)^2 + 2*I*tan(b*x + a) - 1)/(tan(b*x + 
 a)^2 + 1)) - 3*I*d^3*polylog(4, (tan(b*x + a)^2 - 2*I*tan(b*x + a) - 1)/( 
tan(b*x + a)^2 + 1)) - 6*(I*b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c^2*d)*d 
ilog(2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) - 6*(-I*b^2*d^3*x^2 
- 2*I*b^2*c*d^2*x - I*b^2*c^2*d)*dilog(2*(-I*tan(b*x + a) - 1)/(tan(b*x + 
a)^2 + 1) + 1) - 4*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^ 
3)*log(-2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) - 4*(b^3*d^3*x^3 + 3* 
b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*log(-2*(-I*tan(b*x + a) - 1)/(tan 
(b*x + a)^2 + 1)) - 6*(b*d^3*x + b*c*d^2)*polylog(3, (tan(b*x + a)^2 + 2*I 
*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) - 6*(b*d^3*x + b*c*d^2)*polylog(3 
, (tan(b*x + a)^2 - 2*I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)))/b^4
 

Sympy [F]

\[ \int (c+d x)^3 \tan (a+b x) \, dx=\int \left (c + d x\right )^{3} \tan {\left (a + b x \right )}\, dx \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (109) = 218\).

Time = 0.20 (sec) , antiderivative size = 474, normalized size of antiderivative = 3.59 \[ \int (c+d x)^3 \tan (a+b x) \, dx=\frac {12 \, c^{3} \log \left (\sec \left (b x + a\right )\right ) - \frac {36 \, a c^{2} d \log \left (\sec \left (b x + a\right )\right )}{b} + \frac {36 \, a^{2} c d^{2} \log \left (\sec \left (b x + a\right )\right )}{b^{2}} - \frac {12 \, a^{3} d^{3} \log \left (\sec \left (b x + a\right )\right )}{b^{3}} - \frac {-3 i \, {\left (b x + a\right )}^{4} d^{3} - 12 \, {\left (i \, b c d^{2} - i \, a d^{3}\right )} {\left (b x + a\right )}^{3} + 12 i \, d^{3} {\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )}) - 18 \, {\left (i \, b^{2} c^{2} d - 2 i \, a b c d^{2} + i \, a^{2} d^{3}\right )} {\left (b x + a\right )}^{2} - 4 \, {\left (-4 i \, {\left (b x + a\right )}^{3} d^{3} + 9 \, {\left (-i \, b c d^{2} + i \, a d^{3}\right )} {\left (b x + a\right )}^{2} + 9 \, {\left (-i \, b^{2} c^{2} d + 2 i \, a b c d^{2} - i \, a^{2} d^{3}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 6 \, {\left (3 i \, b^{2} c^{2} d - 6 i \, a b c d^{2} + 4 i \, {\left (b x + a\right )}^{2} d^{3} + 3 i \, a^{2} d^{3} + 6 \, {\left (i \, b c d^{2} - i \, a d^{3}\right )} {\left (b x + a\right )}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 \, {\left (4 \, {\left (b x + a\right )}^{3} d^{3} + 9 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}^{2} + 9 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 6 \, {\left (3 \, b c d^{2} + 4 \, {\left (b x + a\right )} d^{3} - 3 \, a d^{3}\right )} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )})}{b^{3}}}{12 \, b} \] Input:

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

Output:

1/12*(12*c^3*log(sec(b*x + a)) - 36*a*c^2*d*log(sec(b*x + a))/b + 36*a^2*c 
*d^2*log(sec(b*x + a))/b^2 - 12*a^3*d^3*log(sec(b*x + a))/b^3 - (-3*I*(b*x 
 + a)^4*d^3 - 12*(I*b*c*d^2 - I*a*d^3)*(b*x + a)^3 + 12*I*d^3*polylog(4, - 
e^(2*I*b*x + 2*I*a)) - 18*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + I*a^2*d^3)*(b*x + 
 a)^2 - 4*(-4*I*(b*x + a)^3*d^3 + 9*(-I*b*c*d^2 + I*a*d^3)*(b*x + a)^2 + 9 
*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 - I*a^2*d^3)*(b*x + a))*arctan2(sin(2*b*x + 
 2*a), cos(2*b*x + 2*a) + 1) - 6*(3*I*b^2*c^2*d - 6*I*a*b*c*d^2 + 4*I*(b*x 
 + a)^2*d^3 + 3*I*a^2*d^3 + 6*(I*b*c*d^2 - I*a*d^3)*(b*x + a))*dilog(-e^(2 
*I*b*x + 2*I*a)) + 2*(4*(b*x + a)^3*d^3 + 9*(b*c*d^2 - a*d^3)*(b*x + a)^2 
+ 9*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x + a))*log(cos(2*b*x + 2*a)^2 
+ sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 6*(3*b*c*d^2 + 4*(b*x + a 
)*d^3 - 3*a*d^3)*polylog(3, -e^(2*I*b*x + 2*I*a)))/b^3)/b
 

Giac [F]

\[ \int (c+d x)^3 \tan (a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \tan \left (b x + a\right ) \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \tan (a+b x) \, dx=\int \mathrm {tan}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^3 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (c+d x)^3 \tan (a+b x) \, dx=\frac {2 \left (\int \tan \left (b x +a \right ) x^{3}d x \right ) b \,d^{3}+6 \left (\int \tan \left (b x +a \right ) x^{2}d x \right ) b c \,d^{2}+6 \left (\int \tan \left (b x +a \right ) x d x \right ) b \,c^{2} d +\mathrm {log}\left (\tan \left (b x +a \right )^{2}+1\right ) c^{3}}{2 b} \] Input:

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

Output:

(2*int(tan(a + b*x)*x**3,x)*b*d**3 + 6*int(tan(a + b*x)*x**2,x)*b*c*d**2 + 
 6*int(tan(a + b*x)*x,x)*b*c**2*d + log(tan(a + b*x)**2 + 1)*c**3)/(2*b)