3.1.29 \(\int (c+d x)^3 \sec (a+b x) \, dx\) [29]

3.1.29.1 Optimal result

Integrand size = 14, antiderivative size = 205 \[ \int (c+d x)^3 \sec (a+b x) \, dx=-\frac {2 i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^4}+\frac {6 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^4} \]

output

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

3.1.29.2 Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.96 \[ \int (c+d x)^3 \sec (a+b x) \, dx=-\frac {i \left (2 b^3 (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )-3 d \left (b^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )+2 i b d (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )-2 d^2 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )\right )+3 d \left (b^2 (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )+2 i b d (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )-2 d^2 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )\right )\right )}{b^4} \]

input

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

output

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

3.1.29.3 Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4669, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

input

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

output

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

3.1.29.3.1 Defintions of rubi rules used

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 4669

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol 
] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Si 
mp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], 
 x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x 
))], x], x]) /; FreeQ[{c, d, e, f}, x] && IntegerQ[2*k] && IGtQ[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]
 

3.1.29.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 684 vs. \(2 (180 ) = 360\).

Time = 1.73 (sec) , antiderivative size = 685, normalized size of antiderivative = 3.34

method	result	size
risch	\(\frac {3 c^{2} d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {3 c^{2} d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {3 i c^{2} d \,\operatorname {Li}_{2}\left (i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 i d^{3} \operatorname {Li}_{2}\left (-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {3 i d^{3} \operatorname {Li}_{2}\left (i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}+\frac {2 i d^{3} a^{3} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 i c^{2} d \,\operatorname {Li}_{2}\left (-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {3 c^{2} d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {6 i d^{2} c \,\operatorname {Li}_{2}\left (i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {6 i d^{3} \operatorname {Li}_{4}\left (i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 i d^{3} \operatorname {Li}_{4}\left (-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 d^{3} \operatorname {Li}_{3}\left (-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {d^{3} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{3}}{b}+\frac {6 d^{3} \operatorname {Li}_{3}\left (i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {a^{3} d^{3} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {a^{3} d^{3} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {d^{3} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{3}}{b}+\frac {6 d^{2} c \,\operatorname {Li}_{3}\left (i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 d^{2} c \,\operatorname {Li}_{3}\left (-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i c^{3} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {6 i d^{2} c \,\operatorname {Li}_{2}\left (-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {6 i c \,d^{2} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 i c^{2} d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {3 a^{2} d^{2} c \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {3 d^{2} c \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {3 a^{2} d^{2} c \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {3 d^{2} c \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {3 c^{2} d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}\)	\(685\)

input

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

output

6*I/b^2*d^2*c*polylog(2,-I*exp(I*(b*x+a)))*x-6*I/b^3*c*d^2*a^2*arctan(exp( 
I*(b*x+a)))+6*I/b^2*c^2*d*a*arctan(exp(I*(b*x+a)))-6*I/b^2*d^2*c*polylog(2 
,I*exp(I*(b*x+a)))*x-3/b^2*c^2*d*ln(1+I*exp(I*(b*x+a)))*a+3/b*c^2*d*ln(1-I 
*exp(I*(b*x+a)))*x+3/b^2*c^2*d*ln(1-I*exp(I*(b*x+a)))*a-3*I/b^2*c^2*d*poly 
log(2,I*exp(I*(b*x+a)))+3*I/b^2*d^3*polylog(2,-I*exp(I*(b*x+a)))*x^2-3*I/b 
^2*d^3*polylog(2,I*exp(I*(b*x+a)))*x^2+2*I/b^4*d^3*a^3*arctan(exp(I*(b*x+a 
)))+3*I/b^2*c^2*d*polylog(2,-I*exp(I*(b*x+a)))-6/b^3*d^3*polylog(3,-I*exp( 
I*(b*x+a)))*x+1/b*d^3*ln(1-I*exp(I*(b*x+a)))*x^3+6/b^3*d^3*polylog(3,I*exp 
(I*(b*x+a)))*x-1/b^4*a^3*d^3*ln(1+I*exp(I*(b*x+a)))+1/b^4*a^3*d^3*ln(1-I*e 
xp(I*(b*x+a)))-1/b*d^3*ln(1+I*exp(I*(b*x+a)))*x^3+6/b^3*d^2*c*polylog(3,I* 
exp(I*(b*x+a)))-6/b^3*d^2*c*polylog(3,-I*exp(I*(b*x+a)))-2*I/b*c^3*arctan( 
exp(I*(b*x+a)))-3/b^3*a^2*d^2*c*ln(1-I*exp(I*(b*x+a)))-3/b*d^2*c*ln(1+I*ex 
p(I*(b*x+a)))*x^2+3/b^3*a^2*d^2*c*ln(1+I*exp(I*(b*x+a)))+3/b*d^2*c*ln(1-I* 
exp(I*(b*x+a)))*x^2-3/b*c^2*d*ln(1+I*exp(I*(b*x+a)))*x+6*I*d^3*polylog(4,I 
*exp(I*(b*x+a)))/b^4-6*I*d^3*polylog(4,-I*exp(I*(b*x+a)))/b^4
 

3.1.29.5 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. 970 vs. \(2 (167) = 334\).

Time = 0.33 (sec) , antiderivative size = 970, normalized size of antiderivative = 4.73 \[ \int (c+d x)^3 \sec (a+b x) \, dx=\text {Too large to display} \]

input

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

output

1/2*(6*I*d^3*polylog(4, I*cos(b*x + a) + sin(b*x + a)) + 6*I*d^3*polylog(4 
, I*cos(b*x + a) - sin(b*x + a)) - 6*I*d^3*polylog(4, -I*cos(b*x + a) + si 
n(b*x + a)) - 6*I*d^3*polylog(4, -I*cos(b*x + a) - sin(b*x + a)) - 3*(I*b^ 
2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c^2*d)*dilog(I*cos(b*x + a) + sin(b*x 
+ a)) - 3*(I*b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c^2*d)*dilog(I*cos(b*x 
+ a) - sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^2*d)* 
dilog(-I*cos(b*x + a) + sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d^2* 
x - I*b^2*c^2*d)*dilog(-I*cos(b*x + a) - sin(b*x + a)) + (b^3*c^3 - 3*a*b^ 
2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cos(b*x + a) + I*sin(b*x + a) + I) 
- (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cos(b*x + a) - I 
*sin(b*x + a) + I) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a* 
b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(I*cos(b*x + a) + sin(b*x + a) + 1 
) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2 
*b*c*d^2 + a^3*d^3)*log(I*cos(b*x + a) - sin(b*x + a) + 1) + (b^3*d^3*x^3 
+ 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^ 
3)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^ 
2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(-I*cos(b* 
x + a) - sin(b*x + a) + 1) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^ 
3*d^3)*log(-cos(b*x + a) + I*sin(b*x + a) + I) - (b^3*c^3 - 3*a*b^2*c^2*d 
+ 3*a^2*b*c*d^2 - a^3*d^3)*log(-cos(b*x + a) - I*sin(b*x + a) + I) - 6*...
 

3.1.29.6 Sympy [F]

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

input

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

output

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

3.1.29.7 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. 722 vs. \(2 (167) = 334\).

Time = 0.48 (sec) , antiderivative size = 722, normalized size of antiderivative = 3.52 \[ \int (c+d x)^3 \sec (a+b x) \, dx=\frac {2 \, c^{3} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right ) - \frac {6 \, a c^{2} d \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b} + \frac {6 \, a^{2} c d^{2} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b^{2}} - \frac {2 \, a^{3} d^{3} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b^{3}} + \frac {12 i \, d^{3} {\rm Li}_{4}(i \, e^{\left (i \, b x + i \, a\right )}) - 12 i \, d^{3} {\rm Li}_{4}(-i \, e^{\left (i \, b x + i \, a\right )}) - 2 \, {\left (i \, {\left (b x + a\right )}^{3} d^{3} + 3 \, {\left (i \, b c d^{2} - i \, a d^{3}\right )} {\left (b x + a\right )}^{2} + 3 \, {\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 (\cos \left (b x + a\right ), \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (i \, {\left (b x + a\right )}^{3} d^{3} + 3 \, {\left (i \, b c d^{2} - i \, a d^{3}\right )} {\left (b x + a\right )}^{2} + 3 \, {\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 (\cos \left (b x + a\right ), -\sin \left (b x + a\right ) + 1\right ) - 6 \, {\left (i \, b^{2} c^{2} d - 2 i \, a b c d^{2} + i \, {\left (b x + a\right )}^{2} d^{3} + i \, a^{2} d^{3} + 2 \, {\left (i \, b c d^{2} - i \, a d^{3}\right )} {\left (b x + a\right )}\right )} {\rm Li}_2\left (i \, e^{\left (i \, b x + i \, a\right )}\right ) - 6 \, {\left (-i \, b^{2} c^{2} d + 2 i \, a b c d^{2} - i \, {\left (b x + a\right )}^{2} d^{3} - i \, a^{2} d^{3} + 2 \, {\left (-i \, b c d^{2} + i \, a d^{3}\right )} {\left (b x + a\right )}\right )} {\rm Li}_2\left (-i \, e^{\left (i \, b x + i \, a\right )}\right ) + {\left ({\left (b x + a\right )}^{3} d^{3} + 3 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}^{2} + 3 \, {\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 (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \sin \left (b x + a\right ) + 1\right ) - {\left ({\left (b x + a\right )}^{3} d^{3} + 3 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}^{2} + 3 \, {\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 (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \sin \left (b x + a\right ) + 1\right ) + 12 \, {\left (b c d^{2} + {\left (b x + a\right )} d^{3} - a d^{3}\right )} {\rm Li}_{3}(i \, e^{\left (i \, b x + i \, a\right )}) - 12 \, {\left (b c d^{2} + {\left (b x + a\right )} d^{3} - a d^{3}\right )} {\rm Li}_{3}(-i \, e^{\left (i \, b x + i \, a\right )})}{b^{3}}}{2 \, b} \]

input

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

output

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

3.1.29.8 Giac [F]

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

input

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

output

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

3.1.29.9 Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \sec (a+b x) \, dx=\int \frac {{\left (c+d\,x\right )}^3}{\cos \left (a+b\,x\right )} \,d x \]

input

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

output

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