\(\int (c+d x)^3 \sec ^3(a+b x) \, dx\) [37]

Optimal result

Integrand size = 16, antiderivative size = 337 \[ \int (c+d x)^3 \sec ^3(a+b x) \, dx=-\frac {6 i d^2 (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^3}-\frac {i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^4}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^4}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}-\frac {3 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^4}+\frac {3 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^4}-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \sec (a+b x) \tan (a+b x)}{2 b} \] Output:

-6*I*d^2*(d*x+c)*arctan(exp(I*(b*x+a)))/b^3-I*(d*x+c)^3*arctan(exp(I*(b*x+ 
a)))/b+3*I*d^3*polylog(2,-I*exp(I*(b*x+a)))/b^4+3/2*I*d*(d*x+c)^2*polylog( 
2,-I*exp(I*(b*x+a)))/b^2-3*I*d^3*polylog(2,I*exp(I*(b*x+a)))/b^4-3/2*I*d*( 
d*x+c)^2*polylog(2,I*exp(I*(b*x+a)))/b^2-3*d^2*(d*x+c)*polylog(3,-I*exp(I* 
(b*x+a)))/b^3+3*d^2*(d*x+c)*polylog(3,I*exp(I*(b*x+a)))/b^3-3*I*d^3*polylo 
g(4,-I*exp(I*(b*x+a)))/b^4+3*I*d^3*polylog(4,I*exp(I*(b*x+a)))/b^4-3/2*d*( 
d*x+c)^2*sec(b*x+a)/b^2+1/2*(d*x+c)^3*sec(b*x+a)*tan(b*x+a)/b
 

Mathematica [A] (verified)

Time = 3.31 (sec) , antiderivative size = 311, normalized size of antiderivative = 0.92 \[ \int (c+d x)^3 \sec ^3(a+b x) \, dx=\frac {-2 i b^3 (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )-6 i d^2 \left (2 b (c+d x) \arctan \left (e^{i (a+b x)}\right )-d \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )+d \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )\right )+3 i d \left (b^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )+2 i b d (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )-2 d^2 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )\right )-3 i d \left (b^2 (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )+2 i b d (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )-2 d^2 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )\right )-3 b^2 d (c+d x)^2 \sec (a+b x)+b^3 (c+d x)^3 \sec (a+b x) \tan (a+b x)}{2 b^4} \] Input:

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

Output:

((-2*I)*b^3*(c + d*x)^3*ArcTan[E^(I*(a + b*x))] - (6*I)*d^2*(2*b*(c + d*x) 
*ArcTan[E^(I*(a + b*x))] - d*PolyLog[2, (-I)*E^(I*(a + b*x))] + d*PolyLog[ 
2, I*E^(I*(a + b*x))]) + (3*I)*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*Po 
lyLog[4, (-I)*E^(I*(a + b*x))]) - (3*I)*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))]) - 3*b^2*d*(c + d*x)^2*Sec[a + b*x] + b^3*( 
c + d*x)^3*Sec[a + b*x]*Tan[a + b*x])/(2*b^4)
 

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3042, 4674, 3042, 4669, 2715, 2838, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4674

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 d^2 \int (c+d x) \csc \left (a+b x+\frac {\pi }{2}\right )dx}{b^2}+\frac {1}{2} \int (c+d x)^3 \csc \left (a+b x+\frac {\pi }{2}\right )dx-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \tan (a+b x) \sec (a+b x)}{2 b}\)

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

(3*d^2*(((-2*I)*(c + d*x)*ArcTan[E^(I*(a + b*x))])/b + (I*d*PolyLog[2, (-I 
)*E^(I*(a + b*x))])/b^2 - (I*d*PolyLog[2, I*E^(I*(a + b*x))])/b^2))/b^2 + 
(((-2*I)*(c + d*x)^3*ArcTan[E^(I*(a + b*x))])/b + (3*d*((I*(c + d*x)^2*Pol 
yLog[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)/2 - (3*d*(c + d*x)^2*Sec[a + b*x])/(2*b^2) + ((c + d*x)^ 
3*Sec[a + b*x]*Tan[a + b*x])/(2*b)
 

Defintions of rubi rules used

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

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 4674

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbo 
l] :> Simp[(-b^2)*(c + d*x)^m*Cot[e + f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n 
- 1))), x] + (-Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^ 
2*(n - 1)*(n - 2))), x] + Simp[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))) 
Int[(c + d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Simp[b^2*((n - 2)/ 
(n - 1))   Int[(c + d*x)^m*(b*Csc[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c 
, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]
 

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. 1126 vs. \(2 (293 ) = 586\).

Time = 1.88 (sec) , antiderivative size = 1127, normalized size of antiderivative = 3.34

method	result	size
risch	\(\text {Expression too large to display}\)	\(1127\)

Input:

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

Output:

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

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. 1315 vs. \(2 (273) = 546\).

Time = 0.16 (sec) , antiderivative size = 1315, normalized size of antiderivative = 3.90 \[ \int (c+d x)^3 \sec ^3(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((c + d*x)**3*sec(a + b*x)**3, 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. 3831 vs. \(2 (273) = 546\).

Time = 1.12 (sec) , antiderivative size = 3831, normalized size of antiderivative = 11.37 \[ \int (c+d x)^3 \sec ^3(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(c^3*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) - log(sin(b*x + a) + 1) + l 
og(sin(b*x + a) - 1)) - 3*a*c^2*d*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) - l 
og(sin(b*x + a) + 1) + log(sin(b*x + a) - 1))/b + 3*a^2*c*d^2*(2*sin(b*x + 
 a)/(sin(b*x + a)^2 - 1) - log(sin(b*x + a) + 1) + log(sin(b*x + a) - 1))/ 
b^2 - a^3*d^3*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) - log(sin(b*x + a) + 1) 
 + log(sin(b*x + a) - 1))/b^3 + 4*(2*((b*x + a)^3*d^3 + 6*b*c*d^2 - 6*a*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 + 
2)*d^3)*(b*x + a) + ((b*x + a)^3*d^3 + 6*b*c*d^2 - 6*a*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 + 2)*d^3)*(b*x + a) 
)*cos(4*b*x + 4*a) + 2*((b*x + a)^3*d^3 + 6*b*c*d^2 - 6*a*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 + 2)*d^3)*(b*x + 
 a))*cos(2*b*x + 2*a) + (I*(b*x + a)^3*d^3 + 6*I*b*c*d^2 - 6*I*a*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 
 + 2*I)*d^3)*(b*x + a))*sin(4*b*x + 4*a) + 2*(I*(b*x + a)^3*d^3 + 6*I*b*c* 
d^2 - 6*I*a*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 + 2*I)*d^3)*(b*x + a))*sin(2*b*x + 2*a))*arctan2(cos 
(b*x + a), sin(b*x + a) + 1) + 2*((b*x + a)^3*d^3 + 6*b*c*d^2 - 6*a*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 + 2)*d 
^3)*(b*x + a) + ((b*x + a)^3*d^3 + 6*b*c*d^2 - 6*a*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 + 2)*d^3)*(b*x + a))...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \sec ^3(a+b x) \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int (c+d x)^3 \sec ^3(a+b x) \, dx=\text {too large to display} \] Input:

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

Output:

(252*cos(a + b*x)*sin(a + b*x)*b**3*c**2*d*x + 252*cos(a + b*x)*sin(a + b* 
x)*b**3*c*d**2*x**2 + 84*cos(a + b*x)*sin(a + b*x)*b**3*d**3*x**3 - 252*co 
s(a + b*x)*sin(a + b*x)*b*c*d**2 + 1044*cos(a + b*x)*sin(a + b*x)*b*d**3*x 
 + 252*cos(a + b*x)*b**2*c**2*d + 1512*cos(a + b*x)*b**2*c*d**2*x + 756*co 
s(a + b*x)*b**2*d**3*x**2 - 216*cos(a + b*x)*d**3 - 224*int(x**3/(tan((a + 
 b*x)/2)**6 - 3*tan((a + b*x)/2)**4 + 3*tan((a + b*x)/2)**2 - 1),x)*sin(a 
+ b*x)**2*b**4*d**3 + 224*int(x**3/(tan((a + b*x)/2)**6 - 3*tan((a + b*x)/ 
2)**4 + 3*tan((a + b*x)/2)**2 - 1),x)*b**4*d**3 - 672*int(x**2/(tan((a + b 
*x)/2)**6 - 3*tan((a + b*x)/2)**4 + 3*tan((a + b*x)/2)**2 - 1),x)*sin(a + 
b*x)**2*b**4*c*d**2 + 672*int(x**2/(tan((a + b*x)/2)**6 - 3*tan((a + b*x)/ 
2)**4 + 3*tan((a + b*x)/2)**2 - 1),x)*b**4*c*d**2 - 2016*int((tan((a + b*x 
)/2)*x**2)/(tan((a + b*x)/2)**6 - 3*tan((a + b*x)/2)**4 + 3*tan((a + b*x)/ 
2)**2 - 1),x)*sin(a + b*x)**2*b**3*d**3 + 2016*int((tan((a + b*x)/2)*x**2) 
/(tan((a + b*x)/2)**6 - 3*tan((a + b*x)/2)**4 + 3*tan((a + b*x)/2)**2 - 1) 
,x)*b**3*d**3 - 4032*int((tan((a + b*x)/2)*x)/(tan((a + b*x)/2)**6 - 3*tan 
((a + b*x)/2)**4 + 3*tan((a + b*x)/2)**2 - 1),x)*sin(a + b*x)**2*b**3*c*d* 
*2 + 4032*int((tan((a + b*x)/2)*x)/(tan((a + b*x)/2)**6 - 3*tan((a + b*x)/ 
2)**4 + 3*tan((a + b*x)/2)**2 - 1),x)*b**3*c*d**2 - 672*int(x/(tan((a + b* 
x)/2)**6 - 3*tan((a + b*x)/2)**4 + 3*tan((a + b*x)/2)**2 - 1),x)*sin(a + b 
*x)**2*b**4*c**2*d - 3456*int(x/(tan((a + b*x)/2)**6 - 3*tan((a + b*x)/...