Integrand size = 16, antiderivative size = 296 \[ \int (c+d x)^3 \text {sech}^3(a+b x) \, dx=-\frac {6 d^2 (c+d x) \arctan \left (e^{a+b x}\right )}{b^3}+\frac {(c+d x)^3 \arctan \left (e^{a+b x}\right )}{b}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^4}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{2 b^2}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^4}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{2 b^2}+\frac {3 i d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}-\frac {3 i d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{a+b x}\right )}{b^3}-\frac {3 i d^3 \operatorname {PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}+\frac {3 i d^3 \operatorname {PolyLog}\left (4,i e^{a+b x}\right )}{b^4}+\frac {3 d (c+d x)^2 \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \text {sech}(a+b x) \tanh (a+b x)}{2 b} \]
-6*d^2*(d*x+c)*arctan(exp(b*x+a))/b^3+(d*x+c)^3*arctan(exp(b*x+a))/b+3*I*d ^3*polylog(2,-I*exp(b*x+a))/b^4-3/2*I*d*(d*x+c)^2*polylog(2,-I*exp(b*x+a)) /b^2-3*I*d^3*polylog(2,I*exp(b*x+a))/b^4+3/2*I*d*(d*x+c)^2*polylog(2,I*exp (b*x+a))/b^2+3*I*d^2*(d*x+c)*polylog(3,-I*exp(b*x+a))/b^3-3*I*d^2*(d*x+c)* polylog(3,I*exp(b*x+a))/b^3-3*I*d^3*polylog(4,-I*exp(b*x+a))/b^4+3*I*d^3*p olylog(4,I*exp(b*x+a))/b^4+3/2*d*(d*x+c)^2*sech(b*x+a)/b^2+1/2*(d*x+c)^3*s ech(b*x+a)*tanh(b*x+a)/b
Time = 6.32 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.54 \[ \int (c+d x)^3 \text {sech}^3(a+b x) \, dx=\frac {i \left (-2 i b^3 c^3 \arctan \left (e^{a+b x}\right )+12 i b c d^2 \arctan \left (e^{a+b x}\right )+3 b^3 c^2 d x \log \left (1-i e^{a+b x}\right )-6 b d^3 x \log \left (1-i e^{a+b x}\right )+3 b^3 c d^2 x^2 \log \left (1-i e^{a+b x}\right )+b^3 d^3 x^3 \log \left (1-i e^{a+b x}\right )-3 b^3 c^2 d x \log \left (1+i e^{a+b x}\right )+6 b d^3 x \log \left (1+i e^{a+b x}\right )-3 b^3 c d^2 x^2 \log \left (1+i e^{a+b x}\right )-b^3 d^3 x^3 \log \left (1+i e^{a+b x}\right )-3 d \left (-2 d^2+b^2 (c+d x)^2\right ) \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )+3 d \left (-2 d^2+b^2 (c+d x)^2\right ) \operatorname {PolyLog}\left (2,i e^{a+b x}\right )+6 b c d^2 \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )+6 b d^3 x \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )-6 b c d^2 \operatorname {PolyLog}\left (3,i e^{a+b x}\right )-6 b d^3 x \operatorname {PolyLog}\left (3,i e^{a+b x}\right )-6 d^3 \operatorname {PolyLog}\left (4,-i e^{a+b x}\right )+6 d^3 \operatorname {PolyLog}\left (4,i e^{a+b x}\right )\right )+b^2 (c+d x)^2 \text {sech}(a+b x) (3 d+b (c+d x) \tanh (a+b x))}{2 b^4} \]
(I*((-2*I)*b^3*c^3*ArcTan[E^(a + b*x)] + (12*I)*b*c*d^2*ArcTan[E^(a + b*x) ] + 3*b^3*c^2*d*x*Log[1 - I*E^(a + b*x)] - 6*b*d^3*x*Log[1 - I*E^(a + b*x) ] + 3*b^3*c*d^2*x^2*Log[1 - I*E^(a + b*x)] + b^3*d^3*x^3*Log[1 - I*E^(a + b*x)] - 3*b^3*c^2*d*x*Log[1 + I*E^(a + b*x)] + 6*b*d^3*x*Log[1 + I*E^(a + b*x)] - 3*b^3*c*d^2*x^2*Log[1 + I*E^(a + b*x)] - b^3*d^3*x^3*Log[1 + I*E^( a + b*x)] - 3*d*(-2*d^2 + b^2*(c + d*x)^2)*PolyLog[2, (-I)*E^(a + b*x)] + 3*d*(-2*d^2 + b^2*(c + d*x)^2)*PolyLog[2, I*E^(a + b*x)] + 6*b*c*d^2*PolyL og[3, (-I)*E^(a + b*x)] + 6*b*d^3*x*PolyLog[3, (-I)*E^(a + b*x)] - 6*b*c*d ^2*PolyLog[3, I*E^(a + b*x)] - 6*b*d^3*x*PolyLog[3, I*E^(a + b*x)] - 6*d^3 *PolyLog[4, (-I)*E^(a + b*x)] + 6*d^3*PolyLog[4, I*E^(a + b*x)]) + b^2*(c + d*x)^2*Sech[a + b*x]*(3*d + b*(c + d*x)*Tanh[a + b*x]))/(2*b^4)
Time = 1.23 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.03, 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, 4668, 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.
| \(\displaystyle \int (c+d x)^3 \text {sech}^3(a+b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d x)^3 \csc \left (i a+i b x+\frac {\pi }{2}\right )^3dx\) |
| \(\Big \downarrow \) 4674 |
| \(\displaystyle -\frac {3 d^2 \int (c+d x) \text {sech}(a+b x)dx}{b^2}+\frac {1}{2} \int (c+d x)^3 \text {sech}(a+b x)dx+\frac {3 d (c+d x)^2 \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \tanh (a+b x) \text {sech}(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 d^2 \int (c+d x) \csc \left (i a+i b x+\frac {\pi }{2}\right )dx}{b^2}+\frac {1}{2} \int (c+d x)^3 \csc \left (i a+i b x+\frac {\pi }{2}\right )dx+\frac {3 d (c+d x)^2 \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \tanh (a+b x) \text {sech}(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 4668 |
| \(\displaystyle -\frac {3 d^2 \left (-\frac {i d \int \log \left (1-i e^{a+b x}\right )dx}{b}+\frac {i d \int \log \left (1+i e^{a+b x}\right )dx}{b}+\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}\right )}{b^2}+\frac {1}{2} \left (-\frac {3 i d \int (c+d x)^2 \log \left (1-i e^{a+b x}\right )dx}{b}+\frac {3 i d \int (c+d x)^2 \log \left (1+i e^{a+b x}\right )dx}{b}+\frac {2 (c+d x)^3 \arctan \left (e^{a+b x}\right )}{b}\right )+\frac {3 d (c+d x)^2 \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \tanh (a+b x) \text {sech}(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle -\frac {3 d^2 \left (-\frac {i d \int e^{-a-b x} \log \left (1-i e^{a+b x}\right )de^{a+b x}}{b^2}+\frac {i d \int e^{-a-b x} \log \left (1+i e^{a+b x}\right )de^{a+b x}}{b^2}+\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}\right )}{b^2}+\frac {1}{2} \left (-\frac {3 i d \int (c+d x)^2 \log \left (1-i e^{a+b x}\right )dx}{b}+\frac {3 i d \int (c+d x)^2 \log \left (1+i e^{a+b x}\right )dx}{b}+\frac {2 (c+d x)^3 \arctan \left (e^{a+b x}\right )}{b}\right )+\frac {3 d (c+d x)^2 \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \tanh (a+b x) \text {sech}(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {1}{2} \left (-\frac {3 i d \int (c+d x)^2 \log \left (1-i e^{a+b x}\right )dx}{b}+\frac {3 i d \int (c+d x)^2 \log \left (1+i e^{a+b x}\right )dx}{b}+\frac {2 (c+d x)^3 \arctan \left (e^{a+b x}\right )}{b}\right )-\frac {3 d^2 \left (\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac {i d \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}\right )}{b^2}+\frac {3 d (c+d x)^2 \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \tanh (a+b x) \text {sech}(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {1}{2} \left (\frac {3 i d \left (\frac {2 d \int (c+d x) \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )dx}{b}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b}\right )}{b}-\frac {3 i d \left (\frac {2 d \int (c+d x) \operatorname {PolyLog}\left (2,i e^{a+b x}\right )dx}{b}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b}\right )}{b}+\frac {2 (c+d x)^3 \arctan \left (e^{a+b x}\right )}{b}\right )-\frac {3 d^2 \left (\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac {i d \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}\right )}{b^2}+\frac {3 d (c+d x)^2 \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \tanh (a+b x) \text {sech}(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {1}{2} \left (\frac {3 i d \left (\frac {2 d \left (\frac {(c+d x) \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )}{b}-\frac {d \int \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )dx}{b}\right )}{b}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b}\right )}{b}-\frac {3 i d \left (\frac {2 d \left (\frac {(c+d x) \operatorname {PolyLog}\left (3,i e^{a+b x}\right )}{b}-\frac {d \int \operatorname {PolyLog}\left (3,i e^{a+b x}\right )dx}{b}\right )}{b}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b}\right )}{b}+\frac {2 (c+d x)^3 \arctan \left (e^{a+b x}\right )}{b}\right )-\frac {3 d^2 \left (\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac {i d \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}\right )}{b^2}+\frac {3 d (c+d x)^2 \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \tanh (a+b x) \text {sech}(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {1}{2} \left (\frac {3 i d \left (\frac {2 d \left (\frac {(c+d x) \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )}{b}-\frac {d \int e^{-a-b x} \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )de^{a+b x}}{b^2}\right )}{b}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b}\right )}{b}-\frac {3 i d \left (\frac {2 d \left (\frac {(c+d x) \operatorname {PolyLog}\left (3,i e^{a+b x}\right )}{b}-\frac {d \int e^{-a-b x} \operatorname {PolyLog}\left (3,i e^{a+b x}\right )de^{a+b x}}{b^2}\right )}{b}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b}\right )}{b}+\frac {2 (c+d x)^3 \arctan \left (e^{a+b x}\right )}{b}\right )-\frac {3 d^2 \left (\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac {i d \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}\right )}{b^2}+\frac {3 d (c+d x)^2 \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \tanh (a+b x) \text {sech}(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle -\frac {3 d^2 \left (\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac {i d \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}\right )}{b^2}+\frac {1}{2} \left (\frac {2 (c+d x)^3 \arctan \left (e^{a+b x}\right )}{b}+\frac {3 i d \left (\frac {2 d \left (\frac {(c+d x) \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )}{b}-\frac {d \operatorname {PolyLog}\left (4,-i e^{a+b x}\right )}{b^2}\right )}{b}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b}\right )}{b}-\frac {3 i d \left (\frac {2 d \left (\frac {(c+d x) \operatorname {PolyLog}\left (3,i e^{a+b x}\right )}{b}-\frac {d \operatorname {PolyLog}\left (4,i e^{a+b x}\right )}{b^2}\right )}{b}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b}\right )}{b}\right )+\frac {3 d (c+d x)^2 \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \tanh (a+b x) \text {sech}(a+b x)}{2 b}\) |
(-3*d^2*((2*(c + d*x)*ArcTan[E^(a + b*x)])/b - (I*d*PolyLog[2, (-I)*E^(a + b*x)])/b^2 + (I*d*PolyLog[2, I*E^(a + b*x)])/b^2))/b^2 + ((2*(c + d*x)^3* ArcTan[E^(a + b*x)])/b + ((3*I)*d*(-(((c + d*x)^2*PolyLog[2, (-I)*E^(a + b *x)])/b) + (2*d*(((c + d*x)*PolyLog[3, (-I)*E^(a + b*x)])/b - (d*PolyLog[4 , (-I)*E^(a + b*x)])/b^2))/b))/b - ((3*I)*d*(-(((c + d*x)^2*PolyLog[2, I*E ^(a + b*x)])/b) + (2*d*(((c + d*x)*PolyLog[3, I*E^(a + b*x)])/b - (d*PolyL og[4, I*E^(a + b*x)])/b^2))/b))/b)/2 + (3*d*(c + d*x)^2*Sech[a + b*x])/(2* b^2) + ((c + d*x)^3*Sech[a + b*x]*Tanh[a + b*x])/(2*b)
3.1.9.3.1 Defintions of rubi rules used
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]
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]]]
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]
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]
Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ ))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I)) Int[(c + d*x)^(m - 1)*Log[ 1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I)) Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c , d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
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]
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]
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]
\[\int \left (d x +c \right )^{3} \operatorname {sech}\left (b x +a \right )^{3}d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 4785 vs. \(2 (242) = 484\).
Time = 0.36 (sec) , antiderivative size = 4785, normalized size of antiderivative = 16.17 \[ \int (c+d x)^3 \text {sech}^3(a+b x) \, dx=\text {Too large to display} \]
1/2*(2*(b^3*d^3*x^3 + b^3*c^3 + 3*b^2*c^2*d + 3*(b^3*c*d^2 + b^2*d^3)*x^2 + 3*(b^3*c^2*d + 2*b^2*c*d^2)*x)*cosh(b*x + a)^3 + 6*(b^3*d^3*x^3 + b^3*c^ 3 + 3*b^2*c^2*d + 3*(b^3*c*d^2 + b^2*d^3)*x^2 + 3*(b^3*c^2*d + 2*b^2*c*d^2 )*x)*cosh(b*x + a)*sinh(b*x + a)^2 + 2*(b^3*d^3*x^3 + b^3*c^3 + 3*b^2*c^2* d + 3*(b^3*c*d^2 + b^2*d^3)*x^2 + 3*(b^3*c^2*d + 2*b^2*c*d^2)*x)*sinh(b*x + a)^3 - 2*(b^3*d^3*x^3 + b^3*c^3 - 3*b^2*c^2*d + 3*(b^3*c*d^2 - b^2*d^3)* x^2 + 3*(b^3*c^2*d - 2*b^2*c*d^2)*x)*cosh(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 + (-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)*cosh(b*x + a)^4 + 4*(-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)*cosh(b*x + a)*sinh(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)*sinh(b*x + a)^4 + 2*I*d^3 + 2*(-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)*cosh(b*x + a)^2 + 2*( -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 + 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)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 4*((-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)*cosh( 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)*co sh(b*x + a))*sinh(b*x + a))*dilog(I*cosh(b*x + a) + I*sinh(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 + (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)*cosh(b*x + a)^4 + 4*(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)*cosh(b*x + a)*sinh(b*x + a)^3 + (I*...
\[ \int (c+d x)^3 \text {sech}^3(a+b x) \, dx=\int \left (c + d x\right )^{3} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \]
\[ \int (c+d x)^3 \text {sech}^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \operatorname {sech}\left (b x + a\right )^{3} \,d x } \]
b^2*d^3*integrate(x^3*e^(b*x + a)/(b^2*e^(2*b*x + 2*a) + b^2), x) + 3*b^2* c*d^2*integrate(x^2*e^(b*x + a)/(b^2*e^(2*b*x + 2*a) + b^2), x) + 3*b^2*c^ 2*d*integrate(x*e^(b*x + a)/(b^2*e^(2*b*x + 2*a) + b^2), x) - c^3*(arctan( e^(-b*x - a))/b - (e^(-b*x - a) - e^(-3*b*x - 3*a))/(b*(2*e^(-2*b*x - 2*a) + e^(-4*b*x - 4*a) + 1))) - 6*d^3*integrate(x*e^(b*x + a)/(b^2*e^(2*b*x + 2*a) + b^2), x) - 6*c*d^2*arctan(e^(b*x + a))/b^3 + ((b*d^3*x^3*e^(3*a) + 3*c^2*d*e^(3*a) + 3*(b*c*d^2 + d^3)*x^2*e^(3*a) + 3*(b*c^2*d + 2*c*d^2)*x *e^(3*a))*e^(3*b*x) - (b*d^3*x^3*e^a - 3*c^2*d*e^a + 3*(b*c*d^2 - d^3)*x^2 *e^a + 3*(b*c^2*d - 2*c*d^2)*x*e^a)*e^(b*x))/(b^2*e^(4*b*x + 4*a) + 2*b^2* e^(2*b*x + 2*a) + b^2)
\[ \int (c+d x)^3 \text {sech}^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \operatorname {sech}\left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int (c+d x)^3 \text {sech}^3(a+b x) \, dx=\int \frac {{\left (c+d\,x\right )}^3}{{\mathrm {cosh}\left (a+b\,x\right )}^3} \,d x \]