3.1.9 \(\int (c+d x)^3 \text {sech}^3(a+b x) \, dx\) [9]
Optimal. Leaf size=296 \[ -\frac {6 d^2 (c+d x) \text {ArcTan}\left (e^{a+b x}\right )}{b^3}+\frac {(c+d x)^3 \text {ArcTan}\left (e^{a+b x}\right )}{b}+\frac {3 i d^3 \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^4}-\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,-i e^{a+b x}\right )}{2 b^2}-\frac {3 i d^3 \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^4}+\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,i e^{a+b x}\right )}{2 b^2}+\frac {3 i d^2 (c+d x) \text {PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}-\frac {3 i d^2 (c+d x) \text {PolyLog}\left (3,i e^{a+b x}\right )}{b^3}-\frac {3 i d^3 \text {PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}+\frac {3 i d^3 \text {PolyLog}\left (4,i e^{a+b x}\right )}{b^4}+\frac {3 d (c+d x)^2 \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \text {sech}(a+b x) \tanh (a+b x)}{2 b} \]
[Out]
-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*e
xp(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*polylog(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*sech(b*x+a)*tanh(b*x+a)/b
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Rubi [A]
time = 0.16, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps
used = 15, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4271, 4265,
2317, 2438, 2611, 6744, 2320, 6724} \begin {gather*} -\frac {6 d^2 (c+d x) \text {ArcTan}\left (e^{a+b x}\right )}{b^3}+\frac {(c+d x)^3 \text {ArcTan}\left (e^{a+b x}\right )}{b}+\frac {3 i d^3 \text {Li}_2\left (-i e^{a+b x}\right )}{b^4}-\frac {3 i d^3 \text {Li}_2\left (i e^{a+b x}\right )}{b^4}-\frac {3 i d^3 \text {Li}_4\left (-i e^{a+b x}\right )}{b^4}+\frac {3 i d^3 \text {Li}_4\left (i e^{a+b x}\right )}{b^4}+\frac {3 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {3 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{2 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} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(c + d*x)^3*Sech[a + b*x]^3,x]
[Out]
(-6*d^2*(c + d*x)*ArcTan[E^(a + b*x)])/b^3 + ((c + d*x)^3*ArcTan[E^(a + b*x)])/b + ((3*I)*d^3*PolyLog[2, (-I)*
E^(a + b*x)])/b^4 - (((3*I)/2)*d*(c + d*x)^2*PolyLog[2, (-I)*E^(a + b*x)])/b^2 - ((3*I)*d^3*PolyLog[2, I*E^(a
+ b*x)])/b^4 + (((3*I)/2)*d*(c + d*x)^2*PolyLog[2, I*E^(a + b*x)])/b^2 + ((3*I)*d^2*(c + d*x)*PolyLog[3, (-I)*
E^(a + b*x)])/b^3 - ((3*I)*d^2*(c + d*x)*PolyLog[3, I*E^(a + b*x)])/b^3 - ((3*I)*d^3*PolyLog[4, (-I)*E^(a + b*
x)])/b^4 + ((3*I)*d^3*PolyLog[4, I*E^(a + b*x)])/b^4 + (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)
Rule 2317
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2320
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 2438
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 2611
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] + Dist[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 4265
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] + (-Dist[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] + Dist[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]
Rule 4271
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-b^2)*(c + d*x)^m*Cot[e
+ f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (Dist[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] + Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)^m*(b*Csc[e + f*x])^
(n - 2), x], 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]) /; Free
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]
Rule 6724
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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 6744
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] - Dist[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]
Rubi steps
\begin {align*} \int (c+d x)^3 \text {sech}^3(a+b x) \, dx &=\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}+\frac {1}{2} \int (c+d x)^3 \text {sech}(a+b x) \, dx-\frac {\left (3 d^2\right ) \int (c+d x) \text {sech}(a+b x) \, dx}{b^2}\\ &=-\frac {6 d^2 (c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {(c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}+\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}-\frac {(3 i d) \int (c+d x)^2 \log \left (1-i e^{a+b x}\right ) \, dx}{2 b}+\frac {(3 i d) \int (c+d x)^2 \log \left (1+i e^{a+b x}\right ) \, dx}{2 b}+\frac {\left (3 i d^3\right ) \int \log \left (1-i e^{a+b x}\right ) \, dx}{b^3}-\frac {\left (3 i d^3\right ) \int \log \left (1+i e^{a+b x}\right ) \, dx}{b^3}\\ &=-\frac {6 d^2 (c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {(c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}+\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}+\frac {\left (3 i d^2\right ) \int (c+d x) \text {Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^2}-\frac {\left (3 i d^2\right ) \int (c+d x) \text {Li}_2\left (i e^{a+b x}\right ) \, dx}{b^2}+\frac {\left (3 i d^3\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac {\left (3 i d^3\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=-\frac {6 d^2 (c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {(c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac {3 i d^3 \text {Li}_2\left (-i e^{a+b x}\right )}{b^4}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}-\frac {3 i d^3 \text {Li}_2\left (i e^{a+b x}\right )}{b^4}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}+\frac {3 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {3 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}+\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}-\frac {\left (3 i d^3\right ) \int \text {Li}_3\left (-i e^{a+b x}\right ) \, dx}{b^3}+\frac {\left (3 i d^3\right ) \int \text {Li}_3\left (i e^{a+b x}\right ) \, dx}{b^3}\\ &=-\frac {6 d^2 (c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {(c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac {3 i d^3 \text {Li}_2\left (-i e^{a+b x}\right )}{b^4}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}-\frac {3 i d^3 \text {Li}_2\left (i e^{a+b x}\right )}{b^4}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}+\frac {3 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {3 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}+\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}-\frac {\left (3 i d^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac {\left (3 i d^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=-\frac {6 d^2 (c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {(c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac {3 i d^3 \text {Li}_2\left (-i e^{a+b x}\right )}{b^4}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}-\frac {3 i d^3 \text {Li}_2\left (i e^{a+b x}\right )}{b^4}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}+\frac {3 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {3 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {3 i d^3 \text {Li}_4\left (-i e^{a+b x}\right )}{b^4}+\frac {3 i d^3 \text {Li}_4\left (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}\\ \end {align*}
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Mathematica [A]
time = 4.02, size = 455, normalized size = 1.54 \begin {gather*} \frac {i \left (-2 i b^3 c^3 \text {ArcTan}\left (e^{a+b x}\right )+12 i b c d^2 \text {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 ) \text {PolyLog}\left (2,-i e^{a+b x}\right )+3 d \left (-2 d^2+b^2 (c+d x)^2\right ) \text {PolyLog}\left (2,i e^{a+b x}\right )+6 b c d^2 \text {PolyLog}\left (3,-i e^{a+b x}\right )+6 b d^3 x \text {PolyLog}\left (3,-i e^{a+b x}\right )-6 b c d^2 \text {PolyLog}\left (3,i e^{a+b x}\right )-6 b d^3 x \text {PolyLog}\left (3,i e^{a+b x}\right )-6 d^3 \text {PolyLog}\left (4,-i e^{a+b x}\right )+6 d^3 \text {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} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)^3*Sech[a + b*x]^3,x]
[Out]
(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*PolyLog[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)
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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \left (d x +c \right )^{3} \mathrm {sech}\left (b x +a \right )^{3}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x+c)^3*sech(b*x+a)^3,x)
[Out]
int((d*x+c)^3*sech(b*x+a)^3,x)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^3*sech(b*x+a)^3,x, algorithm="maxima")
[Out]
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*(arcta
n(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)
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Fricas [B] 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 = 0.47, size = 4785, normalized size = 16.17 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^3*sech(b*x+a)^3,x, algorithm="fricas")
[Out]
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)*cos
h(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)*cosh(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*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)*c
osh(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)*cosh(b*x + a))*sinh(b*x + a))*dilog(-I*cosh(b*x +
a) - I*sinh(b*x + a)) + (I*b^3*c^3 - 3*I*a*b^2*c^2*d + 3*I*(a^2 - 2)*b*c*d^2 + (I*b^3*c^3 - 3*I*a*b^2*c^2*d +
3*I*(a^2 - 2)*b*c*d^2 - I*(a^3 - 6*a)*d^3)*cosh(b*x + a)^4 - 4*(-I*b^3*c^3 + 3*I*a*b^2*c^2*d - 3*I*(a^2 - 2)*
b*c*d^2 + I*(a^3 - 6*a)*d^3)*cosh(b*x + a)*sinh(b*x + a)^3 + (I*b^3*c^3 - 3*I*a*b^2*c^2*d + 3*I*(a^2 - 2)*b*c*
d^2 - I*(a^3 - 6*a)*d^3)*sinh(b*x + a)^4 - I*(a^3 - 6*a)*d^3 - 2*(-I*b^3*c^3 + 3*I*a*b^2*c^2*d - 3*I*(a^2 - 2)
*b*c*d^2 + I*(a^3 - 6*a)*d^3)*cosh(b*x + a)^2 - 2*(-I*b^3*c^3 + 3*I*a*b^2*c^2*d - 3*I*(a^2 - 2)*b*c*d^2 + I*(a
^3 - 6*a)*d^3 + 3*(-I*b^3*c^3 + 3*I*a*b^2*c^2*d - 3*I*(a^2 - 2)*b*c*d^2 + I*(a^3 - 6*a)*d^3)*cosh(b*x + a)^2)*
sinh(b*x + a)^2 - 4*((-I*b^3*c^3 + 3*I*a*b^2*c^2*d - 3*I*(a^2 - 2)*b*c*d^2 + I*(a^3 - 6*a)*d^3)*cosh(b*x + a)^
3 + (-I*b^3*c^3 + 3*I*a*b^2*c^2*d - 3*I*(a^2 - 2)*b*c*d^2 + I*(a^3 - 6*a)*d^3)*cosh(b*x + a))*sinh(b*x + a))*l
og(cosh(b*x + a) + sinh(b*x + a) + I) + (-I*b^3*c^3 + 3*I*a*b^2*c^2*d - 3*I*(a^2 - 2)*b*c*d^2 + (-I*b^3*c^3 +
3*I*a*b^2*c^2*d - 3*I*(a^2 - 2)*b*c*d^2 + I*(a^3 - 6*a)*d^3)*cosh(b*x + a)^4 - 4*(I*b^3*c^3 - 3*I*a*b^2*c^2*d
+ 3*I*(a^2 - 2)*b*c*d^2 - I*(a^3 - 6*a)*d^3)*cosh(b*x + a)*sinh(b*x + a)^3 + (-I*b^3*c^3 + 3*I*a*b^2*c^2*d - 3
*I*(a^2 - 2)*b*c*d^2 + I*(a^3 - 6*a)*d^3)*sinh(b*x + a)^4 + I*(a^3 - 6*a)*d^3 - 2*(I*b^3*c^3 - 3*I*a*b^2*c^2*d
+ 3*I*(a^2 - 2)*b*c*d^2 - I*(a^3 - 6*a)*d^3)*cosh(b*x + a)^2 - 2*(I*b^3*c^3 - 3*I*a*b^2*c^2*d + 3*I*(a^2 - 2)
*b*c*d^2 - I*(a^3 - 6*a)*d^3 + 3*(I*b^3*c^3 - 3*I*a*b^2*c^2*d + 3*I*(a^2 - 2)*b*c*d^2 - I*(a^3 - 6*a)*d^3)*cos
h(b*x + a)^2)*sinh(b*x + a)^2 - 4*((I*b^3*c^3 - 3*I*a*b^2*c^2*d + 3*I*(a^2 - 2)*b*c*d^2 - I*(a^3 - 6*a)*d^3)*c
osh(b*x + a)^3 + (I*b^3*c^3 - 3*I*a*b^2*c^2*d + 3*I*(a^2 - 2)*b*c*d^2 - I*(a^3 - 6*a)*d^3)*cosh(b*x + a))*sinh
(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) - I) + (-I*b^3*d^3*x^3 - 3*I*b^3*c*d^2*x^2 - 3*I*a*b^2*c^2*d + 3*
I*a^2*b*c*d^2 + (-I*b^3*d^3*x^3 - 3*I*b^3*c*d^2*x^2 - 3*I*a*b^2*c^2*d + 3*I*a^2*b*c*d^2 - I*(a^3 - 6*a)*d^3 -
3*I*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a)^4 - 4*(I*b^3*d^3*x^3 + 3*I*b^3*c*d^2*x^2 + 3*I*a*b^2*c^2*d - 3*I*a^
2*b*c*d^2 + I*(a^3 - 6*a)*d^3 + 3*I*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a)*sinh(b*x + a)^3 + (-I*b^3*d^3*x^3 -
3*I*b^3*c*d^2*x^2 - 3*I*a*b^2*c^2*d + 3*I*a^2*b*c*d^2 - I*(a^3 - 6*a)*d^3 - 3*I*(b^3*c^2*d - 2*b*d^3)*x)*sinh
(b*x + a)^4 - I*(a^3 - 6*a)*d^3 - 2*(I*b^3*d^3*x^3 + 3*I*b^3*c*d^2*x^2 + 3*I*a*b^2*c^2*d - 3*I*a^2*b*c*d^2 + I
*(a^3 - 6*a)*d^3 + 3*I*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a)^2 - 2*(I*b^3*d^3*x^3 + 3*I*b^3*c*d^2*x^2 + 3*I*a
*b^2*c^2*d - 3*I*a^2*b*c*d^2 + I*(a^3 - 6*a)*d^3 + 3*(I*b^3*d^3*x^3 + 3*I*b^3*c*d^2*x^2 + 3*I*a*b^2*c^2*d - 3*
I*a^2*b*c*d^2 + I*(a^3 - 6*a)*d^3 + 3*I*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a)^2 + 3*I*(b^3*c^2*d - 2*b*d^3)*x
)*sinh(b*x + a)^2 - 3*I*(b^3*c^2*d - 2*b*d^3)*x - 4*((I*b^3*d^3*x^3 + 3*I*b^3*c*d^2*x^2 + 3*I*a*b^2*c^2*d - 3*
I*a^2*b*c*d^2 + I*(a^3 - 6*a)*d^3 + 3*I*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a)^3 + (I*b^3*d^3*x^3 + 3*I*b^3*c*
d^2*x^2 + 3*I*a*b^2*c^2*d - 3*I*a^2*b*c*d^2 + I...
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c + d x\right )^{3} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)**3*sech(b*x+a)**3,x)
[Out]
Integral((c + d*x)**3*sech(a + b*x)**3, x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^3*sech(b*x+a)^3,x, algorithm="giac")
[Out]
integrate((d*x + c)^3*sech(b*x + a)^3, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^3}{{\mathrm {cosh}\left (a+b\,x\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c + d*x)^3/cosh(a + b*x)^3,x)
[Out]
int((c + d*x)^3/cosh(a + b*x)^3, x)
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