3.26 \(\int (c+d x)^3 \text {sech}(a+b x) \, dx\)

Optimal. Leaf size=179 \[ -\frac {6 i d^3 \text {Li}_4\left (-i e^{a+b x}\right )}{b^4}+\frac {6 i d^3 \text {Li}_4\left (i e^{a+b x}\right )}{b^4}+\frac {6 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {2 (c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4180, 2531, 6609, 2282, 6589} \[ \frac {6 i d^2 (c+d x) \text {PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {PolyLog}\left (3,i e^{a+b x}\right )}{b^3}-\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac {6 i d^3 \text {PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}+\frac {6 i d^3 \text {PolyLog}\left (4,i e^{a+b x}\right )}{b^4}+\frac {2 (c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Sech[a + b*x],x]

[Out]

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

Rule 2282

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 2531

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 4180

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 6589

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 6609

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}(a+b x) \, dx &=\frac {2 (c+d x)^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\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 \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 )}{b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {\left (6 i d^2\right ) \int (c+d x) \text {Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^2}-\frac {\left (6 i d^2\right ) \int (c+d x) \text {Li}_2\left (i e^{a+b x}\right ) \, dx}{b^2}\\ &=\frac {2 (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 )}{b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {6 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {\left (6 i d^3\right ) \int \text {Li}_3\left (-i e^{a+b x}\right ) \, dx}{b^3}+\frac {\left (6 i d^3\right ) \int \text {Li}_3\left (i e^{a+b x}\right ) \, dx}{b^3}\\ &=\frac {2 (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 )}{b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {6 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {\left (6 i d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac {\left (6 i d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=\frac {2 (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 )}{b^2}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac {6 i d^2 (c+d x) \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {6 i d^3 \text {Li}_4\left (-i e^{a+b x}\right )}{b^4}+\frac {6 i d^3 \text {Li}_4\left (i e^{a+b x}\right )}{b^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 2.65, size = 343, normalized size = 1.92 \[ \frac {i \left (-2 i b^3 c^3 \tan ^{-1}\left (e^{a+b x}\right )+3 b^3 c^2 d x \log \left (1-i e^{a+b x}\right )-3 b^3 c^2 d x \log \left (1+i e^{a+b x}\right )+3 b^3 c d^2 x^2 \log \left (1-i e^{a+b x}\right )-3 b^3 c d^2 x^2 \log \left (1+i e^{a+b x}\right )+b^3 d^3 x^3 \log \left (1-i e^{a+b x}\right )-b^3 d^3 x^3 \log \left (1+i e^{a+b x}\right )-3 b^2 d (c+d x)^2 \text {Li}_2\left (-i e^{a+b x}\right )+3 b^2 d (c+d x)^2 \text {Li}_2\left (i e^{a+b x}\right )+6 b c d^2 \text {Li}_3\left (-i e^{a+b x}\right )-6 b c d^2 \text {Li}_3\left (i e^{a+b x}\right )+6 b d^3 x \text {Li}_3\left (-i e^{a+b x}\right )-6 b d^3 x \text {Li}_3\left (i e^{a+b x}\right )-6 d^3 \text {Li}_4\left (-i e^{a+b x}\right )+6 d^3 \text {Li}_4\left (i e^{a+b x}\right )\right )}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [C]  time = 0.65, size = 493, normalized size = 2.75 \[ \frac {6 i \, d^{3} {\rm polylog}\left (4, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 6 i \, d^{3} {\rm polylog}\left (4, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + {\left (3 i \, b^{2} d^{3} x^{2} + 6 i \, b^{2} c d^{2} x + 3 i \, b^{2} c^{2} d\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + {\left (-3 i \, b^{2} d^{3} x^{2} - 6 i \, b^{2} c d^{2} x - 3 i \, b^{2} c^{2} d\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + {\left (i \, b^{3} c^{3} - 3 i \, a b^{2} c^{2} d + 3 i \, a^{2} b c d^{2} - i \, a^{3} d^{3}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + {\left (-i \, b^{3} c^{3} + 3 i \, a b^{2} c^{2} d - 3 i \, a^{2} b c d^{2} + i \, a^{3} d^{3}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + {\left (-i \, b^{3} d^{3} x^{3} - 3 i \, b^{3} c d^{2} x^{2} - 3 i \, b^{3} c^{2} d x - 3 i \, a b^{2} c^{2} d + 3 i \, a^{2} b c d^{2} - i \, a^{3} d^{3}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) + {\left (i \, b^{3} d^{3} x^{3} + 3 i \, b^{3} c d^{2} x^{2} + 3 i \, b^{3} c^{2} d x + 3 i \, a b^{2} c^{2} d - 3 i \, a^{2} b c d^{2} + i \, a^{3} d^{3}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) + {\left (-6 i \, b d^{3} x - 6 i \, b c d^{2}\right )} {\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + {\left (6 i \, b d^{3} x + 6 i \, b c d^{2}\right )} {\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{3} \operatorname {sech}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{3} \mathrm {sech}\left (b x +a \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, c^{3} \arctan \left (e^{\left (-b x - a\right )}\right )}{b} + 2 \, \int \frac {{\left (d^{3} x^{3} e^{a} + 3 \, c d^{2} x^{2} e^{a} + 3 \, c^{2} d x e^{a}\right )} e^{\left (b x\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c^3*arctan(e^(-b*x - a))/b + 2*integrate((d^3*x^3*e^a + 3*c*d^2*x^2*e^a + 3*c^2*d*x*e^a)*e^(b*x)/(e^(2*b*x
+ 2*a) + 1), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,x\right )}^3}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^3/cosh(a + b*x),x)

[Out]

int((c + d*x)^3/cosh(a + b*x), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{3} \operatorname {sech}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________