3.2 \(\int (c+d x)^2 \text{sech}(a+b x) \, dx\)

Optimal. Leaf size=119 \[ -\frac{2 i d (c+d x) \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac{2 i d (c+d x) \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}+\frac{2 i d^2 \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}-\frac{2 i d^2 \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac{2 (c+d x)^2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0820517, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4180, 2531, 2282, 6589} \[ -\frac{2 i d (c+d x) \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac{2 i d (c+d x) \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}+\frac{2 i d^2 \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}-\frac{2 i d^2 \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac{2 (c+d x)^2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sech[a + b*x],x]

[Out]

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

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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.116379, size = 199, normalized size = 1.67 \[ \frac{i \left (-2 b d (c+d x) \text{PolyLog}\left (2,-i e^{a+b x}\right )+2 b d (c+d x) \text{PolyLog}\left (2,i e^{a+b x}\right )+2 d^2 \text{PolyLog}\left (3,-i e^{a+b x}\right )-2 d^2 \text{PolyLog}\left (3,i e^{a+b x}\right )-2 i b^2 c^2 \tan ^{-1}\left (e^{a+b x}\right )+2 b^2 c d x \log \left (1-i e^{a+b x}\right )-2 b^2 c d x \log \left (1+i e^{a+b x}\right )+b^2 d^2 x^2 \log \left (1-i e^{a+b x}\right )-b^2 d^2 x^2 \log \left (1+i e^{a+b x}\right )\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*Sech[a + b*x],x]

[Out]

(I*((-2*I)*b^2*c^2*ArcTan[E^(a + b*x)] + 2*b^2*c*d*x*Log[1 - I*E^(a + b*x)] + b^2*d^2*x^2*Log[1 - I*E^(a + b*x
)] - 2*b^2*c*d*x*Log[1 + I*E^(a + b*x)] - b^2*d^2*x^2*Log[1 + I*E^(a + b*x)] - 2*b*d*(c + d*x)*PolyLog[2, (-I)
*E^(a + b*x)] + 2*b*d*(c + d*x)*PolyLog[2, I*E^(a + b*x)] + 2*d^2*PolyLog[3, (-I)*E^(a + b*x)] - 2*d^2*PolyLog
[3, I*E^(a + b*x)]))/b^3

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Maple [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{2}{\rm sech} \left (bx+a\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*sech(b*x+a),x)

[Out]

int((d*x+c)^2*sech(b*x+a),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, c^{2} \arctan \left (e^{\left (-b x - a\right )}\right )}{b} + 2 \, \int \frac{{\left (d^{2} x^{2} e^{a} + 2 \, c d x e^{a}\right )} e^{\left (b x\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*sech(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 2.30789, size = 845, normalized size = 7.1 \begin{align*} \frac{-2 i \, d^{2}{\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 2 i \, d^{2}{\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) +{\left (2 i \, b d^{2} x + 2 i \, b c d\right )}{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) +{\left (-2 i \, b d^{2} x - 2 i \, b c d\right )}{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) +{\left (i \, b^{2} c^{2} - 2 i \, a b c d + i \, a^{2} d^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) +{\left (-i \, b^{2} c^{2} + 2 i \, a b c d - i \, a^{2} d^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) +{\left (-i \, b^{2} d^{2} x^{2} - 2 i \, b^{2} c d x - 2 i \, a b c d + i \, a^{2} d^{2}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) +{\left (i \, b^{2} d^{2} x^{2} + 2 i \, b^{2} c d x + 2 i \, a b c d - i \, a^{2} d^{2}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*sech(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{2} \operatorname{sech}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*sech(b*x+a),x)

[Out]

Integral((c + d*x)**2*sech(a + b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2} \operatorname{sech}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*sech(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^2*sech(b*x + a), x)