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

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 119, normalized size of antiderivative = 1.00, 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 = {4265, 2611, 2320, 6724} \begin {gather*} \frac {2 (c+d x)^2 \text {ArcTan}\left (e^{a+b x}\right )}{b}+\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}-\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} \end {gather*}

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 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 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 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]

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.07, size = 199, normalized size = 1.67 \begin {gather*} \frac {i \left (-2 i b^2 c^2 \text {ArcTan}\left (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 )-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 )-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 )\right )}{b^3} \end {gather*}

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 = 180.00, size = 0, normalized size = 0.00 \[\int \left (d x +c \right )^{2} \mathrm {sech}\left (b x +a \right )\, dx\]

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 = 0.00, 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)^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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (96) = 192\).
time = 0.37, size = 305, normalized size = 2.56 \begin {gather*} \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 ) - 2 \, {\left (-i \, b d^{2} x - i \, b c d\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 2 \, {\left (i \, b d^{2} x + 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 {gather*}

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 - I*b*c*d)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - 2*(I*b*d^2*x + 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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c + d x\right )^{2} \operatorname {sech}{\left (a + b x \right )}\, dx \end {gather*}

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 = 0.00, 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)^2*sech(b*x+a),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^2}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^2/cosh(a + b*x),x)

[Out]

int((c + d*x)^2/cosh(a + b*x), x)

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