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

Optimal. Leaf size=73 \[ -\frac{d^2 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}-\frac{2 d (c+d x) \log \left (e^{2 (a+b x)}+1\right )}{b^2}+\frac{(c+d x)^2 \tanh (a+b x)}{b}+\frac{(c+d x)^2}{b} \]

[Out]

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

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Rubi [A]  time = 0.132696, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {4184, 3718, 2190, 2279, 2391} \[ -\frac{d^2 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}-\frac{2 d (c+d x) \log \left (e^{2 (a+b x)}+1\right )}{b^2}+\frac{(c+d x)^2 \tanh (a+b x)}{b}+\frac{(c+d x)^2}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sech[a + b*x]^2,x]

[Out]

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

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

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 2391

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]

Rubi steps

\begin{align*} \int (c+d x)^2 \text{sech}^2(a+b x) \, dx &=\frac{(c+d x)^2 \tanh (a+b x)}{b}-\frac{(2 d) \int (c+d x) \tanh (a+b x) \, dx}{b}\\ &=\frac{(c+d x)^2}{b}+\frac{(c+d x)^2 \tanh (a+b x)}{b}-\frac{(4 d) \int \frac{e^{2 (a+b x)} (c+d x)}{1+e^{2 (a+b x)}} \, dx}{b}\\ &=\frac{(c+d x)^2}{b}-\frac{2 d (c+d x) \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac{(c+d x)^2 \tanh (a+b x)}{b}+\frac{\left (2 d^2\right ) \int \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{(c+d x)^2}{b}-\frac{2 d (c+d x) \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac{(c+d x)^2 \tanh (a+b x)}{b}+\frac{d^2 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{b^3}\\ &=\frac{(c+d x)^2}{b}-\frac{2 d (c+d x) \log \left (1+e^{2 (a+b x)}\right )}{b^2}-\frac{d^2 \text{Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}+\frac{(c+d x)^2 \tanh (a+b x)}{b}\\ \end{align*}

Mathematica [C]  time = 6.28069, size = 277, normalized size = 3.79 \[ \frac{d^2 \text{csch}(a) \text{sech}(a) \left (-b^2 x^2 e^{-\tanh ^{-1}(\coth (a))}+\frac{i \coth (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (i \tanh ^{-1}(\coth (a))+i b x\right )}\right )-b x \left (-\pi +2 i \tanh ^{-1}(\coth (a))\right )-2 \left (i \tanh ^{-1}(\coth (a))+i b x\right ) \log \left (1-e^{2 i \left (i \tanh ^{-1}(\coth (a))+i b x\right )}\right )+2 i \tanh ^{-1}(\coth (a)) \log \left (i \sinh \left (\tanh ^{-1}(\coth (a))+b x\right )\right )-\pi \log \left (e^{2 b x}+1\right )+\pi \log (\cosh (b x))\right )}{\sqrt{1-\coth ^2(a)}}\right )}{b^3 \sqrt{\text{csch}^2(a) \left (\sinh ^2(a)-\cosh ^2(a)\right )}}-\frac{2 c d \text{sech}(a) (\cosh (a) \log (\sinh (a) \sinh (b x)+\cosh (a) \cosh (b x))-b x \sinh (a))}{b^2 \left (\cosh ^2(a)-\sinh ^2(a)\right )}+\frac{\text{sech}(a) \text{sech}(a+b x) \left (c^2 \sinh (b x)+2 c d x \sinh (b x)+d^2 x^2 \sinh (b x)\right )}{b} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*c*d*Sech[a]*(Cosh[a]*Log[Cosh[a]*Cosh[b*x] + Sinh[a]*Sinh[b*x]] - b*x*Sinh[a]))/(b^2*(Cosh[a]^2 - Sinh[a]^
2)) + (d^2*Csch[a]*(-((b^2*x^2)/E^ArcTanh[Coth[a]]) + (I*Coth[a]*(-(b*x*(-Pi + (2*I)*ArcTanh[Coth[a]])) - Pi*L
og[1 + E^(2*b*x)] - 2*(I*b*x + I*ArcTanh[Coth[a]])*Log[1 - E^((2*I)*(I*b*x + I*ArcTanh[Coth[a]]))] + Pi*Log[Co
sh[b*x]] + (2*I)*ArcTanh[Coth[a]]*Log[I*Sinh[b*x + ArcTanh[Coth[a]]]] + I*PolyLog[2, E^((2*I)*(I*b*x + I*ArcTa
nh[Coth[a]]))]))/Sqrt[1 - Coth[a]^2])*Sech[a])/(b^3*Sqrt[Csch[a]^2*(-Cosh[a]^2 + Sinh[a]^2)]) + (Sech[a]*Sech[
a + b*x]*(c^2*Sinh[b*x] + 2*c*d*x*Sinh[b*x] + d^2*x^2*Sinh[b*x]))/b

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Maple [B]  time = 0.031, size = 159, normalized size = 2.2 \begin{align*} -2\,{\frac{{d}^{2}{x}^{2}+2\,cdx+{c}^{2}}{b \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }}-2\,{\frac{cd\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{2}}}+4\,{\frac{cd\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+2\,{\frac{{d}^{2}{x}^{2}}{b}}+4\,{\frac{a{d}^{2}x}{{b}^{2}}}+2\,{\frac{{a}^{2}{d}^{2}}{{b}^{3}}}-2\,{\frac{{d}^{2}\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) x}{{b}^{2}}}-{\frac{{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{3}}}-4\,{\frac{a{d}^{2}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(d^2*x^2+2*c*d*x+c^2)/b/(1+exp(2*b*x+2*a))-2*d/b^2*c*ln(1+exp(2*b*x+2*a))+4*d/b^2*c*ln(exp(b*x+a))+2*d^2/b*
x^2+4*d^2/b^2*a*x+2*d^2/b^3*a^2-2*d^2/b^2*ln(1+exp(2*b*x+2*a))*x-d^2*polylog(2,-exp(2*b*x+2*a))/b^3-4*d^2/b^3*
a*ln(exp(b*x+a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.23175, size = 1777, normalized size = 24.34 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2 - (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cosh(b*x + a)^2 - 2*(b^2
*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cosh(b*x + a)*sinh(b*x + a) - (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b
*c*d - a^2*d^2)*sinh(b*x + a)^2 + (d^2*cosh(b*x + a)^2 + 2*d^2*cosh(b*x + a)*sinh(b*x + a) + d^2*sinh(b*x + a)
^2 + d^2)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) + (d^2*cosh(b*x + a)^2 + 2*d^2*cosh(b*x + a)*sinh(b*x + a)
+ d^2*sinh(b*x + a)^2 + d^2)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + (b*c*d - a*d^2 + (b*c*d - a*d^2)*cosh
(b*x + a)^2 + 2*(b*c*d - a*d^2)*cosh(b*x + a)*sinh(b*x + a) + (b*c*d - a*d^2)*sinh(b*x + a)^2)*log(cosh(b*x +
a) + sinh(b*x + a) + I) + (b*c*d - a*d^2 + (b*c*d - a*d^2)*cosh(b*x + a)^2 + 2*(b*c*d - a*d^2)*cosh(b*x + a)*s
inh(b*x + a) + (b*c*d - a*d^2)*sinh(b*x + a)^2)*log(cosh(b*x + a) + sinh(b*x + a) - I) + (b*d^2*x + a*d^2 + (b
*d^2*x + a*d^2)*cosh(b*x + a)^2 + 2*(b*d^2*x + a*d^2)*cosh(b*x + a)*sinh(b*x + a) + (b*d^2*x + a*d^2)*sinh(b*x
+ a)^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) + (b*d^2*x + a*d^2 + (b*d^2*x + a*d^2)*cosh(b*x + a)^2 + 2
*(b*d^2*x + a*d^2)*cosh(b*x + a)*sinh(b*x + a) + (b*d^2*x + a*d^2)*sinh(b*x + a)^2)*log(-I*cosh(b*x + a) - I*s
inh(b*x + a) + 1))/(b^3*cosh(b*x + a)^2 + 2*b^3*cosh(b*x + a)*sinh(b*x + a) + b^3*sinh(b*x + a)^2 + 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}^{2}{\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)**2,x)

[Out]

Integral((c + d*x)**2*sech(a + b*x)**2, 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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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