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

Optimal. Leaf size=73 \[ -\frac {d^2 \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac {2 d (c+d x) \log \left (e^{2 (a+b x)}+1\right )}{b^2}+\frac {(c+d x)^2 \tanh (a+b x)}{b}+\frac {(c+d x)^2}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 73, normalized size of antiderivative = 1.00, 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 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]

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 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]

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 = 6.25, size = 277, normalized size = 3.79 \[ -\frac {2 c d \text {sech}(a) (\cosh (a) \log (\sinh (a) \sinh (b x)+\cosh (a) \cosh (b x))-b x \sinh (a))}{b^2 \left (\cosh ^2(a)-\sinh ^2(a)\right )}-\frac {d^2 \text {csch}(a) \text {sech}(a) \left (b^2 x^2 e^{-\tanh ^{-1}(\coth (a))}-\frac {i \coth (a) \left (i \text {Li}_2\left (e^{2 i \left (i b x+i \tanh ^{-1}(\coth (a))\right )}\right )-b x \left (-\pi +2 i \tanh ^{-1}(\coth (a))\right )-2 \left (i \tanh ^{-1}(\coth (a))+i b x\right ) \log \left (1-e^{2 i \left (i \tanh ^{-1}(\coth (a))+i b x\right )}\right )+2 i \tanh ^{-1}(\coth (a)) \log \left (i \sinh \left (\tanh ^{-1}(\coth (a))+b x\right )\right )-\pi \log \left (e^{2 b x}+1\right )+\pi \log (\cosh (b x))\right )}{\sqrt {1-\coth ^2(a)}}\right )}{b^3 \sqrt {\text {csch}^2(a) \left (\sinh ^2(a)-\cosh ^2(a)\right )}}+\frac {\text {sech}(a) \text {sech}(a+b x) \left (c^2 \sinh (b x)+2 c d x \sinh (b x)+d^2 x^2 \sinh (b x)\right )}{b} \]

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*Log[
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[Cosh[
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*ArcTanh[
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

________________________________________________________________________________________

fricas [C]  time = 0.46, size = 715, normalized size = 9.79 \[ -\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2} - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \cosh \left (b x + a\right )^{2} - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \sinh \left (b x + a\right )^{2} + {\left (d^{2} \cosh \left (b x + a\right )^{2} + 2 \, d^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + d^{2} \sinh \left (b x + a\right )^{2} + d^{2}\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + {\left (d^{2} \cosh \left (b x + a\right )^{2} + 2 \, d^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + d^{2} \sinh \left (b x + a\right )^{2} + d^{2}\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + {\left (b c d - a d^{2} + {\left (b c d - a d^{2}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b c d - a d^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b c d - a d^{2}\right )} \sinh \left (b x + a\right )^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + {\left (b c d - a d^{2} + {\left (b c d - a d^{2}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b c d - a d^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b c d - a d^{2}\right )} \sinh \left (b x + a\right )^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + {\left (b d^{2} x + a d^{2} + {\left (b d^{2} x + a d^{2}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b d^{2} x + a d^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b d^{2} x + a d^{2}\right )} \sinh \left (b x + a\right )^{2}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) + {\left (b d^{2} x + a d^{2} + {\left (b d^{2} x + a d^{2}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b d^{2} x + a d^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b d^{2} x + a d^{2}\right )} \sinh \left (b x + a\right )^{2}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right )\right )}}{b^{3} \cosh \left (b x + a\right )^{2} + 2 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )^{2} + b^{3}} \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \operatorname {sech}\left (b x + a\right )^{2}\,{d x} \]

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)

________________________________________________________________________________________

maple [B]  time = 0.32, size = 159, normalized size = 2.18 \[ -\frac {2 \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}-\frac {2 d c \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b^{2}}+\frac {4 d c \ln \left ({\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {2 d^{2} x^{2}}{b}+\frac {4 d^{2} a x}{b^{2}}+\frac {2 d^{2} a^{2}}{b^{3}}-\frac {2 d^{2} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right ) x}{b^{2}}-\frac {d^{2} \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{3}}-\frac {4 d^{2} a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}} \]

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))

________________________________________________________________________________________

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

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))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^2/cosh(a + b*x)^2,x)

[Out]

int((c + d*x)^2/cosh(a + b*x)^2, x)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________