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

Optimal. Leaf size=29 \[ \frac{(c+d x) \tanh (a+b x)}{b}-\frac{d \log (\cosh (a+b x))}{b^2} \]

[Out]

-((d*Log[Cosh[a + b*x]])/b^2) + ((c + d*x)*Tanh[a + b*x])/b

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Rubi [A]  time = 0.0292901, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4184, 3475} \[ \frac{(c+d x) \tanh (a+b x)}{b}-\frac{d \log (\cosh (a+b x))}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sech[a + b*x]^2,x]

[Out]

-((d*Log[Cosh[a + b*x]])/b^2) + ((c + d*x)*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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int (c+d x) \text{sech}^2(a+b x) \, dx &=\frac{(c+d x) \tanh (a+b x)}{b}-\frac{d \int \tanh (a+b x) \, dx}{b}\\ &=-\frac{d \log (\cosh (a+b x))}{b^2}+\frac{(c+d x) \tanh (a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0898336, size = 51, normalized size = 1.76 \[ -\frac{d \log (\cosh (a+b x))}{b^2}+\frac{c \tanh (a+b x)}{b}+\frac{d x \tanh (a)}{b}+\frac{d x \text{sech}(a) \sinh (b x) \text{sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*Log[Cosh[a + b*x]])/b^2) + (d*x*Sech[a]*Sech[a + b*x]*Sinh[b*x])/b + (d*x*Tanh[a])/b + (c*Tanh[a + b*x])/
b

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Maple [A]  time = 0.023, size = 57, normalized size = 2. \begin{align*} 2\,{\frac{dx}{b}}+2\,{\frac{da}{{b}^{2}}}-2\,{\frac{dx+c}{b \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }}-{\frac{d\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.01154, size = 97, normalized size = 3.34 \begin{align*} 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}{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)*sech(b*x+a)^2,x, algorithm="maxima")

[Out]

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/(b*(e^(-2*b*x
- 2*a) + 1))

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Fricas [B]  time = 2.07024, size = 429, normalized size = 14.79 \begin{align*} \frac{2 \, b d x \cosh \left (b x + a\right )^{2} + 4 \, b d x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 2 \, b d x \sinh \left (b x + a\right )^{2} - 2 \, b c -{\left (d \cosh \left (b x + a\right )^{2} + 2 \, d \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + d \sinh \left (b x + a\right )^{2} + d\right )} \log \left (\frac{2 \, \cosh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right )}{b^{2} \cosh \left (b x + a\right )^{2} + 2 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )^{2} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*d*x*cosh(b*x + a)^2 + 4*b*d*x*cosh(b*x + a)*sinh(b*x + a) + 2*b*d*x*sinh(b*x + a)^2 - 2*b*c - (d*cosh(b*x
+ a)^2 + 2*d*cosh(b*x + a)*sinh(b*x + a) + d*sinh(b*x + a)^2 + d)*log(2*cosh(b*x + a)/(cosh(b*x + a) - sinh(b
*x + a))))/(b^2*cosh(b*x + a)^2 + 2*b^2*cosh(b*x + a)*sinh(b*x + a) + b^2*sinh(b*x + a)^2 + b^2)

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

[Out]

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

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Giac [B]  time = 1.31527, size = 105, normalized size = 3.62 \begin{align*} \frac{2 \, b d x e^{\left (2 \, b x + 2 \, a\right )} - d e^{\left (2 \, b x + 2 \, a\right )} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) - 2 \, b c - d \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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