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*ln(cosh(b*x+a))/b^2+(d*x+c)*tanh(b*x+a)/b

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Rubi [A]  time = 0.03, antiderivative size = 29, normalized size of antiderivative = 1.00, 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 3475

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

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) \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*}

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Mathematica [A]  time = 0.09, 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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fricas [B]  time = 0.76, size = 161, normalized size = 5.55 \[ \frac {2 \, b d x \cosh \left (b x + a\right )^{2} + 4 \, b d x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 2 \, b d x \sinh \left (b x + a\right )^{2} - 2 \, b c - {\left (d \cosh \left (b x + a\right )^{2} + 2 \, d \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + d \sinh \left (b x + a\right )^{2} + d\right )} \log \left (\frac {2 \, \cosh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right )}{b^{2} \cosh \left (b x + a\right )^{2} + 2 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )^{2} + b^{2}} \]

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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giac [B]  time = 0.14, size = 78, normalized size = 2.69 \[ \frac {2 \, b d x e^{\left (2 \, b x + 2 \, a\right )} - d e^{\left (2 \, b x + 2 \, a\right )} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) - 2 \, b c - d \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} \]

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)

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maple [A]  time = 0.18, size = 57, normalized size = 1.97 \[ \frac {2 d x}{b}+\frac {2 d a}{b^{2}}-\frac {2 \left (d x +c \right )}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}-\frac {d \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b^{2}} \]

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 = 0.58, size = 72, normalized size = 2.48 \[ d {\left (\frac {2 \, x e^{\left (2 \, b x + 2 \, a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} - \frac {\log \left ({\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-2 \, a\right )}\right )}{b^{2}}\right )} + \frac {2 \, c}{b {\left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}} \]

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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mupad [B]  time = 0.09, size = 50, normalized size = 1.72 \[ \frac {2\,d\,x}{b}-\frac {2\,\left (c+d\,x\right )}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {d\,\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1\right )}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/cosh(a + b*x)^2,x)

[Out]

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

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

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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