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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0283861, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {4138, 14} \[ \frac{a \log (\cosh (c+d x))}{d}-\frac{b \text{sech}^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
 FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0218577, size = 29, normalized size = 1. \[ \frac{a \log (\cosh (c+d x))}{d}-\frac{b \text{sech}^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 29, normalized size = 1. \begin{align*} -{\frac{b \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{2\,d}}-{\frac{a\ln \left ({\rm sech} \left (dx+c\right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.1496, size = 36, normalized size = 1.24 \begin{align*} \frac{b \tanh \left (d x + c\right )^{2}}{2 \, d} + \frac{a \log \left (\cosh \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*tanh(d*x + c)^2/d + a*log(cosh(d*x + c))/d

________________________________________________________________________________________

Fricas [B]  time = 2.08514, size = 973, normalized size = 33.55 \begin{align*} -\frac{a d x \cosh \left (d x + c\right )^{4} + 4 \, a d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a d x \sinh \left (d x + c\right )^{4} + a d x + 2 \,{\left (a d x + b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, a d x \cosh \left (d x + c\right )^{2} + a d x + b\right )} \sinh \left (d x + c\right )^{2} -{\left (a \cosh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a \sinh \left (d x + c\right )^{4} + 2 \, a \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, a \cosh \left (d x + c\right )^{2} + a\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left (a \cosh \left (d x + c\right )^{3} + a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a\right )} \log \left (\frac{2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 4 \,{\left (a d x \cosh \left (d x + c\right )^{3} +{\left (a d x + b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{d \cosh \left (d x + c\right )^{4} + 4 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + d \sinh \left (d x + c\right )^{4} + 2 \, d \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, d \cosh \left (d x + c\right )^{2} + d\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left (d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*d*x*cosh(d*x + c)^4 + 4*a*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + a*d*x*sinh(d*x + c)^4 + a*d*x + 2*(a*d*x + b
)*cosh(d*x + c)^2 + 2*(3*a*d*x*cosh(d*x + c)^2 + a*d*x + b)*sinh(d*x + c)^2 - (a*cosh(d*x + c)^4 + 4*a*cosh(d*
x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*a*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a)*sinh(d*x + c)^2
 + 4*(a*cosh(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) + a)*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x +
c))) + 4*(a*d*x*cosh(d*x + c)^3 + (a*d*x + b)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^4 + 4*d*cosh(d*x
+ c)*sinh(d*x + c)^3 + d*sinh(d*x + c)^4 + 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^2 +
 4*(d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + d)

________________________________________________________________________________________

Sympy [A]  time = 0.726178, size = 42, normalized size = 1.45 \begin{align*} \begin{cases} a x - \frac{a \log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{d} - \frac{b \operatorname{sech}^{2}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (a + b \operatorname{sech}^{2}{\left (c \right )}\right ) \tanh{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x - a*log(tanh(c + d*x) + 1)/d - b*sech(c + d*x)**2/(2*d), Ne(d, 0)), (x*(a + b*sech(c)**2)*tanh(
c), True))

________________________________________________________________________________________

Giac [B]  time = 1.19476, size = 108, normalized size = 3.72 \begin{align*} -\frac{2 \, a d x - 2 \, a \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) + \frac{3 \, a e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a*d*x - 2*a*log(e^(2*d*x + 2*c) + 1) + (3*a*e^(4*d*x + 4*c) + 6*a*e^(2*d*x + 2*c) + 4*b*e^(2*d*x + 2*c
) + 3*a)/(e^(2*d*x + 2*c) + 1)^2)/d