3.356 \(\int \frac{\text{sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=78 \[ \frac{2 a b \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac{\text{sech}(c+d x) (a-b \sinh (c+d x))}{d \left (a^2+b^2\right )} \]

[Out]

(2*a*b*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(3/2)*d) - (Sech[c + d*x]*(a - b*Sinh[
c + d*x]))/((a^2 + b^2)*d)

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Rubi [A]  time = 0.112229, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2866, 12, 2660, 618, 204} \[ \frac{2 a b \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac{\text{sech}(c+d x) (a-b \sinh (c+d x))}{d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[c + d*x]*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

(2*a*b*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(3/2)*d) - (Sech[c + d*x]*(a - b*Sinh[
c + d*x]))/((a^2 + b^2)*d)

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\text{sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx &=-\frac{\text{sech}(c+d x) (a-b \sinh (c+d x))}{\left (a^2+b^2\right ) d}-\frac{\int \frac{a b}{a+b \sinh (c+d x)} \, dx}{a^2+b^2}\\ &=-\frac{\text{sech}(c+d x) (a-b \sinh (c+d x))}{\left (a^2+b^2\right ) d}-\frac{(a b) \int \frac{1}{a+b \sinh (c+d x)} \, dx}{a^2+b^2}\\ &=-\frac{\text{sech}(c+d x) (a-b \sinh (c+d x))}{\left (a^2+b^2\right ) d}+\frac{(2 i a b) \operatorname{Subst}\left (\int \frac{1}{a-2 i b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{\left (a^2+b^2\right ) d}\\ &=-\frac{\text{sech}(c+d x) (a-b \sinh (c+d x))}{\left (a^2+b^2\right ) d}-\frac{(4 i a b) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,-2 i b+2 a \tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{2 a b \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}-\frac{\text{sech}(c+d x) (a-b \sinh (c+d x))}{\left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 0.192775, size = 104, normalized size = 1.33 \[ -\frac{b \sqrt{-a^2-b^2} \tanh (c+d x)-a \sqrt{-a^2-b^2} \text{sech}(c+d x)-2 a b \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{d \left (-a^2-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[c + d*x]*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

-((-2*a*b*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]] - a*Sqrt[-a^2 - b^2]*Sech[c + d*x] + b*Sqrt[-a^2
- b^2]*Tanh[c + d*x])/((-a^2 - b^2)^(3/2)*d))

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Maple [A]  time = 0.001, size = 100, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -4\,{\frac{ab}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) \sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-2\,{\frac{-\tanh \left ( 1/2\,dx+c/2 \right ) b+a}{ \left ({a}^{2}+{b}^{2} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

1/d*(-4*a*b/(2*a^2+2*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-2/(a^2+b^
2)*(-tanh(1/2*d*x+1/2*c)*b+a)/(tanh(1/2*d*x+1/2*c)^2+1))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.15745, size = 884, normalized size = 11.33 \begin{align*} -\frac{2 \, a^{2} b + 2 \, b^{3} -{\left (a b \cosh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a b \sinh \left (d x + c\right )^{2} + a b\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \,{\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \,{\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) + 2 \,{\left (a^{3} + a b^{2}\right )} \cosh \left (d x + c\right ) + 2 \,{\left (a^{3} + a b^{2}\right )} \sinh \left (d x + c\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \cosh \left (d x + c\right )^{2} + 2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \sinh \left (d x + c\right )^{2} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tanh(c + d*x)*sech(c + d*x)/(a + b*sinh(c + d*x)), x)

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Giac [A]  time = 1.27738, size = 158, normalized size = 2.03 \begin{align*} \frac{\frac{a b \log \left (\frac{{\left | -2 \, b e^{\left (d x + 2 \, c\right )} - 2 \, a e^{c} - 2 \, \sqrt{a^{2} + b^{2}} e^{c} \right |}}{{\left | -2 \, b e^{\left (d x + 2 \, c\right )} - 2 \, a e^{c} + 2 \, \sqrt{a^{2} + b^{2}} e^{c} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac{3}{2}}} - \frac{2 \,{\left (a e^{\left (d x + c\right )} + b\right )}}{{\left (a^{2} + b^{2}\right )}{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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