3.443 \(\int \frac{\text{csch}(c+d x) \text{sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=113 \[ \frac{2 b^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a d \left (a^2+b^2\right )^{3/2}}-\frac{b \text{sech}(c+d x) (a \sinh (c+d x)+b)}{a d \left (a^2+b^2\right )}+\frac{\text{sech}(c+d x)}{a d}-\frac{\tanh ^{-1}(\cosh (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.259943, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2898, 2622, 321, 207, 2696, 12, 2660, 618, 204} \[ \frac{2 b^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a d \left (a^2+b^2\right )^{3/2}}-\frac{b \text{sech}(c+d x) (a \sinh (c+d x)+b)}{a d \left (a^2+b^2\right )}+\frac{\text{sech}(c+d x)}{a d}-\frac{\tanh ^{-1}(\cosh (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2898

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])
, x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/2, 0])

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

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

Rule 2696

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

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

Mathematica [A]  time = 0.264734, size = 171, normalized size = 1.51 \[ -\frac{-a b \sqrt{-a^2-b^2} \tanh (c+d x)+a^2 \sqrt{-a^2-b^2} \text{sech}(c+d x)+b^2 \sqrt{-a^2-b^2} \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )+a^2 \sqrt{-a^2-b^2} \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )-2 b^3 \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{a d \left (-a^2-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.001, size = 136, normalized size = 1.2 \begin{align*}{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{{b}^{3}}{da \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\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}{d \left ({a}^{2}+{b}^{2} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }}+2\,{\frac{a}{d \left ({a}^{2}+{b}^{2} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(csch(d*x+c)*sech(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.11662, size = 1455, normalized size = 12.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^3*b + 2*a*b^3 + (b^3*cosh(d*x + c)^2 + 2*b^3*cosh(d*x + c)*sinh(d*x + c) + b^3*sinh(d*x + c)^2 + b^3)*sqr
t(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^4 + a^2*b^2)*cosh
(d*x + c) - (a^4 + 2*a^2*b^2 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 2*(a^4 + 2*a^2*b^2 + b^4)*cosh(
d*x + c)*sinh(d*x + c) + (a^4 + 2*a^2*b^2 + b^4)*sinh(d*x + c)^2)*log(cosh(d*x + c) + sinh(d*x + c) + 1) + (a^
4 + 2*a^2*b^2 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 2*(a^4 + 2*a^2*b^2 + b^4)*cosh(d*x + c)*sinh(d
*x + c) + (a^4 + 2*a^2*b^2 + b^4)*sinh(d*x + c)^2)*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 2*(a^4 + a^2*b^2)*
sinh(d*x + c))/((a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c)^2 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c)*sinh
(d*x + c) + (a^5 + 2*a^3*b^2 + a*b^4)*d*sinh(d*x + c)^2 + (a^5 + 2*a^3*b^2 + a*b^4)*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)*sech(d*x+c)**2/(a+b*sinh(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.21899, size = 209, normalized size = 1.85 \begin{align*} -\frac{b^{3} \log \left (\frac{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{3} d + a b^{2} d\right )} \sqrt{a^{2} + b^{2}}} + \frac{2 \,{\left (a e^{\left (d x + c\right )} + b\right )}}{{\left (a^{2} d + b^{2} d\right )}{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}} - \frac{\log \left (e^{\left (d x + c\right )} + 1\right )}{a d} + \frac{\log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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