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

Optimal. Leaf size=144 \[ -\frac{2 b^4 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a^2 d \left (a^2+b^2\right )^{3/2}}+\frac{b^2 \text{sech}(c+d x) (a \sinh (c+d x)+b)}{a^2 d \left (a^2+b^2\right )}-\frac{b \text{sech}(c+d x)}{a^2 d}+\frac{b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac{\tanh (c+d x)}{a d}-\frac{\coth (c+d x)}{a d} \]

[Out]

(b*ArcTanh[Cosh[c + d*x]])/(a^2*d) - (2*b^4*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2*(a^2 + b^
2)^(3/2)*d) - Coth[c + d*x]/(a*d) - (b*Sech[c + d*x])/(a^2*d) + (b^2*Sech[c + d*x]*(b + a*Sinh[c + d*x]))/(a^2
*(a^2 + b^2)*d) - Tanh[c + d*x]/(a*d)

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Rubi [A]  time = 0.310752, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 11, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.379, Rules used = {2898, 2622, 321, 207, 2620, 14, 2696, 12, 2660, 618, 204} \[ -\frac{2 b^4 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a^2 d \left (a^2+b^2\right )^{3/2}}+\frac{b^2 \text{sech}(c+d x) (a \sinh (c+d x)+b)}{a^2 d \left (a^2+b^2\right )}-\frac{b \text{sech}(c+d x)}{a^2 d}+\frac{b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac{\tanh (c+d x)}{a d}-\frac{\coth (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTanh[Cosh[c + d*x]])/(a^2*d) - (2*b^4*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2*(a^2 + b^
2)^(3/2)*d) - Coth[c + d*x]/(a*d) - (b*Sech[c + d*x])/(a^2*d) + (b^2*Sech[c + d*x]*(b + a*Sinh[c + d*x]))/(a^2
*(a^2 + b^2)*d) - Tanh[c + d*x]/(a*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 2620

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

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]]

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

Mathematica [A]  time = 2.64522, size = 135, normalized size = 0.94 \[ -\frac{\frac{4 b^4 \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{a^2 \left (-a^2-b^2\right )^{3/2}}+\frac{2 \text{sech}(c+d x) (a \sinh (c+d x)+b)}{a^2+b^2}+\frac{2 b \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{a^2}+\frac{\tanh \left (\frac{1}{2} (c+d x)\right )}{a}+\frac{\coth \left (\frac{1}{2} (c+d x)\right )}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*b^4*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]])/(a^2*(-a^2 - b^2)^(3/2)) + Coth[(c + d*x)/2]/a +
(2*b*Log[Tanh[(c + d*x)/2]])/a^2 + (2*Sech[c + d*x]*(b + a*Sinh[c + d*x]))/(a^2 + b^2) + Tanh[(c + d*x)/2]/a)/
(2*d)

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Maple [A]  time = 0.003, size = 174, normalized size = 1.2 \begin{align*} -{\frac{1}{2\,da}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{b}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+2\,{\frac{{b}^{4}}{d{a}^{2} \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{a\tanh \left ( 1/2\,dx+c/2 \right ) }{d \left ({a}^{2}+{b}^{2} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }}-2\,{\frac{b}{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)^2*sech(d*x+c)^2/(a+b*sinh(d*x+c)),x)

[Out]

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

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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)^2*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 = 3.11544, size = 2483, normalized size = 17.24 \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)^2*sech(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-(4*a^5 + 6*a^3*b^2 + 2*a*b^4 + 2*(a^4*b + a^2*b^3)*cosh(d*x + c)^3 + 2*(a^4*b + a^2*b^3)*sinh(d*x + c)^3 + 2*
(a^3*b^2 + a*b^4)*cosh(d*x + c)^2 + 2*(a^3*b^2 + a*b^4 + 3*(a^4*b + a^2*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 -
(b^4*cosh(d*x + c)^4 + 4*b^4*cosh(d*x + c)^3*sinh(d*x + c) + 6*b^4*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b^4*cos
h(d*x + c)*sinh(d*x + c)^3 + b^4*sinh(d*x + c)^4 - b^4)*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^4*b + a^2*b^3)*cosh(d*x + c) + (a^4*b + 2*a^2*b^3 + b^5 - (a^4*b + 2
*a^2*b^3 + b^5)*cosh(d*x + c)^4 - 4*(a^4*b + 2*a^2*b^3 + b^5)*cosh(d*x + c)^3*sinh(d*x + c) - 6*(a^4*b + 2*a^2
*b^3 + b^5)*cosh(d*x + c)^2*sinh(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3 + b^5)*cosh(d*x + c)*sinh(d*x + c)^3 - (a^4
*b + 2*a^2*b^3 + b^5)*sinh(d*x + c)^4)*log(cosh(d*x + c) + sinh(d*x + c) + 1) - (a^4*b + 2*a^2*b^3 + b^5 - (a^
4*b + 2*a^2*b^3 + b^5)*cosh(d*x + c)^4 - 4*(a^4*b + 2*a^2*b^3 + b^5)*cosh(d*x + c)^3*sinh(d*x + c) - 6*(a^4*b
+ 2*a^2*b^3 + b^5)*cosh(d*x + c)^2*sinh(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3 + b^5)*cosh(d*x + c)*sinh(d*x + c)^3
- (a^4*b + 2*a^2*b^3 + b^5)*sinh(d*x + c)^4)*log(cosh(d*x + c) + sinh(d*x + c) - 1) - 2*(a^4*b + a^2*b^3 - 3*
(a^4*b + a^2*b^3)*cosh(d*x + c)^2 - 2*(a^3*b^2 + a*b^4)*cosh(d*x + c))*sinh(d*x + c))/((a^6 + 2*a^4*b^2 + a^2*
b^4)*d*cosh(d*x + c)^4 + 4*(a^6 + 2*a^4*b^2 + a^2*b^4)*d*cosh(d*x + c)^3*sinh(d*x + c) + 6*(a^6 + 2*a^4*b^2 +
a^2*b^4)*d*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*(a^6 + 2*a^4*b^2 + a^2*b^4)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (
a^6 + 2*a^4*b^2 + a^2*b^4)*d*sinh(d*x + c)^4 - (a^6 + 2*a^4*b^2 + a^2*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)**2*sech(d*x+c)**2/(a+b*sinh(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.72368, size = 262, normalized size = 1.82 \begin{align*} -\frac{b^{4} \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^{4} d + a^{2} b^{2} d\right )} \sqrt{a^{2} + b^{2}}} - \frac{2 \,{\left (a b e^{\left (3 \, d x + 3 \, c\right )} + b^{2} e^{\left (2 \, d x + 2 \, c\right )} - a b e^{\left (d x + c\right )} + 2 \, a^{2} + b^{2}\right )}}{{\left (a^{3} d + a b^{2} d\right )}{\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )}} + \frac{b \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{2} d} - \frac{b \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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