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

Optimal. Leaf size=104 \[ \frac{b^3 \log (a+b \sinh (c+d x))}{a^2 d \left (a^2+b^2\right )}-\frac{a \tan ^{-1}(\sinh (c+d x))}{d \left (a^2+b^2\right )}+\frac{b \log (\cosh (c+d x))}{d \left (a^2+b^2\right )}-\frac{b \log (\sinh (c+d x))}{a^2 d}-\frac{\text{csch}(c+d x)}{a d} \]

[Out]

-((a*ArcTan[Sinh[c + d*x]])/((a^2 + b^2)*d)) - Csch[c + d*x]/(a*d) + (b*Log[Cosh[c + d*x]])/((a^2 + b^2)*d) -
(b*Log[Sinh[c + d*x]])/(a^2*d) + (b^3*Log[a + b*Sinh[c + d*x]])/(a^2*(a^2 + b^2)*d)

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Rubi [A]  time = 0.169645, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2837, 12, 894, 635, 203, 260} \[ \frac{b^3 \log (a+b \sinh (c+d x))}{a^2 d \left (a^2+b^2\right )}-\frac{a \tan ^{-1}(\sinh (c+d x))}{d \left (a^2+b^2\right )}+\frac{b \log (\cosh (c+d x))}{d \left (a^2+b^2\right )}-\frac{b \log (\sinh (c+d x))}{a^2 d}-\frac{\text{csch}(c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*ArcTan[Sinh[c + d*x]])/((a^2 + b^2)*d)) - Csch[c + d*x]/(a*d) + (b*Log[Cosh[c + d*x]])/((a^2 + b^2)*d) -
(b*Log[Sinh[c + d*x]])/(a^2*d) + (b^3*Log[a + b*Sinh[c + d*x]])/(a^2*(a^2 + b^2)*d)

Rule 2837

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

Rule 12

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

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.593664, size = 160, normalized size = 1.54 \[ -\frac{b^3 \left (\frac{\log (\sinh (c+d x))}{a^2 b^2}-\frac{\left (a \sqrt{-b^2}+b^2\right ) \log \left (\sqrt{-b^2}-b \sinh (c+d x)\right )}{2 b^4 \left (a^2+b^2\right )}-\frac{\log (a+b \sinh (c+d x))}{a^2 \left (a^2+b^2\right )}-\frac{\left (\frac{a}{\sqrt{-b^2}}+1\right ) \log \left (\sqrt{-b^2}+b \sinh (c+d x)\right )}{2 b^2 \left (a^2+b^2\right )}+\frac{\text{csch}(c+d x)}{a b^3}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^3*(Csch[c + d*x]/(a*b^3) + Log[Sinh[c + d*x]]/(a^2*b^2) - ((b^2 + a*Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Sinh[c
+ d*x]])/(2*b^4*(a^2 + b^2)) - Log[a + b*Sinh[c + d*x]]/(a^2*(a^2 + b^2)) - ((1 + a/Sqrt[-b^2])*Log[Sqrt[-b^2
] + b*Sinh[c + d*x]])/(2*b^2*(a^2 + b^2))))/d)

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Maple [A]  time = 0.003, size = 159, normalized size = 1.5 \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 ) }+{\frac{{b}^{3}}{d{a}^{2} \left ({a}^{2}+{b}^{2} \right ) }\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) }+{\frac{b}{d \left ({a}^{2}+{b}^{2} \right ) }\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) }-2\,{\frac{a\arctan \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^2*sech(d*x+c)/(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))+1/d*b^3/a^2/(a^2+b^2
)*ln(tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)*b-a)+1/d/(a^2+b^2)*b*ln(tanh(1/2*d*x+1/2*c)^2+1)-2/d/(a^2+b
^2)*a*arctan(tanh(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.74286, size = 234, normalized size = 2.25 \begin{align*} \frac{b^{3} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{4} + a^{2} b^{2}\right )} d} + \frac{2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac{b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac{2 \, e^{\left (-d x - c\right )}}{{\left (a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )} d} - \frac{b \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} - \frac{b \log \left (e^{\left (-d x - c\right )} - 1\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)/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.8194, size = 1098, normalized size = 10.56 \begin{align*} -\frac{2 \,{\left (a^{3} \cosh \left (d x + c\right )^{2} + 2 \, a^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{3} \sinh \left (d x + c\right )^{2} - a^{3}\right )} \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + 2 \,{\left (a^{3} + a b^{2}\right )} \cosh \left (d x + c\right ) -{\left (b^{3} \cosh \left (d x + c\right )^{2} + 2 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{3} \sinh \left (d x + c\right )^{2} - b^{3}\right )} \log \left (\frac{2 \,{\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) -{\left (a^{2} b \cosh \left (d x + c\right )^{2} + 2 \, a^{2} b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2} b \sinh \left (d x + c\right )^{2} - a^{2} b\right )} \log \left (\frac{2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) -{\left (a^{2} b + b^{3} -{\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} - 2 \,{\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) -{\left (a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\frac{2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 2 \,{\left (a^{3} + a b^{2}\right )} \sinh \left (d x + c\right )}{{\left (a^{4} + a^{2} b^{2}\right )} d \cosh \left (d x + c\right )^{2} + 2 \,{\left (a^{4} + a^{2} b^{2}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a^{4} + a^{2} b^{2}\right )} d \sinh \left (d x + c\right )^{2} -{\left (a^{4} + a^{2} b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.31908, size = 284, normalized size = 2.73 \begin{align*} \frac{b^{4} \log \left ({\left | b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{4} b d + a^{2} b^{3} d} - \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} a}{2 \,{\left (a^{2} d + b^{2} d\right )}} + \frac{b \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{2 \,{\left (a^{2} d + b^{2} d\right )}} - \frac{b \log \left ({\left | e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} \right |}\right )}{a^{2} d} + \frac{b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 2 \, a}{a^{2} d{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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