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

[Out]

-a*arctan(sinh(d*x+c))/(a^2+b^2)/d-csch(d*x+c)/a/d+b*ln(cosh(d*x+c))/(a^2+b^2)/d-b*ln(sinh(d*x+c))/a^2/d+b^3*l
n(a+b*sinh(d*x+c))/a^2/(a^2+b^2)/d

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Rubi [A]  time = 0.17, antiderivative size = 104, normalized size of antiderivative = 1.00, 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 12

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

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]

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 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 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]

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*}

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Mathematica [A]  time = 0.60, size = 160, normalized size = 1.54 \[ -\frac {b^3 \left (\frac {\log (\sinh (c+d x))}{a^2 b^2}-\frac {\log (a+b \sinh (c+d x))}{a^2 \left (a^2+b^2\right )}-\frac {\left (\frac {a}{\sqrt {-b^2}}+1\right ) \log \left (\sqrt {-b^2}+b \sinh (c+d x)\right )}{2 b^2 \left (a^2+b^2\right )}-\frac {\left (a \sqrt {-b^2}+b^2\right ) \log \left (\sqrt {-b^2}-b \sinh (c+d x)\right )}{2 b^4 \left (a^2+b^2\right )}+\frac {\text {csch}(c+d x)}{a b^3}\right )}{d} \]

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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fricas [B]  time = 0.70, size = 441, normalized size = 4.24 \[ -\frac {2 \, {\left (a^{3} \cosh \left (d x + c\right )^{2} + 2 \, a^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{3} \sinh \left (d x + c\right )^{2} - a^{3}\right )} \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + 2 \, {\left (a^{3} + a b^{2}\right )} \cosh \left (d x + c\right ) - {\left (b^{3} \cosh \left (d x + c\right )^{2} + 2 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{3} \sinh \left (d x + c\right )^{2} - b^{3}\right )} \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - {\left (a^{2} b \cosh \left (d x + c\right )^{2} + 2 \, a^{2} b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2} b \sinh \left (d x + c\right )^{2} - a^{2} b\right )} \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) - {\left (a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 2 \, {\left (a^{3} + a b^{2}\right )} \sinh \left (d x + c\right )}{{\left (a^{4} + a^{2} b^{2}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{4} + a^{2} b^{2}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{4} + a^{2} b^{2}\right )} d \sinh \left (d x + c\right )^{2} - {\left (a^{4} + a^{2} b^{2}\right )} d} \]

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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giac [A]  time = 0.19, size = 200, normalized size = 1.92 \[ \frac {\frac {2 \, b^{4} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{4} b + a^{2} b^{3}} - \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} a}{a^{2} + b^{2}} + \frac {b \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{a^{2} + b^{2}} - \frac {2 \, b \log \left ({\left | e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} \right |}\right )}{a^{2}} + \frac {2 \, {\left (b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 2 \, a\right )}}{a^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}}{2 \, d} \]

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]

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

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maple [A]  time = 0.00, size = 159, normalized size = 1.53 \[ \frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {b^{3} \ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )}{d \,a^{2} \left (a^{2}+b^{2}\right )}-\frac {1}{2 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {b \ln \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a^{2}+b^{2}\right )}-\frac {2 a \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}+b^{2}\right )} \]

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/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)-1/2/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/(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 = 0.42, size = 173, normalized size = 1.66 \[ \frac {b^{3} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{4} + a^{2} b^{2}\right )} d} + \frac {2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {2 \, e^{\left (-d x - c\right )}}{{\left (a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )} d} - \frac {b \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} - \frac {b \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} d} \]

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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mupad [B]  time = 2.87, size = 142, normalized size = 1.37 \[ \frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}{b\,d+a\,d\,1{}\mathrm {i}}+\frac {b^3\,\ln \left (2\,a\,{\mathrm {e}}^{c+d\,x}-b+b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{d\,a^4+d\,a^2\,b^2}-\frac {2\,{\mathrm {e}}^{c+d\,x}}{a\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {b\,\ln \left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}{a^2\,d}+\frac {\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a\,d+b\,d\,1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(c + d*x)*sinh(c + d*x)^2*(a + b*sinh(c + d*x))),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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