3.3.41 \(\int \frac {\text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx\) [241]

Optimal. Leaf size=64 \[ -\frac {\tanh ^{-1}(\cosh (c+d x))}{a d}+\frac {2 b \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2826, 3855, 2739, 632, 210} \begin {gather*} \frac {2 b \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a d \sqrt {a^2+b^2}}-\frac {\tanh ^{-1}(\cosh (c+d x))}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 210

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

Rule 632

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 2739

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 2826

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 69, normalized size = 1.08 \begin {gather*} \frac {-\frac {2 b \text {ArcTan}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+\log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.00, size = 63, normalized size = 0.98

method	result	size
derivativedivides	\(\frac {-\frac {2 b \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\)	\(63\)
default	\(\frac {-\frac {2 b \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\)	\(63\)
risch	\(\frac {\ln \left ({\mathrm e}^{d x +c}-1\right )}{d a}+\frac {b \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d a}-\frac {b \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d a}-\frac {\ln \left ({\mathrm e}^{d x +c}+1\right )}{d a}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)/(a+b*sinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.47, size = 112, normalized size = 1.75 \begin {gather*} -\frac {b \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a d} - \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (61) = 122\).
time = 0.38, size = 223, normalized size = 3.48 \begin {gather*} \frac {\sqrt {a^{2} + b^{2}} b \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 ) - {\left (a^{2} + b^{2}\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) + {\left (a^{2} + b^{2}\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right )}{{\left (a^{3} + a b^{2}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {csch}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.45, size = 102, normalized size = 1.59 \begin {gather*} -\frac {\frac {b \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 )}{\sqrt {a^{2} + b^{2}} a} + \frac {\log \left (e^{\left (d x + c\right )} + 1\right )}{a} - \frac {\log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.46, size = 347, normalized size = 5.42 \begin {gather*} \frac {\ln \left (32\,a-32\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a\,d}-\frac {\ln \left (32\,a+32\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a\,d}+\frac {b\,\ln \left (128\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-64\,a^2\,b^3-64\,a^4\,b+32\,a\,b^3\,\sqrt {a^2+b^2}+64\,a^3\,b\,\sqrt {a^2+b^2}+160\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}+32\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-96\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{d\,a^3+d\,a\,b^2}-\frac {b\,\ln \left (64\,a^4\,b+64\,a^2\,b^3-128\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\,a\,b^3\,\sqrt {a^2+b^2}+64\,a^3\,b\,\sqrt {a^2+b^2}-160\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}-32\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-96\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{d\,a^3+d\,a\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(32*a - 32*a*exp(d*x)*exp(c))/(a*d) - log(32*a + 32*a*exp(d*x)*exp(c))/(a*d) + (b*log(128*a^5*exp(d*x)*exp(
c) - 64*a^2*b^3 - 64*a^4*b + 32*a*b^3*(a^2 + b^2)^(1/2) + 64*a^3*b*(a^2 + b^2)^(1/2) + 160*a^3*b^2*exp(d*x)*ex
p(c) - 128*a^4*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2) + 32*a*b^4*exp(d*x)*exp(c) - 96*a^2*b^2*exp(d*x)*exp(c)*(a^2
+ b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a^3*d + a*b^2*d) - (b*log(64*a^4*b + 64*a^2*b^3 - 128*a^5*exp(d*x)*exp(c) +
32*a*b^3*(a^2 + b^2)^(1/2) + 64*a^3*b*(a^2 + b^2)^(1/2) - 160*a^3*b^2*exp(d*x)*exp(c) - 128*a^4*exp(d*x)*exp(c
)*(a^2 + b^2)^(1/2) - 32*a*b^4*exp(d*x)*exp(c) - 96*a^2*b^2*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/
2))/(a^3*d + a*b^2*d)

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