3.6.73 \(\int \frac {a+b \text {csch}(x)}{c+d \sinh (x)} \, dx\) [573]

Optimal. Leaf size=58 \[ -\frac {b \tanh ^{-1}(\cosh (x))}{c}-\frac {2 (a c-b d) \tanh ^{-1}\left (\frac {d-c \tanh \left (\frac {x}{2}\right )}{\sqrt {c^2+d^2}}\right )}{c \sqrt {c^2+d^2}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2907, 3080, 3855, 2739, 632, 212} \begin {gather*} -\frac {2 (a c-b d) \tanh ^{-1}\left (\frac {d-c \tanh \left (\frac {x}{2}\right )}{\sqrt {c^2+d^2}}\right )}{c \sqrt {c^2+d^2}}-\frac {b \tanh ^{-1}(\cosh (x))}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 2907

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> In
t[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Int
egerQ[n]

Rule 3080

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 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 {a+b \text {csch}(x)}{c+d \sinh (x)} \, dx &=-\left (i \int \frac {\text {csch}(x) (i b+i a \sinh (x))}{c+d \sinh (x)} \, dx\right )\\ &=\frac {b \int \text {csch}(x) \, dx}{c}+\frac {(a c-b d) \int \frac {1}{c+d \sinh (x)} \, dx}{c}\\ &=-\frac {b \tanh ^{-1}(\cosh (x))}{c}+\frac {(2 (a c-b d)) \text {Subst}\left (\int \frac {1}{c+2 d x-c x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{c}\\ &=-\frac {b \tanh ^{-1}(\cosh (x))}{c}-\frac {(4 (a c-b d)) \text {Subst}\left (\int \frac {1}{4 \left (c^2+d^2\right )-x^2} \, dx,x,2 d-2 c \tanh \left (\frac {x}{2}\right )\right )}{c}\\ &=-\frac {b \tanh ^{-1}(\cosh (x))}{c}-\frac {2 (a c-b d) \tanh ^{-1}\left (\frac {d-c \tanh \left (\frac {x}{2}\right )}{\sqrt {c^2+d^2}}\right )}{c \sqrt {c^2+d^2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 67, normalized size = 1.16 \begin {gather*} \frac {\frac {2 (a c-b d) \text {ArcTan}\left (\frac {d-c \tanh \left (\frac {x}{2}\right )}{\sqrt {-c^2-d^2}}\right )}{\sqrt {-c^2-d^2}}+b \log \left (\tanh \left (\frac {x}{2}\right )\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.98, size = 58, normalized size = 1.00

method	result	size
default	\(\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{c}-\frac {\left (-2 a c +2 b d \right ) \arctanh \left (\frac {2 c \tanh \left (\frac {x}{2}\right )-2 d}{2 \sqrt {c^{2}+d^{2}}}\right )}{c \sqrt {c^{2}+d^{2}}}\)	\(58\)
risch	\(\frac {b \ln \left ({\mathrm e}^{x}-1\right )}{c}-\frac {b \ln \left ({\mathrm e}^{x}+1\right )}{c}+\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c -c^{2}-d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) a}{\sqrt {c^{2}+d^{2}}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c -c^{2}-d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) b d}{\sqrt {c^{2}+d^{2}}\, c}-\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c +c^{2}+d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) a}{\sqrt {c^{2}+d^{2}}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c +c^{2}+d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) b d}{\sqrt {c^{2}+d^{2}}\, c}\)	\(225\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (54) = 108\).
time = 0.48, size = 141, normalized size = 2.43 \begin {gather*} -b {\left (\frac {d \log \left (\frac {d e^{\left (-x\right )} - c - \sqrt {c^{2} + d^{2}}}{d e^{\left (-x\right )} - c + \sqrt {c^{2} + d^{2}}}\right )}{\sqrt {c^{2} + d^{2}} c} + \frac {\log \left (e^{\left (-x\right )} + 1\right )}{c} - \frac {\log \left (e^{\left (-x\right )} - 1\right )}{c}\right )} + \frac {a \log \left (\frac {d e^{\left (-x\right )} - c - \sqrt {c^{2} + d^{2}}}{d e^{\left (-x\right )} - c + \sqrt {c^{2} + d^{2}}}\right )}{\sqrt {c^{2} + d^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (54) = 108\).
time = 0.62, size = 172, normalized size = 2.97 \begin {gather*} -\frac {{\left (a c - b d\right )} \sqrt {c^{2} + d^{2}} \log \left (\frac {d^{2} \cosh \left (x\right )^{2} + d^{2} \sinh \left (x\right )^{2} + 2 \, c d \cosh \left (x\right ) + 2 \, c^{2} + d^{2} + 2 \, {\left (d^{2} \cosh \left (x\right ) + c d\right )} \sinh \left (x\right ) + 2 \, \sqrt {c^{2} + d^{2}} {\left (d \cosh \left (x\right ) + d \sinh \left (x\right ) + c\right )}}{d \cosh \left (x\right )^{2} + d \sinh \left (x\right )^{2} + 2 \, c \cosh \left (x\right ) + 2 \, {\left (d \cosh \left (x\right ) + c\right )} \sinh \left (x\right ) - d}\right ) + {\left (b c^{2} + b d^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - {\left (b c^{2} + b d^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{c^{3} + c d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.41, size = 90, normalized size = 1.55 \begin {gather*} -\frac {b \log \left (e^{x} + 1\right )}{c} + \frac {b \log \left ({\left | e^{x} - 1 \right |}\right )}{c} + \frac {{\left (a c - b d\right )} \log \left (\frac {{\left | 2 \, d e^{x} + 2 \, c - 2 \, \sqrt {c^{2} + d^{2}} \right |}}{{\left | 2 \, d e^{x} + 2 \, c + 2 \, \sqrt {c^{2} + d^{2}} \right |}}\right )}{\sqrt {c^{2} + d^{2}} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.37, size = 539, normalized size = 9.29 \begin {gather*} \frac {b\,\ln \left ({\mathrm {e}}^x-1\right )}{c}-\frac {b\,\ln \left ({\mathrm {e}}^x+1\right )}{c}-\frac {\ln \left (\frac {\left (\frac {32\,\left (a^2\,c^2\,d-2\,a\,b\,c\,d^2-4\,{\mathrm {e}}^x\,b^2\,c^3+2\,b^2\,c^2\,d-3\,{\mathrm {e}}^x\,b^2\,c\,d^2+2\,b^2\,d^3\right )}{d^5}-\frac {\left (a\,c-b\,d\right )\,\left (\frac {32\,c^2\,\left (2\,b\,d^2+4\,a\,c^2\,{\mathrm {e}}^x+a\,d^2\,{\mathrm {e}}^x-2\,a\,c\,d-3\,b\,c\,d\,{\mathrm {e}}^x\right )}{d^5}+\frac {32\,c\,\left (a\,c-b\,d\right )\,\left (-4\,{\mathrm {e}}^x\,c^3+3\,c^2\,d-3\,{\mathrm {e}}^x\,c\,d^2+2\,d^3\right )}{d^5\,\sqrt {c^2+d^2}}\right )}{c\,\sqrt {c^2+d^2}}\right )\,\left (a\,c-b\,d\right )}{c\,\sqrt {c^2+d^2}}+\frac {32\,b\,\left (a\,c-b\,d\right )\,\left (a\,d\,{\mathrm {e}}^x-2\,b\,d+4\,b\,c\,{\mathrm {e}}^x\right )}{d^5}\right )\,\left (a\,c-b\,d\right )\,\sqrt {c^2+d^2}}{c^3+c\,d^2}+\frac {\ln \left (\frac {32\,b\,\left (a\,c-b\,d\right )\,\left (a\,d\,{\mathrm {e}}^x-2\,b\,d+4\,b\,c\,{\mathrm {e}}^x\right )}{d^5}-\frac {\left (\frac {32\,\left (a^2\,c^2\,d-2\,a\,b\,c\,d^2-4\,{\mathrm {e}}^x\,b^2\,c^3+2\,b^2\,c^2\,d-3\,{\mathrm {e}}^x\,b^2\,c\,d^2+2\,b^2\,d^3\right )}{d^5}+\frac {\left (a\,c-b\,d\right )\,\left (\frac {32\,c^2\,\left (2\,b\,d^2+4\,a\,c^2\,{\mathrm {e}}^x+a\,d^2\,{\mathrm {e}}^x-2\,a\,c\,d-3\,b\,c\,d\,{\mathrm {e}}^x\right )}{d^5}-\frac {32\,c\,\left (a\,c-b\,d\right )\,\left (-4\,{\mathrm {e}}^x\,c^3+3\,c^2\,d-3\,{\mathrm {e}}^x\,c\,d^2+2\,d^3\right )}{d^5\,\sqrt {c^2+d^2}}\right )}{c\,\sqrt {c^2+d^2}}\right )\,\left (a\,c-b\,d\right )}{c\,\sqrt {c^2+d^2}}\right )\,\left (a\,c-b\,d\right )\,\sqrt {c^2+d^2}}{c^3+c\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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