3.573 $$\int \frac{a+b \text{csch}(x)}{c+d \sinh (x)} \, dx$$

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

[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])

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Rubi [A]  time = 0.167788, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.4, Rules used = {2828, 3001, 3770, 2660, 618, 206} $-\frac{2 (a c-b d) \tanh ^{-1}\left (\frac{d-c \tanh \left (\frac{x}{2}\right )}{\sqrt{c^2+d^2}}\right )}{c \sqrt{c^2+d^2}}-\frac{b \tanh ^{-1}(\cosh (x))}{c}$

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 2828

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 3001

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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 206

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

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)) \operatorname{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)) \operatorname{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*}

Mathematica [A]  time = 0.136297, size = 67, normalized size = 1.16 $\frac{\frac{2 (a c-b d) \tan ^{-1}\left (\frac{d-c \tanh \left (\frac{x}{2}\right )}{\sqrt{-c^2-d^2}}\right )}{\sqrt{-c^2-d^2}}+b \log \left (\tanh \left (\frac{x}{2}\right )\right )}{c}$

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 = 0.032, size = 86, normalized size = 1.5 \begin{align*}{\frac{b}{c}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+2\,{\frac{a}{\sqrt{{c}^{2}+{d}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,c\tanh \left ( x/2 \right ) -2\,d}{\sqrt{{c}^{2}+{d}^{2}}}} \right ) }-2\,{\frac{bd}{c\sqrt{{c}^{2}+{d}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,c\tanh \left ( x/2 \right ) -2\,d}{\sqrt{{c}^{2}+{d}^{2}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.79355, size = 482, normalized size = 8.31 \begin{align*} -\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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{csch}{\left (x \right )}}{c + d \sinh{\left (x \right )}}\, dx \end{align*}

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 = 1.17117, size = 122, normalized size = 2.1 \begin{align*} -\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{align*}

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)