3.249 \(\int \frac{A+B \tanh (x)}{a+b \sinh (x)} \, dx\)

Optimal. Leaf size=89 \[ -\frac{2 A \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\sqrt{a^2+b^2}}-\frac{a B \log (a+b \sinh (x))}{a^2+b^2}+\frac{b B \tan ^{-1}(\sinh (x))}{a^2+b^2}+\frac{a B \log (\cosh (x))}{a^2+b^2} \]

[Out]

(b*B*ArcTan[Sinh[x]])/(a^2 + b^2) - (2*A*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/Sqrt[a^2 + b^2] + (a*B*Lo
g[Cosh[x]])/(a^2 + b^2) - (a*B*Log[a + b*Sinh[x]])/(a^2 + b^2)

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Rubi [A]  time = 0.177193, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4401, 2660, 618, 206, 2721, 801, 635, 203, 260} \[ -\frac{2 A \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\sqrt{a^2+b^2}}-\frac{a B \log (a+b \sinh (x))}{a^2+b^2}+\frac{b B \tan ^{-1}(\sinh (x))}{a^2+b^2}+\frac{a B \log (\cosh (x))}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tanh[x])/(a + b*Sinh[x]),x]

[Out]

(b*B*ArcTan[Sinh[x]])/(a^2 + b^2) - (2*A*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/Sqrt[a^2 + b^2] + (a*B*Lo
g[Cosh[x]])/(a^2 + b^2) - (a*B*Log[a + b*Sinh[x]])/(a^2 + b^2)

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

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])

Rule 2721

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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{A+B \tanh (x)}{a+b \sinh (x)} \, dx &=\int \left (\frac{A}{a+b \sinh (x)}+\frac{B \tanh (x)}{a+b \sinh (x)}\right ) \, dx\\ &=A \int \frac{1}{a+b \sinh (x)} \, dx+B \int \frac{\tanh (x)}{a+b \sinh (x)} \, dx\\ &=(2 A) \operatorname{Subst}\left (\int \frac{1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )-B \operatorname{Subst}\left (\int \frac{x}{(a+x) \left (-b^2-x^2\right )} \, dx,x,b \sinh (x)\right )\\ &=-\left ((4 A) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac{x}{2}\right )\right )\right )-B \operatorname{Subst}\left (\int \left (\frac{a}{\left (a^2+b^2\right ) (a+x)}+\frac{-b^2-a x}{\left (a^2+b^2\right ) \left (b^2+x^2\right )}\right ) \, dx,x,b \sinh (x)\right )\\ &=-\frac{2 A \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\sqrt{a^2+b^2}}-\frac{a B \log (a+b \sinh (x))}{a^2+b^2}-\frac{B \operatorname{Subst}\left (\int \frac{-b^2-a x}{b^2+x^2} \, dx,x,b \sinh (x)\right )}{a^2+b^2}\\ &=-\frac{2 A \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\sqrt{a^2+b^2}}-\frac{a B \log (a+b \sinh (x))}{a^2+b^2}+\frac{(a B) \operatorname{Subst}\left (\int \frac{x}{b^2+x^2} \, dx,x,b \sinh (x)\right )}{a^2+b^2}+\frac{\left (b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{b^2+x^2} \, dx,x,b \sinh (x)\right )}{a^2+b^2}\\ &=\frac{b B \tan ^{-1}(\sinh (x))}{a^2+b^2}-\frac{2 A \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\sqrt{a^2+b^2}}+\frac{a B \log (\cosh (x))}{a^2+b^2}-\frac{a B \log (a+b \sinh (x))}{a^2+b^2}\\ \end{align*}

Mathematica [A]  time = 0.354621, size = 132, normalized size = 1.48 \[ -\frac{\cosh (x) (A+B \tanh (x)) \left (2 A \left (a^2+b^2\right ) \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )+2 b B \sqrt{-a^2-b^2} \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+a B \sqrt{-a^2-b^2} (\log (\cosh (x))-\log (a+b \sinh (x)))\right )}{\left (-a^2-b^2\right )^{3/2} (A \cosh (x)+B \sinh (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tanh[x])/(a + b*Sinh[x]),x]

[Out]

-((Cosh[x]*(2*b*Sqrt[-a^2 - b^2]*B*ArcTan[Tanh[x/2]] + 2*A*(a^2 + b^2)*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^
2]] + a*Sqrt[-a^2 - b^2]*B*(Log[Cosh[x]] - Log[a + b*Sinh[x]]))*(A + B*Tanh[x]))/((-a^2 - b^2)^(3/2)*(A*Cosh[x
] + B*Sinh[x])))

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Maple [A]  time = 0.032, size = 150, normalized size = 1.7 \begin{align*}{\frac{aB}{{a}^{2}+{b}^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }+2\,{\frac{Bb\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{{a}^{2}+{b}^{2}}}-{\frac{aB}{{a}^{2}+{b}^{2}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }+2\,{\frac{{a}^{2}A}{ \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+2\,{\frac{A{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tanh(x))/(a+b*sinh(x)),x)

[Out]

B/(a^2+b^2)*a*ln(tanh(1/2*x)^2+1)+2*B/(a^2+b^2)*b*arctan(tanh(1/2*x))-1/(a^2+b^2)*a*B*ln(a*tanh(1/2*x)^2-2*tan
h(1/2*x)*b-a)+2/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))*a^2*A+2/(a^2+b^2)^(3/2)*arc
tanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))*A*b^2

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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*tanh(x))/(a+b*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 15.2367, size = 510, normalized size = 5.73 \begin{align*} \frac{2 \, B b \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - B a \log \left (\frac{2 \,{\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + B a \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + \sqrt{a^{2} + b^{2}} A \log \left (\frac{b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \,{\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right )}{a^{2} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tanh(x))/(a+b*sinh(x)),x, algorithm="fricas")

[Out]

(2*B*b*arctan(cosh(x) + sinh(x)) - B*a*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x))) + B*a*log(2*cosh(x)/(cosh(x)
- sinh(x))) + sqrt(a^2 + b^2)*A*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cos
h(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x)
+ 2*(b*cosh(x) + a)*sinh(x) - b)))/(a^2 + b^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \tanh{\left (x \right )}}{a + b \sinh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tanh(x))/(a+b*sinh(x)),x)

[Out]

Integral((A + B*tanh(x))/(a + b*sinh(x)), x)

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Giac [A]  time = 1.17797, size = 166, normalized size = 1.87 \begin{align*} \frac{2 \, B b \arctan \left (e^{x}\right )}{a^{2} + b^{2}} + \frac{B a \log \left (e^{\left (2 \, x\right )} + 1\right )}{a^{2} + b^{2}} - \frac{B a \log \left ({\left | b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b \right |}\right )}{a^{2} + b^{2}} + \frac{A \log \left (\frac{{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tanh(x))/(a+b*sinh(x)),x, algorithm="giac")

[Out]

2*B*b*arctan(e^x)/(a^2 + b^2) + B*a*log(e^(2*x) + 1)/(a^2 + b^2) - B*a*log(abs(b*e^(2*x) + 2*a*e^x - b))/(a^2
+ b^2) + A*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/sqrt(a^2 + b^2)