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

Optimal. Leaf size=51 \[ \frac{B \log (a+b \sinh (x))}{b}-\frac{2 A \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\sqrt{a^2+b^2}} \]

[Out]

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

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Rubi [A]  time = 0.134812, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4401, 2660, 618, 206, 2668, 31} \[ \frac{B \log (a+b \sinh (x))}{b}-\frac{2 A \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\sqrt{a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0762386, size = 59, normalized size = 1.16 \[ \frac{2 A \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}+\frac{B \log (a+b \sinh (x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.025, size = 88, normalized size = 1.7 \begin{align*} -{\frac{B}{b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{B}{b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{B}{b}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }+2\,{\frac{A}{\sqrt{{a}^{2}+{b}^{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*cosh(x))/(a+b*sinh(x)),x)

[Out]

-B/b*ln(tanh(1/2*x)+1)-B/b*ln(tanh(1/2*x)-1)+1/b*B*ln(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)+2/(a^2+b^2)^(1/2)*arc
tanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))*A

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.09345, size = 458, normalized size = 8.98 \begin{align*} \frac{\sqrt{a^{2} + b^{2}} A b \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 ) -{\left (B a^{2} + B b^{2}\right )} x +{\left (B a^{2} + B b^{2}\right )} \log \left (\frac{2 \,{\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b + b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(a^2 + b^2)*A*b*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cosh(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)) - (B*a^2 + B*b^2)*x + (B*a^2 + B*b^2)*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x))))/(a^2*b
+ b^3)

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Sympy [A]  time = 99.5566, size = 695, normalized size = 13.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(A*log(tanh(x/2)) + B*x - 2*B*log(tanh(x/2) + 1) + B*log(tanh(x/2))), Eq(a, 0) & Eq(b, 0)), ((A
*log(tanh(x/2)) + B*x - 2*B*log(tanh(x/2) + 1) + B*log(tanh(x/2)))/b, Eq(a, 0)), (-2*A*tanh(x/2)/(-b*tanh(x/2)
+ I*b) - B*x*tanh(x/2)/(-b*tanh(x/2) + I*b) + I*B*x/(-b*tanh(x/2) + I*b) + 2*B*log(tanh(x/2) + 1)*tanh(x/2)/(
-b*tanh(x/2) + I*b) - 2*I*B*log(tanh(x/2) + 1)/(-b*tanh(x/2) + I*b) - 2*B*log(tanh(x/2) - I)*tanh(x/2)/(-b*tan
h(x/2) + I*b) + 2*I*B*log(tanh(x/2) - I)/(-b*tanh(x/2) + I*b), Eq(a, -I*b)), (2*A*tanh(x/2)/(b*tanh(x/2) + I*b
) + B*x*tanh(x/2)/(b*tanh(x/2) + I*b) + I*B*x/(b*tanh(x/2) + I*b) - 2*B*log(tanh(x/2) + 1)*tanh(x/2)/(b*tanh(x
/2) + I*b) - 2*I*B*log(tanh(x/2) + 1)/(b*tanh(x/2) + I*b) + 2*B*log(tanh(x/2) + I)*tanh(x/2)/(b*tanh(x/2) + I*
b) + 2*I*B*log(tanh(x/2) + I)/(b*tanh(x/2) + I*b), Eq(a, I*b)), ((A*x + B*sinh(x))/a, Eq(b, 0)), (-A*b*sqrt(a*
*2 + b**2)*log(tanh(x/2) - b/a - sqrt(a**2 + b**2)/a)/(a**2*b + b**3) + A*b*sqrt(a**2 + b**2)*log(tanh(x/2) -
b/a + sqrt(a**2 + b**2)/a)/(a**2*b + b**3) + B*a**2*x/(a**2*b + b**3) - 2*B*a**2*log(tanh(x/2) + 1)/(a**2*b +
b**3) + B*a**2*log(tanh(x/2) - b/a - sqrt(a**2 + b**2)/a)/(a**2*b + b**3) + B*a**2*log(tanh(x/2) - b/a + sqrt(
a**2 + b**2)/a)/(a**2*b + b**3) + B*b**2*x/(a**2*b + b**3) - 2*B*b**2*log(tanh(x/2) + 1)/(a**2*b + b**3) + B*b
**2*log(tanh(x/2) - b/a - sqrt(a**2 + b**2)/a)/(a**2*b + b**3) + B*b**2*log(tanh(x/2) - b/a + sqrt(a**2 + b**2
)/a)/(a**2*b + b**3), True))

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Giac [A]  time = 1.13363, size = 117, normalized size = 2.29 \begin{align*} \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}}} - \frac{B x}{b} + \frac{B \log \left ({\left | b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b \right |}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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) - B*x/b +
B*log(abs(b*e^(2*x) + 2*a*e^x - b))/b