3.226 \(\int \frac{\sinh (c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=54 \[ \frac{2 a \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b d \sqrt{a^2+b^2}}+\frac{x}{b} \]

[Out]

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

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Rubi [A]  time = 0.0748817, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2735, 2660, 618, 204} \[ \frac{2 a \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b d \sqrt{a^2+b^2}}+\frac{x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2735

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

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 204

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

Rubi steps

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

Mathematica [A]  time = 0.105599, size = 64, normalized size = 1.19 \[ \frac{-\frac{2 a \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{d \sqrt{-a^2-b^2}}+\frac{c}{d}+x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 87, normalized size = 1.6 \begin{align*}{\frac{1}{bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{a}{bd\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.47331, size = 473, normalized size = 8.76 \begin{align*} \frac{{\left (a^{2} + b^{2}\right )} d x + \sqrt{a^{2} + b^{2}} a \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 + b^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2 + b^2)*d*x + sqrt(a^2 + b^2)*a*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 + b^3)*d)

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Sympy [A]  time = 169.228, size = 371, normalized size = 6.87 \begin{align*} \begin{cases} \tilde{\infty } x & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{x}{b} & \text{for}\: a = 0 \\- \frac{b^{2} d x \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{- b^{3} d \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} + i b^{2} d \sqrt{b^{2}}} + \frac{2 b^{2}}{- b^{3} d \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} + i b^{2} d \sqrt{b^{2}}} + \frac{i b d x \sqrt{b^{2}}}{- b^{3} d \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} + i b^{2} d \sqrt{b^{2}}} & \text{for}\: a = - \sqrt{- b^{2}} \\\frac{b^{2} d x \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{b^{3} d \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} + i b^{2} d \sqrt{b^{2}}} - \frac{2 b^{2}}{b^{3} d \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} + i b^{2} d \sqrt{b^{2}}} + \frac{i b d x \sqrt{b^{2}}}{b^{3} d \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} + i b^{2} d \sqrt{b^{2}}} & \text{for}\: a = \sqrt{- b^{2}} \\\frac{\cosh{\left (c + d x \right )}}{a d} & \text{for}\: b = 0 \\\frac{x \sinh{\left (c \right )}}{a + b \sinh{\left (c \right )}} & \text{for}\: d = 0 \\\frac{a \log{\left (\tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} - \frac{b}{a} - \frac{\sqrt{a^{2} + b^{2}}}{a} \right )}}{b d \sqrt{a^{2} + b^{2}}} - \frac{a \log{\left (\tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} - \frac{b}{a} + \frac{\sqrt{a^{2} + b^{2}}}{a} \right )}}{b d \sqrt{a^{2} + b^{2}}} + \frac{x}{b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (x/b, Eq(a, 0)), (-b**2*d*x*tanh(c/2 + d*x/2)/(-b**3*d*tanh
(c/2 + d*x/2) + I*b**2*d*sqrt(b**2)) + 2*b**2/(-b**3*d*tanh(c/2 + d*x/2) + I*b**2*d*sqrt(b**2)) + I*b*d*x*sqrt
(b**2)/(-b**3*d*tanh(c/2 + d*x/2) + I*b**2*d*sqrt(b**2)), Eq(a, -sqrt(-b**2))), (b**2*d*x*tanh(c/2 + d*x/2)/(b
**3*d*tanh(c/2 + d*x/2) + I*b**2*d*sqrt(b**2)) - 2*b**2/(b**3*d*tanh(c/2 + d*x/2) + I*b**2*d*sqrt(b**2)) + I*b
*d*x*sqrt(b**2)/(b**3*d*tanh(c/2 + d*x/2) + I*b**2*d*sqrt(b**2)), Eq(a, sqrt(-b**2))), (cosh(c + d*x)/(a*d), E
q(b, 0)), (x*sinh(c)/(a + b*sinh(c)), Eq(d, 0)), (a*log(tanh(c/2 + d*x/2) - b/a - sqrt(a**2 + b**2)/a)/(b*d*sq
rt(a**2 + b**2)) - a*log(tanh(c/2 + d*x/2) - b/a + sqrt(a**2 + b**2)/a)/(b*d*sqrt(a**2 + b**2)) + x/b, True))

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Giac [A]  time = 1.29623, size = 115, normalized size = 2.13 \begin{align*} -\frac{a \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}} b d} + \frac{d x + c}{b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*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)*b*d) + (d*x + c)/(b*d)