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(1/2*d*x+1/2*c))/(a^2+b^2)^(1/2))/b/d/(a^2+b^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [B]  time = 0.59, size = 186, normalized size = 3.44 \[ \frac {{\left (a^{2} + b^{2}\right )} d x + \sqrt {a^{2} + b^{2}} a \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right )}{{\left (a^{2} b + b^{3}\right )} d} \]

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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giac [A]  time = 0.39, size = 84, normalized size = 1.56 \[ -\frac {\frac {a \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b} - \frac {d x + c}{b}}{d} \]

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*x + c)/b)/d

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maple [A]  time = 0.00, size = 87, normalized size = 1.61 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d b}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d b}-\frac {2 a \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d b \sqrt {a^{2}+b^{2}}} \]

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)+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))

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maxima [A]  time = 0.56, size = 85, normalized size = 1.57 \[ -\frac {a \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b d} + \frac {d x + c}{b d} \]

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]

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

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mupad [B]  time = 0.82, size = 121, normalized size = 2.24 \[ \frac {x}{b}-\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{b^2}-\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^2\,\sqrt {a^2+b^2}}\right )}{b\,d\,\sqrt {a^2+b^2}}+\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{b^2}+\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^2\,\sqrt {a^2+b^2}}\right )}{b\,d\,\sqrt {a^2+b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 68.10, size = 350, normalized size = 6.48 \[ \begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {\cosh {\left (c + d x \right )}}{a d} & \text {for}\: b = 0 \\\frac {b d x \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b^{2} d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - i b d \sqrt {b^{2}}} - \frac {2 b}{b^{2} d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - i b d \sqrt {b^{2}}} - \frac {i d x \sqrt {b^{2}}}{b^{2} d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - i b d \sqrt {b^{2}}} & \text {for}\: a = - \sqrt {- b^{2}} \\\frac {b d x \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b^{2} d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} + i b d \sqrt {b^{2}}} - \frac {2 b}{b^{2} d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} + i b d \sqrt {b^{2}}} + \frac {i d x \sqrt {b^{2}}}{b^{2} d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} + i b d \sqrt {b^{2}}} & \text {for}\: a = \sqrt {- b^{2}} \\\frac {x \sinh {\relax (c )}}{a + b \sinh {\relax (c )}} & \text {for}\: d = 0 \\\frac {x}{b} & \text {for}\: a = 0 \\\frac {a \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {b}{a} - \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{b d \sqrt {a^{2} + b^{2}}} - \frac {a \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {b}{a} + \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{b d \sqrt {a^{2} + b^{2}}} + \frac {x}{b} & \text {otherwise} \end {cases} \]

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)), (cosh(c + d*x)/(a*d), Eq(b, 0)), (b*d*x*tanh(c/2 + d*x/2)/(
b**2*d*tanh(c/2 + d*x/2) - I*b*d*sqrt(b**2)) - 2*b/(b**2*d*tanh(c/2 + d*x/2) - I*b*d*sqrt(b**2)) - I*d*x*sqrt(
b**2)/(b**2*d*tanh(c/2 + d*x/2) - I*b*d*sqrt(b**2)), Eq(a, -sqrt(-b**2))), (b*d*x*tanh(c/2 + d*x/2)/(b**2*d*ta
nh(c/2 + d*x/2) + I*b*d*sqrt(b**2)) - 2*b/(b**2*d*tanh(c/2 + d*x/2) + I*b*d*sqrt(b**2)) + I*d*x*sqrt(b**2)/(b*
*2*d*tanh(c/2 + d*x/2) + I*b*d*sqrt(b**2)), Eq(a, sqrt(-b**2))), (x*sinh(c)/(a + b*sinh(c)), Eq(d, 0)), (x/b,
Eq(a, 0)), (a*log(tanh(c/2 + d*x/2) - b/a - sqrt(a**2 + b**2)/a)/(b*d*sqrt(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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