3.858 \(\int \frac {1}{a+b \cosh (c+d x) \sinh (c+d x)} \, dx\)

Optimal. Leaf size=44 \[ -\frac {2 \tanh ^{-1}\left (\frac {b-2 a \tanh (c+d x)}{\sqrt {4 a^2+b^2}}\right )}{d \sqrt {4 a^2+b^2}} \]

[Out]

-2*arctanh((b-2*a*tanh(d*x+c))/(4*a^2+b^2)^(1/2))/d/(4*a^2+b^2)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2666, 2660, 618, 204} \[ -\frac {2 \tanh ^{-1}\left (\frac {b-2 a \tanh (c+d x)}{\sqrt {4 a^2+b^2}}\right )}{d \sqrt {4 a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(-1),x]

[Out]

(-2*ArcTanh[(b - 2*a*Tanh[c + d*x])/Sqrt[4*a^2 + b^2]])/(Sqrt[4*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 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 48, normalized size = 1.09 \[ \frac {2 \tan ^{-1}\left (\frac {b-2 a \tanh (c+d x)}{\sqrt {-4 a^2-b^2}}\right )}{d \sqrt {-4 a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(-1),x]

[Out]

(2*ArcTan[(b - 2*a*Tanh[c + d*x])/Sqrt[-4*a^2 - b^2]])/(Sqrt[-4*a^2 - b^2]*d)

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fricas [B]  time = 0.76, size = 299, normalized size = 6.80 \[ \frac {\log \left (\frac {b^{2} \cosh \left (d x + c\right )^{4} + 4 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{2} \sinh \left (d x + c\right )^{4} + 4 \, a b \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} + 2 \, a b\right )} \sinh \left (d x + c\right )^{2} + 8 \, a^{2} + b^{2} + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} + 2 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 2 \, {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + 2 \, a\right )} \sqrt {4 \, a^{2} + b^{2}}}{b \cosh \left (d x + c\right )^{4} + 4 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b \sinh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, b \cosh \left (d x + c\right )^{2} + 2 \, a\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (b \cosh \left (d x + c\right )^{3} + 2 \, a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - b}\right )}{\sqrt {4 \, a^{2} + b^{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c)),x, algorithm="fricas")

[Out]

log((b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 + 4*a*b*cosh(d*x + c)^2 +
2*(3*b^2*cosh(d*x + c)^2 + 2*a*b)*sinh(d*x + c)^2 + 8*a^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 + 2*a*b*cosh(d*x + c
))*sinh(d*x + c) - 2*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + 2*a)*sqrt(4*a^
2 + b^2))/(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 4*a*cosh(d*x + c)^2 + 2
*(3*b*cosh(d*x + c)^2 + 2*a)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 + 2*a*cosh(d*x + c))*sinh(d*x + c) - b))/(
sqrt(4*a^2 + b^2)*d)

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giac [A]  time = 0.41, size = 79, normalized size = 1.80 \[ \frac {\log \left (\frac {{\left | 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a - 2 \, \sqrt {4 \, a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a + 2 \, \sqrt {4 \, a^{2} + b^{2}} \right |}}\right )}{\sqrt {4 \, a^{2} + b^{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c)),x, algorithm="giac")

[Out]

log(abs(2*b*e^(2*d*x + 2*c) + 4*a - 2*sqrt(4*a^2 + b^2))/abs(2*b*e^(2*d*x + 2*c) + 4*a + 2*sqrt(4*a^2 + b^2)))
/(sqrt(4*a^2 + b^2)*d)

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maple [B]  time = 0.66, size = 207, normalized size = 4.70 \[ -\frac {4 a^{2} \ln \left (-\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )}{d \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}}}-\frac {\ln \left (-\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right ) b^{2}}{d \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}}}+\frac {\ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}{d \sqrt {4 a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cosh(d*x+c)*sinh(d*x+c)),x)

[Out]

-4/d*a^2/(4*a^2+b^2)^(3/2)*ln(-tanh(1/2*d*x+1/2*c)^2*a+(4*a^2+b^2)^(1/2)*tanh(1/2*d*x+1/2*c)-tanh(1/2*d*x+1/2*
c)*b-a)-1/d/(4*a^2+b^2)^(3/2)*ln(-tanh(1/2*d*x+1/2*c)^2*a+(4*a^2+b^2)^(1/2)*tanh(1/2*d*x+1/2*c)-tanh(1/2*d*x+1
/2*c)*b-a)*b^2+1/d/(4*a^2+b^2)^(1/2)*ln(tanh(1/2*d*x+1/2*c)^2*a+(4*a^2+b^2)^(1/2)*tanh(1/2*d*x+1/2*c)+tanh(1/2
*d*x+1/2*c)*b+a)

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maxima [A]  time = 0.42, size = 73, normalized size = 1.66 \[ \frac {\log \left (\frac {b e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a - \sqrt {4 \, a^{2} + b^{2}}}{b e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a + \sqrt {4 \, a^{2} + b^{2}}}\right )}{\sqrt {4 \, a^{2} + b^{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c)),x, algorithm="maxima")

[Out]

log((b*e^(-2*d*x - 2*c) - 2*a - sqrt(4*a^2 + b^2))/(b*e^(-2*d*x - 2*c) - 2*a + sqrt(4*a^2 + b^2)))/(sqrt(4*a^2
+ b^2)*d)

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mupad [B]  time = 2.30, size = 343, normalized size = 7.80 \[ \frac {2\,\mathrm {atan}\left (\left (\frac {b^4\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}}{16}+\frac {a^2\,b^2\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}}{4}\right )\,\left (\frac {32\,a\,\left (8\,a^2+b^2\right )}{b^4\,d\,{\left (4\,a^2+b^2\right )}^2}-{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {64\,a\,\left (16\,d\,a^3+4\,d\,a\,b^2\right )}{b^5\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}\,\left (4\,a^2+b^2\right )\,\sqrt {-d^2\,\left (4\,a^2+b^2\right )}}+\frac {16\,\left (8\,a^2+b^2\right )\,\left (8\,a^2\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}+b^2\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}\right )}{b^5\,d\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}\,{\left (4\,a^2+b^2\right )}^2}\right )+\frac {64\,a\,\left (4\,d\,a^2\,b+d\,b^3\right )}{b^5\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}\,\left (4\,a^2+b^2\right )\,\sqrt {-d^2\,\left (4\,a^2+b^2\right )}}\right )\right )}{\sqrt {-4\,a^2\,d^2-b^2\,d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cosh(c + d*x)*sinh(c + d*x)),x)

[Out]

(2*atan(((b^4*(- 4*a^2*d^2 - b^2*d^2)^(1/2))/16 + (a^2*b^2*(- 4*a^2*d^2 - b^2*d^2)^(1/2))/4)*((32*a*(8*a^2 + b
^2))/(b^4*d*(4*a^2 + b^2)^2) - exp(2*c)*exp(2*d*x)*((64*a*(16*a^3*d + 4*a*b^2*d))/(b^5*(- 4*a^2*d^2 - b^2*d^2)
^(1/2)*(4*a^2 + b^2)*(-d^2*(4*a^2 + b^2))^(1/2)) + (16*(8*a^2 + b^2)*(8*a^2*(- 4*a^2*d^2 - b^2*d^2)^(1/2) + b^
2*(- 4*a^2*d^2 - b^2*d^2)^(1/2)))/(b^5*d*(- 4*a^2*d^2 - b^2*d^2)^(1/2)*(4*a^2 + b^2)^2)) + (64*a*(b^3*d + 4*a^
2*b*d))/(b^5*(- 4*a^2*d^2 - b^2*d^2)^(1/2)*(4*a^2 + b^2)*(-d^2*(4*a^2 + b^2))^(1/2)))))/(- 4*a^2*d^2 - b^2*d^2
)^(1/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c)),x)

[Out]

Timed out

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