3.297 \(\int \frac {\cosh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=68 \[ -\frac {2 \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b^2 d}-\frac {a x}{b^2}+\frac {\cosh (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2695, 2735, 2660, 618, 204} \[ -\frac {2 \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b^2 d}-\frac {a x}{b^2}+\frac {\cosh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[c + d*x]^2/(a + b*Sinh[c + d*x]),x]

[Out]

-((a*x)/b^2) - (2*Sqrt[a^2 + b^2]*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(b^2*d) + Cosh[c + d*x]/
(b*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 2695

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(b*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

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

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Mathematica [C]  time = 1.42, size = 458, normalized size = 6.74 \[ \frac {\cosh (c+d x) \left (\sqrt {a+i b} \sqrt {-\frac {b (\sinh (c+d x)-i)}{a+i b}} \left (\sqrt {a-i b} \sqrt {1+i \sinh (c+d x)} \sqrt {-\frac {b (\sinh (c+d x)+i)}{a-i b}}-2 (-1)^{3/4} \sqrt {b} \sin ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a-i b} \sqrt {-\frac {b (\sinh (c+d x)+i)}{a-i b}}}{\sqrt {2} \sqrt {b}}\right )\right )-2 \sqrt {a-i b} \sqrt {a+i b} \sqrt {1+i \sinh (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {-\frac {b (\sinh (c+d x)+i)}{a-i b}}}{\sqrt {-\frac {b (\sinh (c+d x)-i)}{a+i b}}}\right )+2 (a-i b) \sqrt {1+i \sinh (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a-i b} \sqrt {-\frac {b (\sinh (c+d x)+i)}{a-i b}}}{\sqrt {a+i b} \sqrt {-\frac {b (\sinh (c+d x)-i)}{a+i b}}}\right )\right )}{b d \sqrt {a-i b} \sqrt {a+i b} \sqrt {1+i \sinh (c+d x)} \sqrt {-\frac {b (\sinh (c+d x)-i)}{a+i b}} \sqrt {-\frac {b (\sinh (c+d x)+i)}{a-i b}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[c + d*x]^2/(a + b*Sinh[c + d*x]),x]

[Out]

(Cosh[c + d*x]*(-2*Sqrt[a - I*b]*Sqrt[a + I*b]*ArcTanh[Sqrt[-((b*(I + Sinh[c + d*x]))/(a - I*b))]/Sqrt[-((b*(-
I + Sinh[c + d*x]))/(a + I*b))]]*Sqrt[1 + I*Sinh[c + d*x]] + 2*(a - I*b)*ArcTanh[(Sqrt[a - I*b]*Sqrt[-((b*(I +
Sinh[c + d*x]))/(a - I*b))])/(Sqrt[a + I*b]*Sqrt[-((b*(-I + Sinh[c + d*x]))/(a + I*b))])]*Sqrt[1 + I*Sinh[c +
d*x]] + Sqrt[a + I*b]*Sqrt[-((b*(-I + Sinh[c + d*x]))/(a + I*b))]*(-2*(-1)^(3/4)*Sqrt[b]*ArcSin[((-1)^(1/4)*S
qrt[a - I*b]*Sqrt[-((b*(I + Sinh[c + d*x]))/(a - I*b))])/(Sqrt[2]*Sqrt[b])] + Sqrt[a - I*b]*Sqrt[1 + I*Sinh[c
+ d*x]]*Sqrt[-((b*(I + Sinh[c + d*x]))/(a - I*b))])))/(Sqrt[a - I*b]*Sqrt[a + I*b]*b*d*Sqrt[1 + I*Sinh[c + d*x
]]*Sqrt[-((b*(-I + Sinh[c + d*x]))/(a + I*b))]*Sqrt[-((b*(I + Sinh[c + d*x]))/(a - I*b))])

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fricas [B]  time = 0.60, size = 259, normalized size = 3.81 \[ -\frac {2 \, a d x \cosh \left (d x + c\right ) - b \cosh \left (d x + c\right )^{2} - b \sinh \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) + 2 \, {\left (a d x - b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - b}{2 \, {\left (b^{2} d \cosh \left (d x + c\right ) + b^{2} d \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(2*a*d*x*cosh(d*x + c) - b*cosh(d*x + c)^2 - b*sinh(d*x + c)^2 - 2*sqrt(a^2 + b^2)*(cosh(d*x + c) + sinh(
d*x + c))*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)) + 2*(a*d*x - b*cosh(d*x + c
))*sinh(d*x + c) - b)/(b^2*d*cosh(d*x + c) + b^2*d*sinh(d*x + c))

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giac [A]  time = 0.19, size = 110, normalized size = 1.62 \[ -\frac {\frac {2 \, {\left (d x + c\right )} a}{b^{2}} - \frac {e^{\left (d x + c\right )}}{b} - \frac {e^{\left (-d x - c\right )}}{b} - \frac {2 \, \sqrt {a^{2} + b^{2}} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.07, size = 174, normalized size = 2.56 \[ -\frac {1}{d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,b^{2}}+\frac {1}{d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,b^{2}}+\frac {2 a^{2} \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,b^{2} \sqrt {a^{2}+b^{2}}}+\frac {2 \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^2/(a+b*sinh(d*x+c)),x)

[Out]

-1/d/b/(tanh(1/2*d*x+1/2*c)-1)+1/d*a/b^2*ln(tanh(1/2*d*x+1/2*c)-1)+1/d/b/(tanh(1/2*d*x+1/2*c)+1)-1/d*a/b^2*ln(
tanh(1/2*d*x+1/2*c)+1)+2/d*a^2/b^2/(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))+
2/d/(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 = 0.42, size = 116, normalized size = 1.71 \[ -\frac {{\left (d x + c\right )} a}{b^{2} d} + \frac {e^{\left (d x + c\right )}}{2 \, b d} + \frac {e^{\left (-d x - c\right )}}{2 \, b d} + \frac {\sqrt {a^{2} + b^{2}} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.42, size = 121, normalized size = 1.78 \[ \frac {{\mathrm {e}}^{c+d\,x}}{2\,b\,d}-\frac {2\,\mathrm {atan}\left (\frac {a\,\sqrt {-b^4\,d^2}}{b^2\,d\,\sqrt {a^2+b^2}}+\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-b^4\,d^2}}{b\,d\,\sqrt {a^2+b^2}}\right )\,\sqrt {a^2+b^2}}{\sqrt {-b^4\,d^2}}+\frac {{\mathrm {e}}^{-c-d\,x}}{2\,b\,d}-\frac {a\,x}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(c + d*x)^2/(a + b*sinh(c + d*x)),x)

[Out]

exp(c + d*x)/(2*b*d) - (2*atan((a*(-b^4*d^2)^(1/2))/(b^2*d*(a^2 + b^2)^(1/2)) + (exp(d*x)*exp(c)*(-b^4*d^2)^(1
/2))/(b*d*(a^2 + b^2)^(1/2)))*(a^2 + b^2)^(1/2))/(-b^4*d^2)^(1/2) + exp(- c - d*x)/(2*b*d) - (a*x)/b^2

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sympy [A]  time = 156.44, size = 503, normalized size = 7.40 \[ \begin {cases} \frac {\tilde {\infty } x \cosh ^{2}{\relax (c )}}{\sinh {\relax (c )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {- \frac {x \sinh ^{2}{\left (c + d x \right )}}{2} + \frac {x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {\sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d}}{a} & \text {for}\: b = 0 \\\frac {x \cosh ^{2}{\relax (c )}}{a + b \sinh {\relax (c )}} & \text {for}\: d = 0 \\\frac {\frac {\log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )} \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - d} - \frac {\log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )}}{d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - d} - \frac {2}{d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - d}}{b} & \text {for}\: a = 0 \\- \frac {a d x \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b^{2} d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - b^{2} d} + \frac {a d x}{b^{2} d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - b^{2} d} - \frac {2 b}{b^{2} d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - b^{2} d} - \frac {\sqrt {a^{2} + b^{2}} \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {b}{a} - \frac {\sqrt {a^{2} + b^{2}}}{a} \right )} \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b^{2} d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - b^{2} d} + \frac {\sqrt {a^{2} + b^{2}} \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {b}{a} - \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{b^{2} d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - b^{2} d} + \frac {\sqrt {a^{2} + b^{2}} \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {b}{a} + \frac {\sqrt {a^{2} + b^{2}}}{a} \right )} \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b^{2} d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - b^{2} d} - \frac {\sqrt {a^{2} + b^{2}} \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {b}{a} + \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{b^{2} d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - b^{2} d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)**2/(a+b*sinh(d*x+c)),x)

[Out]

Piecewise((zoo*x*cosh(c)**2/sinh(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((-x*sinh(c + d*x)**2/2 + x*cosh(c + d*x
)**2/2 + sinh(c + d*x)*cosh(c + d*x)/(2*d))/a, Eq(b, 0)), (x*cosh(c)**2/(a + b*sinh(c)), Eq(d, 0)), ((log(tanh
(c/2 + d*x/2))*tanh(c/2 + d*x/2)**2/(d*tanh(c/2 + d*x/2)**2 - d) - log(tanh(c/2 + d*x/2))/(d*tanh(c/2 + d*x/2)
**2 - d) - 2/(d*tanh(c/2 + d*x/2)**2 - d))/b, Eq(a, 0)), (-a*d*x*tanh(c/2 + d*x/2)**2/(b**2*d*tanh(c/2 + d*x/2
)**2 - b**2*d) + a*d*x/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d) - 2*b/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d) - s
qrt(a**2 + b**2)*log(tanh(c/2 + d*x/2) - b/a - sqrt(a**2 + b**2)/a)*tanh(c/2 + d*x/2)**2/(b**2*d*tanh(c/2 + d*
x/2)**2 - b**2*d) + sqrt(a**2 + b**2)*log(tanh(c/2 + d*x/2) - b/a - sqrt(a**2 + b**2)/a)/(b**2*d*tanh(c/2 + d*
x/2)**2 - b**2*d) + sqrt(a**2 + b**2)*log(tanh(c/2 + d*x/2) - b/a + sqrt(a**2 + b**2)/a)*tanh(c/2 + d*x/2)**2/
(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d) - sqrt(a**2 + b**2)*log(tanh(c/2 + d*x/2) - b/a + sqrt(a**2 + b**2)/a)/
(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d), True))

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