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) - (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)

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Rubi [A]  time = 0.116962, antiderivative size = 68, normalized size of antiderivative = 1., 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 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]

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{\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*}

Mathematica [C]  time = 2.08619, size = 492, normalized size = 7.24 \[ \frac{\cosh (c+d x) \left (2 \sqrt{b^2} (a-i b) \sqrt{1+i \sinh (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a-i b} \sqrt{-\frac{b (\sinh (c+d x)+i)}{a-i b}}}{\sqrt{a+i b} \sqrt{-\frac{b (\sinh (c+d x)-i)}{a+i b}}}\right )+\sqrt{a+i b} \left (\sqrt{b^2} \sqrt{\frac{b (1+i \sinh (c+d x))}{b-i a}} \left (\sqrt{a-i b} \sqrt{1+i \sinh (c+d x)} \sqrt{-\frac{b (\sinh (c+d x)+i)}{a-i b}}-2 (-1)^{3/4} \sqrt{b} \sin ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{-\frac{b (\sinh (c+d x)+i)}{a-i b}}}{\sqrt{2} \sqrt{b}}\right )\right )-2 i b \sqrt{a-i b} \sqrt{1+i \sinh (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-i b} \sqrt{-\frac{b (\sinh (c+d x)+i)}{a-i b}}}{\sqrt{i b} \sqrt{-\frac{b (\sinh (c+d x)-i)}{a+i b}}}\right )\right )\right )}{b \sqrt{b^2} d \sqrt{a-i b} \sqrt{a+i b} \sqrt{1+i \sinh (c+d x)} \sqrt{-\frac{b (\sinh (c+d x)-i)}{a+i b}} \sqrt{-\frac{b (\sinh (c+d x)+i)}{a-i b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[c + d*x]*(2*(a - I*b)*Sqrt[b^2]*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]*((-2*I)*Sqrt
[a - I*b]*b*ArcTan[(Sqrt[(-I)*b]*Sqrt[-((b*(I + Sinh[c + d*x]))/(a - I*b))])/(Sqrt[I*b]*Sqrt[-((b*(-I + Sinh[c
+ d*x]))/(a + I*b))])]*Sqrt[1 + I*Sinh[c + d*x]] + Sqrt[b^2]*Sqrt[(b*(1 + I*Sinh[c + d*x]))/((-I)*a + b)]*(-2
*(-1)^(3/4)*Sqrt[b]*ArcSin[((-1)^(1/4)*Sqrt[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*Sqrt[b^2]*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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Maple [B]  time = 0.039, size = 174, normalized size = 2.6 \begin{align*}{\frac{1}{bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{a}{d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+2\,{\frac{{a}^{2}}{d{b}^{2}\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 ) }+2\,{\frac{1}{d\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 ) } \end{align*}

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.16567, size = 684, normalized size = 10.06 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.18774, size = 157, normalized size = 2.31 \begin{align*} -\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{{\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} d} \end{align*}

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]

-(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(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)