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

Optimal. Leaf size=73 \[ \frac {2 \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d}-\frac {a x}{b^2}+\frac {\sinh (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2695, 2735, 2659, 205} \[ \frac {2 \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d}-\frac {a x}{b^2}+\frac {\sinh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*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 {\sinh ^2(c+d x)}{a+b \cosh (c+d x)} \, dx &=\frac {\sinh (c+d x)}{b d}+\frac {\int \frac {-b-a \cosh (c+d x)}{a+b \cosh (c+d x)} \, dx}{b}\\ &=-\frac {a x}{b^2}+\frac {\sinh (c+d x)}{b d}-\left (1-\frac {a^2}{b^2}\right ) \int \frac {1}{a+b \cosh (c+d x)} \, dx\\ &=-\frac {a x}{b^2}+\frac {\sinh (c+d x)}{b d}+\frac {\left (2 i \left (1-\frac {a^2}{b^2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{d}\\ &=-\frac {a x}{b^2}+\frac {2 \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d}+\frac {\sinh (c+d x)}{b d}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 69, normalized size = 0.95 \[ \frac {2 \sqrt {b^2-a^2} \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )-a (c+d x)+b \sinh (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.50, size = 415, normalized size = 5.68 \[ \left [-\frac {2 \, a d x \cosh \left (d x + c\right ) - b \cosh \left (d x + c\right )^{2} - b \sinh \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) + b}\right ) + 2 \, {\left (a d x - b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + b}{2 \, {\left (b^{2} d \cosh \left (d x + c\right ) + b^{2} d \sinh \left (d x + c\right )\right )}}, -\frac {2 \, a d x \cosh \left (d x + c\right ) - b \cosh \left (d x + c\right )^{2} - b \sinh \left (d x + c\right )^{2} + 4 \, \sqrt {-a^{2} + b^{2}} {\left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{a^{2} - b^{2}}\right ) + 2 \, {\left (a d x - b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + b}{2 \, {\left (b^{2} d \cosh \left (d x + c\right ) + b^{2} d \sinh \left (d x + c\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2/(a+b*cosh(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)), -1/2*(2*a*d*x*cosh(d*x + c) - b*cosh(d*x +
c)^2 - b*sinh(d*x + c)^2 + 4*sqrt(-a^2 + b^2)*(cosh(d*x + c) + sinh(d*x + c))*arctan(-sqrt(-a^2 + b^2)*(b*cos
h(d*x + c) + b*sinh(d*x + c) + a)/(a^2 - b^2)) + 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 = 0.14, size = 89, normalized size = 1.22 \[ -\frac {\frac {2 \, {\left (d x + c\right )} a}{b^{2}} - \frac {e^{\left (d x + c\right )}}{b} + \frac {e^{\left (-d x - c\right )}}{b} - \frac {4 \, {\left (a^{2} - b^{2}\right )} \arctan \left (\frac {b e^{\left (d x + c\right )} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2/(a+b*cosh(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 - 4*(a^2 - b^2)*arctan((b*e^(d*x + c) + a)/sqrt(-a^2
+ b^2))/(sqrt(-a^2 + b^2)*b^2))/d

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maple [B]  time = 0.08, size = 177, normalized size = 2.42 \[ -\frac {1}{d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,b^{2}}-\frac {1}{d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,b^{2}}+\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) a^{2}}{d \,b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{d \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.12, size = 176, normalized size = 2.41 \[ \frac {{\mathrm {e}}^{c+d\,x}}{2\,b\,d}-\frac {{\mathrm {e}}^{-c-d\,x}}{2\,b\,d}-\frac {a\,x}{b^2}+\frac {\ln \left (-\frac {2\,{\mathrm {e}}^{c+d\,x}\,\left (a^2-b^2\right )}{b^3}-\frac {2\,\sqrt {a+b}\,\sqrt {a-b}\,\left (b+a\,{\mathrm {e}}^{c+d\,x}\right )}{b^3}\right )\,\sqrt {a+b}\,\sqrt {a-b}}{b^2\,d}-\frac {\ln \left (\frac {2\,\sqrt {a+b}\,\sqrt {a-b}\,\left (b+a\,{\mathrm {e}}^{c+d\,x}\right )}{b^3}-\frac {2\,{\mathrm {e}}^{c+d\,x}\,\left (a^2-b^2\right )}{b^3}\right )\,\sqrt {a+b}\,\sqrt {a-b}}{b^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(c + d*x)/(2*b*d) - exp(- c - d*x)/(2*b*d) - (a*x)/b^2 + (log(- (2*exp(c + d*x)*(a^2 - b^2))/b^3 - (2*(a +
b)^(1/2)*(a - b)^(1/2)*(b + a*exp(c + d*x)))/b^3)*(a + b)^(1/2)*(a - b)^(1/2))/(b^2*d) - (log((2*(a + b)^(1/2)
*(a - b)^(1/2)*(b + a*exp(c + d*x)))/b^3 - (2*exp(c + d*x)*(a^2 - b^2))/b^3)*(a + b)^(1/2)*(a - b)^(1/2))/(b^2
*d)

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sympy [A]  time = 122.50, size = 1122, normalized size = 15.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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