3.28 \(\int \frac{\sinh (c+d x)}{a+b \text{sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=47 \[ \frac{\cosh (c+d x)}{a d}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{b}}\right )}{a^{3/2} d} \]

[Out]

-((Sqrt[b]*ArcTan[(Sqrt[a]*Cosh[c + d*x])/Sqrt[b]])/(a^(3/2)*d)) + Cosh[c + d*x]/(a*d)

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Rubi [A]  time = 0.0484901, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4133, 321, 205} \[ \frac{\cosh (c+d x)}{a d}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{b}}\right )}{a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[b]*ArcTan[(Sqrt[a]*Cosh[c + d*x])/Sqrt[b]])/(a^(3/2)*d)) + Cosh[c + d*x]/(a*d)

Rule 4133

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = F
reeFactors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*(ff*x)^n)^p)/(ff*x)^(n*
p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p
]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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]

Rubi steps

\begin{align*} \int \frac{\sinh (c+d x)}{a+b \text{sech}^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{b+a x^2} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{\cosh (c+d x)}{a d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\cosh (c+d x)\right )}{a d}\\ &=-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{b}}\right )}{a^{3/2} d}+\frac{\cosh (c+d x)}{a d}\\ \end{align*}

Mathematica [C]  time = 1.04528, size = 328, normalized size = 6.98 \[ \frac{\text{sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (\frac{a \left (\tan ^{-1}\left (\frac{\sqrt{a}-i \sqrt{a+b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b}}\right )+\tan ^{-1}\left (\frac{\sqrt{a}+i \sqrt{a+b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b}}\right )\right )}{\sqrt{b}}-\frac{(a+4 b) \left (\tan ^{-1}\left (\frac{\sinh (c) \tanh \left (\frac{d x}{2}\right ) \left (\sqrt{a}-i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2}\right )+\cosh (c) \left (\sqrt{a}-i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2} \tanh \left (\frac{d x}{2}\right )\right )}{\sqrt{b}}\right )+\tan ^{-1}\left (\frac{\sinh (c) \tanh \left (\frac{d x}{2}\right ) \left (\sqrt{a}+i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2}\right )+\cosh (c) \left (\sqrt{a}+i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2} \tanh \left (\frac{d x}{2}\right )\right )}{\sqrt{b}}\right )\right )}{\sqrt{b}}+4 \sqrt{a} \cosh (c+d x)\right )}{8 a^{3/2} d \left (a+b \text{sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-(((a + 4*b)*(ArcTan[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*
(Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] + ArcTan[((Sqrt[a] + I*Sqrt[a +
b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[
c])^2]*Tanh[(d*x)/2]))/Sqrt[b]]))/Sqrt[b]) + (a*(ArcTan[(Sqrt[a] - I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]] +
ArcTan[(Sqrt[a] + I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]]))/Sqrt[b] + 4*Sqrt[a]*Cosh[c + d*x])*(a + 2*b + a
*Cosh[2*(c + d*x)])*Sech[c + d*x]^2)/(8*a^(3/2)*d*(a + b*Sech[c + d*x]^2))

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Maple [A]  time = 0.023, size = 44, normalized size = 0.9 \begin{align*}{\frac{1}{da{\rm sech} \left (dx+c\right )}}+{\frac{b}{da}\arctan \left ({b{\rm sech} \left (dx+c\right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )} e^{\left (-d x - c\right )}}{2 \, a d} - \frac{1}{2} \, \int \frac{4 \,{\left (b e^{\left (3 \, d x + 3 \, c\right )} - b e^{\left (d x + c\right )}\right )}}{a^{2} e^{\left (4 \, d x + 4 \, c\right )} + a^{2} + 2 \,{\left (a^{2} e^{\left (2 \, c\right )} + 2 \, a b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e^(2*d*x + 2*c) + 1)*e^(-d*x - c)/(a*d) - 1/2*integrate(4*(b*e^(3*d*x + 3*c) - b*e^(d*x + c))/(a^2*e^(4*d
*x + 4*c) + a^2 + 2*(a^2*e^(2*c) + 2*a*b*e^(2*c))*e^(2*d*x)), x)

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Fricas [B]  time = 2.78275, size = 1643, normalized size = 34.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(-b/a)*(cosh(d*x + c) + sinh(d*x + c))*log((a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 +
a*sinh(d*x + c)^4 + 2*(a - 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a - 2*b)*sinh(d*x + c)^2 + 4*(a*cos
h(d*x + c)^3 + (a - 2*b)*cosh(d*x + c))*sinh(d*x + c) - 4*(a*cosh(d*x + c)^3 + 3*a*cosh(d*x + c)*sinh(d*x + c)
^2 + a*sinh(d*x + c)^3 + a*cosh(d*x + c) + (3*a*cosh(d*x + c)^2 + a)*sinh(d*x + c))*sqrt(-b/a) + a)/(a*cosh(d*
x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d
*x + c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) + c
osh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)/(a*d*cosh(d*x + c) + a*d*sinh(d*x + c)),
1/2*(2*sqrt(b/a)*(cosh(d*x + c) + sinh(d*x + c))*arctan(1/2*(a*cosh(d*x + c)^3 + 3*a*cosh(d*x + c)*sinh(d*x +
c)^2 + a*sinh(d*x + c)^3 + (a + 4*b)*cosh(d*x + c) + (3*a*cosh(d*x + c)^2 + a + 4*b)*sinh(d*x + c))*sqrt(b/a)
/b) - 2*sqrt(b/a)*(cosh(d*x + c) + sinh(d*x + c))*arctan(1/2*(a*cosh(d*x + c) + a*sinh(d*x + c))*sqrt(b/a)/b)
+ cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)/(a*d*cosh(d*x + c) + a*d*sinh(d*x + c
))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh{\left (c + d x \right )}}{a + b \operatorname{sech}^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(c + d*x)/(a + b*sech(c + d*x)**2), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError