### 3.578 $$\int \frac{a+b \text{sech}^2(x)}{c+d \cosh (x)} \, dx$$

Optimal. Leaf size=74 $\frac{2 \left (a c^2+b d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tanh \left (\frac{x}{2}\right )}{\sqrt{c+d}}\right )}{c^2 \sqrt{c-d} \sqrt{c+d}}-\frac{b d \tan ^{-1}(\sinh (x))}{c^2}+\frac{b \tanh (x)}{c}$

[Out]

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

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Rubi [A]  time = 0.245407, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.353, Rules used = {4234, 3056, 3001, 3770, 2659, 208} $\frac{2 \left (a c^2+b d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tanh \left (\frac{x}{2}\right )}{\sqrt{c+d}}\right )}{c^2 \sqrt{c-d} \sqrt{c+d}}-\frac{b d \tan ^{-1}(\sinh (x))}{c^2}+\frac{b \tanh (x)}{c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 4234

Int[(u_)*((A_) + (C_.)*sec[(a_.) + (b_.)*(x_)]^2), x_Symbol] :> Int[(ActivateTrig[u]*(C + A*Cos[a + b*x]^2))/C
os[a + b*x]^2, x] /; FreeQ[{a, b, A, C}, x] && KnownSineIntegrandQ[u, x]

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 208

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

Rubi steps

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

Mathematica [A]  time = 0.217917, size = 127, normalized size = 1.72 $-\frac{2 \text{sech}(x) \left (a \cosh ^2(x)+b\right ) \left (2 \cosh (x) \left (\left (a c^2+b d^2\right ) \tan ^{-1}\left (\frac{(c-d) \tanh \left (\frac{x}{2}\right )}{\sqrt{d^2-c^2}}\right )+b d \sqrt{d^2-c^2} \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )\right )-b c \sqrt{d^2-c^2} \sinh (x)\right )}{c^2 \sqrt{d^2-c^2} (a \cosh (2 x)+a+2 b)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 112, normalized size = 1.5 \begin{align*} 2\,{\frac{\tanh \left ( x/2 \right ) b}{c \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-2\,{\frac{bd\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{{c}^{2}}}+2\,{\frac{a}{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{ \left ( c-d \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) }+2\,{\frac{b{d}^{2}}{{c}^{2}\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{ \left ( c-d \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b/c*tanh(1/2*x)/(tanh(1/2*x)^2+1)-2*b/c^2*d*arctan(tanh(1/2*x))+2/((c+d)*(c-d))^(1/2)*arctanh((c-d)*tanh(1/2
*x)/((c+d)*(c-d))^(1/2))*a+2/c^2/((c+d)*(c-d))^(1/2)*arctanh((c-d)*tanh(1/2*x)/((c+d)*(c-d))^(1/2))*b*d^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((a+b*sech(x)^2)/(c+d*cosh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 6.2375, size = 1480, normalized size = 20. \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*b*c^3 - 2*b*c*d^2 - (a*c^2 + b*d^2 + (a*c^2 + b*d^2)*cosh(x)^2 + 2*(a*c^2 + b*d^2)*cosh(x)*sinh(x) + (a*c
^2 + b*d^2)*sinh(x)^2)*sqrt(c^2 - d^2)*log((d^2*cosh(x)^2 + d^2*sinh(x)^2 + 2*c*d*cosh(x) + 2*c^2 - d^2 + 2*(d
^2*cosh(x) + c*d)*sinh(x) - 2*sqrt(c^2 - d^2)*(d*cosh(x) + d*sinh(x) + c))/(d*cosh(x)^2 + d*sinh(x)^2 + 2*c*co
sh(x) + 2*(d*cosh(x) + c)*sinh(x) + d)) + 2*(b*c^2*d - b*d^3 + (b*c^2*d - b*d^3)*cosh(x)^2 + 2*(b*c^2*d - b*d^
3)*cosh(x)*sinh(x) + (b*c^2*d - b*d^3)*sinh(x)^2)*arctan(cosh(x) + sinh(x)))/(c^4 - c^2*d^2 + (c^4 - c^2*d^2)*
cosh(x)^2 + 2*(c^4 - c^2*d^2)*cosh(x)*sinh(x) + (c^4 - c^2*d^2)*sinh(x)^2), -2*(b*c^3 - b*c*d^2 + (a*c^2 + b*d
^2 + (a*c^2 + b*d^2)*cosh(x)^2 + 2*(a*c^2 + b*d^2)*cosh(x)*sinh(x) + (a*c^2 + b*d^2)*sinh(x)^2)*sqrt(-c^2 + d^
2)*arctan(-sqrt(-c^2 + d^2)*(d*cosh(x) + d*sinh(x) + c)/(c^2 - d^2)) + (b*c^2*d - b*d^3 + (b*c^2*d - b*d^3)*co
sh(x)^2 + 2*(b*c^2*d - b*d^3)*cosh(x)*sinh(x) + (b*c^2*d - b*d^3)*sinh(x)^2)*arctan(cosh(x) + sinh(x)))/(c^4 -
c^2*d^2 + (c^4 - c^2*d^2)*cosh(x)^2 + 2*(c^4 - c^2*d^2)*cosh(x)*sinh(x) + (c^4 - c^2*d^2)*sinh(x)^2)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13716, size = 96, normalized size = 1.3 \begin{align*} -\frac{2 \, b d \arctan \left (e^{x}\right )}{c^{2}} + \frac{2 \,{\left (a c^{2} + b d^{2}\right )} \arctan \left (\frac{d e^{x} + c}{\sqrt{-c^{2} + d^{2}}}\right )}{\sqrt{-c^{2} + d^{2}} c^{2}} - \frac{2 \, b}{c{\left (e^{\left (2 \, x\right )} + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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