3.414 \(\int \frac{\text{sech}(c+d x)}{a+b \sqrt{\sinh (c+d x)}} \, dx\)

Optimal. Leaf size=286 \[ -\frac{2 a b^2 \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{d \left (a^4+b^4\right )}-\frac{b \left (a^2+b^2\right ) \log \left (\sinh (c+d x)-\sqrt{2} \sqrt{\sinh (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^4+b^4\right )}+\frac{b \left (a^2+b^2\right ) \log \left (\sinh (c+d x)+\sqrt{2} \sqrt{\sinh (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^4+b^4\right )}+\frac{a^3 \tan ^{-1}(\sinh (c+d x))}{d \left (a^4+b^4\right )}+\frac{b \left (a^2-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\sinh (c+d x)}\right )}{\sqrt{2} d \left (a^4+b^4\right )}-\frac{b \left (a^2-b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\sinh (c+d x)}+1\right )}{\sqrt{2} d \left (a^4+b^4\right )}+\frac{a b^2 \log (\cosh (c+d x))}{d \left (a^4+b^4\right )} \]

[Out]

(b*(a^2 - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Sinh[c + d*x]]])/(Sqrt[2]*(a^4 + b^4)*d) - (b*(a^2 - b^2)*ArcTan[1 + Sq
rt[2]*Sqrt[Sinh[c + d*x]]])/(Sqrt[2]*(a^4 + b^4)*d) + (a^3*ArcTan[Sinh[c + d*x]])/((a^4 + b^4)*d) + (a*b^2*Log
[Cosh[c + d*x]])/((a^4 + b^4)*d) - (2*a*b^2*Log[a + b*Sqrt[Sinh[c + d*x]]])/((a^4 + b^4)*d) - (b*(a^2 + b^2)*L
og[1 - Sqrt[2]*Sqrt[Sinh[c + d*x]] + Sinh[c + d*x]])/(2*Sqrt[2]*(a^4 + b^4)*d) + (b*(a^2 + b^2)*Log[1 + Sqrt[2
]*Sqrt[Sinh[c + d*x]] + Sinh[c + d*x]])/(2*Sqrt[2]*(a^4 + b^4)*d)

________________________________________________________________________________________

Rubi [A]  time = 0.482466, antiderivative size = 286, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 13, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.565, Rules used = {3223, 6725, 1876, 1168, 1162, 617, 204, 1165, 628, 1248, 635, 203, 260} \[ -\frac{2 a b^2 \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{d \left (a^4+b^4\right )}-\frac{b \left (a^2+b^2\right ) \log \left (\sinh (c+d x)-\sqrt{2} \sqrt{\sinh (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^4+b^4\right )}+\frac{b \left (a^2+b^2\right ) \log \left (\sinh (c+d x)+\sqrt{2} \sqrt{\sinh (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^4+b^4\right )}+\frac{a^3 \tan ^{-1}(\sinh (c+d x))}{d \left (a^4+b^4\right )}+\frac{b \left (a^2-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\sinh (c+d x)}\right )}{\sqrt{2} d \left (a^4+b^4\right )}-\frac{b \left (a^2-b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\sinh (c+d x)}+1\right )}{\sqrt{2} d \left (a^4+b^4\right )}+\frac{a b^2 \log (\cosh (c+d x))}{d \left (a^4+b^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(a^2 - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Sinh[c + d*x]]])/(Sqrt[2]*(a^4 + b^4)*d) - (b*(a^2 - b^2)*ArcTan[1 + Sq
rt[2]*Sqrt[Sinh[c + d*x]]])/(Sqrt[2]*(a^4 + b^4)*d) + (a^3*ArcTan[Sinh[c + d*x]])/((a^4 + b^4)*d) + (a*b^2*Log
[Cosh[c + d*x]])/((a^4 + b^4)*d) - (2*a*b^2*Log[a + b*Sqrt[Sinh[c + d*x]]])/((a^4 + b^4)*d) - (b*(a^2 + b^2)*L
og[1 - Sqrt[2]*Sqrt[Sinh[c + d*x]] + Sinh[c + d*x]])/(2*Sqrt[2]*(a^4 + b^4)*d) + (b*(a^2 + b^2)*Log[1 + Sqrt[2
]*Sqrt[Sinh[c + d*x]] + Sinh[c + d*x]])/(2*Sqrt[2]*(a^4 + b^4)*d)

Rule 3223

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x]
, x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0
] || IGtQ[p, 0] || IntegersQ[m, p])

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 1876

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[(x^ii*(Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii
]*x^(n/2)))/(a + b*x^n), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ
[n/2, 0] && Expon[Pq, x] < n

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
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])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{\text{sech}(c+d x)}{a+b \sqrt{\sinh (c+d x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b \sqrt{x}\right ) \left (1+x^2\right )} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{x}{(a+b x) \left (1+x^4\right )} \, dx,x,\sqrt{\sinh (c+d x)}\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (-\frac{a b^3}{\left (a^4+b^4\right ) (a+b x)}+\frac{b^3+a^3 x-a^2 b x^2+a b^2 x^3}{\left (a^4+b^4\right ) \left (1+x^4\right )}\right ) \, dx,x,\sqrt{\sinh (c+d x)}\right )}{d}\\ &=-\frac{2 a b^2 \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}+\frac{2 \operatorname{Subst}\left (\int \frac{b^3+a^3 x-a^2 b x^2+a b^2 x^3}{1+x^4} \, dx,x,\sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}\\ &=-\frac{2 a b^2 \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}+\frac{2 \operatorname{Subst}\left (\int \left (\frac{b^3-a^2 b x^2}{1+x^4}+\frac{x \left (a^3+a b^2 x^2\right )}{1+x^4}\right ) \, dx,x,\sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}\\ &=-\frac{2 a b^2 \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}+\frac{2 \operatorname{Subst}\left (\int \frac{b^3-a^2 b x^2}{1+x^4} \, dx,x,\sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}+\frac{2 \operatorname{Subst}\left (\int \frac{x \left (a^3+a b^2 x^2\right )}{1+x^4} \, dx,x,\sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}\\ &=-\frac{2 a b^2 \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}+\frac{\operatorname{Subst}\left (\int \frac{a^3+a b^2 x}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{\left (a^4+b^4\right ) d}-\frac{\left (b \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}+\frac{\left (b \left (a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}\\ &=-\frac{2 a b^2 \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}+\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{\left (a^4+b^4\right ) d}+\frac{\left (a b^2\right ) \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{\left (a^4+b^4\right ) d}-\frac{\left (b \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\sinh (c+d x)}\right )}{2 \left (a^4+b^4\right ) d}-\frac{\left (b \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\sinh (c+d x)}\right )}{2 \left (a^4+b^4\right ) d}-\frac{\left (b \left (a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\sinh (c+d x)}\right )}{2 \sqrt{2} \left (a^4+b^4\right ) d}-\frac{\left (b \left (a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\sinh (c+d x)}\right )}{2 \sqrt{2} \left (a^4+b^4\right ) d}\\ &=\frac{a^3 \tan ^{-1}(\sinh (c+d x))}{\left (a^4+b^4\right ) d}+\frac{a b^2 \log (\cosh (c+d x))}{\left (a^4+b^4\right ) d}-\frac{2 a b^2 \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}-\frac{b \left (a^2+b^2\right ) \log \left (1-\sqrt{2} \sqrt{\sinh (c+d x)}+\sinh (c+d x)\right )}{2 \sqrt{2} \left (a^4+b^4\right ) d}+\frac{b \left (a^2+b^2\right ) \log \left (1+\sqrt{2} \sqrt{\sinh (c+d x)}+\sinh (c+d x)\right )}{2 \sqrt{2} \left (a^4+b^4\right ) d}-\frac{\left (b \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\sinh (c+d x)}\right )}{\sqrt{2} \left (a^4+b^4\right ) d}+\frac{\left (b \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\sinh (c+d x)}\right )}{\sqrt{2} \left (a^4+b^4\right ) d}\\ &=\frac{b \left (a^2-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\sinh (c+d x)}\right )}{\sqrt{2} \left (a^4+b^4\right ) d}-\frac{b \left (a^2-b^2\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\sinh (c+d x)}\right )}{\sqrt{2} \left (a^4+b^4\right ) d}+\frac{a^3 \tan ^{-1}(\sinh (c+d x))}{\left (a^4+b^4\right ) d}+\frac{a b^2 \log (\cosh (c+d x))}{\left (a^4+b^4\right ) d}-\frac{2 a b^2 \log \left (a+b \sqrt{\sinh (c+d x)}\right )}{\left (a^4+b^4\right ) d}-\frac{b \left (a^2+b^2\right ) \log \left (1-\sqrt{2} \sqrt{\sinh (c+d x)}+\sinh (c+d x)\right )}{2 \sqrt{2} \left (a^4+b^4\right ) d}+\frac{b \left (a^2+b^2\right ) \log \left (1+\sqrt{2} \sqrt{\sinh (c+d x)}+\sinh (c+d x)\right )}{2 \sqrt{2} \left (a^4+b^4\right ) d}\\ \end{align*}

Mathematica [C]  time = 0.248221, size = 229, normalized size = 0.8 \[ \frac{3 \left (4 a^3 \tan ^{-1}(\sinh (c+d x))-8 a b^2 \log \left (a+b \sqrt{\sinh (c+d x)}\right )+4 a b^2 \log (\cosh (c+d x))-\sqrt{2} b^3 \log \left (\sinh (c+d x)-\sqrt{2} \sqrt{\sinh (c+d x)}+1\right )+\sqrt{2} b^3 \log \left (\sinh (c+d x)+\sqrt{2} \sqrt{\sinh (c+d x)}+1\right )-2 \sqrt{2} b^3 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\sinh (c+d x)}\right )+2 \sqrt{2} b^3 \tan ^{-1}\left (\sqrt{2} \sqrt{\sinh (c+d x)}+1\right )\right )-8 a^2 b \sinh ^{\frac{3}{2}}(c+d x) \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\sinh ^2(c+d x)\right )}{12 d \left (a^4+b^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-2*Sqrt[2]*b^3*ArcTan[1 - Sqrt[2]*Sqrt[Sinh[c + d*x]]] + 2*Sqrt[2]*b^3*ArcTan[1 + Sqrt[2]*Sqrt[Sinh[c + d*
x]]] + 4*a^3*ArcTan[Sinh[c + d*x]] + 4*a*b^2*Log[Cosh[c + d*x]] - 8*a*b^2*Log[a + b*Sqrt[Sinh[c + d*x]]] - Sqr
t[2]*b^3*Log[1 - Sqrt[2]*Sqrt[Sinh[c + d*x]] + Sinh[c + d*x]] + Sqrt[2]*b^3*Log[1 + Sqrt[2]*Sqrt[Sinh[c + d*x]
] + Sinh[c + d*x]]) - 8*a^2*b*Hypergeometric2F1[3/4, 1, 7/4, -Sinh[c + d*x]^2]*Sinh[c + d*x]^(3/2))/(12*(a^4 +
b^4)*d)

________________________________________________________________________________________

Maple [C]  time = 0.108, size = 206, normalized size = 0.7 \begin{align*} -{\frac{a{b}^{2}}{d \left ({a}^{4}+{b}^{4} \right ) }\ln \left ({a}^{2} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,{b}^{2}\tanh \left ( 1/2\,dx+c/2 \right ) -{a}^{2} \right ) }+{\frac{a{b}^{2}}{d \left ({a}^{4}+{b}^{4} \right ) }\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) }+2\,{\frac{{a}^{3}\arctan \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) }{d \left ({a}^{4}+{b}^{4} \right ) }}+{\frac{1}{d}\mbox{{\tt ` int/indef0`}} \left ({\frac{b \left ( -{b}^{2}\sinh \left ( dx+c \right ) +{a}^{2} \right ) }{2\,{a}^{2}{b}^{2}\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{2}-{b}^{4} \left ( \cosh \left ( dx+c \right ) \right ) ^{4}+ \left ( -{a}^{4}+{b}^{4} \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}\sqrt{\sinh \left ( dx+c \right ) }},\sinh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)/(a+b*sinh(d*x+c)^(1/2)),x)

[Out]

-a/d*b^2/(a^4+b^4)*ln(a^2*tanh(1/2*d*x+1/2*c)^2+2*b^2*tanh(1/2*d*x+1/2*c)-a^2)+a/d/(a^4+b^4)*b^2*ln(tanh(1/2*d
*x+1/2*c)^2+1)+2*a^3/d/(a^4+b^4)*arctan(tanh(1/2*d*x+1/2*c))+`int/indef0`(b*sinh(d*x+c)^(1/2)*(-b^2*sinh(d*x+c
)+a^2)/(2*a^2*b^2*sinh(d*x+c)*cosh(d*x+c)^2-b^4*cosh(d*x+c)^4+(-a^4+b^4)*cosh(d*x+c)^2),sinh(d*x+c))/d

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)/(a+b*sinh(d*x+c)^(1/2)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [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(sech(d*x+c)/(a+b*sinh(d*x+c)^(1/2)),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)/(a+b*sinh(d*x+c)**(1/2)),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)/(a+b*sinh(d*x+c)^(1/2)),x, algorithm="giac")

[Out]

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