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

Optimal. Leaf size=384 \[ \frac {2 a b^2}{d \left (a^4+b^4\right ) \left (a+b \sqrt {\sinh (c+d x)}\right )}-\frac {2 b^2 \left (3 a^4-b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{d \left (a^4+b^4\right )^2}+\frac {b^2 \left (3 a^4-b^4\right ) \log (\cosh (c+d x))}{d \left (a^4+b^4\right )^2}+\frac {a^2 \left (a^4-3 b^4\right ) \tan ^{-1}(\sinh (c+d x))}{d \left (a^4+b^4\right )^2}-\frac {a b \left (a^4+2 a^2 b^2-b^4\right ) \log \left (\sinh (c+d x)-\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )}{\sqrt {2} d \left (a^4+b^4\right )^2}+\frac {a b \left (a^4+2 a^2 b^2-b^4\right ) \log \left (\sinh (c+d x)+\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )}{\sqrt {2} d \left (a^4+b^4\right )^2}+\frac {\sqrt {2} a b \left (a^4-2 a^2 b^2-b^4\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\sinh (c+d x)}\right )}{d \left (a^4+b^4\right )^2}-\frac {\sqrt {2} a b \left (a^4-2 a^2 b^2-b^4\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )}{d \left (a^4+b^4\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.63, antiderivative size = 384, normalized size of antiderivative = 1.00, 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} \[ \frac {2 a b^2}{d \left (a^4+b^4\right ) \left (a+b \sqrt {\sinh (c+d x)}\right )}-\frac {a b \left (2 a^2 b^2+a^4-b^4\right ) \log \left (\sinh (c+d x)-\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )}{\sqrt {2} d \left (a^4+b^4\right )^2}+\frac {a b \left (2 a^2 b^2+a^4-b^4\right ) \log \left (\sinh (c+d x)+\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )}{\sqrt {2} d \left (a^4+b^4\right )^2}-\frac {2 b^2 \left (3 a^4-b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{d \left (a^4+b^4\right )^2}+\frac {a^2 \left (a^4-3 b^4\right ) \tan ^{-1}(\sinh (c+d x))}{d \left (a^4+b^4\right )^2}+\frac {\sqrt {2} a b \left (-2 a^2 b^2+a^4-b^4\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\sinh (c+d x)}\right )}{d \left (a^4+b^4\right )^2}-\frac {\sqrt {2} a b \left (-2 a^2 b^2+a^4-b^4\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )}{d \left (a^4+b^4\right )^2}+\frac {b^2 \left (3 a^4-b^4\right ) \log (\cosh (c+d x))}{d \left (a^4+b^4\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*a*b*(a^4 - 2*a^2*b^2 - b^4)*ArcTan[1 - Sqrt[2]*Sqrt[Sinh[c + d*x]]])/((a^4 + b^4)^2*d) - (Sqrt[2]*a*b
*(a^4 - 2*a^2*b^2 - b^4)*ArcTan[1 + Sqrt[2]*Sqrt[Sinh[c + d*x]]])/((a^4 + b^4)^2*d) + (a^2*(a^4 - 3*b^4)*ArcTa
n[Sinh[c + d*x]])/((a^4 + b^4)^2*d) + (b^2*(3*a^4 - b^4)*Log[Cosh[c + d*x]])/((a^4 + b^4)^2*d) - (2*b^2*(3*a^4
- b^4)*Log[a + b*Sqrt[Sinh[c + d*x]]])/((a^4 + b^4)^2*d) - (a*b*(a^4 + 2*a^2*b^2 - b^4)*Log[1 - Sqrt[2]*Sqrt[
Sinh[c + d*x]] + Sinh[c + d*x]])/(Sqrt[2]*(a^4 + b^4)^2*d) + (a*b*(a^4 + 2*a^2*b^2 - b^4)*Log[1 + Sqrt[2]*Sqrt
[Sinh[c + d*x]] + Sinh[c + d*x]])/(Sqrt[2]*(a^4 + b^4)^2*d) + (2*a*b^2)/((a^4 + b^4)*d*(a + b*Sqrt[Sinh[c + d*
x]]))

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 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 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]

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 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 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 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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {\text {sech}(c+d x)}{\left (a+b \sqrt {\sinh (c+d x)}\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b \sqrt {x}\right )^2 \left (1+x^2\right )} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {x}{(a+b x)^2 \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)^2}+\frac {-3 a^4 b^3+b^7}{\left (a^4+b^4\right )^2 (a+b x)}+\frac {4 a^3 b^3+a^2 \left (a^4-3 b^4\right ) x-2 a b \left (a^4-b^4\right ) x^2+b^2 \left (3 a^4-b^4\right ) x^3}{\left (a^4+b^4\right )^2 \left (1+x^4\right )}\right ) \, dx,x,\sqrt {\sinh (c+d x)}\right )}{d}\\ &=-\frac {2 b^2 \left (3 a^4-b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}+\frac {2 a b^2}{\left (a^4+b^4\right ) d \left (a+b \sqrt {\sinh (c+d x)}\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {4 a^3 b^3+a^2 \left (a^4-3 b^4\right ) x-2 a b \left (a^4-b^4\right ) x^2+b^2 \left (3 a^4-b^4\right ) x^3}{1+x^4} \, dx,x,\sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}\\ &=-\frac {2 b^2 \left (3 a^4-b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}+\frac {2 a b^2}{\left (a^4+b^4\right ) d \left (a+b \sqrt {\sinh (c+d x)}\right )}+\frac {2 \operatorname {Subst}\left (\int \left (\frac {4 a^3 b^3-2 a b \left (a^4-b^4\right ) x^2}{1+x^4}+\frac {x \left (a^2 \left (a^4-3 b^4\right )+b^2 \left (3 a^4-b^4\right ) x^2\right )}{1+x^4}\right ) \, dx,x,\sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}\\ &=-\frac {2 b^2 \left (3 a^4-b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}+\frac {2 a b^2}{\left (a^4+b^4\right ) d \left (a+b \sqrt {\sinh (c+d x)}\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {4 a^3 b^3-2 a b \left (a^4-b^4\right ) x^2}{1+x^4} \, dx,x,\sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}+\frac {2 \operatorname {Subst}\left (\int \frac {x \left (a^2 \left (a^4-3 b^4\right )+b^2 \left (3 a^4-b^4\right ) x^2\right )}{1+x^4} \, dx,x,\sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}\\ &=-\frac {2 b^2 \left (3 a^4-b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}+\frac {2 a b^2}{\left (a^4+b^4\right ) d \left (a+b \sqrt {\sinh (c+d x)}\right )}+\frac {\operatorname {Subst}\left (\int \frac {a^2 \left (a^4-3 b^4\right )+b^2 \left (3 a^4-b^4\right ) x}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{\left (a^4+b^4\right )^2 d}-\frac {\left (2 a b \left (a^4-2 a^2 b^2-b^4\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 )^2 d}+\frac {\left (2 a b \left (a^4+2 a^2 b^2-b^4\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 )^2 d}\\ &=-\frac {2 b^2 \left (3 a^4-b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}+\frac {2 a b^2}{\left (a^4+b^4\right ) d \left (a+b \sqrt {\sinh (c+d x)}\right )}+\frac {\left (a^2 \left (a^4-3 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{\left (a^4+b^4\right )^2 d}+\frac {\left (b^2 \left (3 a^4-b^4\right )\right ) \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{\left (a^4+b^4\right )^2 d}-\frac {\left (a b \left (a^4-2 a^2 b^2-b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}-\frac {\left (a b \left (a^4-2 a^2 b^2-b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}-\frac {\left (a b \left (a^4+2 a^2 b^2-b^4\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 )}{\sqrt {2} \left (a^4+b^4\right )^2 d}-\frac {\left (a b \left (a^4+2 a^2 b^2-b^4\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 )}{\sqrt {2} \left (a^4+b^4\right )^2 d}\\ &=\frac {a^2 \left (a^4-3 b^4\right ) \tan ^{-1}(\sinh (c+d x))}{\left (a^4+b^4\right )^2 d}+\frac {b^2 \left (3 a^4-b^4\right ) \log (\cosh (c+d x))}{\left (a^4+b^4\right )^2 d}-\frac {2 b^2 \left (3 a^4-b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}-\frac {a b \left (a^4+2 a^2 b^2-b^4\right ) \log \left (1-\sqrt {2} \sqrt {\sinh (c+d x)}+\sinh (c+d x)\right )}{\sqrt {2} \left (a^4+b^4\right )^2 d}+\frac {a b \left (a^4+2 a^2 b^2-b^4\right ) \log \left (1+\sqrt {2} \sqrt {\sinh (c+d x)}+\sinh (c+d x)\right )}{\sqrt {2} \left (a^4+b^4\right )^2 d}+\frac {2 a b^2}{\left (a^4+b^4\right ) d \left (a+b \sqrt {\sinh (c+d x)}\right )}-\frac {\left (\sqrt {2} a b \left (a^4-2 a^2 b^2-b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}+\frac {\left (\sqrt {2} a b \left (a^4-2 a^2 b^2-b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}\\ &=\frac {\sqrt {2} a b \left (a^4-2 a^2 b^2-b^4\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}-\frac {\sqrt {2} a b \left (a^4-2 a^2 b^2-b^4\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}+\frac {a^2 \left (a^4-3 b^4\right ) \tan ^{-1}(\sinh (c+d x))}{\left (a^4+b^4\right )^2 d}+\frac {b^2 \left (3 a^4-b^4\right ) \log (\cosh (c+d x))}{\left (a^4+b^4\right )^2 d}-\frac {2 b^2 \left (3 a^4-b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}-\frac {a b \left (a^4+2 a^2 b^2-b^4\right ) \log \left (1-\sqrt {2} \sqrt {\sinh (c+d x)}+\sinh (c+d x)\right )}{\sqrt {2} \left (a^4+b^4\right )^2 d}+\frac {a b \left (a^4+2 a^2 b^2-b^4\right ) \log \left (1+\sqrt {2} \sqrt {\sinh (c+d x)}+\sinh (c+d x)\right )}{\sqrt {2} \left (a^4+b^4\right )^2 d}+\frac {2 a b^2}{\left (a^4+b^4\right ) d \left (a+b \sqrt {\sinh (c+d x)}\right )}\\ \end {align*}

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Mathematica [C]  time = 0.72, size = 280, normalized size = 0.73 \[ \frac {-4 a b \left (a^4-b^4\right ) \sinh ^{\frac {3}{2}}(c+d x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\sinh ^2(c+d x)\right )+\frac {6 a b^2 \left (a^4+b^4\right )}{a+b \sqrt {\sinh (c+d x)}}+6 b^2 \left (b^4-3 a^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )-3 b^2 \left (b^4-3 a^4\right ) \log (\cosh (c+d x))-3 \sqrt {2} a^3 b^3 \left (\log \left (\sinh (c+d x)-\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )-\log \left (\sinh (c+d x)+\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )\right )-6 \sqrt {2} a^3 b^3 \left (\tan ^{-1}\left (1-\sqrt {2} \sqrt {\sinh (c+d x)}\right )-\tan ^{-1}\left (\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )\right )+3 a^2 \left (a^4-3 b^4\right ) \tan ^{-1}(\sinh (c+d x))}{3 d \left (a^4+b^4\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*Sqrt[2]*a^3*b^3*(ArcTan[1 - Sqrt[2]*Sqrt[Sinh[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Sinh[c + d*x]]]) + 3*a^
2*(a^4 - 3*b^4)*ArcTan[Sinh[c + d*x]] - 3*b^2*(-3*a^4 + b^4)*Log[Cosh[c + d*x]] + 6*b^2*(-3*a^4 + b^4)*Log[a +
b*Sqrt[Sinh[c + d*x]]] - 3*Sqrt[2]*a^3*b^3*(Log[1 - Sqrt[2]*Sqrt[Sinh[c + d*x]] + Sinh[c + d*x]] - Log[1 + Sq
rt[2]*Sqrt[Sinh[c + d*x]] + Sinh[c + d*x]]) + (6*a*b^2*(a^4 + b^4))/(a + b*Sqrt[Sinh[c + d*x]]) - 4*a*b*(a^4 -
b^4)*Hypergeometric2F1[3/4, 1, 7/4, -Sinh[c + d*x]^2]*Sinh[c + d*x]^(3/2))/(3*(a^4 + b^4)^2*d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Undef/Unsigned Inf encountered in limitLimit: Max order reached or unable to make series expansion
Error: Bad Argument Value

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maple [C]  time = 0.48, size = 567, normalized size = 1.48 \[ -\frac {4 b^{4} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{4}}{d \left (a^{4}+b^{4}\right )^{2} \left (a^{2} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a^{2}\right )}-\frac {4 b^{8} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a^{4}+b^{4}\right )^{2} \left (a^{2} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a^{2}\right )}-\frac {3 b^{2} \ln \left (a^{2} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a^{2}\right ) a^{4}}{d \left (a^{4}+b^{4}\right )^{2}}+\frac {b^{6} \ln \left (a^{2} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a^{2}\right )}{d \left (a^{4}+b^{4}\right )^{2}}+\frac {3 \ln \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{4} b^{2}}{d \left (a^{8}+2 a^{4} b^{4}+b^{8}\right )}-\frac {\ln \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{6}}{d \left (a^{8}+2 a^{4} b^{4}+b^{8}\right )}+\frac {2 \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{6}}{d \left (a^{8}+2 a^{4} b^{4}+b^{8}\right )}-\frac {6 \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b^{4}}{d \left (a^{8}+2 a^{4} b^{4}+b^{8}\right )}+\frac {\mathit {`\,int/indef0`\,}\left (\frac {2 a b \left (\sqrt {\sinh }\left (d x +c \right )\right ) \left (b^{4} \left (\sinh ^{2}\left (d x +c \right )\right )-2 a^{2} b^{2} \sinh \left (d x +c \right )+a^{4}\right )}{4 a^{2} b^{6} \sinh \left (d x +c \right ) \left (\cosh ^{4}\left (d x +c \right )\right )+\left (4 a^{6} b^{2}-4 a^{2} b^{6}\right ) \left (\cosh ^{2}\left (d x +c \right )\right ) \sinh \left (d x +c \right )-b^{8} \left (\cosh ^{6}\left (d x +c \right )\right )+\left (-6 a^{4} b^{4}+2 b^{8}\right ) \left (\cosh ^{4}\left (d x +c \right )\right )+\left (-a^{8}+6 a^{4} b^{4}-b^{8}\right ) \left (\cosh ^{2}\left (d x +c \right )\right )}, \sinh \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/d*b^4/(a^4+b^4)^2*tanh(1/2*d*x+1/2*c)/(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^4-4/d*b^8
/(a^4+b^4)^2*tanh(1/2*d*x+1/2*c)/(a^2*tanh(1/2*d*x+1/2*c)^2+2*b^2*tanh(1/2*d*x+1/2*c)-a^2)-3/d*b^2/(a^4+b^4)^2
*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^4+1/d*b^6/(a^4+b^4)^2*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)+3/d/(a^8+2*a^4*b^4+b^8)*ln(tanh(1/2*d*x+1/2*c)^2+1)*a^4*b^2-1/d/(a^8+2*a^4*
b^4+b^8)*ln(tanh(1/2*d*x+1/2*c)^2+1)*b^6+2/d/(a^8+2*a^4*b^4+b^8)*arctan(tanh(1/2*d*x+1/2*c))*a^6-6/d/(a^8+2*a^
4*b^4+b^8)*arctan(tanh(1/2*d*x+1/2*c))*a^2*b^4+`int/indef0`(2*a*b*sinh(d*x+c)^(1/2)*(b^4*sinh(d*x+c)^2-2*a^2*b
^2*sinh(d*x+c)+a^4)/(4*a^2*b^6*sinh(d*x+c)*cosh(d*x+c)^4+(4*a^6*b^2-4*a^2*b^6)*cosh(d*x+c)^2*sinh(d*x+c)-b^8*c
osh(d*x+c)^6+(-6*a^4*b^4+2*b^8)*cosh(d*x+c)^4+(-a^8+6*a^4*b^4-b^8)*cosh(d*x+c)^2),sinh(d*x+c))/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}\left (d x + c\right )}{{\left (b \sqrt {\sinh \left (d x + c\right )} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{\mathrm {cosh}\left (c+d\,x\right )\,{\left (a+b\,\sqrt {\mathrm {sinh}\left (c+d\,x\right )}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}{\left (c + d x \right )}}{\left (a + b \sqrt {\sinh {\left (c + d x \right )}}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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