\(\int \frac {a+b \text {sech}(c+d \sqrt {x})}{\sqrt {x}} \, dx\) [54]

Optimal result

Integrand size = 20, antiderivative size = 26 \[ \int \frac {a+b \text {sech}\left (c+d \sqrt {x}\right )}{\sqrt {x}} \, dx=2 a \sqrt {x}+\frac {2 b \arctan \left (\sinh \left (c+d \sqrt {x}\right )\right )}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {14, 5544, 3855} \[ \int \frac {a+b \text {sech}\left (c+d \sqrt {x}\right )}{\sqrt {x}} \, dx=2 a \sqrt {x}+\frac {2 b \arctan \left (\sinh \left (c+d \sqrt {x}\right )\right )}{d} \]

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 3855

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

Rule 5544

Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sech[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a}{\sqrt {x}}+\frac {b \text {sech}\left (c+d \sqrt {x}\right )}{\sqrt {x}}\right ) \, dx \\ & = 2 a \sqrt {x}+b \int \frac {\text {sech}\left (c+d \sqrt {x}\right )}{\sqrt {x}} \, dx \\ & = 2 a \sqrt {x}+(2 b) \text {Subst}\left (\int \text {sech}(c+d x) \, dx,x,\sqrt {x}\right ) \\ & = 2 a \sqrt {x}+\frac {2 b \arctan \left (\sinh \left (c+d \sqrt {x}\right )\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \text {sech}\left (c+d \sqrt {x}\right )}{\sqrt {x}} \, dx=2 a \sqrt {x}+\frac {2 b \arctan \left (\sinh \left (c+d \sqrt {x}\right )\right )}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {2 b \arctan \left (\sinh \left (c +d \sqrt {x}\right )\right )}{d}+2 a \sqrt {x}\)	\(23\)
default	\(\frac {2 b \arctan \left (\sinh \left (c +d \sqrt {x}\right )\right )}{d}+2 a \sqrt {x}\)	\(23\)
parts	\(\frac {2 b \arctan \left (\sinh \left (c +d \sqrt {x}\right )\right )}{d}+2 a \sqrt {x}\)	\(23\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {a+b \text {sech}\left (c+d \sqrt {x}\right )}{\sqrt {x}} \, dx=\frac {2 \, {\left (a d \sqrt {x} + 2 \, b \arctan \left (\cosh \left (d \sqrt {x} + c\right ) + \sinh \left (d \sqrt {x} + c\right )\right )\right )}}{d} \]

[In]

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

[Out]

2*(a*d*sqrt(x) + 2*b*arctan(cosh(d*sqrt(x) + c) + sinh(d*sqrt(x) + c)))/d

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {a+b \text {sech}\left (c+d \sqrt {x}\right )}{\sqrt {x}} \, dx=2 a \sqrt {x} + 2 b \left (\begin {cases} \sqrt {x} \operatorname {sech}{\left (c \right )} & \text {for}\: d = 0 \\\frac {2 \operatorname {atan}{\left (\tanh {\left (\frac {c}{2} + \frac {d \sqrt {x}}{2} \right )} \right )}}{d} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

2*a*sqrt(x) + 2*b*Piecewise((sqrt(x)*sech(c), Eq(d, 0)), (2*atan(tanh(c/2 + d*sqrt(x)/2))/d, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \text {sech}\left (c+d \sqrt {x}\right )}{\sqrt {x}} \, dx=2 \, a \sqrt {x} + \frac {2 \, b \arctan \left (\sinh \left (d \sqrt {x} + c\right )\right )}{d} \]

[In]

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

[Out]

2*a*sqrt(x) + 2*b*arctan(sinh(d*sqrt(x) + c))/d

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {a+b \text {sech}\left (c+d \sqrt {x}\right )}{\sqrt {x}} \, dx=\frac {2 \, {\left (d \sqrt {x} + c\right )} a}{d} + \frac {4 \, b \arctan \left (e^{\left (d \sqrt {x} + c\right )}\right )}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.98 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {a+b \text {sech}\left (c+d \sqrt {x}\right )}{\sqrt {x}} \, dx=2\,a\,\sqrt {x}+\frac {4\,\mathrm {atan}\left (\frac {b\,{\mathrm {e}}^{d\,\sqrt {x}}\,{\mathrm {e}}^c\,\sqrt {d^2}}{d\,\sqrt {b^2}}\right )\,\sqrt {b^2}}{\sqrt {d^2}} \]

[In]

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

[Out]

2*a*x^(1/2) + (4*atan((b*exp(d*x^(1/2))*exp(c)*(d^2)^(1/2))/(d*(b^2)^(1/2)))*(b^2)^(1/2))/(d^2)^(1/2)