\(\int \frac {1}{\sqrt {x} (a+b \text {csch}(c+d \sqrt {x}))} \, dx\) [63]

Optimal result

Integrand size = 22, antiderivative size = 63 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2 \sqrt {x}}{a}+\frac {4 b \text {arctanh}\left (\frac {a-b \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d} \]

[Out]

4*b*arctanh((a-b*tanh(1/2*c+1/2*d*x^(1/2)))/(a^2+b^2)^(1/2))/a/d/(a^2+b^2)^(1/2)+2*x^(1/2)/a

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {5545, 3868, 2739, 632, 210} \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\frac {4 b \text {arctanh}\left (\frac {a-b \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a^2+b^2}}\right )}{a d \sqrt {a^2+b^2}}+\frac {2 \sqrt {x}}{a} \]

[In]

Int[1/(Sqrt[x]*(a + b*Csch[c + d*Sqrt[x]])),x]

[Out]

(2*Sqrt[x])/a + (4*b*ArcTanh[(a - b*Tanh[(c + d*Sqrt[x])/2])/Sqrt[a^2 + b^2]])/(a*Sqrt[a^2 + b^2]*d)

Rule 210

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*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 3868

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a/b)*Sin[c
+ d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 5545

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Csch[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}& = 2 \text {Subst}\left (\int \frac {1}{a+b \text {csch}(c+d x)} \, dx,x,\sqrt {x}\right ) \\ & = \frac {2 \sqrt {x}}{a}-\frac {2 \text {Subst}\left (\int \frac {1}{1+\frac {a \sinh (c+d x)}{b}} \, dx,x,\sqrt {x}\right )}{a} \\ & = \frac {2 \sqrt {x}}{a}+\frac {(4 i) \text {Subst}\left (\int \frac {1}{1-\frac {2 i a x}{b}+x^2} \, dx,x,i \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{a d} \\ & = \frac {2 \sqrt {x}}{a}-\frac {(8 i) \text {Subst}\left (\int \frac {1}{-4 \left (1+\frac {a^2}{b^2}\right )-x^2} \, dx,x,-\frac {2 i a}{b}+2 i \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{a d} \\ & = \frac {2 \sqrt {x}}{a}+\frac {4 b \text {arctanh}\left (\frac {b \left (\frac {a}{b}-\tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.16 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2 \left (\frac {c}{d}+\sqrt {x}-\frac {2 b \arctan \left (\frac {a-b \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2} d}\right )}{a} \]

[In]

Integrate[1/(Sqrt[x]*(a + b*Csch[c + d*Sqrt[x]])),x]

[Out]

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

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.41

method	result	size
derivativedivides	\(\frac {-\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )-1\right )}{a}+\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+1\right )}{a}+\frac {4 b \,\operatorname {arctanh}\left (\frac {-2 b \tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}}{d}\)	\(89\)
default	\(\frac {-\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )-1\right )}{a}+\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+1\right )}{a}+\frac {4 b \,\operatorname {arctanh}\left (\frac {-2 b \tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}}{d}\)	\(89\)

[In]

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

[Out]

2/d*(-1/a*ln(tanh(1/2*c+1/2*d*x^(1/2))-1)+1/a*ln(tanh(1/2*c+1/2*d*x^(1/2))+1)+2*b/a/(a^2+b^2)^(1/2)*arctanh(1/
2*(-2*b*tanh(1/2*c+1/2*d*x^(1/2))+2*a)/(a^2+b^2)^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (56) = 112\).

Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.95 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2 \, {\left ({\left (a^{2} + b^{2}\right )} d \sqrt {x} + \sqrt {a^{2} + b^{2}} b \log \left (\frac {a b + {\left (a^{2} + b^{2} + \sqrt {a^{2} + b^{2}} b\right )} \cosh \left (d \sqrt {x} + c\right ) - {\left (b^{2} + \sqrt {a^{2} + b^{2}} b\right )} \sinh \left (d \sqrt {x} + c\right ) + \sqrt {a^{2} + b^{2}} a}{a \sinh \left (d \sqrt {x} + c\right ) + b}\right )\right )}}{{\left (a^{3} + a b^{2}\right )} d} \]

[In]

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

[Out]

2*((a^2 + b^2)*d*sqrt(x) + sqrt(a^2 + b^2)*b*log((a*b + (a^2 + b^2 + sqrt(a^2 + b^2)*b)*cosh(d*sqrt(x) + c) -
(b^2 + sqrt(a^2 + b^2)*b)*sinh(d*sqrt(x) + c) + sqrt(a^2 + b^2)*a)/(a*sinh(d*sqrt(x) + c) + b)))/((a^3 + a*b^2
)*d)

Sympy [F]

\[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\int \frac {1}{\sqrt {x} \left (a + b \operatorname {csch}{\left (c + d \sqrt {x} \right )}\right )}\, dx \]

[In]

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

[Out]

Integral(1/(sqrt(x)*(a + b*csch(c + d*sqrt(x)))), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.46 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=-\frac {2 \, b \log \left (\frac {a e^{\left (-d \sqrt {x} - c\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-d \sqrt {x} - c\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a d} + \frac {2 \, {\left (d \sqrt {x} + c\right )}}{a d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.46 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=-\frac {2 \, b \log \left (\frac {{\left | 2 \, a e^{\left (d \sqrt {x} + c\right )} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{\left (d \sqrt {x} + c\right )} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a d} + \frac {2 \, {\left (d \sqrt {x} + c\right )}}{a d} \]

[In]

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

[Out]

-2*b*log(abs(2*a*e^(d*sqrt(x) + c) + 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*e^(d*sqrt(x) + c) + 2*b + 2*sqrt(a^2 + b
^2)))/(sqrt(a^2 + b^2)*a*d) + 2*(d*sqrt(x) + c)/(a*d)

Mupad [B] (verification not implemented)

Time = 2.57 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.30 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2\,\sqrt {x}}{a}-\frac {2\,b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^{d\,\sqrt {x}}\,{\mathrm {e}}^c}{a^2\,\sqrt {x}}-\frac {2\,b\,\left (a-b\,{\mathrm {e}}^{d\,\sqrt {x}}\,{\mathrm {e}}^c\right )}{a^2\,\sqrt {x}\,\sqrt {a^2+b^2}}\right )}{a\,d\,\sqrt {a^2+b^2}}+\frac {2\,b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^{d\,\sqrt {x}}\,{\mathrm {e}}^c}{a^2\,\sqrt {x}}+\frac {2\,b\,\left (a-b\,{\mathrm {e}}^{d\,\sqrt {x}}\,{\mathrm {e}}^c\right )}{a^2\,\sqrt {x}\,\sqrt {a^2+b^2}}\right )}{a\,d\,\sqrt {a^2+b^2}} \]

[In]

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

[Out]

(2*x^(1/2))/a - (2*b*log((2*b*exp(d*x^(1/2))*exp(c))/(a^2*x^(1/2)) - (2*b*(a - b*exp(d*x^(1/2))*exp(c)))/(a^2*
x^(1/2)*(a^2 + b^2)^(1/2))))/(a*d*(a^2 + b^2)^(1/2)) + (2*b*log((2*b*exp(d*x^(1/2))*exp(c))/(a^2*x^(1/2)) + (2
*b*(a - b*exp(d*x^(1/2))*exp(c)))/(a^2*x^(1/2)*(a^2 + b^2)^(1/2))))/(a*d*(a^2 + b^2)^(1/2))