Integrand size = 22, antiderivative size = 63 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2 \sqrt {x}}{a}+\frac {4 b \text {arctanh}\left (\frac {a-b \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d} \] Output:
2*x^(1/2)/a+4*b*arctanh((a-b*tanh(1/2*c+1/2*d*x^(1/2)))/(a^2+b^2)^(1/2))/a /(a^2+b^2)^(1/2)/d
Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.16 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2 \left (\frac {c}{d}+\sqrt {x}-\frac {2 b \arctan \left (\frac {a-b \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2} d}\right )}{a} \] Input:
Integrate[1/(Sqrt[x]*(a + b*Csch[c + d*Sqrt[x]])),x]
Output:
(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
Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {5960, 3042, 4270, 3042, 3139, 1083, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx\) |
\(\Big \downarrow \) 5960 |
\(\displaystyle 2 \int \frac {1}{a+b \text {csch}\left (c+d \sqrt {x}\right )}d\sqrt {x}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \int \frac {1}{a+i b \csc \left (i c+i d \sqrt {x}\right )}d\sqrt {x}\) |
\(\Big \downarrow \) 4270 |
\(\displaystyle 2 \left (\frac {\sqrt {x}}{a}-\frac {\int \frac {1}{\frac {a \sinh \left (c+d \sqrt {x}\right )}{b}+1}d\sqrt {x}}{a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \left (\frac {\sqrt {x}}{a}-\frac {\int \frac {1}{1-\frac {i a \sin \left (i c+i d \sqrt {x}\right )}{b}}d\sqrt {x}}{a}\right )\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle 2 \left (\frac {\sqrt {x}}{a}+\frac {2 i \int \frac {1}{x+\frac {2 a \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{b}+1}d\left (i \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{a d}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle 2 \left (\frac {\sqrt {x}}{a}-\frac {4 i \int \frac {1}{-4 \left (\frac {a^2}{b^2}+1\right )-x}d\left (2 i \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )-\frac {2 i a}{b}\right )}{a d}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle 2 \left (\frac {\sqrt {x}}{a}-\frac {2 b \text {arctanh}\left (\frac {b \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{2 \sqrt {a^2+b^2}}\right )}{a d \sqrt {a^2+b^2}}\right )\) |
Input:
Int[1/(Sqrt[x]*(a + b*Csch[c + d*Sqrt[x]])),x]
Output:
2*(Sqrt[x]/a - (2*b*ArcTanh[(b*Tanh[(c + d*Sqrt[x])/2])/(2*Sqrt[a^2 + b^2] )])/(a*Sqrt[a^2 + b^2]*d))
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])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[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]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Simp[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]
Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbo l] :> Simp[1/n Subst[Int[x^(Simplify[(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[Simplify[(m + 1)/n], 0] && IntegerQ[p]
Time = 0.35 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.41
method | result | size |
derivativedivides | \(\frac {\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+1\right )}{a}+\frac {4 b \,\operatorname {arctanh}\left (\frac {-2 b \tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}-\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )-1\right )}{a}}{d}\) | \(89\) |
default | \(\frac {\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+1\right )}{a}+\frac {4 b \,\operatorname {arctanh}\left (\frac {-2 b \tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}-\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )-1\right )}{a}}{d}\) | \(89\) |
Input:
int(1/x^(1/2)/(a+b*csch(c+d*x^(1/2))),x,method=_RETURNVERBOSE)
Output:
2/d*(1/a*ln(tanh(1/2*c+1/2*d*x^(1/2))+1)+2/a*b/(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))-1/a*ln(tanh(1/2*c+1 /2*d*x^(1/2))-1))
Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (56) = 112\).
Time = 0.10 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.95 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2 \, {\left ({\left (a^{2} + b^{2}\right )} d \sqrt {x} + \sqrt {a^{2} + b^{2}} b \log \left (\frac {a b + {\left (a^{2} + b^{2} + \sqrt {a^{2} + b^{2}} b\right )} \cosh \left (d \sqrt {x} + c\right ) - {\left (b^{2} + \sqrt {a^{2} + b^{2}} b\right )} \sinh \left (d \sqrt {x} + c\right ) + \sqrt {a^{2} + b^{2}} a}{a \sinh \left (d \sqrt {x} + c\right ) + b}\right )\right )}}{{\left (a^{3} + a b^{2}\right )} d} \] Input:
integrate(1/x^(1/2)/(a+b*csch(c+d*x^(1/2))),x, algorithm="fricas")
Output:
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)
\[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\int \frac {1}{\sqrt {x} \left (a + b \operatorname {csch}{\left (c + d \sqrt {x} \right )}\right )}\, dx \] Input:
integrate(1/x**(1/2)/(a+b*csch(c+d*x**(1/2))),x)
Output:
Integral(1/(sqrt(x)*(a + b*csch(c + d*sqrt(x)))), x)
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.46 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=-\frac {2 \, b \log \left (\frac {a e^{\left (-d \sqrt {x} - c\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-d \sqrt {x} - c\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a d} + \frac {2 \, {\left (d \sqrt {x} + c\right )}}{a d} \] Input:
integrate(1/x^(1/2)/(a+b*csch(c+d*x^(1/2))),x, algorithm="maxima")
Output:
-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)
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.46 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=-\frac {2 \, b \log \left (\frac {{\left | 2 \, a e^{\left (d \sqrt {x} + c\right )} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{\left (d \sqrt {x} + c\right )} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a d} + \frac {2 \, {\left (d \sqrt {x} + c\right )}}{a d} \] Input:
integrate(1/x^(1/2)/(a+b*csch(c+d*x^(1/2))),x, algorithm="giac")
Output:
-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*sqr t(x) + c)/(a*d)
Time = 2.92 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.30 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2\,\sqrt {x}}{a}-\frac {2\,b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^{d\,\sqrt {x}}\,{\mathrm {e}}^c}{a^2\,\sqrt {x}}-\frac {2\,b\,\left (a-b\,{\mathrm {e}}^{d\,\sqrt {x}}\,{\mathrm {e}}^c\right )}{a^2\,\sqrt {x}\,\sqrt {a^2+b^2}}\right )}{a\,d\,\sqrt {a^2+b^2}}+\frac {2\,b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^{d\,\sqrt {x}}\,{\mathrm {e}}^c}{a^2\,\sqrt {x}}+\frac {2\,b\,\left (a-b\,{\mathrm {e}}^{d\,\sqrt {x}}\,{\mathrm {e}}^c\right )}{a^2\,\sqrt {x}\,\sqrt {a^2+b^2}}\right )}{a\,d\,\sqrt {a^2+b^2}} \] Input:
int(1/(x^(1/2)*(a + b/sinh(c + d*x^(1/2)))),x)
Output:
(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))
Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )} \, dx=\frac {-4 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{\sqrt {x}\, d +c} a i +b i}{\sqrt {a^{2}+b^{2}}}\right ) b i +2 \sqrt {x}\, a^{2} d +2 \sqrt {x}\, b^{2} d}{a d \left (a^{2}+b^{2}\right )} \] Input:
int(1/x^(1/2)/(a+b*csch(c+d*x^(1/2))),x)
Output:
(2*( - 2*sqrt(a**2 + b**2)*atan((e**(sqrt(x)*d + c)*a*i + b*i)/sqrt(a**2 + b**2))*b*i + sqrt(x)*a**2*d + sqrt(x)*b**2*d))/(a*d*(a**2 + b**2))