Integrand size = 10, antiderivative size = 62 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^4} \, dx=-\frac {\sqrt {1+x}}{15 x^{5/2}}+\frac {4 \sqrt {1+x}}{45 x^{3/2}}-\frac {8 \sqrt {1+x}}{45 \sqrt {x}}-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{3 x^3} \] Output:
-1/15*(1+x)^(1/2)/x^(5/2)+4/45*(1+x)^(1/2)/x^(3/2)-8/45*(1+x)^(1/2)/x^(1/2 )-1/3*arcsinh(x^(1/2))/x^3
Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.63 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {\sqrt {x} \sqrt {1+x} \left (-3+4 x-8 x^2\right )-15 \text {arcsinh}\left (\sqrt {x}\right )}{45 x^3} \] Input:
Integrate[ArcSinh[Sqrt[x]]/x^4,x]
Output:
(Sqrt[x]*Sqrt[1 + x]*(-3 + 4*x - 8*x^2) - 15*ArcSinh[Sqrt[x]])/(45*x^3)
Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6290, 27, 55, 55, 48}
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.
\(\displaystyle \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 6290 |
\(\displaystyle \frac {1}{3} \int \frac {1}{2 x^{7/2} \sqrt {x+1}}dx-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \int \frac {1}{x^{7/2} \sqrt {x+1}}dx-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {1}{6} \left (-\frac {4}{5} \int \frac {1}{x^{5/2} \sqrt {x+1}}dx-\frac {2 \sqrt {x+1}}{5 x^{5/2}}\right )-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {1}{6} \left (-\frac {4}{5} \left (-\frac {2}{3} \int \frac {1}{x^{3/2} \sqrt {x+1}}dx-\frac {2 \sqrt {x+1}}{3 x^{3/2}}\right )-\frac {2 \sqrt {x+1}}{5 x^{5/2}}\right )-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {1}{6} \left (-\frac {4}{5} \left (\frac {4 \sqrt {x+1}}{3 \sqrt {x}}-\frac {2 \sqrt {x+1}}{3 x^{3/2}}\right )-\frac {2 \sqrt {x+1}}{5 x^{5/2}}\right )-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{3 x^3}\) |
Input:
Int[ArcSinh[Sqrt[x]]/x^4,x]
Output:
((-2*Sqrt[1 + x])/(5*x^(5/2)) - (4*((-2*Sqrt[1 + x])/(3*x^(3/2)) + (4*Sqrt [1 + x])/(3*Sqrt[x])))/5)/6 - ArcSinh[Sqrt[x]]/(3*x^3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Int[((a_.) + ArcSinh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Si mp[(c + d*x)^(m + 1)*((a + b*ArcSinh[u])/(d*(m + 1))), x] - Simp[b/(d*(m + 1)) Int[SimplifyIntegrand[(c + d*x)^(m + 1)*(D[u, x]/Sqrt[1 + u^2]), x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u , x] && !FunctionOfQ[(c + d*x)^(m + 1), u, x] && !FunctionOfExponentialQ[ u, x]
Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(-\frac {\sqrt {1+x}}{15 x^{\frac {5}{2}}}+\frac {4 \sqrt {1+x}}{45 x^{\frac {3}{2}}}-\frac {8 \sqrt {1+x}}{45 \sqrt {x}}-\frac {\operatorname {arcsinh}\left (\sqrt {x}\right )}{3 x^{3}}\) | \(41\) |
default | \(-\frac {\sqrt {1+x}}{15 x^{\frac {5}{2}}}+\frac {4 \sqrt {1+x}}{45 x^{\frac {3}{2}}}-\frac {8 \sqrt {1+x}}{45 \sqrt {x}}-\frac {\operatorname {arcsinh}\left (\sqrt {x}\right )}{3 x^{3}}\) | \(41\) |
parts | \(-\frac {\sqrt {1+x}}{15 x^{\frac {5}{2}}}+\frac {4 \sqrt {1+x}}{45 x^{\frac {3}{2}}}-\frac {8 \sqrt {1+x}}{45 \sqrt {x}}-\frac {\operatorname {arcsinh}\left (\sqrt {x}\right )}{3 x^{3}}\) | \(41\) |
Input:
int(arcsinh(x^(1/2))/x^4,x,method=_RETURNVERBOSE)
Output:
-1/15*(1+x)^(1/2)/x^(5/2)+4/45*(1+x)^(1/2)/x^(3/2)-8/45*(1+x)^(1/2)/x^(1/2 )-1/3*arcsinh(x^(1/2))/x^3
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.60 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^4} \, dx=-\frac {{\left (8 \, x^{2} - 4 \, x + 3\right )} \sqrt {x + 1} \sqrt {x} + 15 \, \log \left (\sqrt {x + 1} + \sqrt {x}\right )}{45 \, x^{3}} \] Input:
integrate(arcsinh(x^(1/2))/x^4,x, algorithm="fricas")
Output:
-1/45*((8*x^2 - 4*x + 3)*sqrt(x + 1)*sqrt(x) + 15*log(sqrt(x + 1) + sqrt(x )))/x^3
\[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^4} \, dx=\int \frac {\operatorname {asinh}{\left (\sqrt {x} \right )}}{x^{4}}\, dx \] Input:
integrate(asinh(x**(1/2))/x**4,x)
Output:
Integral(asinh(sqrt(x))/x**4, x)
Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.65 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^4} \, dx=-\frac {8 \, \sqrt {x + 1}}{45 \, \sqrt {x}} + \frac {4 \, \sqrt {x + 1}}{45 \, x^{\frac {3}{2}}} - \frac {\sqrt {x + 1}}{15 \, x^{\frac {5}{2}}} - \frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{3 \, x^{3}} \] Input:
integrate(arcsinh(x^(1/2))/x^4,x, algorithm="maxima")
Output:
-8/45*sqrt(x + 1)/sqrt(x) + 4/45*sqrt(x + 1)/x^(3/2) - 1/15*sqrt(x + 1)/x^ (5/2) - 1/3*arcsinh(sqrt(x))/x^3
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.08 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^4} \, dx=-\frac {\log \left (\sqrt {x + 1} + \sqrt {x}\right )}{3 \, x^{3}} + \frac {16 \, {\left (10 \, {\left (\sqrt {x + 1} - \sqrt {x}\right )}^{4} - 5 \, {\left (\sqrt {x + 1} - \sqrt {x}\right )}^{2} + 1\right )}}{45 \, {\left ({\left (\sqrt {x + 1} - \sqrt {x}\right )}^{2} - 1\right )}^{5}} \] Input:
integrate(arcsinh(x^(1/2))/x^4,x, algorithm="giac")
Output:
-1/3*log(sqrt(x + 1) + sqrt(x))/x^3 + 16/45*(10*(sqrt(x + 1) - sqrt(x))^4 - 5*(sqrt(x + 1) - sqrt(x))^2 + 1)/((sqrt(x + 1) - sqrt(x))^2 - 1)^5
Timed out. \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^4} \, dx=\int \frac {\mathrm {asinh}\left (\sqrt {x}\right )}{x^4} \,d x \] Input:
int(asinh(x^(1/2))/x^4,x)
Output:
int(asinh(x^(1/2))/x^4, x)
Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.71 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {-15 \mathit {asinh} \left (\sqrt {x}\right )-8 \sqrt {x}\, \sqrt {x +1}\, x^{2}+4 \sqrt {x}\, \sqrt {x +1}\, x -3 \sqrt {x}\, \sqrt {x +1}+8 x^{3}}{45 x^{3}} \] Input:
int(asinh(x^(1/2))/x^4,x)
Output:
( - 15*asinh(sqrt(x)) - 8*sqrt(x)*sqrt(x + 1)*x**2 + 4*sqrt(x)*sqrt(x + 1) *x - 3*sqrt(x)*sqrt(x + 1) + 8*x**3)/(45*x**3)