Integrand size = 10, antiderivative size = 78 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^5} \, dx=-\frac {\sqrt {1+x}}{28 x^{7/2}}+\frac {3 \sqrt {1+x}}{70 x^{5/2}}-\frac {2 \sqrt {1+x}}{35 x^{3/2}}+\frac {4 \sqrt {1+x}}{35 \sqrt {x}}-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{4 x^4} \]
-1/4*arcsinh(x^(1/2))/x^4-1/28*(1+x)^(1/2)/x^(7/2)+3/70*(1+x)^(1/2)/x^(5/2 )-2/35*(1+x)^(1/2)/x^(3/2)+4/35*(1+x)^(1/2)/x^(1/2)
Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.56 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^5} \, dx=\frac {\sqrt {x} \sqrt {1+x} \left (-5+6 x-8 x^2+16 x^3\right )-35 \text {arcsinh}\left (\sqrt {x}\right )}{140 x^4} \]
Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6290, 27, 55, 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.
\(\displaystyle \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^5} \, dx\) |
\(\Big \downarrow \) 6290 |
\(\displaystyle \frac {1}{4} \int \frac {1}{2 x^{9/2} \sqrt {x+1}}dx-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{4 x^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \int \frac {1}{x^{9/2} \sqrt {x+1}}dx-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{4 x^4}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {1}{8} \left (-\frac {6}{7} \int \frac {1}{x^{7/2} \sqrt {x+1}}dx-\frac {2 \sqrt {x+1}}{7 x^{7/2}}\right )-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{4 x^4}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {1}{8} \left (-\frac {6}{7} \left (-\frac {4}{5} \int \frac {1}{x^{5/2} \sqrt {x+1}}dx-\frac {2 \sqrt {x+1}}{5 x^{5/2}}\right )-\frac {2 \sqrt {x+1}}{7 x^{7/2}}\right )-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{4 x^4}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {1}{8} \left (-\frac {6}{7} \left (-\frac {4}{5} \left (-\frac {2}{3} \int \frac {1}{x^{3/2} \sqrt {x+1}}dx-\frac {2 \sqrt {x+1}}{3 x^{3/2}}\right )-\frac {2 \sqrt {x+1}}{5 x^{5/2}}\right )-\frac {2 \sqrt {x+1}}{7 x^{7/2}}\right )-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{4 x^4}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {1}{8} \left (-\frac {6}{7} \left (-\frac {4}{5} \left (\frac {4 \sqrt {x+1}}{3 \sqrt {x}}-\frac {2 \sqrt {x+1}}{3 x^{3/2}}\right )-\frac {2 \sqrt {x+1}}{5 x^{5/2}}\right )-\frac {2 \sqrt {x+1}}{7 x^{7/2}}\right )-\frac {\text {arcsinh}\left (\sqrt {x}\right )}{4 x^4}\) |
((-2*Sqrt[1 + x])/(7*x^(7/2)) - (6*((-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))/7)/8 - ArcSi nh[Sqrt[x]]/(4*x^4)
3.3.99.3.1 Defintions of rubi rules used
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 = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65
method | result | size |
derivativedivides | \(-\frac {\operatorname {arcsinh}\left (\sqrt {x}\right )}{4 x^{4}}-\frac {\sqrt {1+x}}{28 x^{\frac {7}{2}}}+\frac {3 \sqrt {1+x}}{70 x^{\frac {5}{2}}}-\frac {2 \sqrt {1+x}}{35 x^{\frac {3}{2}}}+\frac {4 \sqrt {1+x}}{35 \sqrt {x}}\) | \(51\) |
default | \(-\frac {\operatorname {arcsinh}\left (\sqrt {x}\right )}{4 x^{4}}-\frac {\sqrt {1+x}}{28 x^{\frac {7}{2}}}+\frac {3 \sqrt {1+x}}{70 x^{\frac {5}{2}}}-\frac {2 \sqrt {1+x}}{35 x^{\frac {3}{2}}}+\frac {4 \sqrt {1+x}}{35 \sqrt {x}}\) | \(51\) |
parts | \(-\frac {\operatorname {arcsinh}\left (\sqrt {x}\right )}{4 x^{4}}-\frac {\sqrt {1+x}}{28 x^{\frac {7}{2}}}+\frac {3 \sqrt {1+x}}{70 x^{\frac {5}{2}}}-\frac {2 \sqrt {1+x}}{35 x^{\frac {3}{2}}}+\frac {4 \sqrt {1+x}}{35 \sqrt {x}}\) | \(51\) |
-1/4*arcsinh(x^(1/2))/x^4-1/28*(1+x)^(1/2)/x^(7/2)+3/70*(1+x)^(1/2)/x^(5/2 )-2/35*(1+x)^(1/2)/x^(3/2)+4/35*(1+x)^(1/2)/x^(1/2)
Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.54 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^5} \, dx=\frac {{\left (16 \, x^{3} - 8 \, x^{2} + 6 \, x - 5\right )} \sqrt {x + 1} \sqrt {x} - 35 \, \log \left (\sqrt {x + 1} + \sqrt {x}\right )}{140 \, x^{4}} \]
\[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^5} \, dx=\int \frac {\operatorname {asinh}{\left (\sqrt {x} \right )}}{x^{5}}\, dx \]
Time = 0.40 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.64 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^5} \, dx=\frac {4 \, \sqrt {x + 1}}{35 \, \sqrt {x}} - \frac {2 \, \sqrt {x + 1}}{35 \, x^{\frac {3}{2}}} + \frac {3 \, \sqrt {x + 1}}{70 \, x^{\frac {5}{2}}} - \frac {\sqrt {x + 1}}{28 \, x^{\frac {7}{2}}} - \frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{4 \, x^{4}} \]
4/35*sqrt(x + 1)/sqrt(x) - 2/35*sqrt(x + 1)/x^(3/2) + 3/70*sqrt(x + 1)/x^( 5/2) - 1/28*sqrt(x + 1)/x^(7/2) - 1/4*arcsinh(sqrt(x))/x^4
Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.05 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^5} \, dx=-\frac {\log \left (\sqrt {x + 1} + \sqrt {x}\right )}{4 \, x^{4}} + \frac {8 \, {\left (35 \, {\left (\sqrt {x + 1} - \sqrt {x}\right )}^{6} - 21 \, {\left (\sqrt {x + 1} - \sqrt {x}\right )}^{4} + 7 \, {\left (\sqrt {x + 1} - \sqrt {x}\right )}^{2} - 1\right )}}{35 \, {\left ({\left (\sqrt {x + 1} - \sqrt {x}\right )}^{2} - 1\right )}^{7}} \]
-1/4*log(sqrt(x + 1) + sqrt(x))/x^4 + 8/35*(35*(sqrt(x + 1) - sqrt(x))^6 - 21*(sqrt(x + 1) - sqrt(x))^4 + 7*(sqrt(x + 1) - sqrt(x))^2 - 1)/((sqrt(x + 1) - sqrt(x))^2 - 1)^7
Timed out. \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x^5} \, dx=\int \frac {\mathrm {asinh}\left (\sqrt {x}\right )}{x^5} \,d x \]