Integrand size = 10, antiderivative size = 138 \[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=-\frac {1}{4} \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}+\frac {1}{4} \left (-1+\frac {1}{\sqrt {x}}\right )^{3/2} \left (1+\frac {1}{\sqrt {x}}\right )^{3/2} x^{3/2}-\frac {3}{20} \left (-1+\frac {1}{\sqrt {x}}\right )^{5/2} \left (1+\frac {1}{\sqrt {x}}\right )^{5/2} x^{5/2}+\frac {1}{28} \left (-1+\frac {1}{\sqrt {x}}\right )^{7/2} \left (1+\frac {1}{\sqrt {x}}\right )^{7/2} x^{7/2}+\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right ) \] Output:
-1/4*(-1+1/x^(1/2))^(1/2)*(1/x^(1/2)+1)^(1/2)*x^(1/2)+1/4*(-1+1/x^(1/2))^( 3/2)*(1/x^(1/2)+1)^(3/2)*x^(3/2)-3/20*(-1+1/x^(1/2))^(5/2)*(1/x^(1/2)+1)^( 5/2)*x^(5/2)+1/28*(-1+1/x^(1/2))^(7/2)*(1/x^(1/2)+1)^(7/2)*x^(7/2)+1/4*x^4 *arcsech(x^(1/2))
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.61 \[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=-\frac {1}{140} \sqrt {\frac {1-\sqrt {x}}{1+\sqrt {x}}} \left (16+16 \sqrt {x}+8 x+8 x^{3/2}+6 x^2+6 x^{5/2}+5 x^3+5 x^{7/2}\right )+\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right ) \] Input:
Integrate[x^3*ArcSech[Sqrt[x]],x]
Output:
-1/140*(Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*(16 + 16*Sqrt[x] + 8*x + 8*x^(3/ 2) + 6*x^2 + 6*x^(5/2) + 5*x^3 + 5*x^(7/2))) + (x^4*ArcSech[Sqrt[x]])/4
Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.75, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6899, 27, 53, 2009}
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.
\(\displaystyle \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx\) |
\(\Big \downarrow \) 6899 |
\(\displaystyle \frac {\sqrt {1-x} \int \frac {x^3}{2 \sqrt {1-x}}dx}{4 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}+\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {1-x} \int \frac {x^3}{\sqrt {1-x}}dx}{8 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}+\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {1-x} \int \left (-(1-x)^{5/2}+3 (1-x)^{3/2}-3 \sqrt {1-x}+\frac {1}{\sqrt {1-x}}\right )dx}{8 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}+\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {\left (\frac {2}{7} (1-x)^{7/2}-\frac {6}{5} (1-x)^{5/2}+2 (1-x)^{3/2}-2 \sqrt {1-x}\right ) \sqrt {1-x}}{8 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}\) |
Input:
Int[x^3*ArcSech[Sqrt[x]],x]
Output:
((-2*Sqrt[1 - x] + 2*(1 - x)^(3/2) - (6*(1 - x)^(5/2))/5 + (2*(1 - x)^(7/2 ))/7)*Sqrt[1 - x])/(8*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*Sqrt[x]) + (x^4*ArcSech[Sqrt[x]])/4
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] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[((a_.) + ArcSech[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Si mp[(c + d*x)^(m + 1)*((a + b*ArcSech[u])/(d*(m + 1))), x] + Simp[b*(Sqrt[1 - u^2]/(d*(m + 1)*u*Sqrt[-1 + 1/u]*Sqrt[1 + 1/u])) Int[SimplifyIntegrand[ (c + d*x)^(m + 1)*(D[u, x]/(u*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 = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.39
method | result | size |
derivativedivides | \(\frac {x^{4} \operatorname {arcsech}\left (\sqrt {x}\right )}{4}-\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {x}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (5 x^{3}+6 x^{2}+8 x +16\right )}{140}\) | \(54\) |
default | \(\frac {x^{4} \operatorname {arcsech}\left (\sqrt {x}\right )}{4}-\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {x}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (5 x^{3}+6 x^{2}+8 x +16\right )}{140}\) | \(54\) |
parts | \(\frac {x^{4} \operatorname {arcsech}\left (\sqrt {x}\right )}{4}-\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {x}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (5 x^{3}+6 x^{2}+8 x +16\right )}{140}\) | \(54\) |
Input:
int(x^3*arcsech(x^(1/2)),x,method=_RETURNVERBOSE)
Output:
1/4*x^4*arcsech(x^(1/2))-1/140*(-(x^(1/2)-1)/x^(1/2))^(1/2)*x^(1/2)*((x^(1 /2)+1)/x^(1/2))^(1/2)*(5*x^3+6*x^2+8*x+16)
Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.41 \[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\frac {1}{4} \, x^{4} \log \left (\frac {x \sqrt {-\frac {x - 1}{x}} + \sqrt {x}}{x}\right ) - \frac {1}{140} \, {\left (5 \, x^{3} + 6 \, x^{2} + 8 \, x + 16\right )} \sqrt {x} \sqrt {-\frac {x - 1}{x}} \] Input:
integrate(x^3*arcsech(x^(1/2)),x, algorithm="fricas")
Output:
1/4*x^4*log((x*sqrt(-(x - 1)/x) + sqrt(x))/x) - 1/140*(5*x^3 + 6*x^2 + 8*x + 16)*sqrt(x)*sqrt(-(x - 1)/x)
\[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\int x^{3} \operatorname {asech}{\left (\sqrt {x} \right )}\, dx \] Input:
integrate(x**3*asech(x**(1/2)),x)
Output:
Integral(x**3*asech(sqrt(x)), x)
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.42 \[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\frac {1}{28} \, x^{\frac {7}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {7}{2}} - \frac {3}{20} \, x^{\frac {5}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {5}{2}} + \frac {1}{4} \, x^{4} \operatorname {arsech}\left (\sqrt {x}\right ) + \frac {1}{4} \, x^{\frac {3}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {3}{2}} - \frac {1}{4} \, \sqrt {x} \sqrt {\frac {1}{x} - 1} \] Input:
integrate(x^3*arcsech(x^(1/2)),x, algorithm="maxima")
Output:
1/28*x^(7/2)*(1/x - 1)^(7/2) - 3/20*x^(5/2)*(1/x - 1)^(5/2) + 1/4*x^4*arcs ech(sqrt(x)) + 1/4*x^(3/2)*(1/x - 1)^(3/2) - 1/4*sqrt(x)*sqrt(1/x - 1)
\[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\int { x^{3} \operatorname {arsech}\left (\sqrt {x}\right ) \,d x } \] Input:
integrate(x^3*arcsech(x^(1/2)),x, algorithm="giac")
Output:
integrate(x^3*arcsech(sqrt(x)), x)
Timed out. \[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\int x^3\,\mathrm {acosh}\left (\frac {1}{\sqrt {x}}\right ) \,d x \] Input:
int(x^3*acosh(1/x^(1/2)),x)
Output:
int(x^3*acosh(1/x^(1/2)), x)
\[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\int \mathit {asech} \left (\sqrt {x}\right ) x^{3}d x \] Input:
int(x^3*asech(x^(1/2)),x)
Output:
int(asech(sqrt(x))*x**3,x)