Integrand size = 21, antiderivative size = 94 \[ \int \frac {x^5}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\frac {4 x^2}{105 \sqrt {x^2+\sqrt {1+x^4}}}+\frac {x^6}{7 \sqrt {x^2+\sqrt {1+x^4}}}-\frac {8}{105} \sqrt {x^2+\sqrt {1+x^4}}+\frac {1}{35} x^4 \sqrt {x^2+\sqrt {1+x^4}} \] Output:
4/105*x^2/(x^2+(x^4+1)^(1/2))^(1/2)+1/7*x^6/(x^2+(x^4+1)^(1/2))^(1/2)-8/10 5*(x^2+(x^4+1)^(1/2))^(1/2)+1/35*x^4*(x^2+(x^4+1)^(1/2))^(1/2)
Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.85 \[ \int \frac {x^5}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\frac {-8-49 x^4+21 x^8+84 x^{12}-28 x^2 \sqrt {1+x^4}-21 x^6 \sqrt {1+x^4}+84 x^{10} \sqrt {1+x^4}}{105 \left (x^2+\sqrt {1+x^4}\right )^{7/2}} \] Input:
Integrate[x^5/Sqrt[x^2 + Sqrt[1 + x^4]],x]
Output:
(-8 - 49*x^4 + 21*x^8 + 84*x^12 - 28*x^2*Sqrt[1 + x^4] - 21*x^6*Sqrt[1 + x ^4] + 84*x^10*Sqrt[1 + x^4])/(105*(x^2 + Sqrt[1 + x^4])^(7/2))
Time = 0.52 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {7283, 2544, 355, 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.
\(\displaystyle \int \frac {x^5}{\sqrt {\sqrt {x^4+1}+x^2}} \, dx\) |
\(\Big \downarrow \) 7283 |
\(\displaystyle \frac {1}{2} \int \frac {x^4}{\sqrt {x^2+\sqrt {x^4+1}}}dx^2\) |
\(\Big \downarrow \) 2544 |
\(\displaystyle \frac {1}{16} \int \frac {\left (1-x^4\right )^2 \left (x^4+1\right )}{\left (x^2+\sqrt {x^4+1}\right )^{9/2}}d\left (x^2+\sqrt {x^4+1}\right )\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \frac {1}{16} \int \left (\left (x^2+\sqrt {x^4+1}\right )^{3/2}-\frac {1}{\sqrt {x^2+\sqrt {x^4+1}}}-\frac {1}{\left (x^2+\sqrt {x^4+1}\right )^{5/2}}+\frac {1}{\left (x^2+\sqrt {x^4+1}\right )^{9/2}}\right )d\left (x^2+\sqrt {x^4+1}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{16} \left (\frac {2}{5} \left (\sqrt {x^4+1}+x^2\right )^{5/2}-2 \sqrt {\sqrt {x^4+1}+x^2}+\frac {2}{3 \left (\sqrt {x^4+1}+x^2\right )^{3/2}}-\frac {2}{7 \left (\sqrt {x^4+1}+x^2\right )^{7/2}}\right )\) |
Input:
Int[x^5/Sqrt[x^2 + Sqrt[1 + x^4]],x]
Output:
(-2/(7*(x^2 + Sqrt[1 + x^4])^(7/2)) + 2/(3*(x^2 + Sqrt[1 + x^4])^(3/2)) - 2*Sqrt[x^2 + Sqrt[1 + x^4]] + (2*(x^2 + Sqrt[1 + x^4])^(5/2))/5)/16
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] & & IGtQ[q, 0]
Int[((g_.) + (h_.)*(x_))^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^ 2])^(n_.), x_Symbol] :> Simp[1/(2^(m + 1)*e^(m + 1)) Subst[Int[x^(n - m - 2)*(a*f^2 + x^2)*((-a)*f^2*h + 2*e*g*x + h*x^2)^m, x], x, e*x + f*Sqrt[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && I ntegerQ[m]
Int[(u_)*(x_)^(m_.), x_Symbol] :> With[{lst = PowerVariableExpn[u, m + 1, x ]}, Simp[1/lst[[2]] Subst[Int[NormalizeIntegrand[Simplify[lst[[1]]/x], x] , x], x, (lst[[3]]*x)^lst[[2]]], x] /; !FalseQ[lst] && NeQ[lst[[2]], m + 1 ]] /; IntegerQ[m] && NeQ[m, -1] && NonsumQ[u] && (GtQ[m, 0] || !AlgebraicF unctionQ[u, x])
Result contains higher order function than in optimal. Order 5 vs. order 2.
Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.23
method | result | size |
meijerg | \(\frac {\sqrt {2}\, x^{5} \operatorname {hypergeom}\left (\left [-\frac {5}{4}, \frac {1}{4}, \frac {3}{4}\right ], \left [-\frac {1}{4}, \frac {3}{2}\right ], -\frac {1}{x^{4}}\right )}{10}\) | \(22\) |
Input:
int(x^5/(x^2+(x^4+1)^(1/2))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/10*2^(1/2)*x^5*hypergeom([-5/4,1/4,3/4],[-1/4,3/2],-1/x^4)
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.48 \[ \int \frac {x^5}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=-\frac {1}{105} \, {\left (15 \, x^{8} + x^{4} - {\left (15 \, x^{6} + 4 \, x^{2}\right )} \sqrt {x^{4} + 1} + 8\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} \] Input:
integrate(x^5/(x^2+(x^4+1)^(1/2))^(1/2),x, algorithm="fricas")
Output:
-1/105*(15*x^8 + x^4 - (15*x^6 + 4*x^2)*sqrt(x^4 + 1) + 8)*sqrt(x^2 + sqrt (x^4 + 1))
Leaf count of result is larger than twice the leaf count of optimal. 1168 vs. \(2 (80) = 160\).
Time = 1.50 (sec) , antiderivative size = 1168, normalized size of antiderivative = 12.43 \[ \int \frac {x^5}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\text {Too large to display} \] Input:
integrate(x**5/(x**2+(x**4+1)**(1/2))**(1/2),x)
Output:
15*sqrt(2)*x**10*gamma(1/4)/(840*sqrt(x**4 + 1)*sqrt(sqrt(x**4 + 1) + 1)*g amma(5/4) + 840*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4)) + 216*x**8*(x**4 + 1) **(1/4)*sqrt(sqrt(x**4 + 1) + 1)*cos(7*atan(x**2)/2)*gamma(5/4)/(840*sqrt( x**4 + 1)*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4) + 840*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4)) + 336*x**6*(x**4 + 1)**(3/4)*sqrt(sqrt(x**4 + 1) + 1)*sin(5 *atan(x**2)/2)*gamma(5/4)/(840*sqrt(x**4 + 1)*sqrt(sqrt(x**4 + 1) + 1)*gam ma(5/4) + 840*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4)) + 336*x**6*(x**4 + 1)** (1/4)*sqrt(sqrt(x**4 + 1) + 1)*sin(5*atan(x**2)/2)*gamma(5/4)/(840*sqrt(x* *4 + 1)*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4) + 840*sqrt(sqrt(x**4 + 1) + 1) *gamma(5/4)) + 33*sqrt(2)*x**6*sqrt(x**4 + 1)*gamma(1/4)/(840*sqrt(x**4 + 1)*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4) + 840*sqrt(sqrt(x**4 + 1) + 1)*gamm a(5/4)) + 37*sqrt(2)*x**6*gamma(1/4)/(840*sqrt(x**4 + 1)*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4) + 840*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4)) + 216*x**4*( x**4 + 1)**(3/4)*sqrt(sqrt(x**4 + 1) + 1)*cos(7*atan(x**2)/2)*gamma(5/4)/( 840*sqrt(x**4 + 1)*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4) + 840*sqrt(sqrt(x** 4 + 1) + 1)*gamma(5/4)) + 152*x**4*(x**4 + 1)**(1/4)*sqrt(sqrt(x**4 + 1) + 1)*cos(7*atan(x**2)/2)*gamma(5/4)/(840*sqrt(x**4 + 1)*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4) + 840*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4)) - 224*x**2*(x* *4 + 1)**(3/4)*sqrt(sqrt(x**4 + 1) + 1)*sin(5*atan(x**2)/2)*gamma(5/4)/(84 0*sqrt(x**4 + 1)*sqrt(sqrt(x**4 + 1) + 1)*gamma(5/4) + 840*sqrt(sqrt(x*...
\[ \int \frac {x^5}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\int { \frac {x^{5}}{\sqrt {x^{2} + \sqrt {x^{4} + 1}}} \,d x } \] Input:
integrate(x^5/(x^2+(x^4+1)^(1/2))^(1/2),x, algorithm="maxima")
Output:
integrate(x^5/sqrt(x^2 + sqrt(x^4 + 1)), x)
\[ \int \frac {x^5}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\int { \frac {x^{5}}{\sqrt {x^{2} + \sqrt {x^{4} + 1}}} \,d x } \] Input:
integrate(x^5/(x^2+(x^4+1)^(1/2))^(1/2),x, algorithm="giac")
Output:
integrate(x^5/sqrt(x^2 + sqrt(x^4 + 1)), x)
Timed out. \[ \int \frac {x^5}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\int \frac {x^5}{\sqrt {\sqrt {x^4+1}+x^2}} \,d x \] Input:
int(x^5/((x^4 + 1)^(1/2) + x^2)^(1/2),x)
Output:
int(x^5/((x^4 + 1)^(1/2) + x^2)^(1/2), x)
\[ \int \frac {x^5}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\frac {\sqrt {\sqrt {x^{4}+1}+x^{2}}\, \sqrt {x^{4}+1}\, x^{6}}{8}+\frac {\sqrt {\sqrt {x^{4}+1}+x^{2}}\, \sqrt {x^{4}+1}\, x^{2}}{16}-\frac {\sqrt {\sqrt {x^{4}+1}+x^{2}}}{8}-\frac {9 \left (\int \frac {\sqrt {\sqrt {x^{4}+1}+x^{2}}\, x^{11}}{x^{4}+1}d x \right )}{8}-\frac {19 \left (\int \frac {\sqrt {\sqrt {x^{4}+1}+x^{2}}\, x^{7}}{x^{4}+1}d x \right )}{16}-\frac {\left (\int \frac {\sqrt {\sqrt {x^{4}+1}+x^{2}}\, x^{3}}{x^{4}+1}d x \right )}{16} \] Input:
int(x^5/(x^2+(x^4+1)^(1/2))^(1/2),x)
Output:
(2*sqrt(sqrt(x**4 + 1) + x**2)*sqrt(x**4 + 1)*x**6 + sqrt(sqrt(x**4 + 1) + x**2)*sqrt(x**4 + 1)*x**2 - 2*sqrt(sqrt(x**4 + 1) + x**2) - 18*int((sqrt( sqrt(x**4 + 1) + x**2)*x**11)/(x**4 + 1),x) - 19*int((sqrt(sqrt(x**4 + 1) + x**2)*x**7)/(x**4 + 1),x) - int((sqrt(sqrt(x**4 + 1) + x**2)*x**3)/(x**4 + 1),x))/16