\(\int x^5 \sqrt {1+x^2} \, dx\) [376]

Optimal result

Integrand size = 13, antiderivative size = 40 \[ \int x^5 \sqrt {1+x^2} \, dx=\frac {1}{3} \left (1+x^2\right )^{3/2}-\frac {2}{5} \left (1+x^2\right )^{5/2}+\frac {1}{7} \left (1+x^2\right )^{7/2} \]

[Out]

1/3*(x^2+1)^(3/2)-2/5*(x^2+1)^(5/2)+1/7*(x^2+1)^(7/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int x^5 \sqrt {1+x^2} \, dx=\frac {1}{7} \left (x^2+1\right )^{7/2}-\frac {2}{5} \left (x^2+1\right )^{5/2}+\frac {1}{3} \left (x^2+1\right )^{3/2} \]

[In]

Int[x^5*Sqrt[1 + x^2],x]

[Out]

(1 + x^2)^(3/2)/3 - (2*(1 + x^2)^(5/2))/5 + (1 + x^2)^(7/2)/7

Rule 45

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] && NeQ[b*c - a*d, 0] && 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])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x^2 \sqrt {1+x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\sqrt {1+x}-2 (1+x)^{3/2}+(1+x)^{5/2}\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{3} \left (1+x^2\right )^{3/2}-\frac {2}{5} \left (1+x^2\right )^{5/2}+\frac {1}{7} \left (1+x^2\right )^{7/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.75 \[ \int x^5 \sqrt {1+x^2} \, dx=\frac {1}{105} \sqrt {1+x^2} \left (8-4 x^2+3 x^4+15 x^6\right ) \]

[In]

Integrate[x^5*Sqrt[1 + x^2],x]

[Out]

(Sqrt[1 + x^2]*(8 - 4*x^2 + 3*x^4 + 15*x^6))/105

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.55

method	result	size
gosper	\(\frac {\left (x^{2}+1\right )^{\frac {3}{2}} \left (15 x^{4}-12 x^{2}+8\right )}{105}\)	\(22\)
pseudoelliptic	\(\frac {\left (x^{2}+1\right )^{\frac {3}{2}} \left (15 x^{4}-12 x^{2}+8\right )}{105}\)	\(22\)
trager	\(\left (\frac {1}{7} x^{6}+\frac {1}{35} x^{4}-\frac {4}{105} x^{2}+\frac {8}{105}\right ) \sqrt {x^{2}+1}\)	\(26\)
risch	\(\frac {\left (15 x^{6}+3 x^{4}-4 x^{2}+8\right ) \sqrt {x^{2}+1}}{105}\)	\(27\)
default	\(\frac {x^{4} \left (x^{2}+1\right )^{\frac {3}{2}}}{7}-\frac {4 x^{2} \left (x^{2}+1\right )^{\frac {3}{2}}}{35}+\frac {8 \left (x^{2}+1\right )^{\frac {3}{2}}}{105}\)	\(35\)
meijerg	\(-\frac {\frac {32 \sqrt {\pi }}{105}-\frac {4 \sqrt {\pi }\, \left (x^{2}+1\right )^{\frac {3}{2}} \left (15 x^{4}-12 x^{2}+8\right )}{105}}{4 \sqrt {\pi }}\)	\(36\)

[In]

int(x^5*(x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/105*(x^2+1)^(3/2)*(15*x^4-12*x^2+8)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.65 \[ \int x^5 \sqrt {1+x^2} \, dx=\frac {1}{105} \, {\left (15 \, x^{6} + 3 \, x^{4} - 4 \, x^{2} + 8\right )} \sqrt {x^{2} + 1} \]

[In]

integrate(x^5*(x^2+1)^(1/2),x, algorithm="fricas")

[Out]

1/105*(15*x^6 + 3*x^4 - 4*x^2 + 8)*sqrt(x^2 + 1)

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.32 \[ \int x^5 \sqrt {1+x^2} \, dx=\frac {x^{6} \sqrt {x^{2} + 1}}{7} + \frac {x^{4} \sqrt {x^{2} + 1}}{35} - \frac {4 x^{2} \sqrt {x^{2} + 1}}{105} + \frac {8 \sqrt {x^{2} + 1}}{105} \]

[In]

integrate(x**5*(x**2+1)**(1/2),x)

[Out]

x**6*sqrt(x**2 + 1)/7 + x**4*sqrt(x**2 + 1)/35 - 4*x**2*sqrt(x**2 + 1)/105 + 8*sqrt(x**2 + 1)/105

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int x^5 \sqrt {1+x^2} \, dx=\frac {1}{7} \, {\left (x^{2} + 1\right )}^{\frac {3}{2}} x^{4} - \frac {4}{35} \, {\left (x^{2} + 1\right )}^{\frac {3}{2}} x^{2} + \frac {8}{105} \, {\left (x^{2} + 1\right )}^{\frac {3}{2}} \]

[In]

integrate(x^5*(x^2+1)^(1/2),x, algorithm="maxima")

[Out]

1/7*(x^2 + 1)^(3/2)*x^4 - 4/35*(x^2 + 1)^(3/2)*x^2 + 8/105*(x^2 + 1)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70 \[ \int x^5 \sqrt {1+x^2} \, dx=\frac {1}{7} \, {\left (x^{2} + 1\right )}^{\frac {7}{2}} - \frac {2}{5} \, {\left (x^{2} + 1\right )}^{\frac {5}{2}} + \frac {1}{3} \, {\left (x^{2} + 1\right )}^{\frac {3}{2}} \]

[In]

integrate(x^5*(x^2+1)^(1/2),x, algorithm="giac")

[Out]

1/7*(x^2 + 1)^(7/2) - 2/5*(x^2 + 1)^(5/2) + 1/3*(x^2 + 1)^(3/2)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.62 \[ \int x^5 \sqrt {1+x^2} \, dx=\sqrt {x^2+1}\,\left (\frac {x^6}{7}+\frac {x^4}{35}-\frac {4\,x^2}{105}+\frac {8}{105}\right ) \]

[In]

int(x^5*(x^2 + 1)^(1/2),x)

[Out]

(x^2 + 1)^(1/2)*(x^4/35 - (4*x^2)/105 + x^6/7 + 8/105)