### 3.376 $$\int x^5 \sqrt{1+x^2} \, dx$$

Optimal. Leaf size=40 $\frac{1}{7} \left (x^2+1\right )^{7/2}-\frac{2}{5} \left (x^2+1\right )^{5/2}+\frac{1}{3} \left (x^2+1\right )^{3/2}$

[Out]

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

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Rubi [A]  time = 0.0135769, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.154, Rules used = {266, 43} $\frac{1}{7} \left (x^2+1\right )^{7/2}-\frac{2}{5} \left (x^2+1\right )^{5/2}+\frac{1}{3} \left (x^2+1\right )^{3/2}$

Antiderivative was successfully veriﬁed.

[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 266

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]]

Rule 43

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])

Rubi steps

\begin{align*} \int x^5 \sqrt{1+x^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 \sqrt{1+x} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{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]  time = 0.0074132, size = 25, normalized size = 0.62 $\frac{1}{105} \left (x^2+1\right )^{3/2} \left (15 x^4-12 x^2+8\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^5*Sqrt[1 + x^2],x]

[Out]

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

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Maple [A]  time = 0.005, size = 22, normalized size = 0.6 \begin{align*}{\frac{15\,{x}^{4}-12\,{x}^{2}+8}{105} \left ({x}^{2}+1 \right ) ^{{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.40403, size = 46, normalized size = 1.15 \begin{align*} \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.82029, size = 68, normalized size = 1.7 \begin{align*} \frac{1}{105} \,{\left (15 \, x^{6} + 3 \, x^{4} - 4 \, x^{2} + 8\right )} \sqrt{x^{2} + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 2.02928, size = 53, normalized size = 1.32 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.05398, size = 38, normalized size = 0.95 \begin{align*} \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)