\(\int x^3 \sqrt {1+x^2} \, dx\) [49]

Optimal result

Integrand size = 13, antiderivative size = 27 \[ \int x^3 \sqrt {1+x^2} \, dx=-\frac {1}{3} \left (1+x^2\right )^{3/2}+\frac {1}{5} \left (1+x^2\right )^{5/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, 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} \[ \int x^3 \sqrt {1+x^2} \, dx=\frac {1}{5} \left (x^2+1\right )^{5/2}-\frac {1}{3} \left (x^2+1\right )^{3/2} \]

[In]

Int[x^3*Sqrt[1 + x^2],x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int x^3 \sqrt {1+x^2} \, dx=\frac {1}{15} \left (1+x^2\right )^{3/2} \left (-2+3 x^2\right ) \]

[In]

Integrate[x^3*Sqrt[1 + x^2],x]

[Out]

((1 + x^2)^(3/2)*(-2 + 3*x^2))/15

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63

method	result	size
gosper	\(\frac {\left (x^{2}+1\right )^{\frac {3}{2}} \left (3 x^{2}-2\right )}{15}\)	\(17\)
pseudoelliptic	\(\frac {\left (x^{2}+1\right )^{\frac {3}{2}} \left (3 x^{2}-2\right )}{15}\)	\(17\)
risch	\(\frac {\left (3 x^{4}+x^{2}-2\right ) \sqrt {x^{2}+1}}{15}\)	\(20\)
trager	\(\left (\frac {1}{5} x^{4}+\frac {1}{15} x^{2}-\frac {2}{15}\right ) \sqrt {x^{2}+1}\)	\(21\)
default	\(\frac {x^{2} \left (x^{2}+1\right )^{\frac {3}{2}}}{5}-\frac {2 \left (x^{2}+1\right )^{\frac {3}{2}}}{15}\)	\(23\)
meijerg	\(-\frac {-\frac {8 \sqrt {\pi }}{15}+\frac {4 \sqrt {\pi }\, \left (x^{2}+1\right )^{\frac {3}{2}} \left (-3 x^{2}+2\right )}{15}}{4 \sqrt {\pi }}\)	\(31\)

[In]

int(x^3*(x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15*(x^2+1)^(3/2)*(3*x^2-2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int x^3 \sqrt {1+x^2} \, dx=\frac {1}{15} \, {\left (3 \, x^{4} + x^{2} - 2\right )} \sqrt {x^{2} + 1} \]

[In]

integrate(x^3*(x^2+1)^(1/2),x, algorithm="fricas")

[Out]

1/15*(3*x^4 + x^2 - 2)*sqrt(x^2 + 1)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int x^3 \sqrt {1+x^2} \, dx=\frac {x^{4} \sqrt {x^{2} + 1}}{5} + \frac {x^{2} \sqrt {x^{2} + 1}}{15} - \frac {2 \sqrt {x^{2} + 1}}{15} \]

[In]

integrate(x**3*(x**2+1)**(1/2),x)

[Out]

x**4*sqrt(x**2 + 1)/5 + x**2*sqrt(x**2 + 1)/15 - 2*sqrt(x**2 + 1)/15

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int x^3 \sqrt {1+x^2} \, dx=\frac {1}{5} \, {\left (x^{2} + 1\right )}^{\frac {3}{2}} x^{2} - \frac {2}{15} \, {\left (x^{2} + 1\right )}^{\frac {3}{2}} \]

[In]

integrate(x^3*(x^2+1)^(1/2),x, algorithm="maxima")

[Out]

1/5*(x^2 + 1)^(3/2)*x^2 - 2/15*(x^2 + 1)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int x^3 \sqrt {1+x^2} \, dx=\frac {1}{5} \, {\left (x^{2} + 1\right )}^{\frac {5}{2}} - \frac {1}{3} \, {\left (x^{2} + 1\right )}^{\frac {3}{2}} \]

[In]

integrate(x^3*(x^2+1)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int x^3 \sqrt {1+x^2} \, dx=\sqrt {x^2+1}\,\left (\frac {x^4}{5}+\frac {x^2}{15}-\frac {2}{15}\right ) \]

[In]

int(x^3*(x^2 + 1)^(1/2),x)

[Out]

(x^2 + 1)^(1/2)*(x^2/15 + x^4/5 - 2/15)