\(\int x^3 (1+x^2)^{3/4} \, dx\) [253]

Optimal result

Integrand size = 13, antiderivative size = 25 \[ \int x^3 \left (1+x^2\right )^{3/4} \, dx=\frac {2}{77} \left (1+x^2\right )^{3/4} \left (-4+3 x^2+7 x^4\right ) \]

[Out]

2/77*(x^2+1)^(3/4)*(7*x^4+3*x^2-4)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int x^3 \left (1+x^2\right )^{3/4} \, dx=\frac {2}{11} \left (x^2+1\right )^{11/4}-\frac {2}{7} \left (x^2+1\right )^{7/4} \]

[In]

Int[x^3*(1 + x^2)^(3/4),x]

[Out]

(-2*(1 + x^2)^(7/4))/7 + (2*(1 + x^2)^(11/4))/11

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 (1+x)^{3/4} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-(1+x)^{3/4}+(1+x)^{7/4}\right ) \, dx,x,x^2\right ) \\ & = -\frac {2}{7} \left (1+x^2\right )^{7/4}+\frac {2}{11} \left (1+x^2\right )^{11/4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int x^3 \left (1+x^2\right )^{3/4} \, dx=\frac {2}{77} \left (1+x^2\right )^{7/4} \left (-4+7 x^2\right ) \]

[In]

Integrate[x^3*(1 + x^2)^(3/4),x]

[Out]

(2*(1 + x^2)^(7/4)*(-4 + 7*x^2))/77

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68

method	result	size
gosper	\(\frac {2 \left (x^{2}+1\right )^{\frac {7}{4}} \left (7 x^{2}-4\right )}{77}\)	\(17\)
meijerg	\(\frac {x^{4} \operatorname {hypergeom}\left (\left [-\frac {3}{4}, 2\right ], \left [3\right ], -x^{2}\right )}{4}\)	\(17\)
pseudoelliptic	\(\frac {2 \left (x^{2}+1\right )^{\frac {7}{4}} \left (7 x^{2}-4\right )}{77}\)	\(17\)
trager	\(\left (\frac {2}{11} x^{4}+\frac {6}{77} x^{2}-\frac {8}{77}\right ) \left (x^{2}+1\right )^{\frac {3}{4}}\)	\(21\)
risch	\(\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}} \left (7 x^{4}+3 x^{2}-4\right )}{77}\)	\(22\)

[In]

int(x^3*(x^2+1)^(3/4),x,method=_RETURNVERBOSE)

[Out]

2/77*(x^2+1)^(7/4)*(7*x^2-4)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int x^3 \left (1+x^2\right )^{3/4} \, dx=\frac {2}{77} \, {\left (7 \, x^{4} + 3 \, x^{2} - 4\right )} {\left (x^{2} + 1\right )}^{\frac {3}{4}} \]

[In]

integrate(x^3*(x^2+1)^(3/4),x, algorithm="fricas")

[Out]

2/77*(7*x^4 + 3*x^2 - 4)*(x^2 + 1)^(3/4)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int x^3 \left (1+x^2\right )^{3/4} \, dx=\frac {2 x^{4} \left (x^{2} + 1\right )^{\frac {3}{4}}}{11} + \frac {6 x^{2} \left (x^{2} + 1\right )^{\frac {3}{4}}}{77} - \frac {8 \left (x^{2} + 1\right )^{\frac {3}{4}}}{77} \]

[In]

integrate(x**3*(x**2+1)**(3/4),x)

[Out]

2*x**4*(x**2 + 1)**(3/4)/11 + 6*x**2*(x**2 + 1)**(3/4)/77 - 8*(x**2 + 1)**(3/4)/77

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x^3 \left (1+x^2\right )^{3/4} \, dx=\frac {2}{11} \, {\left (x^{2} + 1\right )}^{\frac {11}{4}} - \frac {2}{7} \, {\left (x^{2} + 1\right )}^{\frac {7}{4}} \]

[In]

integrate(x^3*(x^2+1)^(3/4),x, algorithm="maxima")

[Out]

2/11*(x^2 + 1)^(11/4) - 2/7*(x^2 + 1)^(7/4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x^3 \left (1+x^2\right )^{3/4} \, dx=\frac {2}{11} \, {\left (x^{2} + 1\right )}^{\frac {11}{4}} - \frac {2}{7} \, {\left (x^{2} + 1\right )}^{\frac {7}{4}} \]

[In]

integrate(x^3*(x^2+1)^(3/4),x, algorithm="giac")

[Out]

2/11*(x^2 + 1)^(11/4) - 2/7*(x^2 + 1)^(7/4)

Mupad [B] (verification not implemented)

Time = 5.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int x^3 \left (1+x^2\right )^{3/4} \, dx={\left (x^2+1\right )}^{3/4}\,\left (\frac {2\,x^4}{11}+\frac {6\,x^2}{77}-\frac {8}{77}\right ) \]

[In]

int(x^3*(x^2 + 1)^(3/4),x)

[Out]

(x^2 + 1)^(3/4)*((6*x^2)/77 + (2*x^4)/11 - 8/77)