\(\int x^3 (1+x^2)^{9/14} \, dx\) [466]

Optimal result

Integrand size = 13, antiderivative size = 27 \[ \int x^3 \left (1+x^2\right )^{9/14} \, dx=-\frac {7}{23} \left (1+x^2\right )^{23/14}+\frac {7}{37} \left (1+x^2\right )^{37/14} \]

[Out]

-7/23*(x^2+1)^(23/14)+7/37*(x^2+1)^(37/14)

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} \[ \int x^3 \left (1+x^2\right )^{9/14} \, dx=\frac {7}{37} \left (x^2+1\right )^{37/14}-\frac {7}{23} \left (x^2+1\right )^{23/14} \]

[In]

Int[x^3*(1 + x^2)^(9/14),x]

[Out]

(-7*(1 + x^2)^(23/14))/23 + (7*(1 + x^2)^(37/14))/37

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)^{9/14} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-(1+x)^{9/14}+(1+x)^{23/14}\right ) \, dx,x,x^2\right ) \\ & = -\frac {7}{23} \left (1+x^2\right )^{23/14}+\frac {7}{37} \left (1+x^2\right )^{37/14} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int x^3 \left (1+x^2\right )^{9/14} \, dx=\frac {7}{851} \left (1+x^2\right )^{9/14} \left (-14+9 x^2+23 x^4\right ) \]

[In]

Integrate[x^3*(1 + x^2)^(9/14),x]

[Out]

(7*(1 + x^2)^(9/14)*(-14 + 9*x^2 + 23*x^4))/851

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63

method	result	size
gosper	\(\frac {7 \left (x^{2}+1\right )^{\frac {23}{14}} \left (23 x^{2}-14\right )}{851}\)	\(17\)
meijerg	\(\frac {x^{4} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {9}{14},2;3;-x^{2}\right )}{4}\)	\(17\)
pseudoelliptic	\(\frac {7 \left (x^{2}+1\right )^{\frac {23}{14}} \left (23 x^{2}-14\right )}{851}\)	\(17\)
trager	\(\left (\frac {7}{37} x^{4}+\frac {63}{851} x^{2}-\frac {98}{851}\right ) \left (x^{2}+1\right )^{\frac {9}{14}}\)	\(21\)
risch	\(\frac {7 \left (x^{2}+1\right )^{\frac {9}{14}} \left (23 x^{4}+9 x^{2}-14\right )}{851}\)	\(22\)

[In]

int(x^3*(x^2+1)^(9/14),x,method=_RETURNVERBOSE)

[Out]

7/851*(x^2+1)^(23/14)*(23*x^2-14)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int x^3 \left (1+x^2\right )^{9/14} \, dx=\frac {7}{851} \, {\left (23 \, x^{4} + 9 \, x^{2} - 14\right )} {\left (x^{2} + 1\right )}^{\frac {9}{14}} \]

[In]

integrate(x^3*(x^2+1)^(9/14),x, algorithm="fricas")

[Out]

7/851*(23*x^4 + 9*x^2 - 14)*(x^2 + 1)^(9/14)

Sympy [A] (verification not implemented)

Time = 2.71 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int x^3 \left (1+x^2\right )^{9/14} \, dx=\frac {7 x^{4} \left (x^{2} + 1\right )^{\frac {9}{14}}}{37} + \frac {63 x^{2} \left (x^{2} + 1\right )^{\frac {9}{14}}}{851} - \frac {98 \left (x^{2} + 1\right )^{\frac {9}{14}}}{851} \]

[In]

integrate(x**3*(x**2+1)**(9/14),x)

[Out]

7*x**4*(x**2 + 1)**(9/14)/37 + 63*x**2*(x**2 + 1)**(9/14)/851 - 98*(x**2 + 1)**(9/14)/851

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int x^3 \left (1+x^2\right )^{9/14} \, dx=\frac {7}{37} \, {\left (x^{2} + 1\right )}^{\frac {37}{14}} - \frac {7}{23} \, {\left (x^{2} + 1\right )}^{\frac {23}{14}} \]

[In]

integrate(x^3*(x^2+1)^(9/14),x, algorithm="maxima")

[Out]

7/37*(x^2 + 1)^(37/14) - 7/23*(x^2 + 1)^(23/14)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int x^3 \left (1+x^2\right )^{9/14} \, dx=\frac {7}{37} \, {\left (x^{2} + 1\right )}^{\frac {37}{14}} - \frac {7}{23} \, {\left (x^{2} + 1\right )}^{\frac {23}{14}} \]

[In]

integrate(x^3*(x^2+1)^(9/14),x, algorithm="giac")

[Out]

7/37*(x^2 + 1)^(37/14) - 7/23*(x^2 + 1)^(23/14)

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int x^3 \left (1+x^2\right )^{9/14} \, dx={\left (x^2+1\right )}^{9/14}\,\left (\frac {7\,x^4}{37}+\frac {63\,x^2}{851}-\frac {98}{851}\right ) \]

[In]

int(x^3*(x^2 + 1)^(9/14),x)

[Out]

(x^2 + 1)^(9/14)*((63*x^2)/851 + (7*x^4)/37 - 98/851)