\(\int x^3 (a+b x^2)^p \, dx\) [1041]

Optimal result

Integrand size = 13, antiderivative size = 48 \[ \int x^3 \left (a+b x^2\right )^p \, dx=-\frac {a \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {\left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)} \]

[Out]

-1/2*a*(b*x^2+a)^(p+1)/b^2/(p+1)+1/2*(b*x^2+a)^(2+p)/b^2/(2+p)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, 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} \[ \int x^3 \left (a+b x^2\right )^p \, dx=\frac {\left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}-\frac {a \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)} \]

[In]

Int[x^3*(a + b*x^2)^p,x]

[Out]

-1/2*(a*(a + b*x^2)^(1 + p))/(b^2*(1 + p)) + (a + b*x^2)^(2 + p)/(2*b^2*(2 + p))

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 (a+b x)^p \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {a (a+b x)^p}{b}+\frac {(a+b x)^{1+p}}{b}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {\left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.83 \[ \int x^3 \left (a+b x^2\right )^p \, dx=\frac {\left (a+b x^2\right )^{1+p} \left (-a+b (1+p) x^2\right )}{2 b^2 (1+p) (2+p)} \]

[In]

Integrate[x^3*(a + b*x^2)^p,x]

[Out]

((a + b*x^2)^(1 + p)*(-a + b*(1 + p)*x^2))/(2*b^2*(1 + p)*(2 + p))

Maple [A] (verified)

Time = 2.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88

method	result	size
gosper	\(-\frac {\left (b \,x^{2}+a \right )^{1+p} \left (-x^{2} p b -b \,x^{2}+a \right )}{2 b^{2} \left (p^{2}+3 p +2\right )}\)	\(42\)
risch	\(-\frac {\left (-b^{2} p \,x^{4}-b^{2} x^{4}-a b p \,x^{2}+a^{2}\right ) \left (b \,x^{2}+a \right )^{p}}{2 b^{2} \left (2+p \right ) \left (1+p \right )}\)	\(54\)
norman	\(\frac {x^{4} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{4+2 p}-\frac {a^{2} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 b^{2} \left (p^{2}+3 p +2\right )}+\frac {p a \,x^{2} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 b \left (p^{2}+3 p +2\right )}\)	\(83\)
parallelrisch	\(\frac {x^{4} \left (b \,x^{2}+a \right )^{p} a \,b^{2} p +x^{4} \left (b \,x^{2}+a \right )^{p} a \,b^{2}+x^{2} \left (b \,x^{2}+a \right )^{p} a^{2} b p -\left (b \,x^{2}+a \right )^{p} a^{3}}{2 \left (p^{2}+3 p +2\right ) a \,b^{2}}\)	\(87\)

[In]

int(x^3*(b*x^2+a)^p,x,method=_RETURNVERBOSE)

[Out]

-1/2/b^2*(b*x^2+a)^(1+p)/(p^2+3*p+2)*(-b*p*x^2-b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.21 \[ \int x^3 \left (a+b x^2\right )^p \, dx=\frac {{\left (a b p x^{2} + {\left (b^{2} p + b^{2}\right )} x^{4} - a^{2}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}\right )}} \]

[In]

integrate(x^3*(b*x^2+a)^p,x, algorithm="fricas")

[Out]

1/2*(a*b*p*x^2 + (b^2*p + b^2)*x^4 - a^2)*(b*x^2 + a)^p/(b^2*p^2 + 3*b^2*p + 2*b^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (37) = 74\).

Time = 0.56 (sec) , antiderivative size = 333, normalized size of antiderivative = 6.94 \[ \int x^3 \left (a+b x^2\right )^p \, dx=\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} - \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*(b*x**2+a)**p,x)

[Out]

Piecewise((a**p*x**4/4, Eq(b, 0)), (a*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + a*log(x + sqrt(-a/b))/(2*
a*b**2 + 2*b**3*x**2) + a/(2*a*b**2 + 2*b**3*x**2) + b*x**2*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + b*x
**2*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2), Eq(p, -2)), (-a*log(x - sqrt(-a/b))/(2*b**2) - a*log(x + sqr
t(-a/b))/(2*b**2) + x**2/(2*b), Eq(p, -1)), (-a**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + a*b*p*x
**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*p*x**4*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p +
4*b**2) + b**2*x**4*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98 \[ \int x^3 \left (a+b x^2\right )^p \, dx=\frac {{\left (b^{2} {\left (p + 1\right )} x^{4} + a b p x^{2} - a^{2}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (p^{2} + 3 \, p + 2\right )} b^{2}} \]

[In]

integrate(x^3*(b*x^2+a)^p,x, algorithm="maxima")

[Out]

1/2*(b^2*(p + 1)*x^4 + a*b*p*x^2 - a^2)*(b*x^2 + a)^p/((p^2 + 3*p + 2)*b^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int x^3 \left (a+b x^2\right )^p \, dx=\frac {{\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p}}{2 \, b^{2} {\left (p + 2\right )}} - \frac {{\left (b x^{2} + a\right )}^{p + 1} a}{2 \, b^{2} {\left (p + 1\right )}} \]

[In]

integrate(x^3*(b*x^2+a)^p,x, algorithm="giac")

[Out]

1/2*(b*x^2 + a)^2*(b*x^2 + a)^p/(b^2*(p + 2)) - 1/2*(b*x^2 + a)^(p + 1)*a/(b^2*(p + 1))

Mupad [B] (verification not implemented)

Time = 4.58 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.42 \[ \int x^3 \left (a+b x^2\right )^p \, dx={\left (b\,x^2+a\right )}^p\,\left (\frac {x^4\,\left (p+1\right )}{2\,\left (p^2+3\,p+2\right )}-\frac {a^2}{2\,b^2\,\left (p^2+3\,p+2\right )}+\frac {a\,p\,x^2}{2\,b\,\left (p^2+3\,p+2\right )}\right ) \]

[In]

int(x^3*(a + b*x^2)^p,x)

[Out]

(a + b*x^2)^p*((x^4*(p + 1))/(2*(3*p + p^2 + 2)) - a^2/(2*b^2*(3*p + p^2 + 2)) + (a*p*x^2)/(2*b*(3*p + p^2 + 2
)))