3.1041 \(\int x^3 (a+b x^2)^p \, dx\)

Optimal. Leaf size=48 \[ \frac{\left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}-\frac{a \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0289054, antiderivative size = 48, 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} \[ \frac{\left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}-\frac{a \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)^p,x]

[Out]

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

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^3 \left (a+b x^2\right )^p \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x (a+b x)^p \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{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]  time = 0.0183847, size = 40, normalized size = 0.83 \[ \frac{\left (a+b x^2\right )^{p+1} \left (b (p+1) x^2-a\right )}{2 b^2 (p+1) (p+2)} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.003, size = 42, normalized size = 0.9 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) ^{1+p} \left ( -{x}^{2}pb-b{x}^{2}+a \right ) }{2\,{b}^{2} \left ({p}^{2}+3\,p+2 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.28799, size = 63, normalized size = 1.31 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.5574, size = 115, normalized size = 2.4 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 2.05159, size = 364, normalized size = 7.58 \begin{align*} \begin{cases} \frac{a^{p} x^{4}}{4} & \text{for}\: b = 0 \\\frac{a \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{a \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \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 (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{b x^{2} \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text{for}\: p = -2 \\- \frac{a \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 b^{2}} - \frac{a \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.54358, size = 127, normalized size = 2.65 \begin{align*} \frac{{\left (b x^{2} + a\right )}^{2}{\left (b x^{2} + a\right )}^{p} p -{\left (b x^{2} + a\right )}{\left (b x^{2} + a\right )}^{p} a p +{\left (b x^{2} + a\right )}^{2}{\left (b x^{2} + a\right )}^{p} - 2 \,{\left (b x^{2} + a\right )}{\left (b x^{2} + a\right )}^{p} a}{2 \,{\left (p^{2} + 3 \, p + 2\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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