3.800 \(\int x^3 (a^2+2 a b x^2+b^2 x^4)^p \, dx\)

Optimal. Leaf size=84 \[ \frac{\left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (p+1)}-\frac{a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0566805, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1113, 266, 43} \[ \frac{\left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (p+1)}-\frac{a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

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

Rule 1113

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^2 +
c*x^4)^FracPart[p])/(1 + (2*c*x^2)/b)^(2*FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^2)/b)^(2*p), x], x] /; FreeQ[{
a, b, c, d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[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^2+2 a b x^2+b^2 x^4\right )^p \, dx &=\left (\left (1+\frac{b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \int x^3 \left (1+\frac{b x^2}{a}\right )^{2 p} \, dx\\ &=\frac{1}{2} \left (\left (1+\frac{b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname{Subst}\left (\int x \left (1+\frac{b x}{a}\right )^{2 p} \, dx,x,x^2\right )\\ &=\frac{1}{2} \left (\left (1+\frac{b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname{Subst}\left (\int \left (-\frac{a \left (1+\frac{b x}{a}\right )^{2 p}}{b}+\frac{a \left (1+\frac{b x}{a}\right )^{1+2 p}}{b}\right ) \, dx,x,x^2\right )\\ &=-\frac{a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (1+2 p)}+\frac{\left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.021166, size = 51, normalized size = 0.61 \[ \frac{\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \left (b (2 p+1) x^2-a\right )}{4 b^2 (p+1) (2 p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

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

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Maple [A]  time = 0.048, size = 60, normalized size = 0.7 \begin{align*} -{\frac{ \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{p} \left ( -2\,{x}^{2}pb-b{x}^{2}+a \right ) \left ( b{x}^{2}+a \right ) }{4\,{b}^{2} \left ( 2\,{p}^{2}+3\,p+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b^2*x^4+2*a*b*x^2+a^2)^p,x)

[Out]

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

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Maxima [A]  time = 0.974834, size = 73, normalized size = 0.87 \begin{align*} \frac{{\left (b^{2}{\left (2 \, p + 1\right )} x^{4} + 2 \, a b p x^{2} - a^{2}\right )}{\left (b x^{2} + a\right )}^{2 \, p}}{4 \,{\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.61221, size = 142, normalized size = 1.69 \begin{align*} \frac{{\left (2 \, a b p x^{2} +{\left (2 \, b^{2} p + b^{2}\right )} x^{4} - a^{2}\right )}{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \,{\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b**2*x**4+2*a*b*x**2+a**2)**p,x)

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.32464, size = 178, normalized size = 2.12 \begin{align*} \frac{2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{2} p x^{4} +{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{2} x^{4} + 2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b p x^{2} -{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2}}{4 \,{\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="giac")

[Out]

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