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

Optimal. Leaf size=34 \[ \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{p+1}}{4 b (p+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1247, 629} \[ \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{p+1}}{4 b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 629

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d*(a + b*x + c*x^2)^(p +
1))/(b*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rubi steps

\begin {align*} \int x \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx,x,x^2\right )\\ &=\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{1+p}}{4 b (1+p)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 25, normalized size = 0.74 \[ \frac {\left (\left (a+b x^2\right )^2\right )^{p+1}}{4 b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^2)^(1 + p)/(4*b*(1 + p))

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 47, normalized size = 1.38 \[ \frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \, {\left (b p + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.29, size = 32, normalized size = 0.94 \[ \frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p + 1}}{4 \, b {\left (p + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.00, size = 40, normalized size = 1.18 \[ \frac {\left (b \,x^{2}+a \right )^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 \left (p +1\right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.72, size = 86, normalized size = 2.53 \[ \frac {{\left (b x^{2} + a\right )} {\left (b x^{2} + a\right )}^{2 \, p} a}{2 \, b {\left (2 \, p + 1\right )}} + \frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{4} + 2 \, a b p x^{2} - a^{2}\right )} {\left (b x^{2} + a\right )}^{2 \, p}}{4 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.14, size = 59, normalized size = 1.74 \[ {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p\,\left (\frac {a^2}{4\,b\,\left (p+1\right )}+\frac {a\,x^2}{2\,\left (p+1\right )}+\frac {b\,x^4}{4\,\left (p+1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 9.60, size = 165, normalized size = 4.85 \[ \begin {cases} \frac {x^{2}}{2 a} & \text {for}\: b = 0 \wedge p = -1 \\\frac {a x^{2} \left (a^{2}\right )^{p}}{2} & \text {for}\: b = 0 \\\frac {\log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b} + \frac {\log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b} & \text {for}\: p = -1 \\\frac {a^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{4 b p + 4 b} + \frac {2 a b x^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{4 b p + 4 b} + \frac {b^{2} x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{4 b p + 4 b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________