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]

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

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Rubi [A]  time = 0.06, antiderivative size = 84, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 1.11, size = 70, normalized size = 0.83 \[ \frac {{\left (2 \, a b p x^{2} + {\left (2 \, b^{2} p + b^{2}\right )} x^{4} - a^{2}\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]

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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giac [A]  time = 0.22, size = 132, normalized size = 1.57 \[ \frac {2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{2} p x^{4} + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{2} x^{4} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b p x^{2} - {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2}}{4 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]

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)

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maple [A]  time = 0.01, size = 60, normalized size = 0.71 \[ -\frac {\left (-2 x^{2} p b -b \,x^{2}+a \right ) \left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 \left (2 p^{2}+3 p +1\right ) b^{2}} \]

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 = 1.43, size = 54, normalized size = 0.64 \[ \frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{4} + 2 \, a b p x^{2} - a^{2}\right )} {\left (b x^{2} + a\right )}^{2 \, p}}{4 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} \]

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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mupad [B]  time = 4.26, size = 85, normalized size = 1.01 \[ {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p\,\left (\frac {x^4\,\left (2\,p+1\right )}{4\,\left (2\,p^2+3\,p+1\right )}-\frac {a^2}{4\,b^2\,\left (2\,p^2+3\,p+1\right )}+\frac {a\,p\,x^2}{2\,b\,\left (2\,p^2+3\,p+1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \frac {x^{4} \left (a^{2}\right )^{p}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -1 \\\int \frac {x^{3}}{\sqrt {\left (a + b x^{2}\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\- \frac {a^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 12 b^{2} p + 4 b^{2}} + \frac {2 a b p x^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 12 b^{2} p + 4 b^{2}} + \frac {2 b^{2} p x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 12 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 12 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases} \]

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]

Piecewise((x**4*(a**2)**p/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*sq
rt(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, -1)), (Inte
gral(x**3/sqrt((a + b*x**2)**2), x), Eq(p, -1/2)), (-a**2*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**2*p**2 + 12
*b**2*p + 4*b**2) + 2*a*b*p*x**2*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**2*p**2 + 12*b**2*p + 4*b**2) + 2*b**
2*p*x**4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**2*p**2 + 12*b**2*p + 4*b**2) + b**2*x**4*(a**2 + 2*a*b*x**2
+ b**2*x**4)**p/(8*b**2*p**2 + 12*b**2*p + 4*b**2), True))

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