3.2.34 \(\int x (b+2 c x^2) (-a+b x^2+c x^4)^p \, dx\) [134]

Optimal. Leaf size=27 \[ \frac {\left (-a+b x^2+c x^4\right )^{1+p}}{2 (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1261, 643} \begin {gather*} \frac {\left (-a+b x^2+c x^4\right )^{p+1}}{2 (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 643

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 1261

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

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Mathematica [A]
time = 0.04, size = 27, normalized size = 1.00 \begin {gather*} \frac {\left (-a+b x^2+c x^4\right )^{1+p}}{2 (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 26, normalized size = 0.96

method	result	size
gosper	\(\frac {\left (c \,x^{4}+b \,x^{2}-a \right )^{1+p}}{2+2 p}\)	\(26\)
risch	\(-\frac {\left (-c \,x^{4}-b \,x^{2}+a \right ) \left (c \,x^{4}+b \,x^{2}-a \right )^{p}}{2 \left (1+p \right )}\)	\(38\)
norman	\(-\frac {a \,{\mathrm e}^{p \ln \left (c \,x^{4}+b \,x^{2}-a \right )}}{2 \left (1+p \right )}+\frac {b \,x^{2} {\mathrm e}^{p \ln \left (c \,x^{4}+b \,x^{2}-a \right )}}{2+2 p}+\frac {c \,x^{4} {\mathrm e}^{p \ln \left (c \,x^{4}+b \,x^{2}-a \right )}}{2+2 p}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(2*c*x^2+b)*(c*x^4+b*x^2-a)^p,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.31, size = 37, normalized size = 1.37 \begin {gather*} \frac {{\left (c x^{4} + b x^{2} - a\right )} {\left (c x^{4} + b x^{2} - a\right )}^{p}}{2 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 37, normalized size = 1.37 \begin {gather*} \frac {{\left (c x^{4} + b x^{2} - a\right )} {\left (c x^{4} + b x^{2} - a\right )}^{p}}{2 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (19) = 38\).
time = 130.39, size = 201, normalized size = 7.44 \begin {gather*} \begin {cases} - \frac {a \left (- a + b x^{2} + c x^{4}\right )^{p}}{2 p + 2} + \frac {b x^{2} \left (- a + b x^{2} + c x^{4}\right )^{p}}{2 p + 2} + \frac {c x^{4} \left (- a + b x^{2} + c x^{4}\right )^{p}}{2 p + 2} & \text {for}\: p \neq -1 \\\frac {\log {\left (x - \frac {\sqrt {2} \sqrt {- \frac {b}{c} - \frac {\sqrt {4 a c + b^{2}}}{c}}}{2} \right )}}{2} + \frac {\log {\left (x + \frac {\sqrt {2} \sqrt {- \frac {b}{c} - \frac {\sqrt {4 a c + b^{2}}}{c}}}{2} \right )}}{2} + \frac {\log {\left (x - \frac {\sqrt {2} \sqrt {- \frac {b}{c} + \frac {\sqrt {4 a c + b^{2}}}{c}}}{2} \right )}}{2} + \frac {\log {\left (x + \frac {\sqrt {2} \sqrt {- \frac {b}{c} + \frac {\sqrt {4 a c + b^{2}}}{c}}}{2} \right )}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*(-a + b*x**2 + c*x**4)**p/(2*p + 2) + b*x**2*(-a + b*x**2 + c*x**4)**p/(2*p + 2) + c*x**4*(-a +
b*x**2 + c*x**4)**p/(2*p + 2), Ne(p, -1)), (log(x - sqrt(2)*sqrt(-b/c - sqrt(4*a*c + b**2)/c)/2)/2 + log(x + s
qrt(2)*sqrt(-b/c - sqrt(4*a*c + b**2)/c)/2)/2 + log(x - sqrt(2)*sqrt(-b/c + sqrt(4*a*c + b**2)/c)/2)/2 + log(x
+ sqrt(2)*sqrt(-b/c + sqrt(4*a*c + b**2)/c)/2)/2, True))

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Giac [A]
time = 4.21, size = 25, normalized size = 0.93 \begin {gather*} \frac {{\left (c x^{4} + b x^{2} - a\right )}^{p + 1}}{2 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.05, size = 52, normalized size = 1.93 \begin {gather*} {\left (c\,x^4+b\,x^2-a\right )}^p\,\left (\frac {b\,x^2}{2\,p+2}-\frac {a}{2\,p+2}+\frac {c\,x^4}{2\,p+2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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