3.2.30 \(\int x (b+2 c x^2) (a+b x^2+c x^4)^p \, dx\) [130]

Optimal. Leaf size=25 \[ \frac {\left (a+b x^2+c x^4\right )^{1+p}}{2 (1+p)} \]

[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 = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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 = 25, 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 = 24, normalized size = 0.96

method	result	size
gosper	\(\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{1+p}}{2+2 p}\)	\(24\)
risch	\(\frac {\left (c \,x^{4}+b \,x^{2}+a \right ) \left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{2+2 p}\)	\(34\)
norman	\(\frac {a \,{\mathrm e}^{p \ln \left (c \,x^{4}+b \,x^{2}+a \right )}}{2+2 p}+\frac {b \,x^{2} {\mathrm e}^{p \ln \left (c \,x^{4}+b \,x^{2}+a \right )}}{2+2 p}+\frac {c \,x^{4} {\mathrm e}^{p \ln \left (c \,x^{4}+b \,x^{2}+a \right )}}{2+2 p}\)	\(80\)

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 = 0.32, size = 33, normalized size = 1.32 \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 = 0.38, size = 33, normalized size = 1.32 \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 = 129.16, size = 201, normalized size = 8.04 \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 + 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 + log(x
+ sqrt(2)*sqrt(-b/c + sqrt(-4*a*c + b**2)/c)/2)/2, True))

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Giac [A]
time = 3.59, size = 23, normalized size = 0.92 \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 = 2.09, size = 49, normalized size = 1.96 \begin {gather*} {\left (c\,x^4+b\,x^2+a\right )}^p\,\left (\frac {a}{2\,p+2}+\frac {b\,x^2}{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)*(a + b*x^2 + c*x^4)^p,x)

[Out]

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

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