3.2.38 \(\int x (b+2 c x^2) (b x^2+c x^4)^p \, dx\) [138]

Optimal. Leaf size=24 \[ \frac {\left (b x^2+c x^4\right )^{1+p}}{2 (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1602} \begin {gather*} \frac {\left (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)*(b*x^2 + c*x^4)^p,x]

[Out]

(b*x^2 + c*x^4)^(1 + p)/(2*(1 + p))

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 24, normalized size = 1.00 \begin {gather*} \frac {\left (x^2 \left (b+c x^2\right )\right )^{1+p}}{2 (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(b + c*x^2))^(1 + p)/(2*(1 + p))

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Maple [A]
time = 0.16, size = 31, normalized size = 1.29

method	result	size
gosper	\(\frac {x^{2} \left (c \,x^{2}+b \right ) \left (c \,x^{4}+b \,x^{2}\right )^{p}}{2+2 p}\)	\(31\)
risch	\(\frac {x^{2} \left (c \,x^{2}+b \right ) \left (x^{2} \left (c \,x^{2}+b \right )\right )^{p}}{2+2 p}\)	\(31\)
norman	\(\frac {b \,x^{2} {\mathrm e}^{p \ln \left (c \,x^{4}+b \,x^{2}\right )}}{2+2 p}+\frac {c \,x^{4} {\mathrm e}^{p \ln \left (c \,x^{4}+b \,x^{2}\right )}}{2+2 p}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 35, normalized size = 1.46 \begin {gather*} \frac {{\left (c x^{4} + b x^{2}\right )} e^{\left (p \log \left (c x^{2} + b\right ) + 2 \, p \log \left (x\right )\right )}}{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)^p,x, algorithm="maxima")

[Out]

1/2*(c*x^4 + b*x^2)*e^(p*log(c*x^2 + b) + 2*p*log(x))/(p + 1)

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Fricas [A]
time = 0.34, size = 31, normalized size = 1.29 \begin {gather*} \frac {{\left (c x^{4} + b x^{2}\right )} {\left (c x^{4} + b x^{2}\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)^p,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (17) = 34\).
time = 10.04, size = 75, normalized size = 3.12 \begin {gather*} \begin {cases} \frac {b x^{2} \left (b x^{2} + c x^{4}\right )^{p}}{2 p + 2} + \frac {c x^{4} \left (b x^{2} + c x^{4}\right )^{p}}{2 p + 2} & \text {for}\: p \neq -1 \\\log {\left (x \right )} + \frac {\log {\left (x - \sqrt {- \frac {b}{c}} \right )}}{2} + \frac {\log {\left (x + \sqrt {- \frac {b}{c}} \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)**p,x)

[Out]

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

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Giac [A]
time = 3.91, size = 22, normalized size = 0.92 \begin {gather*} \frac {{\left (c x^{4} + b x^{2}\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)^p,x, algorithm="giac")

[Out]

1/2*(c*x^4 + b*x^2)^(p + 1)/(p + 1)

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Mupad [B]
time = 2.07, size = 31, normalized size = 1.29 \begin {gather*} \frac {x^2\,\left (c\,x^2+b\right )\,{\left (c\,x^4+b\,x^2\right )}^p}{2\,\left (p+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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