3.11.79 \(\int x^{-1+2 (1+p)} (b+c x^2)^p (b+2 c x^2) \, dx\) [1079]

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

[Out]

1/2*x^(2+2*p)*(c*x^2+b)^(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 = 1, number of rules used = 1, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {460} \begin {gather*} \frac {x^{2 (p+1)} \left (b+c x^2\right )^{p+1}}{2 (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*(1 + p))*(b + c*x^2)^p*(b + 2*c*x^2),x]

[Out]

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

Rule 460

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

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 26, normalized size = 0.96 \begin {gather*} \frac {x^{2+2 p} \left (b+c x^2\right )^{1+p}}{2+2 p} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 2*(1 + p))*(b + c*x^2)^p*(b + 2*c*x^2),x]

[Out]

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

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Maple [A]
time = 0.34, size = 26, normalized size = 0.96

method	result	size
gosper	\(\frac {x^{2 p +2} \left (c \,x^{2}+b \right )^{1+p}}{2 p +2}\)	\(26\)
risch	\(\frac {x \left (c \,x^{2}+b \right ) x^{1+2 p} \left (c \,x^{2}+b \right )^{p}}{2 p +2}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1+2*p)*(c*x^2+b)^p*(2*c*x^2+b),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.35, size = 35, normalized size = 1.30 \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^(1+2*p)*(c*x^2+b)^p*(2*c*x^2+b),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 = 1.65, size = 32, normalized size = 1.19 \begin {gather*} \frac {{\left (c x^{3} + b x\right )} {\left (c x^{2} + b\right )}^{p} x^{2 \, p + 1}}{2 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+2*p)*(c*x^2+b)^p*(2*c*x^2+b),x, algorithm="fricas")

[Out]

1/2*(c*x^3 + b*x)*(c*x^2 + b)^p*x^(2*p + 1)/(p + 1)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1+2*p)*(c*x**2+b)**p*(2*c*x**2+b),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (25) = 50\).
time = 0.68, size = 52, normalized size = 1.93 \begin {gather*} \frac {{\left (c x^{2} + b\right )}^{p} c x^{3} e^{\left (2 \, p \log \left (x\right ) + \log \left (x\right )\right )} + {\left (c x^{2} + b\right )}^{p} b x e^{\left (2 \, p \log \left (x\right ) + \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^(1+2*p)*(c*x^2+b)^p*(2*c*x^2+b),x, algorithm="giac")

[Out]

1/2*((c*x^2 + b)^p*c*x^3*e^(2*p*log(x) + log(x)) + (c*x^2 + b)^p*b*x*e^(2*p*log(x) + log(x)))/(p + 1)

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Mupad [B]
time = 4.88, size = 47, normalized size = 1.74 \begin {gather*} {\left (c\,x^2+b\right )}^p\,\left (\frac {c\,x^{2\,p+1}\,x^3}{2\,p+2}+\frac {b\,x\,x^{2\,p+1}}{2\,p+2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2*p + 1)*(b + c*x^2)^p*(b + 2*c*x^2),x)

[Out]

(b + c*x^2)^p*((c*x^(2*p + 1)*x^3)/(2*p + 2) + (b*x*x^(2*p + 1))/(2*p + 2))

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