3.1079 \(\int x^{-1+2 (1+p)} (b+c x^2)^p (b+2 c x^2) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0089422, antiderivative size = 27, normalized size of antiderivative = 1., 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 = {449} \[ \frac{x^{2 (p+1)} \left (b+c x^2\right )^{p+1}}{2 (p+1)} \]

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 449

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*}

Mathematica [C]  time = 0.078724, size = 97, normalized size = 3.59 \[ \frac{x^{2 p+2} \left (b+c x^2\right )^p \left (\frac{c x^2}{b}+1\right )^{-p} \left (2 c (p+1) x^2 \, _2F_1\left (-p,p+2;p+3;-\frac{c x^2}{b}\right )+b (p+2) \, _2F_1\left (-p,p+1;p+2;-\frac{c x^2}{b}\right )\right )}{2 (p+1) (p+2)} \]

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)^p*(b*(2 + p)*Hypergeometric2F1[-p, 1 + p, 2 + p, -((c*x^2)/b)] + 2*c*(1 + p)*x^2*Hype
rgeometric2F1[-p, 2 + p, 3 + p, -((c*x^2)/b)]))/(2*(1 + p)*(2 + p)*(1 + (c*x^2)/b)^p)

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Maple [A]  time = 0.004, size = 26, normalized size = 1. \begin{align*}{\frac{{x}^{2+2\,p} \left ( c{x}^{2}+b \right ) ^{1+p}}{2+2\,p}} \end{align*}

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)

[Out]

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

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Maxima [A]  time = 1.09265, size = 47, normalized size = 1.74 \begin{align*} \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{align*}

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 = 1.06487, size = 72, normalized size = 2.67 \begin{align*} \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{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]  time = 1.10196, size = 70, normalized size = 2.59 \begin{align*} \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{align*}

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)