### 3.138 $$\int x (b+2 c x^2) (b x^2+c x^4)^p \, dx$$

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

[Out]

(b*x^2 + c*x^4)^(1 + p)/(2*(1 + p))

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Rubi [A]  time = 0.0141711, antiderivative size = 24, normalized size of antiderivative = 1., 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 = {1588} $\frac{\left (b x^2+c x^4\right )^{p+1}}{2 (p+1)}$

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 1588

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*}

Mathematica [C]  time = 0.0736556, size = 97, normalized size = 4.04 $\frac{x^2 \left (x^2 \left (b+c x^2\right )\right )^p \left (\frac{c x^2}{b}+1\right )^{-p} \left (2 c (p+1) x^2 \, _2F_1\left (-p,p+2;p+3;-\frac{c x^2}{b}\right )+b (p+2) \, _2F_1\left (-p,p+1;p+2;-\frac{c x^2}{b}\right )\right )}{2 (p+1) (p+2)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(x^2*(x^2*(b + c*x^2))^p*(b*(2 + p)*Hypergeometric2F1[-p, 1 + p, 2 + p, -((c*x^2)/b)] + 2*c*(1 + p)*x^2*Hyperg
eometric2F1[-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 = 0.005, size = 31, normalized size = 1.3 \begin{align*}{\frac{{x}^{2} \left ( c{x}^{2}+b \right ) \left ( c{x}^{4}+b{x}^{2} \right ) ^{p}}{2+2\,p}} \end{align*}

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)

[Out]

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

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Maxima [A]  time = 1.1558, size = 47, normalized size = 1.96 \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*(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 = 1.08402, size = 63, normalized size = 2.62 \begin{align*} \frac{{\left (c x^{4} + b x^{2}\right )}{\left (c x^{4} + b x^{2}\right )}^{p}}{2 \,{\left (p + 1\right )}} \end{align*}

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]  time = 20.1521, size = 85, normalized size = 3.54 \begin{align*} \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 (- i \sqrt{b} \sqrt{\frac{1}{c}} + x \right )}}{2} + \frac{\log{\left (i \sqrt{b} \sqrt{\frac{1}{c}} + x \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

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(-I*sqrt(b)*sqrt(1/c) + x)/2 + log(I*sqrt(b)*sqrt(1/c) + x)/2, True))

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Giac [A]  time = 1.14638, size = 59, normalized size = 2.46 \begin{align*} \frac{{\left (c x^{4} + b x^{2}\right )}^{p} c x^{4} +{\left (c x^{4} + b x^{2}\right )}^{p} b x^{2}}{2 \,{\left (p + 1\right )}} \end{align*}

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