3.1117 \(\int x^3 (a+b x^2+c x^4)^p \, dx\)

Optimal. Leaf size=160 \[ \frac{b 2^{p-1} \left (a+b x^2+c x^4\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}}\right )^{-p-1} \, _2F_1\left (-p,p+1;p+2;\frac{2 c x^2+b+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c (p+1) \sqrt{b^2-4 a c}}+\frac{\left (a+b x^2+c x^4\right )^{p+1}}{4 c (p+1)} \]

[Out]

(a + b*x^2 + c*x^4)^(1 + p)/(4*c*(1 + p)) + (2^(-1 + p)*b*(-((b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/Sqrt[b^2 - 4*a*
c]))^(-1 - p)*(a + b*x^2 + c*x^4)^(1 + p)*Hypergeometric2F1[-p, 1 + p, 2 + p, (b + Sqrt[b^2 - 4*a*c] + 2*c*x^2
)/(2*Sqrt[b^2 - 4*a*c])])/(c*Sqrt[b^2 - 4*a*c]*(1 + p))

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Rubi [A]  time = 0.125484, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1114, 640, 624} \[ \frac{b 2^{p-1} \left (a+b x^2+c x^4\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}}\right )^{-p-1} \, _2F_1\left (-p,p+1;p+2;\frac{2 c x^2+b+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c (p+1) \sqrt{b^2-4 a c}}+\frac{\left (a+b x^2+c x^4\right )^{p+1}}{4 c (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2 + c*x^4)^p,x]

[Out]

(a + b*x^2 + c*x^4)^(1 + p)/(4*c*(1 + p)) + (2^(-1 + p)*b*(-((b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/Sqrt[b^2 - 4*a*
c]))^(-1 - p)*(a + b*x^2 + c*x^4)^(1 + p)*Hypergeometric2F1[-p, 1 + p, 2 + p, (b + Sqrt[b^2 - 4*a*c] + 2*c*x^2
)/(2*Sqrt[b^2 - 4*a*c])])/(c*Sqrt[b^2 - 4*a*c]*(1 + p))

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 624

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, -Simp[((a + b*x + c*
x^2)^(p + 1)*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/(2*q)])/(q*(p + 1)*((q - b - 2*c*x)/(2*q))^(p
+ 1)), x]] /; FreeQ[{a, b, c, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[4*p]

Rubi steps

\begin{align*} \int x^3 \left (a+b x^2+c x^4\right )^p \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int 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}}{4 c (1+p)}-\frac{b \operatorname{Subst}\left (\int \left (a+b x+c x^2\right )^p \, dx,x,x^2\right )}{4 c}\\ &=\frac{\left (a+b x^2+c x^4\right )^{1+p}}{4 c (1+p)}+\frac{2^{-1+p} b \left (-\frac{b-\sqrt{b^2-4 a c}+2 c x^2}{\sqrt{b^2-4 a c}}\right )^{-1-p} \left (a+b x^2+c x^4\right )^{1+p} \, _2F_1\left (-p,1+p;2+p;\frac{b+\sqrt{b^2-4 a c}+2 c x^2}{2 \sqrt{b^2-4 a c}}\right )}{c \sqrt{b^2-4 a c} (1+p)}\\ \end{align*}

Mathematica [C]  time = 0.19605, size = 162, normalized size = 1.01 \[ \frac{1}{4} x^4 \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (2;-p,-p;3;-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*(a + b*x^2 + c*x^4)^p,x]

[Out]

(x^4*(a + b*x^2 + c*x^4)^p*AppellF1[2, -p, -p, 3, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2
- 4*a*c])])/(4*((b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x^
2)/(b + Sqrt[b^2 - 4*a*c]))^p)

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Maple [F]  time = 0.072, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*x^4+b*x^2+a)^p,x)

[Out]

int(x^3*(c*x^4+b*x^2+a)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{4} + b x^{2} + a\right )}^{p} x^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*x^4+b*x^2+a)^p,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2 + a)^p*x^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c x^{4} + b x^{2} + a\right )}^{p} x^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*x^4+b*x^2+a)^p,x, algorithm="fricas")

[Out]

integral((c*x^4 + b*x^2 + a)^p*x^3, x)

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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**3*(c*x**4+b*x**2+a)**p,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{4} + b x^{2} + a\right )}^{p} x^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*x^4+b*x^2+a)^p,x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2 + a)^p*x^3, x)