3.12.17 \(\int x^3 (a+b x^2+c x^4)^p \, dx\) [1117]

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

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 160, normalized size of antiderivative = 1.00, 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 = {1128, 654, 638} \begin {gather*} \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)} \end {gather*}

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 638

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)/(q*(p + 1)*((q - b - 2*c*x)/(2*q))^(p + 1)))*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/
(2*q)], x]] /; FreeQ[{a, b, c, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[4*p]

Rule 654

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 1128

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]

Rubi steps

\begin {align*} \int x^3 \left (a+b x^2+c x^4\right )^p \, dx &=\frac {1}{2} \text {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 \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.25, size = 162, normalized size = 1.01 \begin {gather*} \frac {1}{4} x^4 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}\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}{-b+\sqrt {b^2-4 a c}}\right ) \end {gather*}

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 = 0.01, size = 0, normalized size = 0.00 \[\int x^{3} \left (c \,x^{4}+b \,x^{2}+a \right )^{p}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (a + b x^{2} + c x^{4}\right )^{p}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,{\left (c\,x^4+b\,x^2+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^2 + c*x^4)^p,x)

[Out]

int(x^3*(a + b*x^2 + c*x^4)^p, x)

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