3.404 \(\int \frac{x (d+e x^2)^q}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=198 \[ \frac{c \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(q+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{c \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{(q+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )} \]

[Out]

-((c*(d + e*x^2)^(1 + q)*Hypergeometric2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])
*e)])/(Sqrt[b^2 - 4*a*c]*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(1 + q))) + (c*(d + e*x^2)^(1 + q)*Hypergeometric
2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(Sqrt[b^2 - 4*a*c]*(2*c*d - (b +
Sqrt[b^2 - 4*a*c])*e)*(1 + q))

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Rubi [A]  time = 0.357378, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {1247, 711, 68} \[ \frac{c \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(q+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{c \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{(q+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]

[Out]

-((c*(d + e*x^2)^(1 + q)*Hypergeometric2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])
*e)])/(Sqrt[b^2 - 4*a*c]*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(1 + q))) + (c*(d + e*x^2)^(1 + q)*Hypergeometric
2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(Sqrt[b^2 - 4*a*c]*(2*c*d - (b +
Sqrt[b^2 - 4*a*c])*e)*(1 + q))

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 711

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^
m, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && NeQ[2*c*d - b*e, 0] &&  !IntegerQ[m]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.299574, size = 168, normalized size = 0.85 \[ \frac{c \left (d+e x^2\right )^{q+1} \left (\frac{\, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}-\frac{\, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d+\left (\sqrt{b^2-4 a c}-b\right ) e}\right )}{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}\right )}{(q+1) \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]

[Out]

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

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{x \left ( e{x}^{2}+d \right ) ^{q}}{c{x}^{4}+b{x}^{2}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(e*x^2+d)^q/(c*x^4+b*x^2+a),x)

[Out]

int(x*(e*x^2+d)^q/(c*x^4+b*x^2+a),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)^q*x/(c*x^4 + b*x^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

integral((e*x^2 + d)^q*x/(c*x^4 + b*x^2 + a), 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*(e*x**2+d)**q/(c*x**4+b*x**2+a),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^q*x/(c*x^4 + b*x^2 + a), x)