3.402 \(\int \frac{x^5 (d+e x^2)^q}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=256 \[ \frac{\left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{2 c (q+1) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}+\frac{\left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c (q+1) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{\left (d+e x^2\right )^{q+1}}{2 c e (q+1)} \]

[Out]

(d + e*x^2)^(1 + q)/(2*c*e*(1 + q)) + ((b - (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*(d + e*x^2)^(1 + q)*Hypergeometri
c2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)])/(2*c*(2*c*d - (b - Sqrt[b^2 - 4*
a*c])*e)*(1 + q)) + ((b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*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*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(1 + q))

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Rubi [A]  time = 0.541668, antiderivative size = 256, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {1251, 1628, 68} \[ \frac{\left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{2 c (q+1) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}+\frac{\left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c (q+1) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{\left (d+e x^2\right )^{q+1}}{2 c e (q+1)} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]

[Out]

(d + e*x^2)^(1 + q)/(2*c*e*(1 + q)) + ((b - (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*(d + e*x^2)^(1 + q)*Hypergeometri
c2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)])/(2*c*(2*c*d - (b - Sqrt[b^2 - 4*
a*c])*e)*(1 + q)) + ((b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*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*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(1 + q))

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(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] &&
IntegerQ[(m - 1)/2]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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^5 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2 (d+e x)^q}{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{(d+e x)^q}{c}+\frac{\left (-\frac{b}{c}+\frac{b^2-2 a c}{c \sqrt{b^2-4 a c}}\right ) (d+e x)^q}{b-\sqrt{b^2-4 a c}+2 c x}+\frac{\left (-\frac{b}{c}-\frac{b^2-2 a c}{c \sqrt{b^2-4 a c}}\right ) (d+e x)^q}{b+\sqrt{b^2-4 a c}+2 c x}\right ) \, dx,x,x^2\right )\\ &=\frac{\left (d+e x^2\right )^{1+q}}{2 c e (1+q)}-\frac{\left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{(d+e x)^q}{b-\sqrt{b^2-4 a c}+2 c x} \, dx,x,x^2\right )}{2 c}-\frac{\left (b+\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{(d+e x)^q}{b+\sqrt{b^2-4 a c}+2 c x} \, dx,x,x^2\right )}{2 c}\\ &=\frac{\left (d+e x^2\right )^{1+q}}{2 c e (1+q)}+\frac{\left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \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 )}{2 c \left (2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right ) (1+q)}+\frac{\left (b+\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \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 )}{2 c \left (2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) (1+q)}\\ \end{align*}

Mathematica [A]  time = 0.384033, size = 211, normalized size = 0.82 \[ \frac{\left (d+e x^2\right )^{q+1} \left (\frac{\left (\frac{2 a c-b^2}{\sqrt{b^2-4 a c}}+b\right ) \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d+\left (\sqrt{b^2-4 a c}-b\right ) e}\right )}{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}+\frac{\left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}+\frac{1}{e}\right )}{2 c (q+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]

[Out]

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

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Maple [F]  time = 0.05, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{5} \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^5*(e*x^2+d)^q/(c*x^4+b*x^2+a),x)

[Out]

int(x^5*(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^{5}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)^q*x^5/(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^{5}}{c x^{4} + b x^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

integral((e*x^2 + d)^q*x^5/(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**5*(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^{5}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^q*x^5/(c*x^4 + b*x^2 + a), x)