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]

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

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Rubi [A]  time = 0.54, antiderivative size = 256, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.34, 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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fricas [F]  time = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x^{2} + d\right )}^{q} x^{5}}{c x^{4} + b x^{2} + a}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{q} x^{5}}{c x^{4} + b x^{2} + a}\,{d x} \]

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)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \left (e \,x^{2}+d \right )^{q}}{c \,x^{4}+b \,x^{2}+a}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{q} x^{5}}{c x^{4} + b x^{2} + a}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5\,{\left (e\,x^2+d\right )}^q}{c\,x^4+b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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