3.5.9 \(\int \frac {x^2 (d+e x^2)^q}{a+b x^2+c x^4} \, dx\) [409]

Optimal. Leaf size=162 \[ -\frac {x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{\sqrt {b^2-4 a c}}+\frac {x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{\sqrt {b^2-4 a c}} \]

[Out]

-x*(e*x^2+d)^q*AppellF1(1/2,1,-q,3/2,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-e*x^2/d)/((1+e*x^2/d)^q)/(-4*a*c+b^2)^(1
/2)+x*(e*x^2+d)^q*AppellF1(1/2,1,-q,3/2,-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)),-e*x^2/d)/((1+e*x^2/d)^q)/(-4*a*c+b^2)
^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1317, 441, 440} \begin {gather*} \frac {x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{\sqrt {b^2-4 a c}}-\frac {x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{\sqrt {b^2-4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((x*(d + e*x^2)^q*AppellF1[1/2, 1, -q, 3/2, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), -((e*x^2)/d)])/(Sqrt[b^2 - 4*
a*c]*(1 + (e*x^2)/d)^q)) + (x*(d + e*x^2)^q*AppellF1[1/2, 1, -q, 3/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), -((e
*x^2)/d)])/(Sqrt[b^2 - 4*a*c]*(1 + (e*x^2)/d)^q)

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 1317

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx &=\int \left (\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^q}{b-\sqrt {b^2-4 a c}+2 c x^2}+\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^q}{b+\sqrt {b^2-4 a c}+2 c x^2}\right ) \, dx\\ &=\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {\left (d+e x^2\right )^q}{b-\sqrt {b^2-4 a c}+2 c x^2} \, dx+\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {\left (d+e x^2\right )^q}{b+\sqrt {b^2-4 a c}+2 c x^2} \, dx\\ &=\left (\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q}\right ) \int \frac {\left (1+\frac {e x^2}{d}\right )^q}{b-\sqrt {b^2-4 a c}+2 c x^2} \, dx+\left (\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q}\right ) \int \frac {\left (1+\frac {e x^2}{d}\right )^q}{b+\sqrt {b^2-4 a c}+2 c x^2} \, dx\\ &=-\frac {x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{\sqrt {b^2-4 a c}}+\frac {x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{\sqrt {b^2-4 a c}}\\ \end {align*}

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Mathematica [F]
time = 0.19, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (e \,x^{2}+d \right )^{q}}{c \,x^{4}+b \,x^{2}+a}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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