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

Optimal. Leaf size=218 \[ \frac {x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \left (B-\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{b-\sqrt {b^2-4 a c}}+\frac {x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \left (\frac {b B-2 A c}{\sqrt {b^2-4 a c}}+B\right ) F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{\sqrt {b^2-4 a c}+b} \]

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

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Rubi [A]  time = 0.46, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1692, 430, 429} \[ \frac {x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \left (B-\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{b-\sqrt {b^2-4 a c}}+\frac {x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \left (\frac {b B-2 A c}{\sqrt {b^2-4 a c}}+B\right ) F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{\sqrt {b^2-4 a c}+b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 429

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 430

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 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rubi steps

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

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Mathematica [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 1.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{q}}{c x^{4} + b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((B*x^2+A)*(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 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{q}}{c x^{4} + b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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