3.5.11 \(\int \frac {(d+e x^2)^q}{x^2 (a+b x^2+c x^4)} \, dx\) [411]

Optimal. Leaf size=264 \[ -\frac {c \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{a \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} F_1\left (\frac {1}{2};1,-q;\frac {3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{a \left (b+\sqrt {b^2-4 a c}\right )}-\frac {\left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \, _2F_1\left (-\frac {1}{2},-q;\frac {1}{2};-\frac {e x^2}{d}\right )}{a x} \]

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1317, 372, 371, 1706, 441, 440} \begin {gather*} -\frac {c x \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \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 )}{a \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c x \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \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 )}{a \left (\sqrt {b^2-4 a c}+b\right )}-\frac {\left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \, _2F_1\left (-\frac {1}{2},-q;\frac {1}{2};-\frac {e x^2}{d}\right )}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(1 + b/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)])/(a*(b - Sqrt[b^2 - 4*a*c])*(1 + (e*x^2)/d)^q)) - (c*(1 - b/Sqrt[b^2 - 4*a*c])*x*(d + e*x^2)^q*Ap
pellF1[1/2, 1, -q, 3/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), -((e*x^2)/d)])/(a*(b + Sqrt[b^2 - 4*a*c])*(1 + (e*
x^2)/d)^q) - ((d + e*x^2)^q*Hypergeometric2F1[-1/2, -q, 1/2, -((e*x^2)/d)])/(a*x*(1 + (e*x^2)/d)^q)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

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]

Rule 1706

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

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Mathematica [F]
time = 0.33, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d+e x^2\right )^q}{x^2 \left (a+b x^2+c x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (e \,x^{2}+d \right )^{q}}{x^{2} \left (c \,x^{4}+b \,x^{2}+a \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((x^2*e + d)^q/(c*x^6 + b*x^4 + a*x^2), 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((e*x**2+d)**q/x**2/(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((e*x^2+d)^q/x^2/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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