3.1120 \(\int \frac{(a+b x^2+c x^4)^p}{x^3} \, dx\)

Optimal. Leaf size=166 \[ -\frac{2^{2 p-1} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (1-2 p;-p,-p;2 (1-p);-\frac{b-\sqrt{b^2-4 a c}}{2 c x^2},-\frac{b+\sqrt{b^2-4 a c}}{2 c x^2}\right )}{(1-2 p) x^2} \]

[Out]

-((2^(-1 + 2*p)*(a + b*x^2 + c*x^4)^p*AppellF1[1 - 2*p, -p, -p, 2*(1 - p), -(b - Sqrt[b^2 - 4*a*c])/(2*c*x^2),
-(b + Sqrt[b^2 - 4*a*c])/(2*c*x^2)])/((1 - 2*p)*x^2*((b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(c*x^2))^p*((b + Sqrt[
b^2 - 4*a*c] + 2*c*x^2)/(c*x^2))^p))

________________________________________________________________________________________

Rubi [A]  time = 0.130781, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1114, 758, 133} \[ -\frac{2^{2 p-1} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (1-2 p;-p,-p;2 (1-p);-\frac{b-\sqrt{b^2-4 a c}}{2 c x^2},-\frac{b+\sqrt{b^2-4 a c}}{2 c x^2}\right )}{(1-2 p) x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^p/x^3,x]

[Out]

-((2^(-1 + 2*p)*(a + b*x^2 + c*x^4)^p*AppellF1[1 - 2*p, -p, -p, 2*(1 - p), -(b - Sqrt[b^2 - 4*a*c])/(2*c*x^2),
-(b + Sqrt[b^2 - 4*a*c])/(2*c*x^2)])/((1 - 2*p)*x^2*((b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(c*x^2))^p*((b + Sqrt[
b^2 - 4*a*c] + 2*c*x^2)/(c*x^2))^p))

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 758

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, -Dist[((1/(d + e*x))^(2*p)*(a + b*x + c*x^2)^p)/(e*((e*(b - q + 2*c*x))/(2*c*(d + e*x)))^p*((e*(b + q +
2*c*x))/(2*c*(d + e*x)))^p), Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - (e*(b - q))/(2*c))*x, x]^p*Simp[1 - (d
- (e*(b + q))/(2*c))*x, x]^p, x], x, 1/(d + e*x)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p] && ILtQ[m, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

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

Mathematica [A]  time = 0.210106, size = 163, normalized size = 0.98 \[ \frac{2^{2 p-1} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (1-2 p;-p,-p;2-2 p;-\frac{b+\sqrt{b^2-4 a c}}{2 c x^2},\frac{\sqrt{b^2-4 a c}-b}{2 c x^2}\right )}{(2 p-1) x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^2 + c*x^4)^p/x^3,x]

[Out]

(2^(-1 + 2*p)*(a + b*x^2 + c*x^4)^p*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, -(b + Sqrt[b^2 - 4*a*c])/(2*c*x^2), (-b
+ Sqrt[b^2 - 4*a*c])/(2*c*x^2)])/((-1 + 2*p)*x^2*((b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(c*x^2))^p*((b + Sqrt[b^2
- 4*a*c] + 2*c*x^2)/(c*x^2))^p)

________________________________________________________________________________________

Maple [F]  time = 0.065, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{p}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^p/x^3,x)

[Out]

int((c*x^4+b*x^2+a)^p/x^3,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{2} + a\right )}^{p}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^p/x^3,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2 + a)^p/x^3, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{4} + b x^{2} + a\right )}^{p}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^p/x^3,x, algorithm="fricas")

[Out]

integral((c*x^4 + b*x^2 + a)^p/x^3, x)

________________________________________________________________________________________

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((c*x**4+b*x**2+a)**p/x**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{2} + a\right )}^{p}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^p/x^3,x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2 + a)^p/x^3, x)