3.12.21 \(\int \frac {(a+b x^2+c x^4)^p}{x^5} \, dx\) [1121]

Optimal. Leaf size=164 \[ -\frac {4^{-1+p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{c x^2}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{c x^2}\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (2 (1-p);-p,-p;3-2 p;-\frac {b-\sqrt {b^2-4 a c}}{2 c x^2},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^2}\right )}{(1-p) x^4} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 164, normalized size of antiderivative = 1.00, 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 = {1128, 772, 138} \begin {gather*} -\frac {4^{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 (2 (1-p);-p,-p;3-2 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-p) x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^p/x^5,x]

[Out]

-((4^(-1 + p)*(a + b*x^2 + c*x^4)^p*AppellF1[2*(1 - p), -p, -p, 3 - 2*p, -1/2*(b - Sqrt[b^2 - 4*a*c])/(c*x^2),
-1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x^2)])/((1 - p)*x^4*((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 138

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

Rule 772

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 1128

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2+c x^4\right )^p}{x^5} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^p}{x^3} \, 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 ) \text {Subst}\left (\int x^{3-2 (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 {4^{-1+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 (2 (1-p);-p,-p;3-2 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-p) x^4}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 159, normalized size = 0.97 \begin {gather*} \frac {4^{-1+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 (2-2 p;-p,-p;3-2 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+p) x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)^p/x^5,x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{x^{5}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^p/x^5,x)

[Out]

int((c*x^4+b*x^2+a)^p/x^5,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((c*x^4+b*x^2+a)^p/x^5,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2 + a)^p/x^5, 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((c*x^4+b*x^2+a)^p/x^5,x, algorithm="fricas")

[Out]

integral((c*x^4 + b*x^2 + a)^p/x^5, 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((c*x**4+b*x**2+a)**p/x**5,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((c*x^4+b*x^2+a)^p/x^5,x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2 + a)^p/x^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^4+b\,x^2+a\right )}^p}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)^p/x^5,x)

[Out]

int((a + b*x^2 + c*x^4)^p/x^5, x)

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