3.1121 \(\int \frac {(a+b x^2+c x^4)^p}{x^5} \, dx\)

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

[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 = 0.13, 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 = {1114, 758, 133} \[ -\frac {4^{p-1} \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (2 (1-p);-p,-p;3-2 p;-\frac {b-\sqrt {b^2-4 a c}}{2 c x^2},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^2}\right )}{(1-p) x^4} \]

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, -(b - Sqrt[b^2 - 4*a*c])/(2*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))

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])

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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2+c x^4\right )^p}{x^5} \, dx &=\frac {1}{2} \operatorname {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 ) \operatorname {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 = 0.22, size = 159, normalized size = 0.97 \[ \frac {4^{p-1} \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{c x^2}\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (2-2 p;-p,-p;3-2 p;-\frac {b+\sqrt {b^2-4 a c}}{2 c x^2},\frac {\sqrt {b^2-4 a c}-b}{2 c x^2}\right )}{(p-1) x^4} \]

Warning: Unable to verify antiderivative.

[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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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{4} + b x^{2} + a\right )}^{p}}{x^{5}}, x\right ) \]

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

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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maple [F]  time = 0.04, 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 \[ \int \frac {{\left (c x^{4} + b x^{2} + a\right )}^{p}}{x^{5}}\,{d x} \]

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

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

[Out]

Timed out

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