3.53 \(\int \frac{d+e x^2+f x^4}{x^5 (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=174 \[ -\frac{\log \left (a+b x^2+c x^4\right ) \left (-a b e-a (c d-a f)+b^2 d\right )}{4 a^3}+\frac{\tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right ) \left (2 a^2 c e-a b^2 e-a b (3 c d-a f)+b^3 d\right )}{2 a^3 \sqrt{b^2-4 a c}}+\frac{\log (x) \left (-a b e-a (c d-a f)+b^2 d\right )}{a^3}+\frac{b d-a e}{2 a^2 x^2}-\frac{d}{4 a x^4} \]

[Out]

-d/(4*a*x^4) + (b*d - a*e)/(2*a^2*x^2) + ((b^3*d - a*b^2*e + 2*a^2*c*e - a*b*(3*c*d - a*f))*ArcTanh[(b + 2*c*x
^2)/Sqrt[b^2 - 4*a*c]])/(2*a^3*Sqrt[b^2 - 4*a*c]) + ((b^2*d - a*b*e - a*(c*d - a*f))*Log[x])/a^3 - ((b^2*d - a
*b*e - a*(c*d - a*f))*Log[a + b*x^2 + c*x^4])/(4*a^3)

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Rubi [A]  time = 0.40734, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1663, 1628, 634, 618, 206, 628} \[ -\frac{\log \left (a+b x^2+c x^4\right ) \left (-a b e-a (c d-a f)+b^2 d\right )}{4 a^3}+\frac{\tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right ) \left (2 a^2 c e-a b^2 e-a b (3 c d-a f)+b^3 d\right )}{2 a^3 \sqrt{b^2-4 a c}}+\frac{\log (x) \left (-a b e-a (c d-a f)+b^2 d\right )}{a^3}+\frac{b d-a e}{2 a^2 x^2}-\frac{d}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2 + f*x^4)/(x^5*(a + b*x^2 + c*x^4)),x]

[Out]

-d/(4*a*x^4) + (b*d - a*e)/(2*a^2*x^2) + ((b^3*d - a*b^2*e + 2*a^2*c*e - a*b*(3*c*d - a*f))*ArcTanh[(b + 2*c*x
^2)/Sqrt[b^2 - 4*a*c]])/(2*a^3*Sqrt[b^2 - 4*a*c]) + ((b^2*d - a*b*e - a*(c*d - a*f))*Log[x])/a^3 - ((b^2*d - a
*b*e - a*(c*d - a*f))*Log[a + b*x^2 + c*x^4])/(4*a^3)

Rule 1663

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

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{d+e x^2+f x^4}{x^5 \left (a+b x^2+c x^4\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{d+e x+f x^2}{x^3 \left (a+b x+c x^2\right )} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{d}{a x^3}+\frac{-b d+a e}{a^2 x^2}+\frac{b^2 d-a b e-a (c d-a f)}{a^3 x}+\frac{-b^3 d+a b^2 e-a^2 c e+a b (2 c d-a f)-c \left (b^2 d-a b e-a (c d-a f)\right ) x}{a^3 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{d}{4 a x^4}+\frac{b d-a e}{2 a^2 x^2}+\frac{\left (b^2 d-a b e-a (c d-a f)\right ) \log (x)}{a^3}+\frac{\operatorname{Subst}\left (\int \frac{-b^3 d+a b^2 e-a^2 c e+a b (2 c d-a f)-c \left (b^2 d-a b e-a (c d-a f)\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 a^3}\\ &=-\frac{d}{4 a x^4}+\frac{b d-a e}{2 a^2 x^2}+\frac{\left (b^2 d-a b e-a (c d-a f)\right ) \log (x)}{a^3}-\frac{\left (b^2 d-a b e-a (c d-a f)\right ) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 a^3}-\frac{\left (b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 a^3}\\ &=-\frac{d}{4 a x^4}+\frac{b d-a e}{2 a^2 x^2}+\frac{\left (b^2 d-a b e-a (c d-a f)\right ) \log (x)}{a^3}-\frac{\left (b^2 d-a b e-a (c d-a f)\right ) \log \left (a+b x^2+c x^4\right )}{4 a^3}+\frac{\left (b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 a^3}\\ &=-\frac{d}{4 a x^4}+\frac{b d-a e}{2 a^2 x^2}+\frac{\left (b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^3 \sqrt{b^2-4 a c}}+\frac{\left (b^2 d-a b e-a (c d-a f)\right ) \log (x)}{a^3}-\frac{\left (b^2 d-a b e-a (c d-a f)\right ) \log \left (a+b x^2+c x^4\right )}{4 a^3}\\ \end{align*}

Mathematica [A]  time = 0.35299, size = 314, normalized size = 1.8 \[ -\frac{\frac{a^2 d}{x^4}+\frac{\log \left (-\sqrt{b^2-4 a c}+b+2 c x^2\right ) \left (a b \left (-e \sqrt{b^2-4 a c}+a f-3 c d\right )+a \left (-c d \sqrt{b^2-4 a c}+a f \sqrt{b^2-4 a c}+2 a c e\right )+b^2 \left (d \sqrt{b^2-4 a c}-a e\right )+b^3 d\right )}{\sqrt{b^2-4 a c}}+\frac{\log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right ) \left (-a b \left (e \sqrt{b^2-4 a c}+a f-3 c d\right )+a \left (a f \sqrt{b^2-4 a c}-c \left (d \sqrt{b^2-4 a c}+2 a e\right )\right )+b^2 \left (d \sqrt{b^2-4 a c}+a e\right )+b^3 (-d)\right )}{\sqrt{b^2-4 a c}}-4 \log (x) \left (-a b e+a (a f-c d)+b^2 d\right )+\frac{2 a (a e-b d)}{x^2}}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*d)/x^4 + (2*a*(-(b*d) + a*e))/x^2 - 4*(b^2*d - a*b*e + a*(-(c*d) + a*f))*Log[x] + ((b^3*d + b^2*(Sqrt[b
^2 - 4*a*c]*d - a*e) + a*b*(-3*c*d - Sqrt[b^2 - 4*a*c]*e + a*f) + a*(-(c*Sqrt[b^2 - 4*a*c]*d) + 2*a*c*e + a*Sq
rt[b^2 - 4*a*c]*f))*Log[b - Sqrt[b^2 - 4*a*c] + 2*c*x^2])/Sqrt[b^2 - 4*a*c] + ((-(b^3*d) + b^2*(Sqrt[b^2 - 4*a
*c]*d + a*e) - a*b*(-3*c*d + Sqrt[b^2 - 4*a*c]*e + a*f) + a*(-(c*(Sqrt[b^2 - 4*a*c]*d + 2*a*e)) + a*Sqrt[b^2 -
4*a*c]*f))*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/Sqrt[b^2 - 4*a*c])/(4*a^3)

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Maple [B]  time = 0.011, size = 356, normalized size = 2.1 \begin{align*} -{\frac{d}{4\,a{x}^{4}}}-{\frac{e}{2\,a{x}^{2}}}+{\frac{bd}{2\,{a}^{2}{x}^{2}}}+{\frac{\ln \left ( x \right ) f}{a}}-{\frac{\ln \left ( x \right ) be}{{a}^{2}}}-{\frac{\ln \left ( x \right ) cd}{{a}^{2}}}+{\frac{\ln \left ( x \right ){b}^{2}d}{{a}^{3}}}-{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) f}{4\,a}}+{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) be}{4\,{a}^{2}}}+{\frac{c\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) d}{4\,{a}^{2}}}-{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){b}^{2}d}{4\,{a}^{3}}}-{\frac{bf}{2\,a}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{ce}{a}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{e{b}^{2}}{2\,{a}^{2}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{3\,bcd}{2\,{a}^{2}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{{b}^{3}d}{2\,{a}^{3}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^4+e*x^2+d)/x^5/(c*x^4+b*x^2+a),x)

[Out]

-1/4*d/a/x^4-1/2/a/x^2*e+1/2/a^2/x^2*b*d+1/a*ln(x)*f-1/a^2*ln(x)*b*e-1/a^2*ln(x)*c*d+1/a^3*ln(x)*b^2*d-1/4/a*l
n(c*x^4+b*x^2+a)*f+1/4/a^2*ln(c*x^4+b*x^2+a)*b*e+1/4/a^2*c*ln(c*x^4+b*x^2+a)*d-1/4/a^3*ln(c*x^4+b*x^2+a)*b^2*d
-1/2/a/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b*f-1/a/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4
*a*c-b^2)^(1/2))*c*e+1/2/a^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b^2*e+3/2/a^2/(4*a*c-b^2)
^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b*c*d-1/2/a^3/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1
/2))*b^3*d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^4+e*x^2+d)/x^5/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.26681, size = 1283, normalized size = 7.37 \begin{align*} \left [\frac{{\left (a^{2} b f +{\left (b^{3} - 3 \, a b c\right )} d -{\left (a b^{2} - 2 \, a^{2} c\right )} e\right )} \sqrt{b^{2} - 4 \, a c} x^{4} \log \left (\frac{2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c +{\left (2 \, c x^{2} + b\right )} \sqrt{b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) -{\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d -{\left (a b^{3} - 4 \, a^{2} b c\right )} e +{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} f\right )} x^{4} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \,{\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d -{\left (a b^{3} - 4 \, a^{2} b c\right )} e +{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} f\right )} x^{4} \log \left (x\right ) + 2 \,{\left ({\left (a b^{3} - 4 \, a^{2} b c\right )} d -{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} e\right )} x^{2} -{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d}{4 \,{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x^{4}}, \frac{2 \,{\left (a^{2} b f +{\left (b^{3} - 3 \, a b c\right )} d -{\left (a b^{2} - 2 \, a^{2} c\right )} e\right )} \sqrt{-b^{2} + 4 \, a c} x^{4} \arctan \left (-\frac{{\left (2 \, c x^{2} + b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) -{\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d -{\left (a b^{3} - 4 \, a^{2} b c\right )} e +{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} f\right )} x^{4} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \,{\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d -{\left (a b^{3} - 4 \, a^{2} b c\right )} e +{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} f\right )} x^{4} \log \left (x\right ) + 2 \,{\left ({\left (a b^{3} - 4 \, a^{2} b c\right )} d -{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} e\right )} x^{2} -{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d}{4 \,{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((a^2*b*f + (b^3 - 3*a*b*c)*d - (a*b^2 - 2*a^2*c)*e)*sqrt(b^2 - 4*a*c)*x^4*log((2*c^2*x^4 + 2*b*c*x^2 + b
^2 - 2*a*c + (2*c*x^2 + b)*sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)) - ((b^4 - 5*a*b^2*c + 4*a^2*c^2)*d - (a*b^3
- 4*a^2*b*c)*e + (a^2*b^2 - 4*a^3*c)*f)*x^4*log(c*x^4 + b*x^2 + a) + 4*((b^4 - 5*a*b^2*c + 4*a^2*c^2)*d - (a*
b^3 - 4*a^2*b*c)*e + (a^2*b^2 - 4*a^3*c)*f)*x^4*log(x) + 2*((a*b^3 - 4*a^2*b*c)*d - (a^2*b^2 - 4*a^3*c)*e)*x^2
- (a^2*b^2 - 4*a^3*c)*d)/((a^3*b^2 - 4*a^4*c)*x^4), 1/4*(2*(a^2*b*f + (b^3 - 3*a*b*c)*d - (a*b^2 - 2*a^2*c)*e
)*sqrt(-b^2 + 4*a*c)*x^4*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - ((b^4 - 5*a*b^2*c + 4*a^2*c
^2)*d - (a*b^3 - 4*a^2*b*c)*e + (a^2*b^2 - 4*a^3*c)*f)*x^4*log(c*x^4 + b*x^2 + a) + 4*((b^4 - 5*a*b^2*c + 4*a^
2*c^2)*d - (a*b^3 - 4*a^2*b*c)*e + (a^2*b^2 - 4*a^3*c)*f)*x^4*log(x) + 2*((a*b^3 - 4*a^2*b*c)*d - (a^2*b^2 - 4
*a^3*c)*e)*x^2 - (a^2*b^2 - 4*a^3*c)*d)/((a^3*b^2 - 4*a^4*c)*x^4)]

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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((f*x**4+e*x**2+d)/x**5/(c*x**4+b*x**2+a),x)

[Out]

Timed out

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Giac [A]  time = 1.13726, size = 286, normalized size = 1.64 \begin{align*} -\frac{{\left (b^{2} d - a c d + a^{2} f - a b e\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, a^{3}} + \frac{{\left (b^{2} d - a c d + a^{2} f - a b e\right )} \log \left (x^{2}\right )}{2 \, a^{3}} - \frac{{\left (b^{3} d - 3 \, a b c d + a^{2} b f - a b^{2} e + 2 \, a^{2} c e\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt{-b^{2} + 4 \, a c} a^{3}} - \frac{3 \, b^{2} d x^{4} - 3 \, a c d x^{4} + 3 \, a^{2} f x^{4} - 3 \, a b x^{4} e - 2 \, a b d x^{2} + 2 \, a^{2} x^{2} e + a^{2} d}{4 \, a^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^4+e*x^2+d)/x^5/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

-1/4*(b^2*d - a*c*d + a^2*f - a*b*e)*log(c*x^4 + b*x^2 + a)/a^3 + 1/2*(b^2*d - a*c*d + a^2*f - a*b*e)*log(x^2)
/a^3 - 1/2*(b^3*d - 3*a*b*c*d + a^2*b*f - a*b^2*e + 2*a^2*c*e)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(
-b^2 + 4*a*c)*a^3) - 1/4*(3*b^2*d*x^4 - 3*a*c*d*x^4 + 3*a^2*f*x^4 - 3*a*b*x^4*e - 2*a*b*d*x^2 + 2*a^2*x^2*e +
a^2*d)/(a^3*x^4)