3.313 \(\int \frac{x^5}{a+b x^4+c x^8} \, dx\)

Optimal. Leaf size=159 \[ \frac{\sqrt{\sqrt{b^2-4 a c}+b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x^2}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} \sqrt{c} \sqrt{b^2-4 a c}}-\frac{\sqrt{b-\sqrt{b^2-4 a c}} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x^2}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \sqrt{b^2-4 a c}} \]

[Out]

-(Sqrt[b - Sqrt[b^2 - 4*a*c]]*ArcTan[(Sqrt[2]*Sqrt[c]*x^2)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*Sqrt[c]*Sq
rt[b^2 - 4*a*c]) + (Sqrt[b + Sqrt[b^2 - 4*a*c]]*ArcTan[(Sqrt[2]*Sqrt[c]*x^2)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(2*
Sqrt[2]*Sqrt[c]*Sqrt[b^2 - 4*a*c])

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Rubi [A]  time = 0.126211, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1359, 1130, 205} \[ \frac{\sqrt{\sqrt{b^2-4 a c}+b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x^2}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} \sqrt{c} \sqrt{b^2-4 a c}}-\frac{\sqrt{b-\sqrt{b^2-4 a c}} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x^2}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^4 + c*x^8),x]

[Out]

-(Sqrt[b - Sqrt[b^2 - 4*a*c]]*ArcTan[(Sqrt[2]*Sqrt[c]*x^2)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*Sqrt[c]*Sq
rt[b^2 - 4*a*c]) + (Sqrt[b + Sqrt[b^2 - 4*a*c]]*ArcTan[(Sqrt[2]*Sqrt[c]*x^2)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(2*
Sqrt[2]*Sqrt[c]*Sqrt[b^2 - 4*a*c])

Rule 1359

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[
1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k) + c*x^((2*n)/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b,
c, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 1130

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{x^5}{a+b x^4+c x^8} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{a+b x^2+c x^4} \, dx,x,x^2\right )\\ &=\frac{1}{4} \left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx,x,x^2\right )+\frac{1}{4} \left (1+\frac{b}{\sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{b-\sqrt{b^2-4 a c}} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x^2}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \sqrt{b^2-4 a c}}+\frac{\sqrt{b+\sqrt{b^2-4 a c}} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x^2}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \sqrt{b^2-4 a c}}\\ \end{align*}

Mathematica [A]  time = 0.0926503, size = 171, normalized size = 1.08 \[ \frac{\left (\sqrt{b^2-4 a c}-b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x^2}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )+\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{\sqrt{b^2-4 a c}+b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x^2}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} \sqrt{c} \sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^4 + c*x^8),x]

[Out]

((-b + Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x^2)/Sqrt[b - Sqrt[b^2 - 4*a*c]]] + Sqrt[b - Sqrt[b^2 - 4*a*
c]]*Sqrt[b + Sqrt[b^2 - 4*a*c]]*ArcTan[(Sqrt[2]*Sqrt[c]*x^2)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*Sqrt[c]*
Sqrt[b^2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4*a*c]])

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Maple [A]  time = 0.016, size = 216, normalized size = 1.4 \begin{align*}{\frac{\sqrt{2}}{4}\arctan \left ({c{x}^{2}\sqrt{2}{\frac{1}{\sqrt{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}}+{\frac{\sqrt{2}b}{4}\arctan \left ({c{x}^{2}\sqrt{2}{\frac{1}{\sqrt{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}{\frac{1}{\sqrt{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}}-{\frac{\sqrt{2}}{4}{\it Artanh} \left ({c{x}^{2}\sqrt{2}{\frac{1}{\sqrt{ \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}}+{\frac{\sqrt{2}b}{4}{\it Artanh} \left ({c{x}^{2}\sqrt{2}{\frac{1}{\sqrt{ \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}{\frac{1}{\sqrt{ \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(c*x^8+b*x^4+a),x)

[Out]

1/4*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x^2*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))+1/4/(-4*a*
c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x^2*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*b
-1/4*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctanh(c*x^2*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2))+1/4/(-
4*a*c+b^2)^(1/2)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctanh(c*x^2*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(
1/2))*b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{c x^{8} + b x^{4} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^8+b*x^4+a),x, algorithm="maxima")

[Out]

integrate(x^5/(c*x^8 + b*x^4 + a), x)

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Fricas [B]  time = 1.52664, size = 1214, normalized size = 7.64 \begin{align*} \frac{1}{4} \, \sqrt{\frac{1}{2}} \sqrt{-\frac{b + \frac{b^{2} c - 4 \, a c^{2}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} \log \left (x^{2} + \frac{\sqrt{\frac{1}{2}}{\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt{-\frac{b + \frac{b^{2} c - 4 \, a c^{2}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}\right ) - \frac{1}{4} \, \sqrt{\frac{1}{2}} \sqrt{-\frac{b + \frac{b^{2} c - 4 \, a c^{2}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} \log \left (x^{2} - \frac{\sqrt{\frac{1}{2}}{\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt{-\frac{b + \frac{b^{2} c - 4 \, a c^{2}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}\right ) - \frac{1}{4} \, \sqrt{\frac{1}{2}} \sqrt{-\frac{b - \frac{b^{2} c - 4 \, a c^{2}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} \log \left (x^{2} + \frac{\sqrt{\frac{1}{2}}{\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt{-\frac{b - \frac{b^{2} c - 4 \, a c^{2}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}\right ) + \frac{1}{4} \, \sqrt{\frac{1}{2}} \sqrt{-\frac{b - \frac{b^{2} c - 4 \, a c^{2}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} \log \left (x^{2} - \frac{\sqrt{\frac{1}{2}}{\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt{-\frac{b - \frac{b^{2} c - 4 \, a c^{2}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}}}{\sqrt{b^{2} c^{2} - 4 \, a c^{3}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^8+b*x^4+a),x, algorithm="fricas")

[Out]

1/4*sqrt(1/2)*sqrt(-(b + (b^2*c - 4*a*c^2)/sqrt(b^2*c^2 - 4*a*c^3))/(b^2*c - 4*a*c^2))*log(x^2 + sqrt(1/2)*(b^
2*c - 4*a*c^2)*sqrt(-(b + (b^2*c - 4*a*c^2)/sqrt(b^2*c^2 - 4*a*c^3))/(b^2*c - 4*a*c^2))/sqrt(b^2*c^2 - 4*a*c^3
)) - 1/4*sqrt(1/2)*sqrt(-(b + (b^2*c - 4*a*c^2)/sqrt(b^2*c^2 - 4*a*c^3))/(b^2*c - 4*a*c^2))*log(x^2 - sqrt(1/2
)*(b^2*c - 4*a*c^2)*sqrt(-(b + (b^2*c - 4*a*c^2)/sqrt(b^2*c^2 - 4*a*c^3))/(b^2*c - 4*a*c^2))/sqrt(b^2*c^2 - 4*
a*c^3)) - 1/4*sqrt(1/2)*sqrt(-(b - (b^2*c - 4*a*c^2)/sqrt(b^2*c^2 - 4*a*c^3))/(b^2*c - 4*a*c^2))*log(x^2 + sqr
t(1/2)*(b^2*c - 4*a*c^2)*sqrt(-(b - (b^2*c - 4*a*c^2)/sqrt(b^2*c^2 - 4*a*c^3))/(b^2*c - 4*a*c^2))/sqrt(b^2*c^2
- 4*a*c^3)) + 1/4*sqrt(1/2)*sqrt(-(b - (b^2*c - 4*a*c^2)/sqrt(b^2*c^2 - 4*a*c^3))/(b^2*c - 4*a*c^2))*log(x^2
- sqrt(1/2)*(b^2*c - 4*a*c^2)*sqrt(-(b - (b^2*c - 4*a*c^2)/sqrt(b^2*c^2 - 4*a*c^3))/(b^2*c - 4*a*c^2))/sqrt(b^
2*c^2 - 4*a*c^3))

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Sympy [A]  time = 2.21134, size = 76, normalized size = 0.48 \begin{align*} \operatorname{RootSum}{\left (t^{4} \left (4096 a^{2} c^{3} - 2048 a b^{2} c^{2} + 256 b^{4} c\right ) + t^{2} \left (- 64 a b c + 16 b^{3}\right ) + a, \left ( t \mapsto t \log{\left (512 t^{3} a c^{2} - 128 t^{3} b^{2} c - 4 t b + x^{2} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(c*x**8+b*x**4+a),x)

[Out]

RootSum(_t**4*(4096*a**2*c**3 - 2048*a*b**2*c**2 + 256*b**4*c) + _t**2*(-64*a*b*c + 16*b**3) + a, Lambda(_t, _
t*log(512*_t**3*a*c**2 - 128*_t**3*b**2*c - 4*_t*b + x**2)))

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Giac [C]  time = 7.82648, size = 1403, normalized size = 8.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^8+b*x^4+a),x, algorithm="giac")

[Out]

1/4*(((a*c^3)^(1/4)*b^2 - 4*(a*c^3)^(1/4)*a*c + (a*c^3)^(1/4)*sqrt(b^2 - 4*a*c)*b)*x^4*cosh(1/2*imag_part(arcs
in(1/2*sqrt(a*c)*b/(a*abs(c)))))*sin(5/4*pi + 1/2*real_part(arcsin(1/2*sqrt(a*c)*b/(a*abs(c))))) - ((a*c^3)^(1
/4)*b^2 - 4*(a*c^3)^(1/4)*a*c + (a*c^3)^(1/4)*sqrt(b^2 - 4*a*c)*b)*x^4*sin(5/4*pi + 1/2*real_part(arcsin(1/2*s
qrt(a*c)*b/(a*abs(c)))))*sinh(1/2*imag_part(arcsin(1/2*sqrt(a*c)*b/(a*abs(c))))))*arctan((x^2 - (a/c)^(1/4)*co
s(5/4*pi + 1/2*arcsin(1/2*sqrt(a*c)*b/(a*abs(c)))))/((a/c)^(1/4)*sin(5/4*pi + 1/2*arcsin(1/2*sqrt(a*c)*b/(a*ab
s(c))))))/(a*b^2*c - 4*a^2*c^2) + 1/4*(((a*c^3)^(1/4)*b^2 - 4*(a*c^3)^(1/4)*a*c + (a*c^3)^(1/4)*sqrt(b^2 - 4*a
*c)*b)*x^4*cosh(1/2*imag_part(arcsin(1/2*sqrt(a*c)*b/(a*abs(c)))))*sin(1/4*pi + 1/2*real_part(arcsin(1/2*sqrt(
a*c)*b/(a*abs(c))))) - ((a*c^3)^(1/4)*b^2 - 4*(a*c^3)^(1/4)*a*c + (a*c^3)^(1/4)*sqrt(b^2 - 4*a*c)*b)*x^4*sin(1
/4*pi + 1/2*real_part(arcsin(1/2*sqrt(a*c)*b/(a*abs(c)))))*sinh(1/2*imag_part(arcsin(1/2*sqrt(a*c)*b/(a*abs(c)
)))))*arctan((x^2 - (a/c)^(1/4)*cos(1/4*pi + 1/2*arcsin(1/2*sqrt(a*c)*b/(a*abs(c)))))/((a/c)^(1/4)*sin(1/4*pi
+ 1/2*arcsin(1/2*sqrt(a*c)*b/(a*abs(c))))))/(a*b^2*c - 4*a^2*c^2) - 1/8*(((a*c^3)^(1/4)*b^2 - 4*(a*c^3)^(1/4)*
a*c + (a*c^3)^(1/4)*sqrt(b^2 - 4*a*c)*b)*x^4*cos(5/4*pi + 1/2*real_part(arcsin(1/2*sqrt(a*c)*b/(a*abs(c)))))*c
osh(1/2*imag_part(arcsin(1/2*sqrt(a*c)*b/(a*abs(c))))) - ((a*c^3)^(1/4)*b^2 - 4*(a*c^3)^(1/4)*a*c + (a*c^3)^(1
/4)*sqrt(b^2 - 4*a*c)*b)*x^4*cos(5/4*pi + 1/2*real_part(arcsin(1/2*sqrt(a*c)*b/(a*abs(c)))))*sinh(1/2*imag_par
t(arcsin(1/2*sqrt(a*c)*b/(a*abs(c))))))*log(x^4 - 2*x^2*(a/c)^(1/4)*cos(5/4*pi + 1/2*arcsin(1/2*sqrt(a*c)*b/(a
*abs(c)))) + sqrt(a/c))/(a*b^2*c - 4*a^2*c^2) - 1/8*(((a*c^3)^(1/4)*b^2 - 4*(a*c^3)^(1/4)*a*c + (a*c^3)^(1/4)*
sqrt(b^2 - 4*a*c)*b)*x^4*cos(1/4*pi + 1/2*real_part(arcsin(1/2*sqrt(a*c)*b/(a*abs(c)))))*cosh(1/2*imag_part(ar
csin(1/2*sqrt(a*c)*b/(a*abs(c))))) - ((a*c^3)^(1/4)*b^2 - 4*(a*c^3)^(1/4)*a*c + (a*c^3)^(1/4)*sqrt(b^2 - 4*a*c
)*b)*x^4*cos(1/4*pi + 1/2*real_part(arcsin(1/2*sqrt(a*c)*b/(a*abs(c)))))*sinh(1/2*imag_part(arcsin(1/2*sqrt(a*
c)*b/(a*abs(c))))))*log(x^4 - 2*x^2*(a/c)^(1/4)*cos(1/4*pi + 1/2*arcsin(1/2*sqrt(a*c)*b/(a*abs(c)))) + sqrt(a/
c))/(a*b^2*c - 4*a^2*c^2)