3.1453 \(\int \frac {x^5}{a+b x^8} \, dx\)

Optimal. Leaf size=193 \[ -\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{3/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} \sqrt [4]{a} b^{3/4}}+\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {a}+\sqrt {b} x^4\right )}{8 \sqrt {2} \sqrt [4]{a} b^{3/4}}-\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {a}+\sqrt {b} x^4\right )}{8 \sqrt {2} \sqrt [4]{a} b^{3/4}} \]

[Out]

1/8*arctan(-1+b^(1/4)*x^2*2^(1/2)/a^(1/4))/a^(1/4)/b^(3/4)*2^(1/2)+1/8*arctan(1+b^(1/4)*x^2*2^(1/2)/a^(1/4))/a
^(1/4)/b^(3/4)*2^(1/2)+1/16*ln(-a^(1/4)*b^(1/4)*x^2*2^(1/2)+a^(1/2)+x^4*b^(1/2))/a^(1/4)/b^(3/4)*2^(1/2)-1/16*
ln(a^(1/4)*b^(1/4)*x^2*2^(1/2)+a^(1/2)+x^4*b^(1/2))/a^(1/4)/b^(3/4)*2^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {275, 297, 1162, 617, 204, 1165, 628} \[ \frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {a}+\sqrt {b} x^4\right )}{8 \sqrt {2} \sqrt [4]{a} b^{3/4}}-\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {a}+\sqrt {b} x^4\right )}{8 \sqrt {2} \sqrt [4]{a} b^{3/4}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{3/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} \sqrt [4]{a} b^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^8),x]

[Out]

-ArcTan[1 - (Sqrt[2]*b^(1/4)*x^2)/a^(1/4)]/(4*Sqrt[2]*a^(1/4)*b^(3/4)) + ArcTan[1 + (Sqrt[2]*b^(1/4)*x^2)/a^(1
/4)]/(4*Sqrt[2]*a^(1/4)*b^(3/4)) + Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x^2 + Sqrt[b]*x^4]/(8*Sqrt[2]*a^(1/4)
*b^(3/4)) - Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x^2 + Sqrt[b]*x^4]/(8*Sqrt[2]*a^(1/4)*b^(3/4))

Rule 204

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

Rule 275

Int[(x_)^(m_.)*((a_) + (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))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {x^5}{a+b x^8} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,x^2\right )\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,x^2\right )}{4 \sqrt {b}}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,x^2\right )}{4 \sqrt {b}}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,x^2\right )}{8 b}+\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,x^2\right )}{8 b}+\frac {\operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,x^2\right )}{8 \sqrt {2} \sqrt [4]{a} b^{3/4}}+\frac {\operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,x^2\right )}{8 \sqrt {2} \sqrt [4]{a} b^{3/4}}\\ &=\frac {\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} \sqrt [4]{a} b^{3/4}}-\frac {\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} \sqrt [4]{a} b^{3/4}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{3/4}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{3/4}}\\ &=-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{3/4}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{3/4}}+\frac {\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} \sqrt [4]{a} b^{3/4}}-\frac {\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} \sqrt [4]{a} b^{3/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 279, normalized size = 1.45 \[ -\frac {\log \left (-2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac {\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )+\log \left (2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac {\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )-\log \left (-2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac {\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )-\log \left (2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac {\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )-2 \tan ^{-1}\left (\frac {\sqrt [8]{b} x \sec \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}-\tan \left (\frac {\pi }{8}\right )\right )+2 \tan ^{-1}\left (\frac {\sqrt [8]{b} x \sec \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}+\tan \left (\frac {\pi }{8}\right )\right )+2 \tan ^{-1}\left (\cot \left (\frac {\pi }{8}\right )-\frac {\sqrt [8]{b} x \csc \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}\right )+2 \tan ^{-1}\left (\frac {\sqrt [8]{b} x \csc \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}+\cot \left (\frac {\pi }{8}\right )\right )}{8 \sqrt {2} \sqrt [4]{a} b^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^8),x]

[Out]

-1/8*(2*ArcTan[Cot[Pi/8] - (b^(1/8)*x*Csc[Pi/8])/a^(1/8)] + 2*ArcTan[Cot[Pi/8] + (b^(1/8)*x*Csc[Pi/8])/a^(1/8)
] - 2*ArcTan[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) - Tan[Pi/8]] + 2*ArcTan[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) + Tan[Pi/8]]
- Log[a^(1/4) + b^(1/4)*x^2 - 2*a^(1/8)*b^(1/8)*x*Cos[Pi/8]] - Log[a^(1/4) + b^(1/4)*x^2 + 2*a^(1/8)*b^(1/8)*x
*Cos[Pi/8]] + Log[a^(1/4) + b^(1/4)*x^2 - 2*a^(1/8)*b^(1/8)*x*Sin[Pi/8]] + Log[a^(1/4) + b^(1/4)*x^2 + 2*a^(1/
8)*b^(1/8)*x*Sin[Pi/8]])/(Sqrt[2]*a^(1/4)*b^(3/4))

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 128, normalized size = 0.66 \[ -\frac {1}{2} \, \left (-\frac {1}{a b^{3}}\right )^{\frac {1}{4}} \arctan \left (-b x^{2} \left (-\frac {1}{a b^{3}}\right )^{\frac {1}{4}} + \sqrt {x^{4} - a b \sqrt {-\frac {1}{a b^{3}}}} b \left (-\frac {1}{a b^{3}}\right )^{\frac {1}{4}}\right ) + \frac {1}{8} \, \left (-\frac {1}{a b^{3}}\right )^{\frac {1}{4}} \log \left (a b^{2} \left (-\frac {1}{a b^{3}}\right )^{\frac {3}{4}} + x^{2}\right ) - \frac {1}{8} \, \left (-\frac {1}{a b^{3}}\right )^{\frac {1}{4}} \log \left (-a b^{2} \left (-\frac {1}{a b^{3}}\right )^{\frac {3}{4}} + x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-1/(a*b^3))^(1/4)*arctan(-b*x^2*(-1/(a*b^3))^(1/4) + sqrt(x^4 - a*b*sqrt(-1/(a*b^3)))*b*(-1/(a*b^3))^(1/
4)) + 1/8*(-1/(a*b^3))^(1/4)*log(a*b^2*(-1/(a*b^3))^(3/4) + x^2) - 1/8*(-1/(a*b^3))^(1/4)*log(-a*b^2*(-1/(a*b^
3))^(3/4) + x^2)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 199, normalized size = 1.03 \[ \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} x^{4} \arctan \left (\frac {\sqrt {2} {\left (2 \, x^{2} + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} x^{4} \arctan \left (\frac {\sqrt {2} {\left (2 \, x^{2} - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} x^{4} \log \left (x^{4} + \sqrt {2} x^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{16 \, a b} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} x^{4} \log \left (x^{4} - \sqrt {2} x^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{16 \, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*(a*b^3)^(1/4)*x^4*arctan(1/2*sqrt(2)*(2*x^2 + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a*b) + 1/8*sqrt(2
)*(a*b^3)^(1/4)*x^4*arctan(1/2*sqrt(2)*(2*x^2 - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a*b) + 1/16*sqrt(2)*(a*b^3)
^(1/4)*x^4*log(x^4 + sqrt(2)*x^2*(a/b)^(1/4) + sqrt(a/b))/(a*b) - 1/16*sqrt(2)*(a*b^3)^(1/4)*x^4*log(x^4 - sqr
t(2)*x^2*(a/b)^(1/4) + sqrt(a/b))/(a*b)

________________________________________________________________________________________

maple [A]  time = 0.00, size = 136, normalized size = 0.70 \[ \frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x^{2}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x^{2}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\sqrt {2}\, \ln \left (\frac {x^{4}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x^{2}+\sqrt {\frac {a}{b}}}{x^{4}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x^{2}+\sqrt {\frac {a}{b}}}\right )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/b/(a/b)^(1/4)*2^(1/2)*ln((x^4-(a/b)^(1/4)*2^(1/2)*x^2+(a/b)^(1/2))/(x^4+(a/b)^(1/4)*2^(1/2)*x^2+(a/b)^(1/
2)))+1/8/b/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^2+1)+1/8/b/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b
)^(1/4)*x^2-1)

________________________________________________________________________________________

maxima [A]  time = 2.28, size = 177, normalized size = 0.92 \[ \frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{8 \, \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{8 \, \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{4} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x^{2} + \sqrt {a}\right )}{16 \, a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{4} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x^{2} + \sqrt {a}\right )}{16 \, a^{\frac {1}{4}} b^{\frac {3}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x^2 + sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*
sqrt(b))*sqrt(b)) + 1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt
(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - 1/16*sqrt(2)*log(sqrt(b)*x^4 + sqrt(2)*a^(1/4)*b^(1/4)*x^2 + sqrt(a))/
(a^(1/4)*b^(3/4)) + 1/16*sqrt(2)*log(sqrt(b)*x^4 - sqrt(2)*a^(1/4)*b^(1/4)*x^2 + sqrt(a))/(a^(1/4)*b^(3/4))

________________________________________________________________________________________

mupad [B]  time = 1.11, size = 39, normalized size = 0.20 \[ \frac {\mathrm {atan}\left (\frac {b^{1/4}\,x^2}{{\left (-a\right )}^{1/4}}\right )-\mathrm {atanh}\left (\frac {b^{1/4}\,x^2}{{\left (-a\right )}^{1/4}}\right )}{4\,{\left (-a\right )}^{1/4}\,b^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^8),x)

[Out]

(atan((b^(1/4)*x^2)/(-a)^(1/4)) - atanh((b^(1/4)*x^2)/(-a)^(1/4)))/(4*(-a)^(1/4)*b^(3/4))

________________________________________________________________________________________

sympy [A]  time = 0.28, size = 27, normalized size = 0.14 \[ \operatorname {RootSum} {\left (4096 t^{4} a b^{3} + 1, \left (t \mapsto t \log {\left (512 t^{3} a b^{2} + x^{2} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(4096*_t**4*a*b**3 + 1, Lambda(_t, _t*log(512*_t**3*a*b**2 + x**2)))

________________________________________________________________________________________