3.18.26 \(\int \frac {b+a x^4}{\sqrt [4]{-b+a x^4} (b+c x^4+a x^8)} \, dx\) [1726]

Optimal. Leaf size=116 \[ \frac {1}{4} \text {RootSum}\left [a^2+a b+a c-2 a \text {$\#$1}^4-c \text {$\#$1}^4+\text {$\#$1}^8\& ,\frac {2 a \log (x)-2 a \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+\log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{2 a \text {$\#$1}+c \text {$\#$1}-2 \text {$\#$1}^5}\& \right ] \]

[Out]

Unintegrable

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(539\) vs. \(2(116)=232\).
time = 0.84, antiderivative size = 539, normalized size of antiderivative = 4.65, number of steps used = 10, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6860, 385, 218, 214, 211} \begin {gather*} \frac {a^{3/4} \left (\frac {2 b-c}{\sqrt {c^2-4 a b}}+1\right ) \text {ArcTan}\left (\frac {\sqrt [4]{a} x \sqrt [4]{-\sqrt {c^2-4 a b}+2 b+c}}{\sqrt [4]{c-\sqrt {c^2-4 a b}} \sqrt [4]{a x^4-b}}\right )}{2 \left (c-\sqrt {c^2-4 a b}\right )^{3/4} \sqrt [4]{-\sqrt {c^2-4 a b}+2 b+c}}+\frac {a^{3/4} \left (1-\frac {2 b-c}{\sqrt {c^2-4 a b}}\right ) \text {ArcTan}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {c^2-4 a b}+2 b+c}}{\sqrt [4]{\sqrt {c^2-4 a b}+c} \sqrt [4]{a x^4-b}}\right )}{2 \left (\sqrt {c^2-4 a b}+c\right )^{3/4} \sqrt [4]{\sqrt {c^2-4 a b}+2 b+c}}+\frac {a^{3/4} \left (\frac {2 b-c}{\sqrt {c^2-4 a b}}+1\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x \sqrt [4]{-\sqrt {c^2-4 a b}+2 b+c}}{\sqrt [4]{c-\sqrt {c^2-4 a b}} \sqrt [4]{a x^4-b}}\right )}{2 \left (c-\sqrt {c^2-4 a b}\right )^{3/4} \sqrt [4]{-\sqrt {c^2-4 a b}+2 b+c}}+\frac {a^{3/4} \left (1-\frac {2 b-c}{\sqrt {c^2-4 a b}}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {c^2-4 a b}+2 b+c}}{\sqrt [4]{\sqrt {c^2-4 a b}+c} \sqrt [4]{a x^4-b}}\right )}{2 \left (\sqrt {c^2-4 a b}+c\right )^{3/4} \sqrt [4]{\sqrt {c^2-4 a b}+2 b+c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b + a*x^4)/((-b + a*x^4)^(1/4)*(b + c*x^4 + a*x^8)),x]

[Out]

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

Rule 211

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

Rule 214

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

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 117, normalized size = 1.01 \begin {gather*} \frac {1}{4} \text {RootSum}\left [a^2+a b+a c-2 a \text {$\#$1}^4-c \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-2 a \log (x)+2 a \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-2 a \text {$\#$1}-c \text {$\#$1}+2 \text {$\#$1}^5}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x^4)/((-b + a*x^4)^(1/4)*(b + c*x^4 + a*x^8)),x]

[Out]

RootSum[a^2 + a*b + a*c - 2*a*#1^4 - c*#1^4 + #1^8 & , (-2*a*Log[x] + 2*a*Log[(-b + a*x^4)^(1/4) - x*#1] + Log
[x]*#1^4 - Log[(-b + a*x^4)^(1/4) - x*#1]*#1^4)/(-2*a*#1 - c*#1 + 2*#1^5) & ]/4

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{4}+b}{\left (a \,x^{4}-b \right )^{\frac {1}{4}} \left (a \,x^{8}+c \,x^{4}+b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4+b)/(a*x^4-b)^(1/4)/(a*x^8+c*x^4+b),x)

[Out]

int((a*x^4+b)/(a*x^4-b)^(1/4)/(a*x^8+c*x^4+b),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((a*x^4+b)/(a*x^4-b)^(1/4)/(a*x^8+c*x^4+b),x, algorithm="maxima")

[Out]

integrate((a*x^4 + b)/((a*x^8 + c*x^4 + b)*(a*x^4 - b)^(1/4)), x)

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Fricas [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((a*x^4+b)/(a*x^4-b)^(1/4)/(a*x^8+c*x^4+b),x, algorithm="fricas")

[Out]

Timed out

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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((a*x**4+b)/(a*x**4-b)**(1/4)/(a*x**8+c*x**4+b),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((a*x^4+b)/(a*x^4-b)^(1/4)/(a*x^8+c*x^4+b),x, algorithm="giac")

[Out]

integrate((a*x^4 + b)/((a*x^8 + c*x^4 + b)*(a*x^4 - b)^(1/4)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a\,x^4+b}{{\left (a\,x^4-b\right )}^{1/4}\,\left (a\,x^8+c\,x^4+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b + a*x^4)/((a*x^4 - b)^(1/4)*(b + a*x^8 + c*x^4)),x)

[Out]

int((b + a*x^4)/((a*x^4 - b)^(1/4)*(b + a*x^8 + c*x^4)), x)

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