3.18.38 \(\int \frac {-b+a x^6}{x^6 \sqrt [4]{-b x+a x^4}} \, dx\)

Optimal. Leaf size=117 \[ \frac {2}{3} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \left (a x^4-b x\right )^{3/4}}{a x^3-b}\right )+\frac {2}{3} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \left (a x^4-b x\right )^{3/4}}{a x^3-b}\right )-\frac {4 \left (a x^4-b x\right )^{3/4} \left (4 a x^3+3 b\right )}{63 b x^6} \]

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 179, normalized size of antiderivative = 1.53, number of steps used = 11, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {2052, 2011, 329, 275, 240, 212, 206, 203, 2016, 2014} \begin {gather*} \frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{a x^3-b} \tan ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3-b}}\right )}{3 \sqrt [4]{a x^4-b x}}+\frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{a x^3-b} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3-b}}\right )}{3 \sqrt [4]{a x^4-b x}}-\frac {4 \left (a x^4-b x\right )^{3/4}}{21 x^6}-\frac {16 a \left (a x^4-b x\right )^{3/4}}{63 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-b + a*x^6)/(x^6*(-(b*x) + a*x^4)^(1/4)),x]

[Out]

(-4*(-(b*x) + a*x^4)^(3/4))/(21*x^6) - (16*a*(-(b*x) + a*x^4)^(3/4))/(63*b*x^3) + (2*a^(3/4)*x^(1/4)*(-b + a*x
^3)^(1/4)*ArcTan[(a^(1/4)*x^(3/4))/(-b + a*x^3)^(1/4)])/(3*(-(b*x) + a*x^4)^(1/4)) + (2*a^(3/4)*x^(1/4)*(-b +
a*x^3)^(1/4)*ArcTanh[(a^(1/4)*x^(3/4))/(-b + a*x^3)^(1/4)])/(3*(-(b*x) + a*x^4)^(1/4))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 212

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] &&
!GtQ[a/b, 0]

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2052

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps

\begin {align*} \int \frac {-b+a x^6}{x^6 \sqrt [4]{-b x+a x^4}} \, dx &=\int \left (\frac {a}{\sqrt [4]{-b x+a x^4}}-\frac {b}{x^6 \sqrt [4]{-b x+a x^4}}\right ) \, dx\\ &=a \int \frac {1}{\sqrt [4]{-b x+a x^4}} \, dx-b \int \frac {1}{x^6 \sqrt [4]{-b x+a x^4}} \, dx\\ &=-\frac {4 \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {1}{7} (4 a) \int \frac {1}{x^3 \sqrt [4]{-b x+a x^4}} \, dx+\frac {\left (a \sqrt [4]{x} \sqrt [4]{-b+a x^3}\right ) \int \frac {1}{\sqrt [4]{x} \sqrt [4]{-b+a x^3}} \, dx}{\sqrt [4]{-b x+a x^4}}\\ &=-\frac {4 \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {16 a \left (-b x+a x^4\right )^{3/4}}{63 b x^3}+\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{-b+a x^3}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{-b+a x^{12}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x+a x^4}}\\ &=-\frac {4 \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {16 a \left (-b x+a x^4\right )^{3/4}}{63 b x^3}+\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{-b+a x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-b+a x^4}} \, dx,x,x^{3/4}\right )}{3 \sqrt [4]{-b x+a x^4}}\\ &=-\frac {4 \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {16 a \left (-b x+a x^4\right )^{3/4}}{63 b x^3}+\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{-b+a x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )}{3 \sqrt [4]{-b x+a x^4}}\\ &=-\frac {4 \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {16 a \left (-b x+a x^4\right )^{3/4}}{63 b x^3}+\frac {\left (2 a \sqrt [4]{x} \sqrt [4]{-b+a x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )}{3 \sqrt [4]{-b x+a x^4}}+\frac {\left (2 a \sqrt [4]{x} \sqrt [4]{-b+a x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )}{3 \sqrt [4]{-b x+a x^4}}\\ &=-\frac {4 \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {16 a \left (-b x+a x^4\right )^{3/4}}{63 b x^3}+\frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{-b+a x^3} \tan ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )}{3 \sqrt [4]{-b x+a x^4}}+\frac {2 a^{3/4} \sqrt [4]{x} \sqrt [4]{-b+a x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )}{3 \sqrt [4]{-b x+a x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 147, normalized size = 1.26 \begin {gather*} \frac {2 \left (21 a^{3/4} b x^{21/4} \sqrt [4]{a x^3-b} \tan ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3-b}}\right )+21 a^{3/4} b x^{21/4} \sqrt [4]{a x^3-b} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3-b}}\right )-8 a^2 x^6+2 a b x^3+6 b^2\right )}{63 b x^5 \sqrt [4]{a x^4-b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-b + a*x^6)/(x^6*(-(b*x) + a*x^4)^(1/4)),x]

[Out]

(2*(6*b^2 + 2*a*b*x^3 - 8*a^2*x^6 + 21*a^(3/4)*b*x^(21/4)*(-b + a*x^3)^(1/4)*ArcTan[(a^(1/4)*x^(3/4))/(-b + a*
x^3)^(1/4)] + 21*a^(3/4)*b*x^(21/4)*(-b + a*x^3)^(1/4)*ArcTanh[(a^(1/4)*x^(3/4))/(-b + a*x^3)^(1/4)]))/(63*b*x
^5*(-(b*x) + a*x^4)^(1/4))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.71, size = 117, normalized size = 1.00 \begin {gather*} -\frac {4 \left (3 b+4 a x^3\right ) \left (-b x+a x^4\right )^{3/4}}{63 b x^6}+\frac {2}{3} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )+\frac {2}{3} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-b + a*x^6)/(x^6*(-(b*x) + a*x^4)^(1/4)),x]

[Out]

(-4*(3*b + 4*a*x^3)*(-(b*x) + a*x^4)^(3/4))/(63*b*x^6) + (2*a^(3/4)*ArcTan[(a^(1/4)*(-(b*x) + a*x^4)^(3/4))/(-
b + a*x^3)])/3 + (2*a^(3/4)*ArcTanh[(a^(1/4)*(-(b*x) + a*x^4)^(3/4))/(-b + a*x^3)])/3

________________________________________________________________________________________

fricas [F(-1)]  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^6-b)/x^6/(a*x^4-b*x)^(1/4),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [B]  time = 0.35, size = 217, normalized size = 1.85 \begin {gather*} \frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right ) + \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right ) + \frac {4 \, {\left (3 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {7}{4}} b^{6} - 7 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {3}{4}} a b^{6}\right )}}{63 \, b^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^6-b)/x^6/(a*x^4-b*x)^(1/4),x, algorithm="giac")

[Out]

1/3*sqrt(2)*(-a)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a - b/x^3)^(1/4))/(-a)^(1/4)) + 1/3*sqrt(2)
*(-a)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a - b/x^3)^(1/4))/(-a)^(1/4)) - 1/6*sqrt(2)*(-a)^(3/4
)*log(sqrt(2)*(-a)^(1/4)*(a - b/x^3)^(1/4) + sqrt(-a) + sqrt(a - b/x^3)) + 1/6*sqrt(2)*(-a)^(3/4)*log(-sqrt(2)
*(-a)^(1/4)*(a - b/x^3)^(1/4) + sqrt(-a) + sqrt(a - b/x^3)) + 4/63*(3*(a - b/x^3)^(7/4)*b^6 - 7*(a - b/x^3)^(3
/4)*a*b^6)/b^7

________________________________________________________________________________________

maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{6}-b}{x^{6} \left (a \,x^{4}-b x \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^6-b)/x^6/(a*x^4-b*x)^(1/4),x)

[Out]

int((a*x^6-b)/x^6/(a*x^4-b*x)^(1/4),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{6} - b}{{\left (a x^{4} - b x\right )}^{\frac {1}{4}} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^6-b)/x^6/(a*x^4-b*x)^(1/4),x, algorithm="maxima")

[Out]

integrate((a*x^6 - b)/((a*x^4 - b*x)^(1/4)*x^6), x)

________________________________________________________________________________________

mupad [B]  time = 1.37, size = 73, normalized size = 0.62 \begin {gather*} \frac {4\,a\,x\,{\left (1-\frac {a\,x^3}{b}\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ \frac {a\,x^3}{b}\right )}{3\,{\left (a\,x^4-b\,x\right )}^{1/4}}-\frac {4\,{\left (a\,x^4-b\,x\right )}^{3/4}\,\left (4\,a\,x^3+3\,b\right )}{63\,b\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(b - a*x^6)/(x^6*(a*x^4 - b*x)^(1/4)),x)

[Out]

(4*a*x*(1 - (a*x^3)/b)^(1/4)*hypergeom([1/4, 1/4], 5/4, (a*x^3)/b))/(3*(a*x^4 - b*x)^(1/4)) - (4*(a*x^4 - b*x)
^(3/4)*(3*b + 4*a*x^3))/(63*b*x^6)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{6} - b}{x^{6} \sqrt [4]{x \left (a x^{3} - b\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**6-b)/x**6/(a*x**4-b*x)**(1/4),x)

[Out]

Integral((a*x**6 - b)/(x**6*(x*(a*x**3 - b))**(1/4)), x)

________________________________________________________________________________________