3.18.2 \(\int \frac {(-b+a x^3) (-b+2 a x^3)}{x^6 \sqrt [4]{-b x+a x^4}} \, dx\)

Optimal. Leaf size=114 \[ \frac {4}{3} a^{7/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \left (a x^4-b x\right )^{3/4}}{a x^3-b}\right )+\frac {4}{3} a^{7/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \left (a x^4-b x\right )^{3/4}}{a x^3-b}\right )+\frac {4 \left (3 b-17 a x^3\right ) \left (a x^4-b x\right )^{3/4}}{63 x^6} \]

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Rubi [A]  time = 0.40, antiderivative size = 177, normalized size of antiderivative = 1.55, number of steps used = 12, number of rules used = 10, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {2052, 2011, 329, 275, 240, 212, 206, 203, 2016, 2014} \begin {gather*} \frac {4 a^{7/4} \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 {4 a^{7/4} \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 b \left (a x^4-b x\right )^{3/4}}{21 x^6}-\frac {68 a \left (a x^4-b x\right )^{3/4}}{63 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-b + a*x^3)*(-b + 2*a*x^3))/(x^6*(-(b*x) + a*x^4)^(1/4)),x]

[Out]

(4*b*(-(b*x) + a*x^4)^(3/4))/(21*x^6) - (68*a*(-(b*x) + a*x^4)^(3/4))/(63*x^3) + (4*a^(7/4)*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)) + (4*a^(7/4)*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 {\left (-b+a x^3\right ) \left (-b+2 a x^3\right )}{x^6 \sqrt [4]{-b x+a x^4}} \, dx &=\int \left (\frac {2 a^2}{\sqrt [4]{-b x+a x^4}}+\frac {b^2}{x^6 \sqrt [4]{-b x+a x^4}}-\frac {3 a b}{x^3 \sqrt [4]{-b x+a x^4}}\right ) \, dx\\ &=\left (2 a^2\right ) \int \frac {1}{\sqrt [4]{-b x+a x^4}} \, dx-(3 a b) \int \frac {1}{x^3 \sqrt [4]{-b x+a x^4}} \, dx+b^2 \int \frac {1}{x^6 \sqrt [4]{-b x+a x^4}} \, dx\\ &=\frac {4 b \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {4 a \left (-b x+a x^4\right )^{3/4}}{3 x^3}+\frac {1}{7} (4 a b) \int \frac {1}{x^3 \sqrt [4]{-b x+a x^4}} \, dx+\frac {\left (2 a^2 \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 b \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {68 a \left (-b x+a x^4\right )^{3/4}}{63 x^3}+\frac {\left (8 a^2 \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 b \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {68 a \left (-b x+a x^4\right )^{3/4}}{63 x^3}+\frac {\left (8 a^2 \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 b \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {68 a \left (-b x+a x^4\right )^{3/4}}{63 x^3}+\frac {\left (8 a^2 \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 b \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {68 a \left (-b x+a x^4\right )^{3/4}}{63 x^3}+\frac {\left (4 a^2 \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 (4 a^2 \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 b \left (-b x+a x^4\right )^{3/4}}{21 x^6}-\frac {68 a \left (-b x+a x^4\right )^{3/4}}{63 x^3}+\frac {4 a^{7/4} \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 {4 a^{7/4} \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*}

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Mathematica [C]  time = 0.09, size = 71, normalized size = 0.62 \begin {gather*} \frac {4 \left (a x^4-b x\right )^{3/4} \left (3 \left (b-a x^3\right )-\frac {14 a x^3 \, _2F_1\left (-\frac {3}{4},-\frac {3}{4};\frac {1}{4};\frac {a x^3}{b}\right )}{\left (1-\frac {a x^3}{b}\right )^{3/4}}\right )}{63 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-b + a*x^3)*(-b + 2*a*x^3))/(x^6*(-(b*x) + a*x^4)^(1/4)),x]

[Out]

(4*(-(b*x) + a*x^4)^(3/4)*(3*(b - a*x^3) - (14*a*x^3*Hypergeometric2F1[-3/4, -3/4, 1/4, (a*x^3)/b])/(1 - (a*x^
3)/b)^(3/4)))/(63*x^6)

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IntegrateAlgebraic [A]  time = 0.43, size = 114, normalized size = 1.00 \begin {gather*} \frac {4 \left (3 b-17 a x^3\right ) \left (-b x+a x^4\right )^{3/4}}{63 x^6}+\frac {4}{3} a^{7/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )+\frac {4}{3} a^{7/4} \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^3)*(-b + 2*a*x^3))/(x^6*(-(b*x) + a*x^4)^(1/4)),x]

[Out]

(4*(3*b - 17*a*x^3)*(-(b*x) + a*x^4)^(3/4))/(63*x^6) + (4*a^(7/4)*ArcTan[(a^(1/4)*(-(b*x) + a*x^4)^(3/4))/(-b
+ a*x^3)])/3 + (4*a^(7/4)*ArcTanh[(a^(1/4)*(-(b*x) + a*x^4)^(3/4))/(-b + a*x^3)])/3

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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^3-b)*(2*a*x^3-b)/x^6/(a*x^4-b*x)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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giac [B]  time = 0.30, size = 209, normalized size = 1.83 \begin {gather*} \frac {2}{3} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} a \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 {2}{3} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} a \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}} a \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}{3} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} a \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}{21} \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {7}{4}} - \frac {8}{9} \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {3}{4}} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^3-b)*(2*a*x^3-b)/x^6/(a*x^4-b*x)^(1/4),x, algorithm="giac")

[Out]

2/3*sqrt(2)*(-a)^(3/4)*a*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a - b/x^3)^(1/4))/(-a)^(1/4)) + 2/3*sqrt(
2)*(-a)^(3/4)*a*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)*a*log(sqrt(2)*(-a)^(1/4)*(a - b/x^3)^(1/4) + sqrt(-a) + sqrt(a - b/x^3)) + 1/3*sqrt(2)*(-a)^(3/4)*a*log(
-sqrt(2)*(-a)^(1/4)*(a - b/x^3)^(1/4) + sqrt(-a) + sqrt(a - b/x^3)) - 4/21*(a - b/x^3)^(7/4) - 8/9*(a - b/x^3)
^(3/4)*a

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{3}-b \right ) \left (2 a \,x^{3}-b \right )}{x^{6} \left (a \,x^{4}-b x \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^3-b)*(2*a*x^3-b)/x^6/(a*x^4-b*x)^(1/4),x)

[Out]

int((a*x^3-b)*(2*a*x^3-b)/x^6/(a*x^4-b*x)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, a x^{3} - b\right )} {\left (a x^{3} - b\right )}}{{\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^3-b)*(2*a*x^3-b)/x^6/(a*x^4-b*x)^(1/4),x, algorithm="maxima")

[Out]

integrate((2*a*x^3 - b)*(a*x^3 - b)/((a*x^4 - b*x)^(1/4)*x^6), x)

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mupad [B]  time = 1.88, size = 72, normalized size = 0.63 \begin {gather*} -\frac {4\,\left (3\,b^2+17\,a^2\,x^6-20\,a\,b\,x^3-42\,a^2\,x^6\,{\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 )\right )}{63\,x^5\,{\left (a\,x^4-b\,x\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b - a*x^3)*(b - 2*a*x^3))/(x^6*(a*x^4 - b*x)^(1/4)),x)

[Out]

-(4*(3*b^2 + 17*a^2*x^6 - 20*a*b*x^3 - 42*a^2*x^6*(1 - (a*x^3)/b)^(1/4)*hypergeom([1/4, 1/4], 5/4, (a*x^3)/b))
)/(63*x^5*(a*x^4 - b*x)^(1/4))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a x^{3} - b\right ) \left (2 a x^{3} - b\right )}{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**3-b)*(2*a*x**3-b)/x**6/(a*x**4-b*x)**(1/4),x)

[Out]

Integral((a*x**3 - b)*(2*a*x**3 - b)/(x**6*(x*(a*x**3 - b))**(1/4)), x)

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