3.22.34 \(\int \frac {\sqrt [4]{-b+a x^4} (-8 b+a x^8)}{x^{10} (b+a x^4)} \, dx\)

Optimal. Leaf size=155 \[ \frac {\left (8 a^{9/4}-a^{5/4} b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2^{3/4} b^2}-\frac {\left (8 a^{9/4}-a^{5/4} b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2^{3/4} b^2}+\frac {\sqrt [4]{a x^4-b} \left (80 a^2 x^8-9 a b x^8-16 a b x^4+8 b^2\right )}{9 b^2 x^9} \]

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Rubi [C]  time = 0.70, antiderivative size = 232, normalized size of antiderivative = 1.50, number of steps used = 12, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6725, 271, 264, 277, 331, 298, 203, 206, 511, 510} \begin {gather*} \frac {a^{5/4} (8 a-b) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 b^2}-\frac {a^{5/4} (8 a-b) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 b^2}+\frac {a^2 x^3 (8 a-b) \sqrt [4]{a x^4-b} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {a x^4}{b},\frac {a x^4}{b}\right )}{3 b^3 \sqrt [4]{1-\frac {a x^4}{b}}}+\frac {a (8 a-b) \sqrt [4]{a x^4-b}}{b^2 x}+\frac {8 a \left (a x^4-b\right )^{5/4}}{9 b^2 x^5}-\frac {8 \left (a x^4-b\right )^{5/4}}{9 b x^9} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[((-b + a*x^4)^(1/4)*(-8*b + a*x^8))/(x^10*(b + a*x^4)),x]

[Out]

(a*(8*a - b)*(-b + a*x^4)^(1/4))/(b^2*x) - (8*(-b + a*x^4)^(5/4))/(9*b*x^9) + (8*a*(-b + a*x^4)^(5/4))/(9*b^2*
x^5) + (a^2*(8*a - b)*x^3*(-b + a*x^4)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, -((a*x^4)/b), (a*x^4)/b])/(3*b^3*(1 -
(a*x^4)/b)^(1/4)) + (a^(5/4)*(8*a - b)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(2*b^2) - (a^(5/4)*(8*a - b)*A
rcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(2*b^2)

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 264

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

Rule 271

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

Rule 277

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

Rule 298

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

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 6725

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

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{-b+a x^4} \left (-8 b+a x^8\right )}{x^{10} \left (b+a x^4\right )} \, dx &=\int \left (-\frac {8 \sqrt [4]{-b+a x^4}}{x^{10}}+\frac {8 a \sqrt [4]{-b+a x^4}}{b x^6}-\frac {a (8 a-b) \sqrt [4]{-b+a x^4}}{b^2 x^2}+\frac {a^2 (8 a-b) x^2 \sqrt [4]{-b+a x^4}}{b^2 \left (b+a x^4\right )}\right ) \, dx\\ &=-\left (8 \int \frac {\sqrt [4]{-b+a x^4}}{x^{10}} \, dx\right )-\frac {(a (8 a-b)) \int \frac {\sqrt [4]{-b+a x^4}}{x^2} \, dx}{b^2}+\frac {\left (a^2 (8 a-b)\right ) \int \frac {x^2 \sqrt [4]{-b+a x^4}}{b+a x^4} \, dx}{b^2}+\frac {(8 a) \int \frac {\sqrt [4]{-b+a x^4}}{x^6} \, dx}{b}\\ &=\frac {a (8 a-b) \sqrt [4]{-b+a x^4}}{b^2 x}-\frac {8 \left (-b+a x^4\right )^{5/4}}{9 b x^9}+\frac {8 a \left (-b+a x^4\right )^{5/4}}{5 b^2 x^5}-\frac {\left (a^2 (8 a-b)\right ) \int \frac {x^2}{\left (-b+a x^4\right )^{3/4}} \, dx}{b^2}-\frac {(32 a) \int \frac {\sqrt [4]{-b+a x^4}}{x^6} \, dx}{9 b}+\frac {\left (a^2 (8 a-b) \sqrt [4]{-b+a x^4}\right ) \int \frac {x^2 \sqrt [4]{1-\frac {a x^4}{b}}}{b+a x^4} \, dx}{b^2 \sqrt [4]{1-\frac {a x^4}{b}}}\\ &=\frac {a (8 a-b) \sqrt [4]{-b+a x^4}}{b^2 x}-\frac {8 \left (-b+a x^4\right )^{5/4}}{9 b x^9}+\frac {8 a \left (-b+a x^4\right )^{5/4}}{9 b^2 x^5}+\frac {a^2 (8 a-b) x^3 \sqrt [4]{-b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {a x^4}{b},\frac {a x^4}{b}\right )}{3 b^3 \sqrt [4]{1-\frac {a x^4}{b}}}-\frac {\left (a^2 (8 a-b)\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{b^2}\\ &=\frac {a (8 a-b) \sqrt [4]{-b+a x^4}}{b^2 x}-\frac {8 \left (-b+a x^4\right )^{5/4}}{9 b x^9}+\frac {8 a \left (-b+a x^4\right )^{5/4}}{9 b^2 x^5}+\frac {a^2 (8 a-b) x^3 \sqrt [4]{-b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {a x^4}{b},\frac {a x^4}{b}\right )}{3 b^3 \sqrt [4]{1-\frac {a x^4}{b}}}-\frac {\left (a^{3/2} (8 a-b)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{2 b^2}+\frac {\left (a^{3/2} (8 a-b)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{2 b^2}\\ &=\frac {a (8 a-b) \sqrt [4]{-b+a x^4}}{b^2 x}-\frac {8 \left (-b+a x^4\right )^{5/4}}{9 b x^9}+\frac {8 a \left (-b+a x^4\right )^{5/4}}{9 b^2 x^5}+\frac {a^2 (8 a-b) x^3 \sqrt [4]{-b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {a x^4}{b},\frac {a x^4}{b}\right )}{3 b^3 \sqrt [4]{1-\frac {a x^4}{b}}}+\frac {a^{5/4} (8 a-b) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 b^2}-\frac {a^{5/4} (8 a-b) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 b^2}\\ \end {align*}

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Mathematica [C]  time = 0.33, size = 129, normalized size = 0.83 \begin {gather*} \frac {\left (a x^4-b\right ) \left (80 a^2 x^8-a b \left (9 x^4+16\right ) x^4+8 b^2\right )+\frac {6 a^2 x^{12} (b-8 a) \left (1-\frac {a x^4}{b}\right )^{3/4} \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};\frac {2 a x^4}{a x^4+b}\right )}{\left (\frac {a x^4}{b}+1\right )^{3/4}}}{9 b^2 x^9 \left (a x^4-b\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((-b + a*x^4)^(1/4)*(-8*b + a*x^8))/(x^10*(b + a*x^4)),x]

[Out]

((-b + a*x^4)*(8*b^2 + 80*a^2*x^8 - a*b*x^4*(16 + 9*x^4)) + (6*a^2*(-8*a + b)*x^12*(1 - (a*x^4)/b)^(3/4)*Hyper
geometric2F1[3/4, 3/4, 7/4, (2*a*x^4)/(b + a*x^4)])/(1 + (a*x^4)/b)^(3/4))/(9*b^2*x^9*(-b + a*x^4)^(3/4))

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IntegrateAlgebraic [A]  time = 0.50, size = 155, normalized size = 1.00 \begin {gather*} \frac {\sqrt [4]{-b+a x^4} \left (8 b^2-16 a b x^4+80 a^2 x^8-9 a b x^8\right )}{9 b^2 x^9}+\frac {\left (8 a^{9/4}-a^{5/4} b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2^{3/4} b^2}-\frac {\left (8 a^{9/4}-a^{5/4} b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2^{3/4} b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-b + a*x^4)^(1/4)*(-8*b + a*x^8))/(x^10*(b + a*x^4)),x]

[Out]

((-b + a*x^4)^(1/4)*(8*b^2 - 16*a*b*x^4 + 80*a^2*x^8 - 9*a*b*x^8))/(9*b^2*x^9) + ((8*a^(9/4) - a^(5/4)*b)*ArcT
an[(2^(1/4)*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(2^(3/4)*b^2) - ((8*a^(9/4) - a^(5/4)*b)*ArcTanh[(2^(1/4)*a^(1/4)*
x)/(-b + a*x^4)^(1/4)])/(2^(3/4)*b^2)

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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^4-b)^(1/4)*(a*x^8-8*b)/x^10/(a*x^4+b),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a x^{8} - 8 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{{\left (a x^{4} + b\right )} x^{10}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)^(1/4)*(a*x^8-8*b)/x^10/(a*x^4+b),x, algorithm="giac")

[Out]

integrate((a*x^8 - 8*b)*(a*x^4 - b)^(1/4)/((a*x^4 + b)*x^10), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}} \left (a \,x^{8}-8 b \right )}{x^{10} \left (a \,x^{4}+b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4-b)^(1/4)*(a*x^8-8*b)/x^10/(a*x^4+b),x)

[Out]

int((a*x^4-b)^(1/4)*(a*x^8-8*b)/x^10/(a*x^4+b),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a x^{8} - 8 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{{\left (a x^{4} + b\right )} x^{10}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)^(1/4)*(a*x^8-8*b)/x^10/(a*x^4+b),x, algorithm="maxima")

[Out]

integrate((a*x^8 - 8*b)*(a*x^4 - b)^(1/4)/((a*x^4 + b)*x^10), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {{\left (a\,x^4-b\right )}^{1/4}\,\left (8\,b-a\,x^8\right )}{x^{10}\,\left (a\,x^4+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((a*x^4 - b)^(1/4)*(8*b - a*x^8))/(x^10*(b + a*x^4)),x)

[Out]

int(-((a*x^4 - b)^(1/4)*(8*b - a*x^8))/(x^10*(b + a*x^4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{a x^{4} - b} \left (a x^{8} - 8 b\right )}{x^{10} \left (a x^{4} + b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**4-b)**(1/4)*(a*x**8-8*b)/x**10/(a*x**4+b),x)

[Out]

Integral((a*x**4 - b)**(1/4)*(a*x**8 - 8*b)/(x**10*(a*x**4 + b)), x)

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