3.23.36 \(\int \frac {1}{x^7 \sqrt [4]{-b+a x^3}} \, dx\)

Optimal. Leaf size=167 \[ -\frac {5 a^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}{\sqrt {a x^3-b}-\sqrt {b}}\right )}{48 \sqrt {2} b^{9/4}}-\frac {5 a^2 \tanh ^{-1}\left (\frac {\frac {\sqrt {a x^3-b}}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b}}{\sqrt {2}}}{\sqrt [4]{a x^3-b}}\right )}{48 \sqrt {2} b^{9/4}}+\frac {\left (a x^3-b\right )^{3/4} \left (5 a x^3+4 b\right )}{24 b^2 x^6} \]

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Rubi [A]  time = 0.27, antiderivative size = 260, normalized size of antiderivative = 1.56, number of steps used = 13, number of rules used = 9, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {266, 51, 63, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {5 a^2 \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{96 \sqrt {2} b^{9/4}}-\frac {5 a^2 \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{96 \sqrt {2} b^{9/4}}-\frac {5 a^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{48 \sqrt {2} b^{9/4}}+\frac {5 a^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{48 \sqrt {2} b^{9/4}}+\frac {5 a \left (a x^3-b\right )^{3/4}}{24 b^2 x^3}+\frac {\left (a x^3-b\right )^{3/4}}{6 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(-b + a*x^3)^(1/4)),x]

[Out]

(-b + a*x^3)^(3/4)/(6*b*x^6) + (5*a*(-b + a*x^3)^(3/4))/(24*b^2*x^3) - (5*a^2*ArcTan[1 - (Sqrt[2]*(-b + a*x^3)
^(1/4))/b^(1/4)])/(48*Sqrt[2]*b^(9/4)) + (5*a^2*ArcTan[1 + (Sqrt[2]*(-b + a*x^3)^(1/4))/b^(1/4)])/(48*Sqrt[2]*
b^(9/4)) + (5*a^2*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4) + Sqrt[-b + a*x^3]])/(96*Sqrt[2]*b^(9/4)) -
(5*a^2*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4) + Sqrt[-b + a*x^3]])/(96*Sqrt[2]*b^(9/4))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 266

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

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 {1}{x^7 \sqrt [4]{-b+a x^3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt [4]{-b+a x}} \, dx,x,x^3\right )\\ &=\frac {\left (-b+a x^3\right )^{3/4}}{6 b x^6}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt [4]{-b+a x}} \, dx,x,x^3\right )}{24 b}\\ &=\frac {\left (-b+a x^3\right )^{3/4}}{6 b x^6}+\frac {5 a \left (-b+a x^3\right )^{3/4}}{24 b^2 x^3}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt [4]{-b+a x}} \, dx,x,x^3\right )}{96 b^2}\\ &=\frac {\left (-b+a x^3\right )^{3/4}}{6 b x^6}+\frac {5 a \left (-b+a x^3\right )^{3/4}}{24 b^2 x^3}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^3}\right )}{24 b^2}\\ &=\frac {\left (-b+a x^3\right )^{3/4}}{6 b x^6}+\frac {5 a \left (-b+a x^3\right )^{3/4}}{24 b^2 x^3}-\frac {(5 a) \operatorname {Subst}\left (\int \frac {\sqrt {b}-x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^3}\right )}{48 b^2}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {\sqrt {b}+x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^3}\right )}{48 b^2}\\ &=\frac {\left (-b+a x^3\right )^{3/4}}{6 b x^6}+\frac {5 a \left (-b+a x^3\right )^{3/4}}{24 b^2 x^3}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{b}+2 x}{-\sqrt {b}-\sqrt {2} \sqrt [4]{b} x-x^2} \, dx,x,\sqrt [4]{-b+a x^3}\right )}{96 \sqrt {2} b^{9/4}}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{b}-2 x}{-\sqrt {b}+\sqrt {2} \sqrt [4]{b} x-x^2} \, dx,x,\sqrt [4]{-b+a x^3}\right )}{96 \sqrt {2} b^{9/4}}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}-\sqrt {2} \sqrt [4]{b} x+x^2} \, dx,x,\sqrt [4]{-b+a x^3}\right )}{96 b^2}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}+\sqrt {2} \sqrt [4]{b} x+x^2} \, dx,x,\sqrt [4]{-b+a x^3}\right )}{96 b^2}\\ &=\frac {\left (-b+a x^3\right )^{3/4}}{6 b x^6}+\frac {5 a \left (-b+a x^3\right )^{3/4}}{24 b^2 x^3}+\frac {5 a^2 \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{96 \sqrt {2} b^{9/4}}-\frac {5 a^2 \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{96 \sqrt {2} b^{9/4}}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{-b+a x^3}}{\sqrt [4]{b}}\right )}{48 \sqrt {2} b^{9/4}}-\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{-b+a x^3}}{\sqrt [4]{b}}\right )}{48 \sqrt {2} b^{9/4}}\\ &=\frac {\left (-b+a x^3\right )^{3/4}}{6 b x^6}+\frac {5 a \left (-b+a x^3\right )^{3/4}}{24 b^2 x^3}-\frac {5 a^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{-b+a x^3}}{\sqrt [4]{b}}\right )}{48 \sqrt {2} b^{9/4}}+\frac {5 a^2 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{-b+a x^3}}{\sqrt [4]{b}}\right )}{48 \sqrt {2} b^{9/4}}+\frac {5 a^2 \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{96 \sqrt {2} b^{9/4}}-\frac {5 a^2 \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{96 \sqrt {2} b^{9/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 42, normalized size = 0.25 \begin {gather*} \frac {4 a^2 \left (a x^3-b\right )^{3/4} \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};1-\frac {a x^3}{b}\right )}{9 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(-b + a*x^3)^(1/4)),x]

[Out]

(4*a^2*(-b + a*x^3)^(3/4)*Hypergeometric2F1[3/4, 3, 7/4, 1 - (a*x^3)/b])/(9*b^3)

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IntegrateAlgebraic [A]  time = 0.29, size = 166, normalized size = 0.99 \begin {gather*} \frac {\left (-b+a x^3\right )^{3/4} \left (4 b+5 a x^3\right )}{24 b^2 x^6}+\frac {5 a^2 \tan ^{-1}\left (\frac {-\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^3}}\right )}{48 \sqrt {2} b^{9/4}}-\frac {5 a^2 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{\sqrt {b}+\sqrt {-b+a x^3}}\right )}{48 \sqrt {2} b^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^7*(-b + a*x^3)^(1/4)),x]

[Out]

((-b + a*x^3)^(3/4)*(4*b + 5*a*x^3))/(24*b^2*x^6) + (5*a^2*ArcTan[(-(b^(1/4)/Sqrt[2]) + Sqrt[-b + a*x^3]/(Sqrt
[2]*b^(1/4)))/(-b + a*x^3)^(1/4)])/(48*Sqrt[2]*b^(9/4)) - (5*a^2*ArcTanh[(Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4))/
(Sqrt[b] + Sqrt[-b + a*x^3])])/(48*Sqrt[2]*b^(9/4))

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fricas [A]  time = 0.85, size = 240, normalized size = 1.44 \begin {gather*} -\frac {20 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {125 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{6} b^{2} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} - \sqrt {-15625 \, a^{8} b^{5} \sqrt {-\frac {a^{8}}{b^{9}}} + 15625 \, \sqrt {a x^{3} - b} a^{12}} b^{2} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}}}{125 \, a^{8}}\right ) - 5 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (125 \, b^{7} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{6}\right ) + 5 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-125 \, b^{7} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{6}\right ) - 4 \, {\left (5 \, a x^{3} + 4 \, b\right )} {\left (a x^{3} - b\right )}^{\frac {3}{4}}}{96 \, b^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(a*x^3-b)^(1/4),x, algorithm="fricas")

[Out]

-1/96*(20*b^2*x^6*(-a^8/b^9)^(1/4)*arctan(-1/125*(125*(a*x^3 - b)^(1/4)*a^6*b^2*(-a^8/b^9)^(1/4) - sqrt(-15625
*a^8*b^5*sqrt(-a^8/b^9) + 15625*sqrt(a*x^3 - b)*a^12)*b^2*(-a^8/b^9)^(1/4))/a^8) - 5*b^2*x^6*(-a^8/b^9)^(1/4)*
log(125*b^7*(-a^8/b^9)^(3/4) + 125*(a*x^3 - b)^(1/4)*a^6) + 5*b^2*x^6*(-a^8/b^9)^(1/4)*log(-125*b^7*(-a^8/b^9)
^(3/4) + 125*(a*x^3 - b)^(1/4)*a^6) - 4*(5*a*x^3 + 4*b)*(a*x^3 - b)^(3/4))/(b^2*x^6)

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giac [A]  time = 0.33, size = 224, normalized size = 1.34 \begin {gather*} \frac {\frac {10 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {9}{4}}} + \frac {10 \, \sqrt {2} a^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {9}{4}}} - \frac {5 \, \sqrt {2} a^{3} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {9}{4}}} + \frac {5 \, \sqrt {2} a^{3} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {9}{4}}} + \frac {8 \, {\left (5 \, {\left (a x^{3} - b\right )}^{\frac {7}{4}} a^{3} + 9 \, {\left (a x^{3} - b\right )}^{\frac {3}{4}} a^{3} b\right )}}{a^{2} b^{2} x^{6}}}{192 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(a*x^3-b)^(1/4),x, algorithm="giac")

[Out]

1/192*(10*sqrt(2)*a^3*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^3 - b)^(1/4))/b^(1/4))/b^(9/4) + 10*sqrt(2)
*a^3*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^3 - b)^(1/4))/b^(1/4))/b^(9/4) - 5*sqrt(2)*a^3*log(sqrt(2)*
(a*x^3 - b)^(1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b))/b^(9/4) + 5*sqrt(2)*a^3*log(-sqrt(2)*(a*x^3 - b)^(1/4)*
b^(1/4) + sqrt(a*x^3 - b) + sqrt(b))/b^(9/4) + 8*(5*(a*x^3 - b)^(7/4)*a^3 + 9*(a*x^3 - b)^(3/4)*a^3*b)/(a^2*b^
2*x^6))/a

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maple [F]  time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{7} \left (a \,x^{3}-b \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(a*x^3-b)^(1/4),x)

[Out]

int(1/x^7/(a*x^3-b)^(1/4),x)

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maxima [A]  time = 0.43, size = 241, normalized size = 1.44 \begin {gather*} \frac {5 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}}\right )} a^{2}}{192 \, b^{2}} + \frac {5 \, {\left (a x^{3} - b\right )}^{\frac {7}{4}} a^{2} + 9 \, {\left (a x^{3} - b\right )}^{\frac {3}{4}} a^{2} b}{24 \, {\left ({\left (a x^{3} - b\right )}^{2} b^{2} + 2 \, {\left (a x^{3} - b\right )} b^{3} + b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(a*x^3-b)^(1/4),x, algorithm="maxima")

[Out]

5/192*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^3 - b)^(1/4))/b^(1/4))/b^(1/4) + 2*sqrt(2)*arcta
n(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^3 - b)^(1/4))/b^(1/4))/b^(1/4) - sqrt(2)*log(sqrt(2)*(a*x^3 - b)^(1/4
)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b))/b^(1/4) + sqrt(2)*log(-sqrt(2)*(a*x^3 - b)^(1/4)*b^(1/4) + sqrt(a*x^3 -
b) + sqrt(b))/b^(1/4))*a^2/b^2 + 1/24*(5*(a*x^3 - b)^(7/4)*a^2 + 9*(a*x^3 - b)^(3/4)*a^2*b)/((a*x^3 - b)^2*b^
2 + 2*(a*x^3 - b)*b^3 + b^4)

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mupad [B]  time = 1.32, size = 98, normalized size = 0.59 \begin {gather*} \frac {3\,{\left (a\,x^3-b\right )}^{3/4}}{8\,b\,x^6}+\frac {5\,{\left (a\,x^3-b\right )}^{7/4}}{24\,b^2\,x^6}+\frac {5\,a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{48\,{\left (-b\right )}^{9/4}}+\frac {a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,5{}\mathrm {i}}{48\,{\left (-b\right )}^{9/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(a*x^3 - b)^(1/4)),x)

[Out]

(3*(a*x^3 - b)^(3/4))/(8*b*x^6) + (5*(a*x^3 - b)^(7/4))/(24*b^2*x^6) + (5*a^2*atan((a*x^3 - b)^(1/4)/(-b)^(1/4
)))/(48*(-b)^(9/4)) + (a^2*atan(((a*x^3 - b)^(1/4)*1i)/(-b)^(1/4))*5i)/(48*(-b)^(9/4))

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sympy [C]  time = 1.44, size = 42, normalized size = 0.25 \begin {gather*} - \frac {\Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 \sqrt [4]{a} x^{\frac {27}{4}} \Gamma \left (\frac {13}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(a*x**3-b)**(1/4),x)

[Out]

-gamma(9/4)*hyper((1/4, 9/4), (13/4,), b*exp_polar(2*I*pi)/(a*x**3))/(3*a**(1/4)*x**(27/4)*gamma(13/4))

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