3.23.35 \(\int \frac {1}{x^7 (-b+a x^3)^{3/4}} \, dx\)

Optimal. Leaf size=167 \[ -\frac {7 a^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}{\sqrt {a x^3-b}-\sqrt {b}}\right )}{16 \sqrt {2} b^{11/4}}+\frac {7 a^2 \tanh ^{-1}\left (\frac {\frac {\sqrt {a x^3-b}}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b}}{\sqrt {2}}}{\sqrt [4]{a x^3-b}}\right )}{16 \sqrt {2} b^{11/4}}+\frac {\sqrt [4]{a x^3-b} \left (7 a x^3+4 b\right )}{24 b^2 x^6} \]

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Rubi [A]  time = 0.26, 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, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {7 a^2 \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{32 \sqrt {2} b^{11/4}}+\frac {7 a^2 \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{32 \sqrt {2} b^{11/4}}-\frac {7 a^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{16 \sqrt {2} b^{11/4}}+\frac {7 a^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{16 \sqrt {2} b^{11/4}}+\frac {7 a \sqrt [4]{a x^3-b}}{24 b^2 x^3}+\frac {\sqrt [4]{a x^3-b}}{6 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(-b + a*x^3)^(3/4)),x]

[Out]

(-b + a*x^3)^(1/4)/(6*b*x^6) + (7*a*(-b + a*x^3)^(1/4))/(24*b^2*x^3) - (7*a^2*ArcTan[1 - (Sqrt[2]*(-b + a*x^3)
^(1/4))/b^(1/4)])/(16*Sqrt[2]*b^(11/4)) + (7*a^2*ArcTan[1 + (Sqrt[2]*(-b + a*x^3)^(1/4))/b^(1/4)])/(16*Sqrt[2]
*b^(11/4)) - (7*a^2*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4) + Sqrt[-b + a*x^3]])/(32*Sqrt[2]*b^(11/4)
) + (7*a^2*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4) + Sqrt[-b + a*x^3]])/(32*Sqrt[2]*b^(11/4))

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 211

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

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 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 \left (-b+a x^3\right )^{3/4}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x^3 (-b+a x)^{3/4}} \, dx,x,x^3\right )\\ &=\frac {\sqrt [4]{-b+a x^3}}{6 b x^6}+\frac {(7 a) \operatorname {Subst}\left (\int \frac {1}{x^2 (-b+a x)^{3/4}} \, dx,x,x^3\right )}{24 b}\\ &=\frac {\sqrt [4]{-b+a x^3}}{6 b x^6}+\frac {7 a \sqrt [4]{-b+a x^3}}{24 b^2 x^3}+\frac {\left (7 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x (-b+a x)^{3/4}} \, dx,x,x^3\right )}{32 b^2}\\ &=\frac {\sqrt [4]{-b+a x^3}}{6 b x^6}+\frac {7 a \sqrt [4]{-b+a x^3}}{24 b^2 x^3}+\frac {(7 a) \operatorname {Subst}\left (\int \frac {1}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^3}\right )}{8 b^2}\\ &=\frac {\sqrt [4]{-b+a x^3}}{6 b x^6}+\frac {7 a \sqrt [4]{-b+a x^3}}{24 b^2 x^3}+\frac {(7 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 )}{16 b^{5/2}}+\frac {(7 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 )}{16 b^{5/2}}\\ &=\frac {\sqrt [4]{-b+a x^3}}{6 b x^6}+\frac {7 a \sqrt [4]{-b+a x^3}}{24 b^2 x^3}-\frac {\left (7 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 )}{32 \sqrt {2} b^{11/4}}-\frac {\left (7 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 )}{32 \sqrt {2} b^{11/4}}+\frac {\left (7 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 )}{32 b^{5/2}}+\frac {\left (7 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 )}{32 b^{5/2}}\\ &=\frac {\sqrt [4]{-b+a x^3}}{6 b x^6}+\frac {7 a \sqrt [4]{-b+a x^3}}{24 b^2 x^3}-\frac {7 a^2 \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{32 \sqrt {2} b^{11/4}}+\frac {7 a^2 \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{32 \sqrt {2} b^{11/4}}+\frac {\left (7 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 )}{16 \sqrt {2} b^{11/4}}-\frac {\left (7 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 )}{16 \sqrt {2} b^{11/4}}\\ &=\frac {\sqrt [4]{-b+a x^3}}{6 b x^6}+\frac {7 a \sqrt [4]{-b+a x^3}}{24 b^2 x^3}-\frac {7 a^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{-b+a x^3}}{\sqrt [4]{b}}\right )}{16 \sqrt {2} b^{11/4}}+\frac {7 a^2 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{-b+a x^3}}{\sqrt [4]{b}}\right )}{16 \sqrt {2} b^{11/4}}-\frac {7 a^2 \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{32 \sqrt {2} b^{11/4}}+\frac {7 a^2 \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{32 \sqrt {2} b^{11/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 42, normalized size = 0.25 \begin {gather*} \frac {4 a^2 \sqrt [4]{a x^3-b} \, _2F_1\left (\frac {1}{4},3;\frac {5}{4};1-\frac {a x^3}{b}\right )}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(-b + a*x^3)^(1/4)*Hypergeometric2F1[1/4, 3, 5/4, 1 - (a*x^3)/b])/(3*b^3)

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IntegrateAlgebraic [A]  time = 0.25, size = 166, normalized size = 0.99 \begin {gather*} \frac {\sqrt [4]{-b+a x^3} \left (4 b+7 a x^3\right )}{24 b^2 x^6}+\frac {7 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 )}{16 \sqrt {2} b^{11/4}}+\frac {7 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 )}{16 \sqrt {2} b^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-b + a*x^3)^(1/4)*(4*b + 7*a*x^3))/(24*b^2*x^6) + (7*a^2*ArcTan[(-(b^(1/4)/Sqrt[2]) + Sqrt[-b + a*x^3]/(Sqrt
[2]*b^(1/4)))/(-b + a*x^3)^(1/4)])/(16*Sqrt[2]*b^(11/4)) + (7*a^2*ArcTanh[(Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4))
/(Sqrt[b] + Sqrt[-b + a*x^3])])/(16*Sqrt[2]*b^(11/4))

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fricas [A]  time = 0.94, size = 234, normalized size = 1.40 \begin {gather*} \frac {84 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2} b^{8} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {3}{4}} - \sqrt {b^{6} \sqrt {-\frac {a^{8}}{b^{11}}} + \sqrt {a x^{3} - b} a^{4}} b^{8} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {3}{4}}}{a^{8}}\right ) + 21 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (7 \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2}\right ) - 21 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-7 \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2}\right ) + 4 \, {\left (7 \, a x^{3} + 4 \, b\right )} {\left (a x^{3} - b\right )}^{\frac {1}{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)^(3/4),x, algorithm="fricas")

[Out]

1/96*(84*b^2*x^6*(-a^8/b^11)^(1/4)*arctan(-((a*x^3 - b)^(1/4)*a^2*b^8*(-a^8/b^11)^(3/4) - sqrt(b^6*sqrt(-a^8/b
^11) + sqrt(a*x^3 - b)*a^4)*b^8*(-a^8/b^11)^(3/4))/a^8) + 21*b^2*x^6*(-a^8/b^11)^(1/4)*log(7*b^3*(-a^8/b^11)^(
1/4) + 7*(a*x^3 - b)^(1/4)*a^2) - 21*b^2*x^6*(-a^8/b^11)^(1/4)*log(-7*b^3*(-a^8/b^11)^(1/4) + 7*(a*x^3 - b)^(1
/4)*a^2) + 4*(7*a*x^3 + 4*b)*(a*x^3 - b)^(1/4))/(b^2*x^6)

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giac [A]  time = 0.19, size = 224, normalized size = 1.34 \begin {gather*} \frac {\frac {42 \, \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 {11}{4}}} + \frac {42 \, \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 {11}{4}}} + \frac {21 \, \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 {11}{4}}} - \frac {21 \, \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 {11}{4}}} + \frac {8 \, {\left (7 \, {\left (a x^{3} - b\right )}^{\frac {5}{4}} a^{3} + 11 \, {\left (a x^{3} - b\right )}^{\frac {1}{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)^(3/4),x, algorithm="giac")

[Out]

1/192*(42*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^(11/4) + 42*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^(11/4) + 21*sqrt(2)*a^3*log(sqrt(
2)*(a*x^3 - b)^(1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b))/b^(11/4) - 21*sqrt(2)*a^3*log(-sqrt(2)*(a*x^3 - b)^(
1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b))/b^(11/4) + 8*(7*(a*x^3 - b)^(5/4)*a^3 + 11*(a*x^3 - b)^(1/4)*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 \[\int \frac {1}{x^{7} \left (a \,x^{3}-b \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 250, normalized size = 1.50 \begin {gather*} \frac {7 \, {\left (a x^{3} - b\right )}^{\frac {5}{4}} a^{2} + 11 \, {\left (a x^{3} - b\right )}^{\frac {1}{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 )}} + \frac {7 \, {\left (\frac {2 \, \sqrt {2} a^{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 {3}{4}}} + \frac {2 \, \sqrt {2} a^{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 {3}{4}}} + \frac {\sqrt {2} a^{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 {3}{4}}} - \frac {\sqrt {2} a^{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 {3}{4}}}\right )}}{64 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(7*(a*x^3 - b)^(5/4)*a^2 + 11*(a*x^3 - b)^(1/4)*a^2*b)/((a*x^3 - b)^2*b^2 + 2*(a*x^3 - b)*b^3 + b^4) + 7/
64*(2*sqrt(2)*a^2*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^3 - b)^(1/4))/b^(1/4))/b^(3/4) + 2*sqrt(2)*a^2*
arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^3 - b)^(1/4))/b^(1/4))/b^(3/4) + sqrt(2)*a^2*log(sqrt(2)*(a*x^3
- b)^(1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b))/b^(3/4) - sqrt(2)*a^2*log(-sqrt(2)*(a*x^3 - b)^(1/4)*b^(1/4) +
sqrt(a*x^3 - b) + sqrt(b))/b^(3/4))/b^2

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mupad [B]  time = 1.40, size = 98, normalized size = 0.59 \begin {gather*} \frac {11\,{\left (a\,x^3-b\right )}^{1/4}}{24\,b\,x^6}+\frac {7\,{\left (a\,x^3-b\right )}^{5/4}}{24\,b^2\,x^6}-\frac {7\,a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{16\,{\left (-b\right )}^{11/4}}+\frac {a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,7{}\mathrm {i}}{16\,{\left (-b\right )}^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(a*x^3 - b)^(3/4)),x)

[Out]

(11*(a*x^3 - b)^(1/4))/(24*b*x^6) + (7*(a*x^3 - b)^(5/4))/(24*b^2*x^6) - (7*a^2*atan((a*x^3 - b)^(1/4)/(-b)^(1
/4)))/(16*(-b)^(11/4)) + (a^2*atan(((a*x^3 - b)^(1/4)*1i)/(-b)^(1/4))*7i)/(16*(-b)^(11/4))

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sympy [C]  time = 1.69, size = 42, normalized size = 0.25 \begin {gather*} - \frac {\Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 a^{\frac {3}{4}} x^{\frac {33}{4}} \Gamma \left (\frac {15}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(11/4)*hyper((3/4, 11/4), (15/4,), b*exp_polar(2*I*pi)/(a*x**3))/(3*a**(3/4)*x**(33/4)*gamma(15/4))

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