\(\int \frac {\sqrt [4]{-b+a x^3}}{x^7} \, dx\) [2226]

Optimal result

Integrand size = 17, antiderivative size = 166 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^7} \, dx=\frac {\left (-4 b+a x^3\right ) \sqrt [4]{-b+a x^3}}{24 b x^6}-\frac {a^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{-\sqrt {b}+\sqrt {-b+a x^3}}\right )}{16 \sqrt {2} b^{7/4}}+\frac {a^2 \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^3}}\right )}{16 \sqrt {2} b^{7/4}} \]

[Out]

1/24*(a*x^3-4*b)*(a*x^3-b)^(1/4)/b/x^6-1/32*a^2*arctan(2^(1/2)*b^(1/4)*(a*x^3-b)^(1/4)/(-b^(1/2)+(a*x^3-b)^(1/
2)))*2^(1/2)/b^(7/4)+1/32*a^2*arctanh((1/2*b^(1/4)*2^(1/2)+1/2*(a*x^3-b)^(1/2)*2^(1/2)/b^(1/4))/(a*x^3-b)^(1/4
))*2^(1/2)/b^(7/4)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.55, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.588, Rules used = {272, 43, 44, 65, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^7} \, dx=-\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{16 \sqrt {2} b^{7/4}}+\frac {a^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{16 \sqrt {2} b^{7/4}}-\frac {a^2 \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{32 \sqrt {2} b^{7/4}}+\frac {a^2 \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{32 \sqrt {2} b^{7/4}}+\frac {a \sqrt [4]{a x^3-b}}{24 b x^3}-\frac {\sqrt [4]{a x^3-b}}{6 x^6} \]

[In]

Int[(-b + a*x^3)^(1/4)/x^7,x]

[Out]

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

Rule 43

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

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 210

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

Rule 217

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 272

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 631

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 642

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 1176

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 1179

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^7} \, dx=\frac {4 b^{3/4} \left (-4 b+a x^3\right ) \sqrt [4]{-b+a x^3}+3 \sqrt {2} a^2 x^6 \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}\right )+3 \sqrt {2} a^2 x^6 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{\sqrt {b}+\sqrt {-b+a x^3}}\right )}{96 b^{7/4} x^6} \]

[In]

Integrate[(-b + a*x^3)^(1/4)/x^7,x]

[Out]

(4*b^(3/4)*(-4*b + a*x^3)*(-b + a*x^3)^(1/4) + 3*Sqrt[2]*a^2*x^6*ArcTan[(-Sqrt[b] + Sqrt[-b + a*x^3])/(Sqrt[2]
*b^(1/4)*(-b + a*x^3)^(1/4))] + 3*Sqrt[2]*a^2*x^6*ArcTanh[(Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4))/(Sqrt[b] + Sqrt
[-b + a*x^3])])/(96*b^(7/4)*x^6)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.25

method	result	size
pseudoelliptic	\(\frac {3 \ln \left (\frac {-b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{3}-b}-\sqrt {b}}{b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{3}-b}-\sqrt {b}}\right ) \sqrt {2}\, a^{2} x^{6}+6 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a^{2} x^{6}-6 \arctan \left (\frac {-\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a^{2} x^{6}+8 a \,x^{3} \left (a \,x^{3}-b \right )^{\frac {1}{4}} b^{\frac {3}{4}}-32 b^{\frac {7}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}}}{192 b^{\frac {7}{4}} x^{6}}\)	\(207\)

[In]

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

[Out]

1/192/b^(7/4)*(3*ln((-b^(1/4)*(a*x^3-b)^(1/4)*2^(1/2)-(a*x^3-b)^(1/2)-b^(1/2))/(b^(1/4)*(a*x^3-b)^(1/4)*2^(1/2
)-(a*x^3-b)^(1/2)-b^(1/2)))*2^(1/2)*a^2*x^6+6*arctan((2^(1/2)*(a*x^3-b)^(1/4)+b^(1/4))/b^(1/4))*2^(1/2)*a^2*x^
6-6*arctan((-2^(1/2)*(a*x^3-b)^(1/4)+b^(1/4))/b^(1/4))*2^(1/2)*a^2*x^6+8*a*x^3*(a*x^3-b)^(1/4)*b^(3/4)-32*b^(7
/4)*(a*x^3-b)^(1/4))/x^6

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.34 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^7} \, dx=\frac {3 \, \left (-\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b x^{6} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2} + \left (-\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b^{2}\right ) + 3 i \, \left (-\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b x^{6} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2} + i \, \left (-\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b^{2}\right ) - 3 i \, \left (-\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b x^{6} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2} - i \, \left (-\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b^{2}\right ) - 3 \, \left (-\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b x^{6} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2} - \left (-\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b^{2}\right ) + 4 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} {\left (a x^{3} - 4 \, b\right )}}{96 \, b x^{6}} \]

[In]

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

[Out]

1/96*(3*(-a^8/b^7)^(1/4)*b*x^6*log((a*x^3 - b)^(1/4)*a^2 + (-a^8/b^7)^(1/4)*b^2) + 3*I*(-a^8/b^7)^(1/4)*b*x^6*
log((a*x^3 - b)^(1/4)*a^2 + I*(-a^8/b^7)^(1/4)*b^2) - 3*I*(-a^8/b^7)^(1/4)*b*x^6*log((a*x^3 - b)^(1/4)*a^2 - I
*(-a^8/b^7)^(1/4)*b^2) - 3*(-a^8/b^7)^(1/4)*b*x^6*log((a*x^3 - b)^(1/4)*a^2 - (-a^8/b^7)^(1/4)*b^2) + 4*(a*x^3
- b)^(1/4)*(a*x^3 - 4*b))/(b*x^6)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.46 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^7} \, dx=- \frac {\sqrt [4]{a} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 x^{\frac {21}{4}} \Gamma \left (\frac {11}{4}\right )} \]

[In]

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

[Out]

-a**(1/4)*gamma(7/4)*hyper((-1/4, 7/4), (11/4,), b*exp_polar(2*I*pi)/(a*x**3))/(3*x**(21/4)*gamma(11/4))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.49 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^7} \, dx=\frac {{\left (a x^{3} - b\right )}^{\frac {5}{4}} a^{2} - 3 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2} b}{24 \, {\left ({\left (a x^{3} - b\right )}^{2} b + 2 \, {\left (a x^{3} - b\right )} b^{2} + b^{3}\right )}} + \frac {\frac {2 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {\sqrt {2} a^{2} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}}}{64 \, b} \]

[In]

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

[Out]

1/24*((a*x^3 - b)^(5/4)*a^2 - 3*(a*x^3 - b)^(1/4)*a^2*b)/((a*x^3 - b)^2*b + 2*(a*x^3 - b)*b^2 + b^3) + 1/64*(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*arcta
n(-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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.34 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^7} \, dx=\frac {\frac {6 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {7}{4}}} + \frac {6 \, \sqrt {2} a^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {7}{4}}} + \frac {3 \, \sqrt {2} a^{3} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {7}{4}}} - \frac {3 \, \sqrt {2} a^{3} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {7}{4}}} + \frac {8 \, {\left ({\left (a x^{3} - b\right )}^{\frac {5}{4}} a^{3} - 3 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{3} b\right )}}{a^{2} b x^{6}}}{192 \, a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.52 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^7} \, dx=\frac {{\left (a\,x^3-b\right )}^{5/4}}{24\,b\,x^6}-\frac {{\left (a\,x^3-b\right )}^{1/4}}{8\,x^6}+\frac {a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{16\,{\left (-b\right )}^{7/4}}-\frac {a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,1{}\mathrm {i}}{16\,{\left (-b\right )}^{7/4}} \]

[In]

int((a*x^3 - b)^(1/4)/x^7,x)

[Out]

(a*x^3 - b)^(5/4)/(24*b*x^6) - (a*x^3 - b)^(1/4)/(8*x^6) + (a^2*atan((a*x^3 - b)^(1/4)/(-b)^(1/4)))/(16*(-b)^(
7/4)) - (a^2*atan(((a*x^3 - b)^(1/4)*1i)/(-b)^(1/4))*1i)/(16*(-b)^(7/4))