\(\int \frac {1}{x^7 \sqrt [4]{-b+a x^3}} \, dx\) [2236]

Optimal result

Integrand size = 17, antiderivative size = 167 \[ \int \frac {1}{x^7 \sqrt [4]{-b+a x^3}} \, dx=\frac {\left (-b+a x^3\right )^{3/4} \left (4 b+5 a x^3\right )}{24 b^2 x^6}-\frac {5 a^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{-\sqrt {b}+\sqrt {-b+a x^3}}\right )}{48 \sqrt {2} b^{9/4}}-\frac {5 a^2 \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^3}}\right )}{48 \sqrt {2} b^{9/4}} \]

[Out]

1/24*(a*x^3-b)^(3/4)*(5*a*x^3+4*b)/b^2/x^6-5/96*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^(9/4)-5/96*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^(9/4)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.56, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {272, 44, 65, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {1}{x^7 \sqrt [4]{-b+a x^3}} \, dx=-\frac {5 a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{48 \sqrt {2} b^{9/4}}+\frac {5 a^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{48 \sqrt {2} b^{9/4}}+\frac {5 a^2 \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{96 \sqrt {2} b^{9/4}}-\frac {5 a^2 \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{96 \sqrt {2} b^{9/4}}+\frac {5 a \left (a x^3-b\right )^{3/4}}{24 b^2 x^3}+\frac {\left (a x^3-b\right )^{3/4}}{6 b x^6} \]

[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 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 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 303

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 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 {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) \text {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 ) \text {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) \text {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) \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 )}{48 b^2}+\frac {(5 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 )}{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 ) \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 )}{96 \sqrt {2} b^{9/4}}+\frac {\left (5 a^2\right ) \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 )}{96 \sqrt {2} b^{9/4}}+\frac {\left (5 a^2\right ) \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 )}{96 b^2}+\frac {\left (5 a^2\right ) \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 )}{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 ) \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 )}{48 \sqrt {2} b^{9/4}}-\frac {\left (5 a^2\right ) \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 )}{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 \arctan \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 \arctan \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*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^7 \sqrt [4]{-b+a x^3}} \, dx=\frac {4 \sqrt [4]{b} \left (-b+a x^3\right )^{3/4} \left (4 b+5 a x^3\right )+5 \sqrt {2} a^2 x^6 \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}\right )-5 \sqrt {2} a^2 x^6 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{\sqrt {b}+\sqrt {-b+a x^3}}\right )}{96 b^{9/4} x^6} \]

[In]

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

[Out]

(4*b^(1/4)*(-b + a*x^3)^(3/4)*(4*b + 5*a*x^3) + 5*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))] - 5*Sqrt[2]*a^2*x^6*ArcTanh[(Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4))/(Sqrt[b] + Sqr
t[-b + a*x^3])])/(96*b^(9/4)*x^6)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.19

method	result	size
pseudoelliptic	\(\frac {5 \sqrt {2}\, \ln \left (\frac {\sqrt {a \,x^{3}-b}-b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}+\sqrt {b}}{\sqrt {a \,x^{3}-b}+b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}+\sqrt {b}}\right ) a^{2} x^{6}+10 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a^{2} x^{6}-10 \arctan \left (\frac {-\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a^{2} x^{6}+40 a \,x^{3} b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {3}{4}}+32 b^{\frac {5}{4}} \left (a \,x^{3}-b \right )^{\frac {3}{4}}}{192 b^{\frac {9}{4}} x^{6}}\)	\(199\)

[In]

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

[Out]

1/192/b^(9/4)*(5*2^(1/2)*ln(((a*x^3-b)^(1/2)-b^(1/4)*(a*x^3-b)^(1/4)*2^(1/2)+b^(1/2))/((a*x^3-b)^(1/2)+b^(1/4)
*(a*x^3-b)^(1/4)*2^(1/2)+b^(1/2)))*a^2*x^6+10*arctan((2^(1/2)*(a*x^3-b)^(1/4)+b^(1/4))/b^(1/4))*2^(1/2)*a^2*x^
6-10*arctan((-2^(1/2)*(a*x^3-b)^(1/4)+b^(1/4))/b^(1/4))*2^(1/2)*a^2*x^6+40*a*x^3*b^(1/4)*(a*x^3-b)^(3/4)+32*b^
(5/4)*(a*x^3-b)^(3/4))/x^6

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.41 \[ \int \frac {1}{x^7 \sqrt [4]{-b+a x^3}} \, dx=\frac {5 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (125 \, b^{7} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{6}\right ) - 5 i \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (125 i \, b^{7} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{6}\right ) + 5 i \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-125 i \, b^{7} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{6}\right ) - 5 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-125 \, b^{7} \left (-\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{6}\right ) + 4 \, {\left (5 \, a x^{3} + 4 \, b\right )} {\left (a x^{3} - b\right )}^{\frac {3}{4}}}{96 \, b^{2} x^{6}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.79 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.25 \[ \int \frac {1}{x^7 \sqrt [4]{-b+a x^3}} \, dx=- \frac {\Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 \sqrt [4]{a} x^{\frac {27}{4}} \Gamma \left (\frac {13}{4}\right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.44 \[ \int \frac {1}{x^7 \sqrt [4]{-b+a x^3}} \, dx=\frac {5 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}}\right )} a^{2}}{192 \, b^{2}} + \frac {5 \, {\left (a x^{3} - b\right )}^{\frac {7}{4}} a^{2} + 9 \, {\left (a x^{3} - b\right )}^{\frac {3}{4}} a^{2} b}{24 \, {\left ({\left (a x^{3} - b\right )}^{2} b^{2} + 2 \, {\left (a x^{3} - b\right )} b^{3} + b^{4}\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x^7 \sqrt [4]{-b+a x^3}} \, dx=\frac {\frac {10 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {9}{4}}} + \frac {10 \, \sqrt {2} a^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {9}{4}}} - \frac {5 \, \sqrt {2} a^{3} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {9}{4}}} + \frac {5 \, \sqrt {2} a^{3} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {9}{4}}} + \frac {8 \, {\left (5 \, {\left (a x^{3} - b\right )}^{\frac {7}{4}} a^{3} + 9 \, {\left (a x^{3} - b\right )}^{\frac {3}{4}} a^{3} b\right )}}{a^{2} b^{2} x^{6}}}{192 \, a} \]

[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

Mupad [B] (verification not implemented)

Time = 6.60 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^7 \sqrt [4]{-b+a x^3}} \, dx=\frac {3\,{\left (a\,x^3-b\right )}^{3/4}}{8\,b\,x^6}+\frac {5\,{\left (a\,x^3-b\right )}^{7/4}}{24\,b^2\,x^6}+\frac {5\,a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{48\,{\left (-b\right )}^{9/4}}+\frac {a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,5{}\mathrm {i}}{48\,{\left (-b\right )}^{9/4}} \]

[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))