\(\int \frac {1}{x^{11} (-b+a x^5)^{3/4}} \, dx\) [2241]

Optimal result

Integrand size = 17, antiderivative size = 167 \[ \int \frac {1}{x^{11} \left (-b+a x^5\right )^{3/4}} \, dx=\frac {\sqrt [4]{-b+a x^5} \left (4 b+7 a x^5\right )}{40 b^2 x^{10}}-\frac {21 a^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}}{-\sqrt {b}+\sqrt {-b+a x^5}}\right )}{80 \sqrt {2} b^{11/4}}+\frac {21 a^2 \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^5}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^5}}\right )}{80 \sqrt {2} b^{11/4}} \]

[Out]

1/40*(a*x^5-b)^(1/4)*(7*a*x^5+4*b)/b^2/x^10-21/160*a^2*arctan(2^(1/2)*b^(1/4)*(a*x^5-b)^(1/4)/(-b^(1/2)+(a*x^5
-b)^(1/2)))*2^(1/2)/b^(11/4)+21/160*a^2*arctanh((1/2*b^(1/4)*2^(1/2)+1/2*(a*x^5-b)^(1/2)*2^(1/2)/b^(1/4))/(a*x
^5-b)^(1/4))*2^(1/2)/b^(11/4)

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, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{x^{11} \left (-b+a x^5\right )^{3/4}} \, dx=-\frac {21 a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^5-b}}{\sqrt [4]{b}}\right )}{80 \sqrt {2} b^{11/4}}+\frac {21 a^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^5-b}}{\sqrt [4]{b}}+1\right )}{80 \sqrt {2} b^{11/4}}-\frac {21 a^2 \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^5-b}+\sqrt {a x^5-b}+\sqrt {b}\right )}{160 \sqrt {2} b^{11/4}}+\frac {21 a^2 \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^5-b}+\sqrt {a x^5-b}+\sqrt {b}\right )}{160 \sqrt {2} b^{11/4}}+\frac {7 a \sqrt [4]{a x^5-b}}{40 b^2 x^5}+\frac {\sqrt [4]{a x^5-b}}{10 b x^{10}} \]

[In]

Int[1/(x^11*(-b + a*x^5)^(3/4)),x]

[Out]

(-b + a*x^5)^(1/4)/(10*b*x^10) + (7*a*(-b + a*x^5)^(1/4))/(40*b^2*x^5) - (21*a^2*ArcTan[1 - (Sqrt[2]*(-b + a*x
^5)^(1/4))/b^(1/4)])/(80*Sqrt[2]*b^(11/4)) + (21*a^2*ArcTan[1 + (Sqrt[2]*(-b + a*x^5)^(1/4))/b^(1/4)])/(80*Sqr
t[2]*b^(11/4)) - (21*a^2*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*(-b + a*x^5)^(1/4) + Sqrt[-b + a*x^5]])/(160*Sqrt[2]*b^
(11/4)) + (21*a^2*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*(-b + a*x^5)^(1/4) + Sqrt[-b + a*x^5]])/(160*Sqrt[2]*b^(11/4))

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 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}{5} \text {Subst}\left (\int \frac {1}{x^3 (-b+a x)^{3/4}} \, dx,x,x^5\right ) \\ & = \frac {\sqrt [4]{-b+a x^5}}{10 b x^{10}}+\frac {(7 a) \text {Subst}\left (\int \frac {1}{x^2 (-b+a x)^{3/4}} \, dx,x,x^5\right )}{40 b} \\ & = \frac {\sqrt [4]{-b+a x^5}}{10 b x^{10}}+\frac {7 a \sqrt [4]{-b+a x^5}}{40 b^2 x^5}+\frac {\left (21 a^2\right ) \text {Subst}\left (\int \frac {1}{x (-b+a x)^{3/4}} \, dx,x,x^5\right )}{160 b^2} \\ & = \frac {\sqrt [4]{-b+a x^5}}{10 b x^{10}}+\frac {7 a \sqrt [4]{-b+a x^5}}{40 b^2 x^5}+\frac {(21 a) \text {Subst}\left (\int \frac {1}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^5}\right )}{40 b^2} \\ & = \frac {\sqrt [4]{-b+a x^5}}{10 b x^{10}}+\frac {7 a \sqrt [4]{-b+a x^5}}{40 b^2 x^5}+\frac {(21 a) \text {Subst}\left (\int \frac {\sqrt {b}-x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^5}\right )}{80 b^{5/2}}+\frac {(21 a) \text {Subst}\left (\int \frac {\sqrt {b}+x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^5}\right )}{80 b^{5/2}} \\ & = \frac {\sqrt [4]{-b+a x^5}}{10 b x^{10}}+\frac {7 a \sqrt [4]{-b+a x^5}}{40 b^2 x^5}-\frac {\left (21 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^5}\right )}{160 \sqrt {2} b^{11/4}}-\frac {\left (21 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^5}\right )}{160 \sqrt {2} b^{11/4}}+\frac {\left (21 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^5}\right )}{160 b^{5/2}}+\frac {\left (21 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^5}\right )}{160 b^{5/2}} \\ & = \frac {\sqrt [4]{-b+a x^5}}{10 b x^{10}}+\frac {7 a \sqrt [4]{-b+a x^5}}{40 b^2 x^5}-\frac {21 a^2 \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}+\sqrt {-b+a x^5}\right )}{160 \sqrt {2} b^{11/4}}+\frac {21 a^2 \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}+\sqrt {-b+a x^5}\right )}{160 \sqrt {2} b^{11/4}}+\frac {\left (21 a^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{-b+a x^5}}{\sqrt [4]{b}}\right )}{80 \sqrt {2} b^{11/4}}-\frac {\left (21 a^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{-b+a x^5}}{\sqrt [4]{b}}\right )}{80 \sqrt {2} b^{11/4}} \\ & = \frac {\sqrt [4]{-b+a x^5}}{10 b x^{10}}+\frac {7 a \sqrt [4]{-b+a x^5}}{40 b^2 x^5}-\frac {21 a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-b+a x^5}}{\sqrt [4]{b}}\right )}{80 \sqrt {2} b^{11/4}}+\frac {21 a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-b+a x^5}}{\sqrt [4]{b}}\right )}{80 \sqrt {2} b^{11/4}}-\frac {21 a^2 \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}+\sqrt {-b+a x^5}\right )}{160 \sqrt {2} b^{11/4}}+\frac {21 a^2 \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}+\sqrt {-b+a x^5}\right )}{160 \sqrt {2} b^{11/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^{11} \left (-b+a x^5\right )^{3/4}} \, dx=\frac {4 b^{3/4} \sqrt [4]{-b+a x^5} \left (4 b+7 a x^5\right )+21 \sqrt {2} a^2 x^{10} \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^5}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}}\right )+21 \sqrt {2} a^2 x^{10} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^5}}{\sqrt {b}+\sqrt {-b+a x^5}}\right )}{160 b^{11/4} x^{10}} \]

[In]

Integrate[1/(x^11*(-b + a*x^5)^(3/4)),x]

[Out]

(4*b^(3/4)*(-b + a*x^5)^(1/4)*(4*b + 7*a*x^5) + 21*Sqrt[2]*a^2*x^10*ArcTan[(-Sqrt[b] + Sqrt[-b + a*x^5])/(Sqrt
[2]*b^(1/4)*(-b + a*x^5)^(1/4))] + 21*Sqrt[2]*a^2*x^10*ArcTanh[(Sqrt[2]*b^(1/4)*(-b + a*x^5)^(1/4))/(Sqrt[b] +
Sqrt[-b + a*x^5])])/(160*b^(11/4)*x^10)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.24

method	result	size
pseudoelliptic	\(\frac {21 \sqrt {2}\, \ln \left (\frac {-b^{\frac {1}{4}} \left (a \,x^{5}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{5}-b}-\sqrt {b}}{b^{\frac {1}{4}} \left (a \,x^{5}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{5}-b}-\sqrt {b}}\right ) a^{2} x^{10}+42 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{5}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) a^{2} x^{10}-42 \sqrt {2}\, \arctan \left (\frac {-\sqrt {2}\, \left (a \,x^{5}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) a^{2} x^{10}+56 a \,x^{5} \left (a \,x^{5}-b \right )^{\frac {1}{4}} b^{\frac {3}{4}}+32 b^{\frac {7}{4}} \left (a \,x^{5}-b \right )^{\frac {1}{4}}}{320 b^{\frac {11}{4}} x^{10}}\)	\(207\)

[In]

int(1/x^11/(a*x^5-b)^(3/4),x,method=_RETURNVERBOSE)

[Out]

1/320/b^(11/4)*(21*2^(1/2)*ln((-b^(1/4)*(a*x^5-b)^(1/4)*2^(1/2)-(a*x^5-b)^(1/2)-b^(1/2))/(b^(1/4)*(a*x^5-b)^(1
/4)*2^(1/2)-(a*x^5-b)^(1/2)-b^(1/2)))*a^2*x^10+42*2^(1/2)*arctan((2^(1/2)*(a*x^5-b)^(1/4)+b^(1/4))/b^(1/4))*a^
2*x^10-42*2^(1/2)*arctan((-2^(1/2)*(a*x^5-b)^(1/4)+b^(1/4))/b^(1/4))*a^2*x^10+56*a*x^5*(a*x^5-b)^(1/4)*b^(3/4)
+32*b^(7/4)*(a*x^5-b)^(1/4))/x^10

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.41 \[ \int \frac {1}{x^{11} \left (-b+a x^5\right )^{3/4}} \, dx=\frac {21 \, b^{2} x^{10} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (21 \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}} a^{2}\right ) + 21 i \, b^{2} x^{10} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (21 i \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}} a^{2}\right ) - 21 i \, b^{2} x^{10} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-21 i \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}} a^{2}\right ) - 21 \, b^{2} x^{10} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-21 \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}} a^{2}\right ) + 4 \, {\left (7 \, a x^{5} + 4 \, b\right )} {\left (a x^{5} - b\right )}^{\frac {1}{4}}}{160 \, b^{2} x^{10}} \]

[In]

integrate(1/x^11/(a*x^5-b)^(3/4),x, algorithm="fricas")

[Out]

1/160*(21*b^2*x^10*(-a^8/b^11)^(1/4)*log(21*b^3*(-a^8/b^11)^(1/4) + 21*(a*x^5 - b)^(1/4)*a^2) + 21*I*b^2*x^10*
(-a^8/b^11)^(1/4)*log(21*I*b^3*(-a^8/b^11)^(1/4) + 21*(a*x^5 - b)^(1/4)*a^2) - 21*I*b^2*x^10*(-a^8/b^11)^(1/4)
*log(-21*I*b^3*(-a^8/b^11)^(1/4) + 21*(a*x^5 - b)^(1/4)*a^2) - 21*b^2*x^10*(-a^8/b^11)^(1/4)*log(-21*b^3*(-a^8
/b^11)^(1/4) + 21*(a*x^5 - b)^(1/4)*a^2) + 4*(7*a*x^5 + 4*b)*(a*x^5 - b)^(1/4))/(b^2*x^10)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.25 \[ \int \frac {1}{x^{11} \left (-b+a x^5\right )^{3/4}} \, dx=- \frac {\Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{5}}} \right )}}{5 a^{\frac {3}{4}} x^{\frac {55}{4}} \Gamma \left (\frac {15}{4}\right )} \]

[In]

integrate(1/x**11/(a*x**5-b)**(3/4),x)

[Out]

-gamma(11/4)*hyper((3/4, 11/4), (15/4,), b*exp_polar(2*I*pi)/(a*x**5))/(5*a**(3/4)*x**(55/4)*gamma(15/4))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.50 \[ \int \frac {1}{x^{11} \left (-b+a x^5\right )^{3/4}} \, dx=\frac {7 \, {\left (a x^{5} - b\right )}^{\frac {5}{4}} a^{2} + 11 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}} a^{2} b}{40 \, {\left ({\left (a x^{5} - b\right )}^{2} b^{2} + 2 \, {\left (a x^{5} - b\right )} b^{3} + b^{4}\right )}} + \frac {21 \, {\left (\frac {2 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {\sqrt {2} a^{2} \log \left (\sqrt {2} {\left (a x^{5} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{5} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{5} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{5} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}}\right )}}{320 \, b^{2}} \]

[In]

integrate(1/x^11/(a*x^5-b)^(3/4),x, algorithm="maxima")

[Out]

1/40*(7*(a*x^5 - b)^(5/4)*a^2 + 11*(a*x^5 - b)^(1/4)*a^2*b)/((a*x^5 - b)^2*b^2 + 2*(a*x^5 - b)*b^3 + b^4) + 21
/320*(2*sqrt(2)*a^2*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^5 - 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^5 - b)^(1/4))/b^(1/4))/b^(3/4) + sqrt(2)*a^2*log(sqrt(2)*(a*x^
5 - b)^(1/4)*b^(1/4) + sqrt(a*x^5 - b) + sqrt(b))/b^(3/4) - sqrt(2)*a^2*log(-sqrt(2)*(a*x^5 - b)^(1/4)*b^(1/4)
+ sqrt(a*x^5 - b) + sqrt(b))/b^(3/4))/b^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x^{11} \left (-b+a x^5\right )^{3/4}} \, dx=\frac {\frac {42 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {11}{4}}} + \frac {42 \, \sqrt {2} a^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {11}{4}}} + \frac {21 \, \sqrt {2} a^{3} \log \left (\sqrt {2} {\left (a x^{5} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{5} - b} + \sqrt {b}\right )}{b^{\frac {11}{4}}} - \frac {21 \, \sqrt {2} a^{3} \log \left (-\sqrt {2} {\left (a x^{5} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{5} - b} + \sqrt {b}\right )}{b^{\frac {11}{4}}} + \frac {8 \, {\left (7 \, {\left (a x^{5} - b\right )}^{\frac {5}{4}} a^{3} + 11 \, {\left (a x^{5} - b\right )}^{\frac {1}{4}} a^{3} b\right )}}{a^{2} b^{2} x^{10}}}{320 \, a} \]

[In]

integrate(1/x^11/(a*x^5-b)^(3/4),x, algorithm="giac")

[Out]

1/320*(42*sqrt(2)*a^3*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^5 - 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^5 - b)^(1/4))/b^(1/4))/b^(11/4) + 21*sqrt(2)*a^3*log(sqrt(
2)*(a*x^5 - b)^(1/4)*b^(1/4) + sqrt(a*x^5 - b) + sqrt(b))/b^(11/4) - 21*sqrt(2)*a^3*log(-sqrt(2)*(a*x^5 - b)^(
1/4)*b^(1/4) + sqrt(a*x^5 - b) + sqrt(b))/b^(11/4) + 8*(7*(a*x^5 - b)^(5/4)*a^3 + 11*(a*x^5 - b)^(1/4)*a^3*b)/
(a^2*b^2*x^10))/a

Mupad [B] (verification not implemented)

Time = 6.01 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^{11} \left (-b+a x^5\right )^{3/4}} \, dx=\frac {11\,{\left (a\,x^5-b\right )}^{1/4}}{40\,b\,x^{10}}+\frac {7\,{\left (a\,x^5-b\right )}^{5/4}}{40\,b^2\,x^{10}}-\frac {21\,a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^5-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{80\,{\left (-b\right )}^{11/4}}+\frac {a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^5-b\right )}^{1/4}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,21{}\mathrm {i}}{80\,{\left (-b\right )}^{11/4}} \]

[In]

int(1/(x^11*(a*x^5 - b)^(3/4)),x)

[Out]

(11*(a*x^5 - b)^(1/4))/(40*b*x^10) + (7*(a*x^5 - b)^(5/4))/(40*b^2*x^10) - (21*a^2*atan((a*x^5 - b)^(1/4)/(-b)
^(1/4)))/(80*(-b)^(11/4)) + (a^2*atan(((a*x^5 - b)^(1/4)*1i)/(-b)^(1/4))*21i)/(80*(-b)^(11/4))