Integrand size = 17, antiderivative size = 167 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {\sqrt [4]{-b+a x^3} \left (4 b+7 a x^3\right )}{24 b^2 x^6}-\frac {7 a^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{-\sqrt {b}+\sqrt {-b+a x^3}}\right )}{16 \sqrt {2} b^{11/4}}+\frac {7 a^2 \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^3}}\right )}{16 \sqrt {2} b^{11/4}} \]
1/24*(a*x^3-b)^(1/4)*(7*a*x^3+4*b)/b^2/x^6-7/32*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^(11/4)+7/32*a^2*arc tanh((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^(11/4)
Time = 0.26 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {4 b^{3/4} \sqrt [4]{-b+a x^3} \left (4 b+7 a x^3\right )+21 \sqrt {2} a^2 x^6 \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}\right )+21 \sqrt {2} a^2 x^6 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{\sqrt {b}+\sqrt {-b+a x^3}}\right )}{96 b^{11/4} x^6} \]
(4*b^(3/4)*(-b + a*x^3)^(1/4)*(4*b + 7*a*x^3) + 21*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))] + 21*S qrt[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^(11/4)*x^6)
Time = 0.40 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.56, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.765, Rules used = {798, 52, 52, 73, 755, 27, 1476, 1082, 217, 1479, 25, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{x^7 \left (a x^3-b\right )^{3/4}} \, dx\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle \frac {1}{3} \int \frac {1}{x^9 \left (a x^3-b\right )^{3/4}}dx^3\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \int \frac {1}{x^6 \left (a x^3-b\right )^{3/4}}dx^3}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \left (\frac {3 a \int \frac {1}{x^3 \left (a x^3-b\right )^{3/4}}dx^3}{4 b}+\frac {\sqrt [4]{a x^3-b}}{b x^3}\right )}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \left (\frac {3 \int \frac {1}{\frac {x^{12}}{a}+\frac {b}{a}}d\sqrt [4]{a x^3-b}}{b}+\frac {\sqrt [4]{a x^3-b}}{b x^3}\right )}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \left (\frac {3 \left (\frac {\int \frac {a \left (\sqrt {b}-x^6\right )}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}+\frac {\int \frac {a \left (x^6+\sqrt {b}\right )}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^3-b}}{b x^3}\right )}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \left (\frac {3 \left (\frac {a \int \frac {\sqrt {b}-x^6}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}+\frac {a \int \frac {x^6+\sqrt {b}}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^3-b}}{b x^3}\right )}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \left (\frac {3 \left (\frac {a \left (\frac {1}{2} \int \frac {1}{x^6+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}+\frac {1}{2} \int \frac {1}{x^6+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}\right )}{2 \sqrt {b}}+\frac {a \int \frac {\sqrt {b}-x^6}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^3-b}}{b x^3}\right )}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \left (\frac {3 \left (\frac {a \left (\frac {\int \frac {1}{-x^6-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\int \frac {1}{-x^6-1}d\left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \int \frac {\sqrt {b}-x^6}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^3-b}}{b x^3}\right )}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \left (\frac {3 \left (\frac {a \int \frac {\sqrt {b}-x^6}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^3-b}}{b x^3}\right )}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \left (\frac {3 \left (\frac {a \left (-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^3-b}}{x^6+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt {2} \sqrt [4]{b}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^3-b}\right )}{x^6+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^3-b}}{b x^3}\right )}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \left (\frac {3 \left (\frac {a \left (\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^3-b}}{x^6+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt {2} \sqrt [4]{b}}+\frac {\int \frac {\sqrt {2} \left (\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^3-b}\right )}{x^6+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^3-b}}{b x^3}\right )}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \left (\frac {3 \left (\frac {a \left (\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^3-b}}{x^6+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt {2} \sqrt [4]{b}}+\frac {\int \frac {\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^3-b}}{x^6+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^3-b}}{b x^3}\right )}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{3} \left (\frac {7 a \left (\frac {3 \left (\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {b}+x^6\right )}{2 \sqrt {2} \sqrt [4]{b}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {b}+x^6\right )}{2 \sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^3-b}}{b x^3}\right )}{8 b}+\frac {\sqrt [4]{a x^3-b}}{2 b x^6}\right )\) |
((-b + a*x^3)^(1/4)/(2*b*x^6) + (7*a*((-b + a*x^3)^(1/4)/(b*x^3) + (3*((a* (-(ArcTan[1 - (Sqrt[2]*(-b + a*x^3)^(1/4))/b^(1/4)]/(Sqrt[2]*b^(1/4))) + A rcTan[1 + (Sqrt[2]*(-b + a*x^3)^(1/4))/b^(1/4)]/(Sqrt[2]*b^(1/4))))/(2*Sqr t[b]) + (a*(-1/2*Log[Sqrt[b] + x^6 - Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4)]/( Sqrt[2]*b^(1/4)) + Log[Sqrt[b] + x^6 + Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4)] /(2*Sqrt[2]*b^(1/4))))/(2*Sqrt[b])))/b))/(8*b))/3
3.23.35.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[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]]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Time = 0.27 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.24
| method | result | size |
| pseudoelliptic | \(\frac {21 \ln \left (\frac {-b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{3}-b}-\sqrt {b}}{b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{3}-b}-\sqrt {b}}\right ) \sqrt {2}\, a^{2} x^{6}+42 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a^{2} x^{6}-42 \arctan \left (\frac {-\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a^{2} x^{6}+56 a \,x^{3} \left (a \,x^{3}-b \right )^{\frac {1}{4}} b^{\frac {3}{4}}+32 b^{\frac {7}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}}}{192 b^{\frac {11}{4}} x^{6}}\) | \(207\) |
1/192/b^(11/4)*(21*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+42*arctan((2^(1/2)*(a*x^3-b)^(1/4)+b^(1/4))/b^(1/4))*2^(1/2)*a^2*x ^6-42*arctan((-2^(1/2)*(a*x^3-b)^(1/4)+b^(1/4))/b^(1/4))*2^(1/2)*a^2*x^6+5 6*a*x^3*(a*x^3-b)^(1/4)*b^(3/4)+32*b^(7/4)*(a*x^3-b)^(1/4))/x^6
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.41 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {21 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (7 \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2}\right ) + 21 i \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (7 i \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2}\right ) - 21 i \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-7 i \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2}\right ) - 21 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-7 \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2}\right ) + 4 \, {\left (7 \, a x^{3} + 4 \, b\right )} {\left (a x^{3} - b\right )}^{\frac {1}{4}}}{96 \, b^{2} x^{6}} \]
1/96*(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*I*b^2*x^6*(-a^8/b^11)^(1/4)*log(7*I*b^3*(-a^8/b^11)^( 1/4) + 7*(a*x^3 - b)^(1/4)*a^2) - 21*I*b^2*x^6*(-a^8/b^11)^(1/4)*log(-7*I* 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)
Result contains complex when optimal does not.
Time = 1.82 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.25 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=- \frac {\Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 a^{\frac {3}{4}} x^{\frac {33}{4}} \Gamma \left (\frac {15}{4}\right )} \]
-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))
Time = 0.28 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.50 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {7 \, {\left (a x^{3} - b\right )}^{\frac {5}{4}} a^{2} + 11 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2} b}{24 \, {\left ({\left (a x^{3} - b\right )}^{2} b^{2} + 2 \, {\left (a x^{3} - b\right )} b^{3} + b^{4}\right )}} + \frac {7 \, {\left (\frac {2 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {\sqrt {2} a^{2} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}}\right )}}{64 \, b^{2}} \]
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*ar ctan(-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) + s qrt(b))/b^(3/4) - sqrt(2)*a^2*log(-sqrt(2)*(a*x^3 - b)^(1/4)*b^(1/4) + sqr t(a*x^3 - b) + sqrt(b))/b^(3/4))/b^2
Time = 0.27 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {\frac {42 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {11}{4}}} + \frac {42 \, \sqrt {2} a^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {11}{4}}} + \frac {21 \, \sqrt {2} a^{3} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {11}{4}}} - \frac {21 \, \sqrt {2} a^{3} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {11}{4}}} + \frac {8 \, {\left (7 \, {\left (a x^{3} - b\right )}^{\frac {5}{4}} a^{3} + 11 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{3} b\right )}}{a^{2} b^{2} x^{6}}}{192 \, a} \]
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
Time = 6.78 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {11\,{\left (a\,x^3-b\right )}^{1/4}}{24\,b\,x^6}+\frac {7\,{\left (a\,x^3-b\right )}^{5/4}}{24\,b^2\,x^6}-\frac {7\,a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{16\,{\left (-b\right )}^{11/4}}+\frac {a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,7{}\mathrm {i}}{16\,{\left (-b\right )}^{11/4}} \]