Integrand size = 17, antiderivative size = 167 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {\sqrt [4]{-b+a x^3} \left (4 b+7 a x^3\right )}{24 b^2 x^6}-\frac {7 a^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{-\sqrt {b}+\sqrt {-b+a x^3}}\right )}{16 \sqrt {2} b^{11/4}}+\frac {7 a^2 \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^3}}\right )}{16 \sqrt {2} b^{11/4}} \]
[Out]
Time = 0.16 (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} \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=-\frac {7 a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{16 \sqrt {2} b^{11/4}}+\frac {7 a^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{16 \sqrt {2} b^{11/4}}-\frac {7 a^2 \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{32 \sqrt {2} b^{11/4}}+\frac {7 a^2 \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{32 \sqrt {2} b^{11/4}}+\frac {7 a \sqrt [4]{a x^3-b}}{24 b^2 x^3}+\frac {\sqrt [4]{a x^3-b}}{6 b x^6} \]
[In]
[Out]
Rule 44
Rule 65
Rule 210
Rule 217
Rule 272
Rule 631
Rule 642
Rule 1176
Rule 1179
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {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) \text {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 ) \text {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) \text {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) \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^{5/2}}+\frac {(7 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^{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 ) \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^{11/4}}-\frac {\left (7 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 )}{32 \sqrt {2} b^{11/4}}+\frac {\left (7 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 )}{32 b^{5/2}}+\frac {\left (7 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 )}{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 ) \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^{11/4}}-\frac {\left (7 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 )}{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 \arctan \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 \arctan \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*}
Time = 0.26 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {4 b^{3/4} \sqrt [4]{-b+a x^3} \left (4 b+7 a x^3\right )+21 \sqrt {2} a^2 x^6 \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}\right )+21 \sqrt {2} a^2 x^6 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{\sqrt {b}+\sqrt {-b+a x^3}}\right )}{96 b^{11/4} x^6} \]
[In]
[Out]
Time = 0.27 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.24
method | result | size |
pseudoelliptic | \(\frac {21 \ln \left (\frac {-b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{3}-b}-\sqrt {b}}{b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{3}-b}-\sqrt {b}}\right ) \sqrt {2}\, a^{2} x^{6}+42 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a^{2} x^{6}-42 \arctan \left (\frac {-\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a^{2} x^{6}+56 a \,x^{3} \left (a \,x^{3}-b \right )^{\frac {1}{4}} b^{\frac {3}{4}}+32 b^{\frac {7}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}}}{192 b^{\frac {11}{4}} x^{6}}\) | \(207\) |
[In]
[Out]
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.41 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {21 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (7 \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2}\right ) + 21 i \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (7 i \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2}\right ) - 21 i \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-7 i \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2}\right ) - 21 \, b^{2} x^{6} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-7 \, b^{3} \left (-\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2}\right ) + 4 \, {\left (7 \, a x^{3} + 4 \, b\right )} {\left (a x^{3} - b\right )}^{\frac {1}{4}}}{96 \, b^{2} x^{6}} \]
[In]
[Out]
Result contains complex when optimal does not.
Time = 1.82 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.25 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=- \frac {\Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 a^{\frac {3}{4}} x^{\frac {33}{4}} \Gamma \left (\frac {15}{4}\right )} \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.50 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {7 \, {\left (a x^{3} - b\right )}^{\frac {5}{4}} a^{2} + 11 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{2} b}{24 \, {\left ({\left (a x^{3} - b\right )}^{2} b^{2} + 2 \, {\left (a x^{3} - b\right )} b^{3} + b^{4}\right )}} + \frac {7 \, {\left (\frac {2 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {\sqrt {2} a^{2} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}}\right )}}{64 \, b^{2}} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {\frac {42 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {11}{4}}} + \frac {42 \, \sqrt {2} a^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {11}{4}}} + \frac {21 \, \sqrt {2} a^{3} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {11}{4}}} - \frac {21 \, \sqrt {2} a^{3} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {11}{4}}} + \frac {8 \, {\left (7 \, {\left (a x^{3} - b\right )}^{\frac {5}{4}} a^{3} + 11 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a^{3} b\right )}}{a^{2} b^{2} x^{6}}}{192 \, a} \]
[In]
[Out]
Time = 6.78 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^7 \left (-b+a x^3\right )^{3/4}} \, dx=\frac {11\,{\left (a\,x^3-b\right )}^{1/4}}{24\,b\,x^6}+\frac {7\,{\left (a\,x^3-b\right )}^{5/4}}{24\,b^2\,x^6}-\frac {7\,a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{16\,{\left (-b\right )}^{11/4}}+\frac {a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,7{}\mathrm {i}}{16\,{\left (-b\right )}^{11/4}} \]
[In]
[Out]