Integrand size = 17, antiderivative size = 139 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=\frac {(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {5 (3-x)^2 (1+x)}{64 \left (9+3 x-5 x^2+x^3\right )^{3/2}}-\frac {15 (3-x)^3 (1+x)}{256 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {15 (3-x)^3 (1+x)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+x}}{2}\right )}{512 \left (9+3 x-5 x^2+x^3\right )^{3/2}} \]
[Out]
Time = 0.09 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2092, 2089, 44, 53, 65, 212} \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=\frac {15 (x+1)^{3/2} (3-x)^3 \text {arctanh}\left (\frac {\sqrt {x+1}}{2}\right )}{512 \left (x^3-5 x^2+3 x+9\right )^{3/2}}-\frac {15 (x+1) (3-x)^3}{256 \left (x^3-5 x^2+3 x+9\right )^{3/2}}+\frac {5 (x+1) (3-x)^2}{64 \left (x^3-5 x^2+3 x+9\right )^{3/2}}+\frac {(x+1) (3-x)}{8 \left (x^3-5 x^2+3 x+9\right )^{3/2}} \]
[In]
[Out]
Rule 44
Rule 53
Rule 65
Rule 212
Rule 2089
Rule 2092
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{\left (\frac {128}{27}-\frac {16 x}{3}+x^3\right )^{3/2}} \, dx,x,-\frac {5}{3}+x\right ) \\ & = \frac {\left (2097152 (3-x)^3 (1+x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {128}{9}-\frac {32 x}{3}\right )^3 \left (\frac {128}{9}+\frac {16 x}{3}\right )^{3/2}} \, dx,x,-\frac {5}{3}+x\right )}{81 \sqrt {3} \left (9+3 x-5 x^2+x^3\right )^{3/2}} \\ & = \frac {(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {\left (20480 (3-x)^3 (1+x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {128}{9}-\frac {32 x}{3}\right )^2 \left (\frac {128}{9}+\frac {16 x}{3}\right )^{3/2}} \, dx,x,-\frac {5}{3}+x\right )}{27 \sqrt {3} \left (9+3 x-5 x^2+x^3\right )^{3/2}} \\ & = \frac {(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {5 (3-x)^2 (1+x)}{64 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {\left (80 (3-x)^3 (1+x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {128}{9}-\frac {32 x}{3}\right ) \left (\frac {128}{9}+\frac {16 x}{3}\right )^{3/2}} \, dx,x,-\frac {5}{3}+x\right )}{3 \sqrt {3} \left (9+3 x-5 x^2+x^3\right )^{3/2}} \\ & = \frac {(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {5 (3-x)^2 (1+x)}{64 \left (9+3 x-5 x^2+x^3\right )^{3/2}}-\frac {15 (3-x)^3 (1+x)}{256 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {\left (5 (3-x)^3 (1+x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {128}{9}-\frac {32 x}{3}\right ) \sqrt {\frac {128}{9}+\frac {16 x}{3}}} \, dx,x,-\frac {5}{3}+x\right )}{4 \sqrt {3} \left (9+3 x-5 x^2+x^3\right )^{3/2}} \\ & = \frac {(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {5 (3-x)^2 (1+x)}{64 \left (9+3 x-5 x^2+x^3\right )^{3/2}}-\frac {15 (3-x)^3 (1+x)}{256 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {\left (5 \sqrt {3} (3-x)^3 (1+x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\frac {128}{3}-2 x^2} \, dx,x,\frac {4 \sqrt {1+x}}{\sqrt {3}}\right )}{32 \left (9+3 x-5 x^2+x^3\right )^{3/2}} \\ & = \frac {(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {5 (3-x)^2 (1+x)}{64 \left (9+3 x-5 x^2+x^3\right )^{3/2}}-\frac {15 (3-x)^3 (1+x)}{256 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {15 (3-x)^3 (1+x)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+x}}{2}\right )}{512 \left (9+3 x-5 x^2+x^3\right )^{3/2}} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=\frac {86-140 x+30 x^2-15 (-3+x)^2 \sqrt {1+x} \text {arctanh}\left (\frac {\sqrt {1+x}}{2}\right )}{512 (-3+x) \sqrt {(-3+x)^2 (1+x)}} \]
[In]
[Out]
Time = 0.16 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.51
method | result | size |
risch | \(\frac {15 x^{2}-70 x +43}{256 \left (-3+x \right ) \sqrt {\left (1+x \right ) \left (-3+x \right )^{2}}}+\frac {\left (\frac {15 \ln \left (\sqrt {1+x}-2\right )}{1024}-\frac {15 \ln \left (\sqrt {1+x}+2\right )}{1024}\right ) \sqrt {1+x}\, \left (-3+x \right )}{\sqrt {\left (1+x \right ) \left (-3+x \right )^{2}}}\) | \(71\) |
trager | \(\frac {\left (15 x^{2}-70 x +43\right ) \sqrt {x^{3}-5 x^{2}+3 x +9}}{256 \left (-3+x \right )^{3} \left (1+x \right )}+\frac {15 \ln \left (\frac {-x^{2}+4 \sqrt {x^{3}-5 x^{2}+3 x +9}-2 x +15}{\left (-3+x \right )^{2}}\right )}{1024}\) | \(75\) |
default | \(\frac {\left (-3+x \right )^{3} \left (1+x \right ) \left (15 \left (1+x \right )^{\frac {5}{2}} \ln \left (\sqrt {1+x}-2\right )-15 \left (1+x \right )^{\frac {5}{2}} \ln \left (\sqrt {1+x}+2\right )-120 \left (1+x \right )^{\frac {3}{2}} \ln \left (\sqrt {1+x}-2\right )+120 \left (1+x \right )^{\frac {3}{2}} \ln \left (\sqrt {1+x}+2\right )+240 \ln \left (\sqrt {1+x}-2\right ) \sqrt {1+x}-240 \ln \left (\sqrt {1+x}+2\right ) \sqrt {1+x}+60 x^{2}-280 x +172\right )}{1024 \left (\sqrt {1+x}+2\right )^{2} \left (\sqrt {1+x}-2\right )^{2} \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {3}{2}}}\) | \(144\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=-\frac {15 \, {\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )} \log \left (\frac {2 \, x + \sqrt {x^{3} - 5 \, x^{2} + 3 \, x + 9} - 6}{x - 3}\right ) - 15 \, {\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )} \log \left (-\frac {2 \, x - \sqrt {x^{3} - 5 \, x^{2} + 3 \, x + 9} - 6}{x - 3}\right ) - 4 \, \sqrt {x^{3} - 5 \, x^{2} + 3 \, x + 9} {\left (15 \, x^{2} - 70 \, x + 43\right )}}{1024 \, {\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )}} \]
[In]
[Out]
\[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=\int \frac {1}{\left (x^{3} - 5 x^{2} + 3 x + 9\right )^{\frac {3}{2}}}\, dx \]
[In]
[Out]
\[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {3}{2}}} \,d x } \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=-\frac {15 \, \log \left (\sqrt {x + 1} + 2\right )}{1024 \, \mathrm {sgn}\left (x - 3\right )} + \frac {15 \, \log \left ({\left | \sqrt {x + 1} - 2 \right |}\right )}{1024 \, \mathrm {sgn}\left (x - 3\right )} + \frac {1}{32 \, \sqrt {x + 1} \mathrm {sgn}\left (x - 3\right )} + \frac {7 \, {\left (x + 1\right )}^{\frac {3}{2}} - 36 \, \sqrt {x + 1}}{256 \, {\left (x - 3\right )}^{2} \mathrm {sgn}\left (x - 3\right )} \]
[In]
[Out]
Timed out. \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=\int \frac {1}{{\left (x^3-5\,x^2+3\,x+9\right )}^{3/2}} \,d x \]
[In]
[Out]