Integrand size = 24, antiderivative size = 114 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^3 \left (a+b x^2\right )} \, dx=-\frac {c \sqrt {c+d x^2}}{2 a x^2}+\frac {\sqrt {c} (2 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2}-\frac {(b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 \sqrt {b}} \]
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Time = 0.11 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {457, 100, 162, 65, 214} \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^3 \left (a+b x^2\right )} \, dx=-\frac {(b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 \sqrt {b}}+\frac {\sqrt {c} (2 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2}-\frac {c \sqrt {c+d x^2}}{2 a x^2} \]
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Rule 65
Rule 100
Rule 162
Rule 214
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{x^2 (a+b x)} \, dx,x,x^2\right ) \\ & = -\frac {c \sqrt {c+d x^2}}{2 a x^2}-\frac {\text {Subst}\left (\int \frac {\frac {1}{2} c (2 b c-3 a d)+\frac {1}{2} d (b c-2 a d) x}{x (a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 a} \\ & = -\frac {c \sqrt {c+d x^2}}{2 a x^2}-\frac {(c (2 b c-3 a d)) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{4 a^2}+\frac {(b c-a d)^2 \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 a^2} \\ & = -\frac {c \sqrt {c+d x^2}}{2 a x^2}-\frac {(c (2 b c-3 a d)) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 a^2 d}+\frac {(b c-a d)^2 \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{a^2 d} \\ & = -\frac {c \sqrt {c+d x^2}}{2 a x^2}+\frac {\sqrt {c} (2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2}-\frac {(b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 \sqrt {b}} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.95 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^3 \left (a+b x^2\right )} \, dx=\frac {-\frac {a c \sqrt {c+d x^2}}{x^2}+\frac {2 (-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{\sqrt {b}}+\sqrt {c} (2 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2} \]
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Time = 2.99 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.03
method | result | size |
pseudoelliptic | \(\frac {\left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) x^{2}+\sqrt {\left (a d -b c \right ) b}\, \left (x^{2} \left (c^{\frac {3}{2}} b -\frac {3 a d \sqrt {c}}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )-\frac {\sqrt {d \,x^{2}+c}\, c a}{2}\right )}{\sqrt {\left (a d -b c \right ) b}\, a^{2} x^{2}}\) | \(117\) |
risch | \(-\frac {c \sqrt {d \,x^{2}+c}}{2 a \,x^{2}}+\frac {-\frac {\sqrt {c}\, \left (3 a d -2 b c \right ) \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{a}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{a b \sqrt {-\frac {a d -b c}{b}}}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{a b \sqrt {-\frac {a d -b c}{b}}}}{2 a}\) | \(412\) |
default | \(\text {Expression too large to display}\) | \(1377\) |
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Time = 0.45 (sec) , antiderivative size = 732, normalized size of antiderivative = 6.42 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^3 \left (a+b x^2\right )} \, dx=\left [-\frac {{\left (b c - a d\right )} x^{2} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + {\left (2 \, b c - 3 \, a d\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, \sqrt {d x^{2} + c} a c}{4 \, a^{2} x^{2}}, -\frac {2 \, {\left (2 \, b c - 3 \, a d\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (b c - a d\right )} x^{2} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, \sqrt {d x^{2} + c} a c}{4 \, a^{2} x^{2}}, -\frac {2 \, {\left (b c - a d\right )} x^{2} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + {\left (2 \, b c - 3 \, a d\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, \sqrt {d x^{2} + c} a c}{4 \, a^{2} x^{2}}, -\frac {{\left (b c - a d\right )} x^{2} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + {\left (2 \, b c - 3 \, a d\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + \sqrt {d x^{2} + c} a c}{2 \, a^{2} x^{2}}\right ] \]
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\[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^3 \left (a+b x^2\right )} \, dx=\int \frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{x^{3} \left (a + b x^{2}\right )}\, dx \]
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\[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^3 \left (a+b x^2\right )} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}}}{{\left (b x^{2} + a\right )} x^{3}} \,d x } \]
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Time = 0.31 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.05 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^3 \left (a+b x^2\right )} \, dx=\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{2}} - \frac {{\left (2 \, b c^{2} - 3 \, a c d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, a^{2} \sqrt {-c}} - \frac {\sqrt {d x^{2} + c} c}{2 \, a x^{2}} \]
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Time = 5.86 (sec) , antiderivative size = 560, normalized size of antiderivative = 4.91 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x^3 \left (a+b x^2\right )} \, dx=-\frac {c\,\sqrt {d\,x^2+c}}{2\,a\,x^2}-\frac {\sqrt {c}\,\mathrm {atanh}\left (\frac {29\,b^2\,c^{3/2}\,d^6\,\sqrt {d\,x^2+c}}{4\,\left (\frac {29\,b^2\,c^2\,d^6}{4}-3\,a\,b\,c\,d^7-\frac {23\,b^3\,c^3\,d^5}{4\,a}+\frac {3\,b^4\,c^4\,d^4}{2\,a^2}\right )}+\frac {23\,b^3\,c^{5/2}\,d^5\,\sqrt {d\,x^2+c}}{4\,\left (\frac {23\,b^3\,c^3\,d^5}{4}-\frac {29\,a\,b^2\,c^2\,d^6}{4}-\frac {3\,b^4\,c^4\,d^4}{2\,a}+3\,a^2\,b\,c\,d^7\right )}+\frac {3\,b^4\,c^{7/2}\,d^4\,\sqrt {d\,x^2+c}}{2\,\left (-3\,a^3\,b\,c\,d^7+\frac {29\,a^2\,b^2\,c^2\,d^6}{4}-\frac {23\,a\,b^3\,c^3\,d^5}{4}+\frac {3\,b^4\,c^4\,d^4}{2}\right )}-\frac {3\,a\,b\,\sqrt {c}\,d^7\,\sqrt {d\,x^2+c}}{\frac {29\,b^2\,c^2\,d^6}{4}-3\,a\,b\,c\,d^7-\frac {23\,b^3\,c^3\,d^5}{4\,a}+\frac {3\,b^4\,c^4\,d^4}{2\,a^2}}\right )\,\left (3\,a\,d-2\,b\,c\right )}{2\,a^2}-\frac {\mathrm {atanh}\left (\frac {3\,b^2\,c^2\,d^4\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b\,d^3+3\,a^2\,b^2\,c\,d^2-3\,a\,b^3\,c^2\,d+b^4\,c^3}}{2\,\left (-2\,a^3\,b\,c\,d^7+\frac {11\,a^2\,b^2\,c^2\,d^6}{2}-5\,a\,b^3\,c^3\,d^5+\frac {3\,b^4\,c^4\,d^4}{2}\right )}+\frac {2\,b\,c\,d^5\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b\,d^3+3\,a^2\,b^2\,c\,d^2-3\,a\,b^3\,c^2\,d+b^4\,c^3}}{5\,b^3\,c^3\,d^5-\frac {11\,a\,b^2\,c^2\,d^6}{2}-\frac {3\,b^4\,c^4\,d^4}{2\,a}+2\,a^2\,b\,c\,d^7}\right )\,\sqrt {-b\,{\left (a\,d-b\,c\right )}^3}}{a^2\,b} \]
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