Integrand size = 24, antiderivative size = 129 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )^2} \, dx=\frac {(b c-a d) \sqrt {c+d x^2}}{2 a b \left (a+b x^2\right )}-\frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a^2}+\frac {\sqrt {b c-a d} (2 b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^2 b^{3/2}} \]
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Time = 0.11 (sec) , antiderivative size = 129, 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} \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )^2} \, dx=\frac {\sqrt {b c-a d} (a d+2 b c) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^2 b^{3/2}}-\frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a^2}+\frac {\sqrt {c+d x^2} (b c-a d)}{2 a b \left (a+b x^2\right )} \]
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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 (a+b x)^2} \, dx,x,x^2\right ) \\ & = \frac {(b c-a d) \sqrt {c+d x^2}}{2 a b \left (a+b x^2\right )}+\frac {\text {Subst}\left (\int \frac {b c^2+\frac {1}{2} d (b c+a d) x}{x (a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 a b} \\ & = \frac {(b c-a d) \sqrt {c+d x^2}}{2 a b \left (a+b x^2\right )}+\frac {c^2 \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{2 a^2}-\frac {((b c-a d) (2 b c+a d)) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 a^2 b} \\ & = \frac {(b c-a d) \sqrt {c+d x^2}}{2 a b \left (a+b x^2\right )}+\frac {c^2 \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{a^2 d}-\frac {((b c-a d) (2 b c+a d)) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 a^2 b d} \\ & = \frac {(b c-a d) \sqrt {c+d x^2}}{2 a b \left (a+b x^2\right )}-\frac {c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a^2}+\frac {\sqrt {b c-a d} (2 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^2 b^{3/2}} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )^2} \, dx=\frac {\frac {a (b c-a d) \sqrt {c+d x^2}}{b \left (a+b x^2\right )}+\frac {\sqrt {-b c+a d} (2 b c+a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{b^{3/2}}-2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2} \]
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Time = 3.05 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.09
method | result | size |
pseudoelliptic | \(-\frac {-\left (b \,x^{2}+a \right ) \left (a d +2 b c \right ) \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\sqrt {\left (a d -b c \right ) b}\, \left (2 b \,c^{\frac {3}{2}} \left (b \,x^{2}+a \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )+a \sqrt {d \,x^{2}+c}\, \left (a d -b c \right )\right )}{2 \sqrt {\left (a d -b c \right ) b}\, a^{2} b \left (b \,x^{2}+a \right )}\) | \(140\) |
default | \(\text {Expression too large to display}\) | \(3437\) |
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Time = 0.44 (sec) , antiderivative size = 883, normalized size of antiderivative = 6.84 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )^2} \, dx=\left [\frac {{\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x^{2}\right )} \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 ) + 4 \, {\left (b^{2} c x^{2} + a b c\right )} \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 4 \, {\left (a b c - a^{2} d\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a^{2} b^{2} x^{2} + a^{3} b\right )}}, \frac {8 \, {\left (b^{2} c x^{2} + a b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x^{2}\right )} \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 ) + 4 \, {\left (a b c - a^{2} d\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a^{2} b^{2} x^{2} + a^{3} b\right )}}, \frac {{\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x^{2}\right )} \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 ) + 2 \, {\left (b^{2} c x^{2} + a b c\right )} \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (a b c - a^{2} d\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a^{2} b^{2} x^{2} + a^{3} b\right )}}, \frac {{\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x^{2}\right )} \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 ) + 4 \, {\left (b^{2} c x^{2} + a b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + 2 \, {\left (a b c - a^{2} d\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a^{2} b^{2} x^{2} + a^{3} b\right )}}\right ] \]
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\[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )^2} \, dx=\int \frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{x \left (a + b x^{2}\right )^{2}}\, dx \]
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\[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{2} x} \,d x } \]
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Time = 0.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.19 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )^2} \, dx=\frac {c^{2} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c}} - \frac {{\left (2 \, b^{2} c^{2} - a b c d - a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} a^{2} b} + \frac {\sqrt {d x^{2} + c} b c d - \sqrt {d x^{2} + c} a d^{2}}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} a b} \]
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Time = 7.19 (sec) , antiderivative size = 488, normalized size of antiderivative = 3.78 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )^2} \, dx=-\frac {\mathrm {atanh}\left (\frac {d^6\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{2\,\left (\frac {c^2\,d^6}{2}+\frac {b\,c^3\,d^5}{a}-\frac {3\,b^2\,c^4\,d^4}{2\,a^2}\right )}+\frac {c\,d^5\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{c^3\,d^5+\frac {a\,c^2\,d^6}{2\,b}-\frac {3\,b\,c^4\,d^4}{2\,a}}-\frac {3\,b\,c^2\,d^4\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{2\,\left (a\,c^3\,d^5-\frac {3\,b\,c^4\,d^4}{2}+\frac {a^2\,c^2\,d^6}{2\,b}\right )}\right )\,\sqrt {c^3}}{a^2}-\frac {\mathrm {atanh}\left (\frac {5\,c^2\,d^5\,\sqrt {d\,x^2+c}\,\sqrt {b^4\,c-a\,b^3\,d}}{4\,\left (\frac {a^2\,c\,d^7}{4}+\frac {b^2\,c^3\,d^5}{4}-\frac {3\,b^3\,c^4\,d^4}{2\,a}+a\,b\,c^2\,d^6\right )}+\frac {3\,c^3\,d^4\,\sqrt {d\,x^2+c}\,\sqrt {b^4\,c-a\,b^3\,d}}{2\,\left (a^2\,c^2\,d^6-\frac {3\,b^2\,c^4\,d^4}{2}+\frac {a^3\,c\,d^7}{4\,b}+\frac {a\,b\,c^3\,d^5}{4}\right )}+\frac {c\,d^6\,\sqrt {d\,x^2+c}\,\sqrt {b^4\,c-a\,b^3\,d}}{4\,\left (b^2\,c^2\,d^6+\frac {a\,b\,c\,d^7}{4}+\frac {b^3\,c^3\,d^5}{4\,a}-\frac {3\,b^4\,c^4\,d^4}{2\,a^2}\right )}\right )\,\left (a\,d+2\,b\,c\right )\,\sqrt {-b^3\,\left (a\,d-b\,c\right )}}{2\,a^2\,b^3}-\frac {d\,\sqrt {d\,x^2+c}\,\left (a\,d-b\,c\right )}{2\,a\,b\,\left (b\,\left (d\,x^2+c\right )+a\,d-b\,c\right )} \]
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