Optimal. Leaf size=166 \[ \frac {2}{a x^3 \sqrt {a x^2+b x^3}}-\frac {9 \sqrt {a x^2+b x^3}}{4 a^2 x^5}+\frac {21 b \sqrt {a x^2+b x^3}}{8 a^3 x^4}-\frac {105 b^2 \sqrt {a x^2+b x^3}}{32 a^4 x^3}+\frac {315 b^3 \sqrt {a x^2+b x^3}}{64 a^5 x^2}-\frac {315 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{64 a^{11/2}} \]
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Rubi [A]
time = 0.28, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2048, 2050,
2033, 212} \begin {gather*} -\frac {315 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{64 a^{11/2}}+\frac {315 b^3 \sqrt {a x^2+b x^3}}{64 a^5 x^2}-\frac {105 b^2 \sqrt {a x^2+b x^3}}{32 a^4 x^3}+\frac {21 b \sqrt {a x^2+b x^3}}{8 a^3 x^4}-\frac {9 \sqrt {a x^2+b x^3}}{4 a^2 x^5}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 2033
Rule 2048
Rule 2050
Rubi steps
\begin {align*} \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx &=\frac {2}{a x^3 \sqrt {a x^2+b x^3}}+\frac {9 \int \frac {1}{x^4 \sqrt {a x^2+b x^3}} \, dx}{a}\\ &=\frac {2}{a x^3 \sqrt {a x^2+b x^3}}-\frac {9 \sqrt {a x^2+b x^3}}{4 a^2 x^5}-\frac {(63 b) \int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx}{8 a^2}\\ &=\frac {2}{a x^3 \sqrt {a x^2+b x^3}}-\frac {9 \sqrt {a x^2+b x^3}}{4 a^2 x^5}+\frac {21 b \sqrt {a x^2+b x^3}}{8 a^3 x^4}+\frac {\left (105 b^2\right ) \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx}{16 a^3}\\ &=\frac {2}{a x^3 \sqrt {a x^2+b x^3}}-\frac {9 \sqrt {a x^2+b x^3}}{4 a^2 x^5}+\frac {21 b \sqrt {a x^2+b x^3}}{8 a^3 x^4}-\frac {105 b^2 \sqrt {a x^2+b x^3}}{32 a^4 x^3}-\frac {\left (315 b^3\right ) \int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx}{64 a^4}\\ &=\frac {2}{a x^3 \sqrt {a x^2+b x^3}}-\frac {9 \sqrt {a x^2+b x^3}}{4 a^2 x^5}+\frac {21 b \sqrt {a x^2+b x^3}}{8 a^3 x^4}-\frac {105 b^2 \sqrt {a x^2+b x^3}}{32 a^4 x^3}+\frac {315 b^3 \sqrt {a x^2+b x^3}}{64 a^5 x^2}+\frac {\left (315 b^4\right ) \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx}{128 a^5}\\ &=\frac {2}{a x^3 \sqrt {a x^2+b x^3}}-\frac {9 \sqrt {a x^2+b x^3}}{4 a^2 x^5}+\frac {21 b \sqrt {a x^2+b x^3}}{8 a^3 x^4}-\frac {105 b^2 \sqrt {a x^2+b x^3}}{32 a^4 x^3}+\frac {315 b^3 \sqrt {a x^2+b x^3}}{64 a^5 x^2}-\frac {\left (315 b^4\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x}{\sqrt {a x^2+b x^3}}\right )}{64 a^5}\\ &=\frac {2}{a x^3 \sqrt {a x^2+b x^3}}-\frac {9 \sqrt {a x^2+b x^3}}{4 a^2 x^5}+\frac {21 b \sqrt {a x^2+b x^3}}{8 a^3 x^4}-\frac {105 b^2 \sqrt {a x^2+b x^3}}{32 a^4 x^3}+\frac {315 b^3 \sqrt {a x^2+b x^3}}{64 a^5 x^2}-\frac {315 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{64 a^{11/2}}\\ \end {align*}
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Mathematica [A]
time = 0.15, size = 106, normalized size = 0.64 \begin {gather*} \frac {\sqrt {a} \left (-16 a^4+24 a^3 b x-42 a^2 b^2 x^2+105 a b^3 x^3+315 b^4 x^4\right )-315 b^4 x^4 \sqrt {a+b x} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{11/2} x^3 \sqrt {x^2 (a+b x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.46, size = 100, normalized size = 0.60
method | result | size |
default | \(-\frac {\left (b x +a \right ) \left (315 \sqrt {b x +a}\, \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{4} x^{4}-24 a^{\frac {7}{2}} b x +42 a^{\frac {5}{2}} b^{2} x^{2}-105 a^{\frac {3}{2}} b^{3} x^{3}-315 b^{4} x^{4} \sqrt {a}+16 a^{\frac {9}{2}}\right )}{64 x \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {11}{2}}}\) | \(100\) |
risch | \(-\frac {\left (b x +a \right ) \left (-187 b^{3} x^{3}+82 a \,b^{2} x^{2}-40 a^{2} b x +16 a^{3}\right )}{64 a^{5} x^{3} \sqrt {x^{2} \left (b x +a \right )}}+\frac {b^{4} \left (-\frac {630 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {256}{\sqrt {b x +a}}\right ) \sqrt {b x +a}\, x}{128 a^{5} \sqrt {x^{2} \left (b x +a \right )}}\) | \(110\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.60, size = 263, normalized size = 1.58 \begin {gather*} \left [\frac {315 \, {\left (b^{5} x^{6} + a b^{4} x^{5}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (315 \, a b^{4} x^{4} + 105 \, a^{2} b^{3} x^{3} - 42 \, a^{3} b^{2} x^{2} + 24 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{128 \, {\left (a^{6} b x^{6} + a^{7} x^{5}\right )}}, \frac {315 \, {\left (b^{5} x^{6} + a b^{4} x^{5}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (315 \, a b^{4} x^{4} + 105 \, a^{2} b^{3} x^{3} - 42 \, a^{3} b^{2} x^{2} + 24 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{64 \, {\left (a^{6} b x^{6} + a^{7} x^{5}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 1.13, size = 122, normalized size = 0.73 \begin {gather*} \frac {315 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{64 \, \sqrt {-a} a^{5} \mathrm {sgn}\left (x\right )} + \frac {2 \, b^{4}}{\sqrt {b x + a} a^{5} \mathrm {sgn}\left (x\right )} + \frac {187 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{4} - 643 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{4} + 765 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{4} - 325 \, \sqrt {b x + a} a^{3} b^{4}}{64 \, a^{5} b^{4} x^{4} \mathrm {sgn}\left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.68, size = 44, normalized size = 0.27 \begin {gather*} -\frac {2\,{\left (\frac {a}{b\,x}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {11}{2};\ \frac {13}{2};\ -\frac {a}{b\,x}\right )}{11\,x\,{\left (b\,x^3+a\,x^2\right )}^{3/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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