Optimal. Leaf size=75 \[ \frac{(d x)^{m+1} \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )}{d (m+1)}-\frac{2 b c d (d x)^{m-1} \text{Hypergeometric2F1}\left (1,\frac{1-m}{4},\frac{5-m}{4},\frac{c^2}{x^4}\right )}{1-m^2} \]
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Rubi [A] time = 0.0557798, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6097, 16, 339, 364} \[ \frac{(d x)^{m+1} \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )}{d (m+1)}-\frac{2 b c d (d x)^{m-1} \, _2F_1\left (1,\frac{1-m}{4};\frac{5-m}{4};\frac{c^2}{x^4}\right )}{1-m^2} \]
Antiderivative was successfully verified.
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Rule 6097
Rule 16
Rule 339
Rule 364
Rubi steps
\begin{align*} \int (d x)^m \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right ) \, dx &=\frac{(d x)^{1+m} \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )}{d (1+m)}+\frac{(2 b c) \int \frac{(d x)^{1+m}}{\left (1-\frac{c^2}{x^4}\right ) x^3} \, dx}{d (1+m)}\\ &=\frac{(d x)^{1+m} \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )}{d (1+m)}+\frac{\left (2 b c d^2\right ) \int \frac{(d x)^{-2+m}}{1-\frac{c^2}{x^4}} \, dx}{1+m}\\ &=\frac{(d x)^{1+m} \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )}{d (1+m)}-\frac{\left (2 b c d \left (\frac{1}{x}\right )^{-1+m} (d x)^{-1+m}\right ) \operatorname{Subst}\left (\int \frac{x^{-m}}{1-c^2 x^4} \, dx,x,\frac{1}{x}\right )}{1+m}\\ &=\frac{(d x)^{1+m} \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )}{d (1+m)}-\frac{2 b c d (d x)^{-1+m} \, _2F_1\left (1,\frac{1-m}{4};\frac{5-m}{4};\frac{c^2}{x^4}\right )}{1-m^2}\\ \end{align*}
Mathematica [A] time = 0.0673567, size = 68, normalized size = 0.91 \[ \frac{(d x)^m \left (2 b c \text{Hypergeometric2F1}\left (1,\frac{1}{4}-\frac{m}{4},\frac{5}{4}-\frac{m}{4},\frac{c^2}{x^4}\right )+(m-1) x^2 \left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )\right )}{(m-1) (m+1) x} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.294, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{m} \left ( a+b{\it Artanh} \left ({\frac{c}{{x}^{2}}} \right ) \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \operatorname{artanh}\left (\frac{c}{x^{2}}\right ) + a\right )} \left (d x\right )^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{artanh}\left (\frac{c}{x^{2}}\right ) + a\right )} \left (d x\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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