Optimal. Leaf size=82 \[ -\frac {2 e f p (f x)^{-1+m} \, _2F_1\left (1,\frac {1-m}{2};\frac {3-m}{2};-\frac {e}{d x^2}\right )}{d \left (1-m^2\right )}+\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{x^2}\right )^p\right )}{f (1+m)} \]
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Rubi [A]
time = 0.04, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2505, 16, 346,
371} \begin {gather*} \frac {(f x)^{m+1} \log \left (c \left (d+\frac {e}{x^2}\right )^p\right )}{f (m+1)}-\frac {2 e f p (f x)^{m-1} \, _2F_1\left (1,\frac {1-m}{2};\frac {3-m}{2};-\frac {e}{d x^2}\right )}{d \left (1-m^2\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 16
Rule 346
Rule 371
Rule 2505
Rubi steps
\begin {align*} \int (f x)^m \log \left (c \left (d+\frac {e}{x^2}\right )^p\right ) \, dx &=\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{x^2}\right )^p\right )}{f (1+m)}+\frac {(2 e p) \int \frac {(f x)^{1+m}}{\left (d+\frac {e}{x^2}\right ) x^3} \, dx}{f (1+m)}\\ &=\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{x^2}\right )^p\right )}{f (1+m)}+\frac {\left (2 e f^2 p\right ) \int \frac {(f x)^{-2+m}}{d+\frac {e}{x^2}} \, dx}{1+m}\\ &=\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{x^2}\right )^p\right )}{f (1+m)}-\frac {\left (2 e f p \left (\frac {1}{x}\right )^{-1+m} (f x)^{-1+m}\right ) \text {Subst}\left (\int \frac {x^{-m}}{d+e x^2} \, dx,x,\frac {1}{x}\right )}{1+m}\\ &=-\frac {2 e f p (f x)^{-1+m} \, _2F_1\left (1,\frac {1-m}{2};\frac {3-m}{2};-\frac {e}{d x^2}\right )}{d \left (1-m^2\right )}+\frac {(f x)^{1+m} \log \left (c \left (d+\frac {e}{x^2}\right )^p\right )}{f (1+m)}\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 76, normalized size = 0.93 \begin {gather*} \frac {(f x)^m \left (2 e p \, _2F_1\left (1,\frac {1}{2}-\frac {m}{2};\frac {3}{2}-\frac {m}{2};-\frac {e}{d x^2}\right )+d (-1+m) x^2 \log \left (c \left (d+\frac {e}{x^2}\right )^p\right )\right )}{d (-1+m) (1+m) x} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \left (f x \right )^{m} \ln \left (c \left (d +\frac {e}{x^{2}}\right )^{p}\right )\, dx\]
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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 32.79, size = 369, normalized size = 4.50 \begin {gather*} 2 e p \left (\begin {cases} - \frac {0^{m} \sqrt {- \frac {1}{d e}} \log {\left (- e \sqrt {- \frac {1}{d e}} + x \right )}}{2} + \frac {0^{m} \sqrt {- \frac {1}{d e}} \log {\left (e \sqrt {- \frac {1}{d e}} + x \right )}}{2} & \text {for}\: \left (f = 0 \wedge m \neq -1\right ) \vee f = 0 \\\frac {f f^{m} m x^{m} \Phi \left (\frac {e e^{i \pi }}{d x^{2}}, 1, \frac {1}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {1}{2} - \frac {m}{2}\right )}{4 d f m x \Gamma \left (\frac {3}{2} - \frac {m}{2}\right ) + 4 d f x \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )} - \frac {f f^{m} x^{m} \Phi \left (\frac {e e^{i \pi }}{d x^{2}}, 1, \frac {1}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {1}{2} - \frac {m}{2}\right )}{4 d f m x \Gamma \left (\frac {3}{2} - \frac {m}{2}\right ) + 4 d f x \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )} & \text {for}\: m > -\infty \wedge m < \infty \wedge m \neq -1 \\\frac {\begin {cases} \frac {\operatorname {Li}_{2}\left (\frac {e e^{i \pi }}{d x^{2}}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} + \frac {\operatorname {Li}_{2}\left (\frac {e e^{i \pi }}{d x^{2}}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} + \frac {\operatorname {Li}_{2}\left (\frac {e e^{i \pi }}{d x^{2}}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + \frac {\operatorname {Li}_{2}\left (\frac {e e^{i \pi }}{d x^{2}}\right )}{2} & \text {otherwise} \end {cases}}{2 e f} - \frac {\log {\left (f x \right )} \log {\left (d + \frac {e}{x^{2}} \right )}}{2 e f} & \text {otherwise} \end {cases}\right ) + \left (\begin {cases} 0^{m} x & \text {for}\: f = 0 \\\frac {\begin {cases} \frac {\left (f x\right )^{m + 1}}{m + 1} & \text {for}\: m \neq -1 \\\log {\left (f x \right )} & \text {otherwise} \end {cases}}{f} & \text {otherwise} \end {cases}\right ) \log {\left (c \left (d + \frac {e}{x^{2}}\right )^{p} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \ln \left (c\,{\left (d+\frac {e}{x^2}\right )}^p\right )\,{\left (f\,x\right )}^m \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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