Optimal. Leaf size=168 \[ -\frac {6 b^2 e p^2 \text {Li}_2\left (\frac {e}{d (f+g x)}+1\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )}{d g}-\frac {3 b e p \log \left (-\frac {e}{d (f+g x)}\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2}{d g}+\frac {(d (f+g x)+e) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^3}{d g}+\frac {6 b^3 e p^3 \text {Li}_3\left (\frac {e}{d (f+g x)}+1\right )}{d g} \]
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Rubi [A] time = 0.18, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {2483, 2449, 2454, 2396, 2433, 2374, 6589} \[ -\frac {6 b^2 e p^2 \text {PolyLog}\left (2,\frac {e}{d (f+g x)}+1\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )}{d g}+\frac {6 b^3 e p^3 \text {PolyLog}\left (3,\frac {e}{d (f+g x)}+1\right )}{d g}-\frac {3 b e p \log \left (-\frac {e}{d (f+g x)}\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2}{d g}+\frac {(d (f+g x)+e) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^3}{d g} \]
Antiderivative was successfully verified.
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Rule 2374
Rule 2396
Rule 2433
Rule 2449
Rule 2454
Rule 2483
Rule 6589
Rubi steps
\begin {align*} \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b \log \left (c \left (d+\frac {e}{x}\right )^p\right )\right )^3 \, dx,x,f+g x\right )}{g}\\ &=\frac {(e+d (f+g x)) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^3}{d g}+\frac {(3 b e p) \operatorname {Subst}\left (\int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x}\right )^p\right )\right )^2}{x} \, dx,x,f+g x\right )}{d g}\\ &=\frac {(e+d (f+g x)) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^3}{d g}-\frac {(3 b e p) \operatorname {Subst}\left (\int \frac {\left (a+b \log \left (c (d+e x)^p\right )\right )^2}{x} \, dx,x,\frac {1}{f+g x}\right )}{d g}\\ &=-\frac {3 b e p \log \left (-\frac {e}{d (f+g x)}\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2}{d g}+\frac {(e+d (f+g x)) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^3}{d g}+\frac {\left (6 b^2 e^2 p^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^p\right )\right )}{d+e x} \, dx,x,\frac {1}{f+g x}\right )}{d g}\\ &=-\frac {3 b e p \log \left (-\frac {e}{d (f+g x)}\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2}{d g}+\frac {(e+d (f+g x)) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^3}{d g}+\frac {\left (6 b^2 e p^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \log \left (c x^p\right )\right ) \log \left (-\frac {e \left (-\frac {d}{e}+\frac {x}{e}\right )}{d}\right )}{x} \, dx,x,d+\frac {e}{f+g x}\right )}{d g}\\ &=-\frac {3 b e p \log \left (-\frac {e}{d (f+g x)}\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2}{d g}+\frac {(e+d (f+g x)) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^3}{d g}-\frac {6 b^2 e p^2 \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right ) \text {Li}_2\left (\frac {d+\frac {e}{f+g x}}{d}\right )}{d g}+\frac {\left (6 b^3 e p^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{d}\right )}{x} \, dx,x,d+\frac {e}{f+g x}\right )}{d g}\\ &=-\frac {3 b e p \log \left (-\frac {e}{d (f+g x)}\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^2}{d g}+\frac {(e+d (f+g x)) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right )^3}{d g}-\frac {6 b^2 e p^2 \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right ) \text {Li}_2\left (\frac {d+\frac {e}{f+g x}}{d}\right )}{d g}+\frac {6 b^3 e p^3 \text {Li}_3\left (\frac {d+\frac {e}{f+g x}}{d}\right )}{d g}\\ \end {align*}
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Mathematica [B] time = 0.72, size = 415, normalized size = 2.47 \[ \frac {3 b^2 p^2 \left (2 e \text {Li}_2\left (\frac {d (f+g x)}{e}+1\right )+d (f+g x) \log ^2\left (d+\frac {e}{f+g x}\right )+e \left (2 \log \left (-\frac {d (f+g x)}{e}\right )-\log (d f+d g x+e)+2 \log \left (d+\frac {e}{f+g x}\right )\right ) \log (d (f+g x)+e)\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )-b p \log \left (d+\frac {e}{f+g x}\right )\right )+3 b d p (f+g x) \log \left (d+\frac {e}{f+g x}\right ) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )-b p \log \left (d+\frac {e}{f+g x}\right )\right )^2+3 b e p \log (d (f+g x)+e) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )-b p \log \left (d+\frac {e}{f+g x}\right )\right )^2+d (f+g x) \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )-b p \log \left (d+\frac {e}{f+g x}\right )\right )^3+b^3 p^3 \left (6 e \text {Li}_3\left (\frac {e}{d f+d g x}+1\right )-6 e \text {Li}_2\left (\frac {e}{d f+d g x}+1\right ) \log \left (d+\frac {e}{f+g x}\right )+\left ((d f+d g x+e) \log \left (d+\frac {e}{f+g x}\right )-3 e \log \left (-\frac {e}{d f+d g x}\right )\right ) \log ^2\left (d+\frac {e}{f+g x}\right )\right )}{d g} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{3} \log \left (c \left (\frac {d g x + d f + e}{g x + f}\right )^{p}\right )^{3} + 3 \, a b^{2} \log \left (c \left (\frac {d g x + d f + e}{g x + f}\right )^{p}\right )^{2} + 3 \, a^{2} b \log \left (c \left (\frac {d g x + d f + e}{g x + f}\right )^{p}\right ) + a^{3}, x\right ) \]
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 \[ \int {\left (b \log \left (c {\left (d + \frac {e}{g x + f}\right )}^{p}\right ) + a\right )}^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.10, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \left (d +\frac {e}{g x +f}\right )^{p}\right )+a \right )^{3}\, dx \]
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 \[ -3 \, a^{2} b e g p {\left (\frac {f \log \left (g x + f\right )}{e g^{2}} - \frac {{\left (d f + e\right )} \log \left (d g x + d f + e\right )}{d e g^{2}}\right )} + 3 \, a^{2} b x \log \left (c {\left (d + \frac {e}{g x + f}\right )}^{p}\right ) + a^{3} x + \frac {b^{3} d g x \log \left ({\left (d g x + d f + e\right )}^{p}\right )^{3} - 3 \, {\left (b^{3} d f p \log \left (g x + f\right ) + b^{3} d g x \log \left ({\left (g x + f\right )}^{p}\right ) - {\left (d f p + e p\right )} b^{3} \log \left (d g x + d f + e\right ) - {\left (b^{3} d g \log \relax (c) + a b^{2} d g\right )} x\right )} \log \left ({\left (d g x + d f + e\right )}^{p}\right )^{2}}{d g} + \int \frac {{\left (d f + e\right )} b^{3} \log \relax (c)^{3} + 3 \, {\left (d f + e\right )} a b^{2} \log \relax (c)^{2} - {\left (b^{3} d g x + {\left (d f + e\right )} b^{3}\right )} \log \left ({\left (g x + f\right )}^{p}\right )^{3} + 3 \, {\left ({\left (d f + e\right )} b^{3} \log \relax (c) + {\left (d f + e\right )} a b^{2} + {\left (b^{3} d g \log \relax (c) + a b^{2} d g\right )} x\right )} \log \left ({\left (g x + f\right )}^{p}\right )^{2} + {\left (b^{3} d g \log \relax (c)^{3} + 3 \, a b^{2} d g \log \relax (c)^{2}\right )} x + 3 \, {\left (2 \, b^{3} d f p^{2} \log \left (g x + f\right ) + {\left (d f + e\right )} b^{3} \log \relax (c)^{2} - 2 \, {\left (d f p^{2} + e p^{2}\right )} b^{3} \log \left (d g x + d f + e\right ) + 2 \, {\left (d f + e\right )} a b^{2} \log \relax (c) + {\left (b^{3} d g x + {\left (d f + e\right )} b^{3}\right )} \log \left ({\left (g x + f\right )}^{p}\right )^{2} - {\left (2 \, {\left (d g p - d g \log \relax (c)\right )} a b^{2} + {\left (2 \, d g p \log \relax (c) - d g \log \relax (c)^{2}\right )} b^{3}\right )} x - 2 \, {\left ({\left (d f + e\right )} b^{3} \log \relax (c) + {\left (d f + e\right )} a b^{2} + {\left (a b^{2} d g - {\left (d g p - d g \log \relax (c)\right )} b^{3}\right )} x\right )} \log \left ({\left (g x + f\right )}^{p}\right )\right )} \log \left ({\left (d g x + d f + e\right )}^{p}\right ) - 3 \, {\left ({\left (d f + e\right )} b^{3} \log \relax (c)^{2} + 2 \, {\left (d f + e\right )} a b^{2} \log \relax (c) + {\left (b^{3} d g \log \relax (c)^{2} + 2 \, a b^{2} d g \log \relax (c)\right )} x\right )} \log \left ({\left (g x + f\right )}^{p}\right )}{d g x + d f + e}\,{d x} \]
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 \[ \int {\left (a+b\,\ln \left (c\,{\left (d+\frac {e}{f+g\,x}\right )}^p\right )\right )}^3 \,d x \]
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 \[ \int \left (a + b \log {\left (c \left (d + \frac {e}{f + g x}\right )^{p} \right )}\right )^{3}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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